FOCUSING ON FUNDAMENTAL CONCEPTS The revision of Physics by Giambattista/Richardson/Richardson incorporates great new pedagogical tools to help students understand that physics is based on a few basic principles, enabling them to draw connections between physics concepts.

È ÒConnectionsÓ identify areas in each chapter where important concepts are revisited. Marginal ÒConnectionsÓ notes help students easily recognize that a previously introduced concept is being applied to the current discussion. Concepts are being applied; they are not being newly introduced.

È Checkpoint questions have been added to applicable sections, allowing students to quickly test their understanding of the concept within the current section.

È The exercises in the Review & Synthesis sections have been revised and expanded. With the exercises in the Review & Synthesis sections, students can test their collective understanding of concepts within groups of chapters, which helps them to better prepare for cumulative exams. These sections also contain MCAT¨ Review exercisesÑ actual reading passages and questions written for the MCAT¨ exam.

È Some of the more detailed coverage and derivations have been moved to the textÕs website. Thus, students can focus on fundamental, core concepts in the text, and then proceed online where noted for expanded coverage and explanation of topics of interest.

Online Homework and Resources McGraw-HillÕs Physics website offers online electronic homework and a myriad of resources for both instructors and students:

È Instructors can create homework with easy-to-assign algorithmically generated problems PhysicsÕ end-of-chapter problems and Review & Synthesis exercises appear in the online homework system in diverse formats and with various tools. The online homework system incorporates new and exciting interactive tools and problem types: ranking problems, a graphing tool, a free-body diagram drawing tool, symbolic entry, and a math palette.

È Instructor resources include PowerPoint lecture outlines, an InstructorÕs Resource Guide with solutions, suggested demonstrations, electronic images from the text, and clicker questions. Both instructors and students have access to quizzes, interactive simulations, tutorials, selected solutions for the textÕs problems, and more.

È Go to www.mhhe.com/grr to learn more and to register!

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from the text and the simplicity of automatic grading and reporting.

Revised Pages

SECOND EDITION

Physics

Alan Giambattista Cornell University

Betty McCarthy Richardson Cornell University

Robert C. Richardson Cornell University

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PHYSICS, SECOND EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved. Previous edition © 2008. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 VNH/VNH 0 9 ISBN 978–0–07–340453–0 MHID 0–07–340453–5 Publisher: Thomas D. Timp Sponsoring Editor: Debra B. Hash Vice-President New Product Launches: Michael Lange Senior Developmental Editor: Mary E. Hurley Senior Marketing Manager: Lisa Nicks Senior Project Manager: Gloria G. Schiesl Senior Production Supervisor: Laura Fuller Senior Media Project Manager: Tammy Juran Senior Designer: David W. Hash Cover/Interior Designer: Rokusek Design (USE) Cover Image: ©Kazuya Shiota/Aflo/Jupiterimages Lead Photo Research Coordinator: Carrie K. Burger Photo Research: Danny Meldung/Photo Affairs, Inc Supplement Producer: Mary Jane Lampe Compositor: Laserwords Private Limited Typeface: 10/12 Times Printer: R. R. Donnelley, Jefferson City, MO The credits section for this book begins on page C-1 and is considered an extension of the copyright page. MCAT® is a registered trademark of the Association of American Medical Colleges. MCAT exam material included is printed with permission of the AAMC. The AAMC does not endorse this book. Library of Congress Cataloging-in-Publication Data Giambattista, Alan. Physics / Alan Giambattista, Betty McCarthy Richardson, Robert C. Richardson.—2nd ed. p. cm. Includes index. ISBN 978–0–07–340453–0 — ISBN 0–07–340453–5 (hard copy : alk. paper) 1. Physics–Textbooks. I. Richardson, Betty McCarthy. II. Richardson, Robert C. III. Title. QC21.3.G537 2010 530–dc22 2008034667

www.mhhe.com

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About the Authors Alan Giambattista grew up in Nutley, New Jersey. In his junior year at Brigham Young University he decided to pursue a physics major, after having explored math, music, and psychology. He did his graduate studies at Cornell University and has taught introductory college physics for over 20 years. When not found at the computer keyboard working on Physics, he can often be found at the keyboard of a harpsichord or piano. He is a member of the Cayuga Chamber Orchestra and has given performances of the Bach harpsichord concerti at several regional Bach festivals. He met his wife Marion in a singing group. They live in an 1824 parsonage built for an abolitionist minister, which is now surrounded by an organic dairy farm. Besides making music and taking care of the house, gardens, and fruit trees, they love to travel together.

Betty McCarthy Richardson was born and grew up in Marblehead, Massachusetts, and tried to avoid taking any science classes after eighth grade but managed to avoid only ninth grade science. After discovering that physics tells how things work, she decided to become a physicist. She attended Wellesley College and did graduate work at Duke University. While at Duke, Betty met and married fellow graduate student Bob Richardson and had two daughters, Jennifer and Pamela. Betty began teaching physics at Cornell in 1977. Many years later, she is still teaching the same course, Physics 101/102, an algebra-based course with all teaching done one-on-one in a Learning Center. From her own early experience of math and science avoidance, Betty has empathy with students who are apprehensive about learning physics. Betty’s hobbies include collecting old children’s books, reading, enjoying music, travel, and dining with royalty. A highlight for Betty during the Nobel Prize festivities in 1996 was being escorted to dinner on the arm of King Carl XVI Gustav of Sweden. Currently she is spending spare time enjoying grandsons Jasper (the 1-m child in Chapter 1), Dashiell and Oliver (the twins of Chapter 12), and Quintin, the newest arrival.

Robert C. Richardson was born in Washington, D.C., attended Virginia Polytechnic Institute, spent time in the United States Army, and then returned to graduate school in physics at Duke University where his thesis work involved NMR studies of solid helium-3. In the fall of 1966 Bob began work at Cornell University in the laboratory of David M. Lee. Their research goal was to observe the nuclear magnetic phase transition in solid 3He that could be predicted from Richardson’s thesis work with Professor Horst Meyer at Duke. In collaboration with graduate student Douglas D. Osheroff, they worked on cooling techniques and NMR instrumentation for studying low-temperature helium liquids and solids. In the fall of 1971, they made the accidental discovery that liquid 3He undergoes a pairing transition similar to that of superconductors. The three were awarded the Nobel Prize for that work in 1996. Bob is currently the F. R. Newman Professor of Physics and the Senior Science Advisor at Cornell. In his spare time he enjoys gardening and photography. In loving memory of Dad and of my niece, Natalie Alan In memory of our daughter Pamela, and for Quintin, Oliver, Dashiell, Jasper, Jennifer, and Jim Merlis Bob and Betty iii

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Brief Contents Chapter 1

PART ONE

2 3 4 5 6 7 8 9 10 11 12

Electric Forces and Fields 561 Electric Potential 601 Electric Current and Circuits 640 Magnetic Forces and Fields 693 Electromagnetic Induction 741 Alternating Current 780

Electromagnetic Waves and Optics Chapter 22 Chapter 23 Chapter 24 Chapter 25

PART FIVE

Temperature and Ideal Gas 457 Heat 489 Thermodynamics 527

Electromagnetism Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21

PART FOUR

Motion Along a Line 25 Motion in a Plane 55 Force and Newton’s Laws of Motion 87 Circular Motion 146 Conservation of Energy 186 Linear Momentum 225 Torque and Angular Momentum 260 Fluids 316 Elasticity and Oscillations 356 Waves 392 Sound 420

Thermal Physics Chapter 13 Chapter 14 Chapter 15

PART THREE

1

Mechanics Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter

PART TWO

Introduction

Electromagnetic Waves 811 Reflection and Refraction of Light 848 Optical Instruments 891 Interference and Diffraction 922

Quantum and Particle Physics and Relativity Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30

Relativity 967 Early Quantum Physics and the Photon 997 Quantum Physics 1030 Nuclear Physics 1065 Particle Physics 1105

Appendix A

Mathematics Review A–1

Appendix B

Table of Selected Nuclides A-15

iv

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Contents List of Selected Applications Preface xiii To the Student xxii Acknowledgments xxx

Chapter 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

Chapter 4

x

Introduction

4.1 4.2

1

4.3

Why Study Physics? 2 Talking Physics 2 The Use of Mathematics 3 Scientific Notation and Significant Figures Units 8 Dimensional Analysis 11 Problem-Solving Techniques 13 Approximation 14 Graphs 15

4.4 5

PART ONE

2.1 2.2 2.3 2.4 2.5 2.6

3.3 3.4 3.5 3.6

25

Position and Displacement 26 Velocity: Rate of Change of Position 28 Acceleration: Rate of Change of Velocity 33 Motion Along a Line with Constant Acceleration 37 Visualizing Motion Along a Line with Constant Acceleration 40 Free Fall 43

Chapter 3 3.1 3.2

Motion Along a Line

Force 88 Inertia and Equilibrium: Newton’s First Law of Motion 92 Net Force, Mass, and Acceleration: Newtons’s Second Law of Motion 96 Interaction Pairs: Newton’s Third Law of Motion 97 Gravitational Forces 99 Contact Forces 102 Tension 109 Applying Newton’s Second Law 113 Reference Frames 122 Apparent Weight 123 Air Resistance 126 Fundamental Forces 126

Chapter 5

Mechanics Chapter 2

4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12

Motion in a Plane 55

Graphical Addition and Subtraction of Vectors 56 Vector Addition and Subtraction Using Components 59 Velocity 63 Acceleration 64 Motion in a Plane with Constant Acceleration 67 Velocity Is Relative; Reference Frames 73

5.1 5.2 5.3 5.4 5.5 5.6 5.7

Force and Newton’s Laws of Motion 87

Circular Motion 146

Description of Uniform Circular Motion 147 Radial Acceleration 152 Unbanked and Banked Curves 157 Circular Orbits of Satellites and Planets 160 Nonuniform Circular Motion 164 Tangential and Angular Acceleration 168 Apparent Weight and Artificial Gravity 170

Review & Synthesis: Chapters 1–5

Chapter 6 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

182

Conservation of Energy 186

The Law of Conservation of Energy 187 Work Done by a Constant Force 188 Kinetic Energy 195 Gravitational Potential Energy (1) 197 Gravitational Potential Energy (2) 202 Work Done by Variable Forces: Hooke’s Law 205 Elastic Potential Energy 208 Power 212

v

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vi

CONTENTS

Chapter 7 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

Conservation Law for a Vector Quantity 226 Momentum 226 The Impulse-Momentum Theorem 228 Conservation of Momentum 234 Center of Mass 237 Motion of the Center of Mass 240 Collisions in One Dimension 242 Collisions in Two Dimensions 247

Chapter 8 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

Linear Momentum 225

Torque and Angular Momentum 260

Rotational Kinetic Energy and Rotational Inertia 261 Torque 266 Calculating Work Done from a Torque 271 Rotational Equilibrium 273 Equilibrium in the Human Body 281 Rotational Form of Newton’s Second Law 285 The Motion of Rolling Objects 286 Angular Momentum 289 The Vector Nature of Angular Momentum 293

Review & Synthesis: Chapters 6–8

Chapter 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11

311

Fluids 316

States of Matter 317 Pressure 317 Pascal’s Principle 320 The Effect of Gravity on Fluid Pressure 321 Measuring Pressure 324 327 The Buoyant Force Fluid Flow 332 Bernoulli’s Equation 334 Viscosity 338 Viscous Drag 341 Surface Tension 343

Chapter 10 Elasticity and Oscillations 356 10.1 10.2

Elastic Deformations of Solids 357 Hooke’s Law for Tensile and Compressive Forces 357

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10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10

Beyond Hooke’s Law 360 Shear and Volume Deformations 363 Simple Harmonic Motion 367 The Period and Frequency for SHM 370 Graphical Analysis of SHM 374 The Pendulum 376 Damped Oscillations 380 Forced Oscillations and Resonance 380

Chapter 11 Waves 392 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10

Waves and Energy Transport 393 Transverse and Longitudinal Waves 395 Speed of Transverse Waves on a String 397 Periodic Waves 398 Mathematical Description of a Wave 400 Graphing Waves 401 Principle of Superposition 403 Reflection and Refraction 404 Interference and Diffraction 406 Standing Waves 409

Chapter 12 Sound 420 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9

Sound Waves 421 The Speed of Sound Waves 423 Amplitude and Intensity of Sound Waves 425 Standing Sound Waves 429 Timbre 433 The Human Ear 435 Beats 437 The Doppler Effect 439 Echolocation and Medical Imaging 443

Review & Synthesis: Chapters 9–12

453

PART TWO Thermal Physics Chapter 13 Temperature and the Ideal Gas 457 13.1 13.2 13.3

Temperature and Thermal Equilibrium 458 Temperature Scales 459 Thermal Expansion of Solids and Liquids 460

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vii

CONTENTS

13.4 13.5 13.6 13.7 13.8

Molecular Picture of a Gas 464 Absolute Temperature and the Ideal Gas Law 466 Kinetic Theory of the Ideal Gas 471 Temperature and Reaction Rates 475 Diffusion 477

Chapter 14 Heat 489 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8

Internal Energy 490 Heat 492 Heat Capacity and Specific Heat 494 Specific Heat of Ideal Gases 498 Phase Transitions 500 Thermal Conduction 506 Thermal Convection 510 Thermal Radiation 511

Chapter 15 Thermodynamics 527 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9

The First Law of Thermodynamics 528 Thermodynamic Processes 530 Thermodynamic Processes for an Ideal Gas 533 Reversible and Irreversible Processes 536 Heat Engines 537 Refrigerators and Heat Pumps 540 Reversible Engines and Heat Pumps 542 Entropy 546 The Third Law of Thermodynamics 549

Review & Synthesis: Chapters 13–15

557

PART THREE Electromagnetism Chapter 16 Electric Forces and Fields 561 16.1 16.2 16.3 16.4 16.5 16.6 16.7

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Electric Charge 562 Electric Conductors and Insulators 565 Coulomb’s Law 570 The Electric Field 573 Motion of a Point Charge in a Uniform Electric Field 581 Conductors in Electrostatic Equilibrium 584 Gauss’s Law for Electric Fields 587

Chapter 17 Electric Potential 601 17.1 17.2 17.3 17.4 17.5 17.6 17.7

Electric Potential Energy 602 Electric Potential 605 The Relationship Between Electric Field and Potential 612 Conservation of Energy for Moving Charges 616 Capacitors 617 Dielectrics 621 Energy Stored in a Capacitor 626

Chapter 18 Electric Current and Circuits 640 18.1 18.2 18.3 18.4 18.5 18.6 18.7 18.8 18.9 18.10 18.11

Electric Current 641 Emf and Circuits 643 Microscopic View of Current in a Metal: The Free-Electron Model 645 Resistance and Resistivity 648 Kirchhoff’s Rules 654 Series and Parallel Circuits 655 Circuit Analysis Using Kirchhoff’s Rules 661 Power and Energy in Circuits 664 Measuring Currents and Voltages 666 RC Circuits 668 Electrical Safety 672

Review & Synthesis: Chapters 16–18

688

Chapter 19 Magnetic Forces and Fields 693 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.8 19.9 19.10

Magnetic Fields 694 Magnetic Force on a Point Charge 697 Charged Particle Moving Perpendicularly to a Uniform Magnetic Field 703 Motion of a Charged Particle in a Uniform Magnetic Field: General 707 ⃗ A Charged Particle in Crossed E ⃗ Fields 708 and B Magnetic Force on a Current-Carrying Wire 713 Torque on a Current Loop 715 Magnetic Field due to an Electric Current 718 Ampère’s Law 723 Magnetic Materials 725

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viii

CONTENTS

Chapter 20 Electromagnetic Induction 741 20.1 20.2 20.3 20.4 20.5 20.6 20.7 20.8 20.9 20.10

Chapter 23 Reflection and Refraction of Light 848 23.1

Motional Emf 742 Electric Generators 745 Faraday’s Law 748 Lenz’s Law 753 Back Emf in a Motor 756 Transformers 756 Eddy Currents 758 Induced Electric Fields 759 Inductance 761 LR Circuits 765

23.2 23.3 23.4 23.5 23.6 23.7 23.8 23.9

Chapter 21 Alternating Current 21.1 21.2 21.3 21.4 21.5 21.6 21.7

780

Sinusoidal Currents and Voltages; Resistors in ac Circuits 781 Electricity in the Home 783 Capacitors in ac Circuits 785 Inductors in ac Circuits 788 RLC Series Circuits 790 Resonance in an RLC Circuit 794 Converting ac to dc; Filters 796

Review & Synthesis: Chapters 19–21

Wavefronts, Rays, and Huygens’s Principle 849 The Reflection of Light 852 The Refraction of Light: Snell’s Law 853 Total Internal Reflection 858 Polarization by Reflection 864 The Formation of Images Through Reflection or Refraction 865 Plane Mirrors 867 Spherical Mirrors 869 Thin Lenses 876

807

Chapter 24 Optical Instruments 891 24.1 24.2 24.3 24.4 24.5 24.6 24.7

Lenses in Combination 892 Cameras 895 The Eye 898 Angular Magnification and the Simple Magnifier 903 Compound Microscopes 905 Telescopes 907 Aberrations of Lenses and Mirrors 911

Chapter 25 Interference and Diffraction 922

PART FOUR

25.1

Electromagnetic Waves and Optics Chapter 22 Electromagnetic Waves 811 22.1 22.2 22.3 22.4 22.5 22.6 22.7 22.8

Maxwell’s Equations and Electromagnetic Waves 812 Antennas 813 The Electromagnetic Spectrum 816 Speed of EM Waves in Vacuum and in Matter 821 Characteristics of Traveling Electromagnetic Waves in Vacuum 824 Energy Transport by EM Waves 827 Polarization 830 The Doppler Effect for EM Waves 838

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25.2 25.3 25.4 25.5 25.6 25.7 25.8 25.9 25.10

Constructive and Destructive Interference 923 The Michelson Interferometer 927 Thin Films 929 Young’s Double-Slit Experiment 935 Gratings 939 Diffraction and Huygens’s Principle 942 Diffraction by a Single Slit 945 Diffraction and the Resolution of Optical Instruments 947 X-Ray Diffraction 950 Holography 952

Review & Synthesis: Chapters 22–25

963

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ix

CONTENTS

Chapter 29 Nuclear Physics 1065

PART FIVE

29.1 29.2 29.3 29.4

Quantum and Particle Physics and Relativity Chapter 26 Relativity 26.1 26.2 26.3 26.4 26.5 26.6 26.7 26.8

967

Postulates of Relativity 968 Simultaneity and Ideal Observers 971 Time Dilation 974 Length Contraction 977 Velocities in Different Reference Frames Relativistic Momentum 981 Mass and Energy 983 Relativistic Kinetic Energy 985

29.5 29.6 29.7 29.8 979

Chapter 27 Early Quantum Physics and the Photon 997 27.1 27.2 27.3 27.4 27.5 27.6 27.7 27.8

Quantization 998 Blackbody Radiation 998 The Photoelectric Effect 1000 X-Ray Production 1005 Compton Scattering 1006 Spectroscopy and Early Models of the Atom 1009 The Bohr Model of the Hydrogen Atom; Atomic Energy Levels 1012 Pair Annihilation and Pair Production 1020

Chapter 28 Quantum Physics 28.1 28.2 28.3 28.4 28.5 28.6 28.7

28.8 28.9 28.10

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Nuclear Structure 1066 Binding Energy 1069 Radioactivity 1073 Radioactive Decay Rates and Half-Lives 1079 Biological Effects of Radiation 1085 Induced Nuclear Reactions 1090 Fission 1091 Fusion 1096

1030

The Wave-Particle Duality 1031 Matter Waves 1032 Electron Microscopes 1036 The Uncertainty Principle 1037 Wave Functions for a Confined Particle 1040 The Hydrogen Atom: Wave Functions and Quantum Numbers 1042 The Exclusion Principle; Electron Configurations for Atoms Other than Hydrogen 1044 Electron Energy Levels in a Solid 1048 Lasers 1049 Tunneling 1053

Chapter 30 Particle Physics 1105 30.1 30.2 30.3 30.4 30.5

Fundamental Particles 1106 Fundamental Interactions 1108 Unification 1111 Particle Accelerators 1113 Twenty-First-Century Particle Physics

Review & Synthesis: Chapters 26–30

1114

1118

Appendix A Mathematics Review A–1 Appendix B Table of Selected Nuclides A–15 Answers to Selected Questions and Problems AP–1 Credits

C–1

Index I–1

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List of Selected Applications Note: Within the Problems, (#) = Chapter Number; CQ = Conceptual Question; MC = Multiple-Choice Question; P = Problem; R&S = Review & Synthesis.

Biology/Life Science Number of cells in the body, Ex. 1.9, p. 14 Can the lion catch the buffalo? Sec. 2.3, p. 34 Gull dropping a clam Sec. 3.5, p. 73 Tensile forces in the body, Sec. 4.7, p. 111 Speed in centrifuge, Ex. 5.2, p. 150 The centrifuge, Sec. 5.7, p. 171 Energy conversion in animal jumping, Ex. 6.12, p. 210 Energy transformation in a jumping flea, Sec. 6.7, p. 210 Conditions for equilibrium in the human body, Sec. 8.5, p. 281 Flexor versus extensor muscles, Sec. 8.5, p. 281 Force to hold arm horizontal, Ex. 8.10, p. 281 Forces on the human spine during heavy lifting, Sec. 8.5, p. 283 Sphygmomanometer and blood pressure, Sec. 9.5, p. 327 Specific gravity measurements in medicine, Sec. 9.6, p. 329 Floating and sinking of fish and animals, Ex. 9.8, p. 331 Speed of blood flow, Ex. 9.9, p. 334 Plaque buildup and narrowed arteries, Sec. 9.8, p. 337 Narrowing arteries and high blood pressure, Sec. 9.9, p. 340 Arterial blockage, Ex. 9.12, p. 341 How insects can walk on the surface of a pond, Sec. 9.11, p. 343 Surface tension of alveoli in the lungs, Ex. 9.14, p. 344 Surfactant in the lungs, Sec. 9.11, p. 344 Tension and compression in bone, Ex. 10.2, p. 359 Osteoporosis, Sec. 10.3, p. 361 Size limitations on organisms, Sec. 10.3, p. 363 Comparison of walking speeds for various creatures, Ex. 10.10, p. 378 Sensitivity of the human ear, Sec. 11.1, p. 394 Sound waves from a songbird, Ex. 12.2, p. 425 Human ear, Sec. 12.6, p. 432 Echolocation of bats and dolphins, Sec. 12.10, p. 443 Ultrasound and ultrasonic imaging, Sec. 12.10, p. 444 Temperature conversion, Ex. 13.1, p. 460 Warm-blooded vs.cold-blooded animals, Ex. 13.8, Sec. 13.7, p. 476 Diffusion of O2 into the bloodstream, Sec. 13.8, Ex. 13.9, pp. 478-479 Using ice to protect buds from freezing, Sec. 14.5, p. 500 Temperature regulation in the human body, Sec. 14.7, p. 510 Thermal radiation, Sec. 14.8, p. 515 Thermal radiation from the human body, Ex. 14.15, p. 515 Electrolocation in fish, Sec. 16.4, p. 581 Electrocardiogram and electroencephalogram, Sec. 17.2, p. 612 Neuron capacitance, Ex. 17.11, p. 624 Defibrillator, Ex. 17.12, Sec. 18.11, pp. 627, 672 RC circuits in neurons, Sec. 18.10, p. 671 Defibrillator, Sec. 18.11, p. 672 Magnetotactic bacteria, Sec. 19.1, p. 697 Medical uses of cyclotrons, Sec. 19.3, p. 705 Electromagnetic blood flowmeter, Sec. 19.5, p. 710 Magnetic resonance imaging, Sec. 19.7, p. 723 Magnetoencephalography, Sec. 20.3, p. 753 Fluorescence, Sec. 22.3, p. 818 Thermograms of the human body, Sec. 22.3, p. 818 X-rays in medicine and dentistry, CAT scans, Sec. 22.3, p. 820 Navigation of bees, Sec. 22.7, p. 837 Endoscope, Sec. 23.4, p. 863 Kingfisher looking for prey, Ex. 23.4, p. 866 Human eye, Sec. 24.3, p. 898 Correcting myopia, Sec. 24.3, Ex. 24.4, pp. 900-901 Correcting hyperopia, Ex. 24.5, p. 901 Iridescent colors in butterfly wings, Sec. 25.3, p. 934 Resolution of the human eye, Sec. 25.8, p. 950 Positron emission tomography, Sec. 27.9, p. 1021 Electron microscopes, Sec. 28.3, p. 1036 Lasers in medicine, Sec. 28.9, p. 1052 Radiocarbon dating, Sec. 29.4, p. 1082 Dating archeological sites, Ex. 29.9, p. 1083

Biological effect of radiation, Sec. 29.5, p. 1085 Radioactive tracers in medical diagnosis, Sec. 29.5, p. 1088 Gamma knife radio surgery, Sec. 29.5, p. 1089 Radiation therapy, Sec. 29.5, p. 1089 Problems (1) P: 49, 51-52, 59-60, 66, 69, 87. (2) P: 58, 73. (4) P: 81, 123. (5) P: 11, 53, 54, 77, 83; R&S: 4, 18. (6) P: 50, 57, 73, 103, 108. (7) CQ: 8; P: 87. (8) CQ: 9-11, 16; P: 39-46, 85, 99, 105, 107, 112-113; MCAT: 6-17. (9) CQ: 7, 12, 14; P: 8, 13, 18, 20, 22, 26, 36, 38, 42, 55, 59, 63, 70, 82, 85-88, 100, 101. (10) P: 2, 6, 8, 11, 15, 38, 76, 86-88, 91, 96. (11) CQ: 2; P: 16. (12) CQ: 4, 5, 8; P: 1, 2, 11, 19, 42, 49-52, 56-58, 60, 67, 68; R&S: 4, 15. (13) P: 31, 51, 57-58, 77, 89, 90, 102-103, 105. (14) CQ: 3; P: 14-15, 20, 41-42, 46, 53-55, 58, 63-64, 68, 70-71, 76-80, 90-91, 93, 97. (15) P: 55, 60, 67, 75. (16) P: 26, 53, 75. (17) P: 36, 60, 72, 84-85, 102, 108, 107, 111, 114. (18) CQ: 11, 17, 19; P: 24, 25, 86, 95-96, 103, 118. (19) P: 41-42, 97, 106. (20) CQ: 8. (21) P: 69. (22) P: 23. (23) CQ: 20. (24) CQ: 10, 13, 14, 17, 18; MC: 4-5; P: 22-28, 39, 72, 75, 76. (25) CQ: 16; P: 10, 54, 57, 61, 71. (26) CQ: 5. (27) CQ: 3, 21; P: 59, 62, 63. (28) P: 14-15. (29) CQ: 9, 11-12, 14; MC: 10; P: 35, 36, 41, 46-48, 69, 75, 79. (30) R&S: 10-13.

Chemistry Collision between krypton atom and water molecule, Ex. 7.9, p. 242 Why reaction rates increase with temperature, Sec. 13.7, Ex. 13.8, p. 474 Polarization of charge in water, Sec. 16.1, p. 565 Current in neon signs and fluorescent lights, Sec. 18.1, p. 643 Spectroscopic analysis of elements, Sec. 27.6, p. 1010 Fluorescence, phosphorescence, and chemiluminescence, Sec. 27.7, p. 1018 Electronic configurations of atoms, Sec. 28.7, p. 1045 Understanding the periodic table, Sec. 28.7, p. 1046 Problems (7) P: 39. (13) CQ: 13-14; P: 27-30, 32-39, 76, 79, 86, 87. (14) P: 10. (16) P: 17, 92. (17) P: 2, 3, 49, 92. (18) MC: 1; P: 7; R&S: 11. (19) P: 28-32, 88, 96. (26) P: 43, 86. (27) P: 5, 6, 9-10, 31-44, 46-48, 51, 52, 64, 66, 68, 71-72, 74, 76, 78-79, 81-82, 85. (28) CQ: 12-18; MC: 4; P: 6, 11, 20, 30, 36, 40-41, 43-45, 47, 57, 64, 65. (29) P: 4-17, 19, 21-32, 37-40, 42, 51-57, 59-67. (30) R&S: 11-12, 16, 17, 27; MCAT: 1-2, 6-8.

Geology/Earth Science Angular speed of Earth, Ex. 5.1, p. 149 Hidden depths of an iceberg, Ex. 9.7, p. 330 Why ocean waves approach shore nearly head on, Sec. 11.8, p. 406 Resonance and damage caused by earthquakes, Sec. 11.10, p. 411 Ocean currents and global warming, Sec. 14.7, p. 511 Global climate change, Sec. 14.8, p. 516 Second law and evolution, Sec. 15.8, p. 548 Second law and the “energy crisis”, Sec. 15.8, p. 548 Electric potential energy in a thundercloud, Ex. 17.1, p. 604 Thunderclouds and lightning, Sec. 17.6, p. 624 Earth’s magnetic field, Sec. 19.1, p. 696 Deflection of cosmic rays, Ex. 19.1, p. 700 Magnetic force on an ion in the air, Ex. 19.2, p. 701 Intensity of sunlight reaching the Earth, Ex. 22.6, p. 829 Colors of the sky during the day and at sunset, Sec. 22.7, p. 836 Rainbows, Sec. 23.3, p. 858 Cosmic rays, Ex. 26.2, 26.4, pp. 979-982 Radioactive dating of geologic formations, Sec. 29.4, p. 1084 Neutron activation analysis, Sec. 29.6, p. 1095 Angular momentum of hurricanes and pulsars, Sec. 8.8, p. 291 Problems (1) P: 55. (5) P: 70. (8) CQ 21. (9) CQ: 9; P: 46, 67, 80, 98-99. (11) CQ: 9; P: 61-62, 66, 68-70. (12) P: 4, 5, 46-48. (13) P: 49. (14) CQ: 4, 6; P: 89, 92, 94. (15) MCAT: 2-3. (16) P: 61, 75, 80. (17) CQ: 19; P: 66, 77, 86, 87. (18) P: 125, 126. (22) CQ: 6, 7, 15; P: 53, 54, 63. (29) CQ: 6; P: 70.

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Astronomy/Space Science Mars Climate Orbiter failure, Sec. 1.5, p. 9 Orbiting satellite, Ex. 4.5, p. 98 Circular orbits, Sec. 5.4, p. 160 Speed of Hubble Telescope orbiting Earth, Ex. 5.8, p. 161 Geostationary orbits, Sec. 5.4, p. 162 Kepler’s laws of planetary motion, Sec. s 5.4, 8.8, pp. 162, 292 Orbit of geostationary satellite, Ex. 5.9, p. 163 Orbiting satellites, Ex. 5.10, p. 164 Apparent weightlessness of orbiting astronauts, Sec. 5.7, p. 170 Artificial gravity and the human body, Sec. 5.7, p. 171 Elliptical orbits, Sec. 6.2, p. 191 Escape speed from Earth, Ex. 6.8, p. 204 Orbital speed of Mercury, Ex. 6.7, p. 204 Center of mass of a binary star system, Ex. 7.7, p. 239 Motion of an exploding model rocket, Ex. 7.8, p. 241 Orbital speed of Earth, Ex. 8.15, p. 292 Composition of planetary atmospheres, Sec. 13.6, p. 475 Temperature of the Sun, Ex. 14.13, p. 514 Aurorae on Earth, Jupiter, and Saturn, Sec. 19.4, p. 708 Cosmic microwave background radiation, Sec. 22.3, p. 819 Light from a supernova, Ex. 22.2, p. 822 Doppler radar and the expanding universe, Sec. 22.8, p. 839 Telescopes, Sec. 24.6, p. 907 Hubble Space Telescope, Sec. 24.6, p. 910 Radio telescopes, Sec. 24.7, p. 911 Observing active galactic nuclei, Sec. 26.2, p. 973 Aging of astronauts during space voyages, Ex. 26.1, p. 976 Nuclear fusion in stars, Sec. 29.8, p. 1097 Problems (1) P: 9, 58, 78, 85. (2) P: 59, 66, 69. (4) MC 12; P 42, 44-51, 132-135, 149. (5) R&S: 11, 16, 25, 36, 37. (6) P: 19, 37-38, 40-46, 81, 98. (8) CQ: 17; P: 7, 70; R&S: 14, 83, 94, 104. (9) CQ 5. (10) P: 21. (11) P: 1, 5. (13) P: 71. (14) MC: 1-3. (15) R&S: 3, 9. (16) P: 91. (19) P: 17, 102. (21) R&S: 5. (22) P: 16, 34, 36-37, 39, 40, 44, 56-57, 64, 78. (24) CQ: 5, 12; MC: 6; P: 45-50, 63, 71, 63, 73, 79. (25) CQ: 3-4; P: 52, 55, 63, 73, 79; R&S: 16, 22; MCAT: 3-6. (26) CQ: 8, 12; MC: 3-4; P: 3, 5, 8-9, 13-19, 22, 35, 40-41, 46, 61, 65-66, 68-69, 72, 74, 78-79, 82, 84. (27) CQ: 5; P: 58. (29) P: 3. (30) P: 22.

Architecture Cantilever building construction, Sec. 8.4, p. 275 Strength of building materials, Sec. 10.3, p. 361 Vibration of bridges and buildings, Sec. 10.10, p. 381 Expansion joints in bridges and buildings, Sec. 13.3, p. 461 Heat transfer through window glass, Ex. 14.10, 14.11, pp. 508-509 Building heating systems, Sec. 14.7, p. 510 Problems (9) CQ: 4. (10) CQ: 5, 12; P: 1, 17, 89. (13) P: 12, 19, 84. (14) P: 62, 72, 81. (15) CQ: 12; R&S: 10.

Technology/Machines Catapults and projectile motion, Ex. 3.7, p. 70 Advantages of a pulley, Sec. 4.7, p. 112 Mercury manometer, Ex. 9.5, p. 225 Products to protect the human body from injury, Ex. 7.2, p. 229 Safety features in a modern car, Sec. 7.3, p. 230 Recoil of a rifle, Sec. 7.4, p. 236 Atwood’s machine, Ex. 8.2, p. 265 Angular momentum of a gyroscope, Sec. 8.9, p. 294 Hydraulic lifts, brakes, and controls, Sec. 9.3, p. 320 Hydraulic lift, Ex. 9.2, p. 321 Hot air balloons, Sec. 9.6, p. 331 Venturi meter, Ex. 9.11, p. 337 Sedimentation velocity and the centrifuge, Sec. 9.10, p. 343 Operation of sonar and radar, Sec. 12.10, p. 444 Bimetallic strip in a thermostat, Sec. 13.3, p. 462 Volume expansion in thermometers, Sec. 13.3, p. 464 Heat engines, Sec. 15.5, p. 538 Internal combustion engine, Sec. 15.5, p. 538 Refrigerators and heat pumps, Sec. 15.6, p. 540 Photocopiers and laser printers, Sec. 16.2, p. 569 Cathode ray tube, Ex. 16.8, p. 582

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Oscilloscope, Sec. 16.5, p. 582 Electrostatic shielding, Sec. 16.6, p. 585 Electrostatic precipitator, Sec. 16.6, p. 586 Lightning rods, Sec. 16.6, p. 586 Battery-powered lantern, Ex. 17.3, p. 607 van de Graaff generator, Sec. 17.2, p. 610 Transmission of nerve impulses, Sec. 17.2, p. 611 Computer keyboard, Ex. 17.9, p. 619 Camera flash attachments, Sec. 17.5 18.10, pp. 620-670 Condenser microphone, Sec. 17.5, p. 620 Oscilloscopes, Sec. 17.5, p. 620 Random-access memory (RAM) chips, Sec. 17.5, p. 620 Electron drift velocity in household wiring, Ex. 18.2, p. 647 Resistance thermometer, Sec. 18.4, p. 651 Battery connection in a flashlight, Sec. 18.6, p. 656 Resistive heating, Sec. 18.10, p. 665 Camera flash, Sec. 18.10, p. 670 Electric fence, Sec. 18.11, p. 672 Household wiring, Sec. 18.11, 21.2, pp. 673, 787 Magnetic compass, Sec. 19.1, p. 694 Bubble chamber, Sec. 19.3, p. 703 Mass spectrometer, Sec. 19.3, p. 704 Cyclotrons, Ex. 19.5, p. 706 Velocity selector, Sec. 19.5, p. 709 The Hall effect, Sec. 19.5, p. 711 Electric motor, Sec. 19.7, p. 716 Galvanometer, Sec. 19.7, p. 717 Audio speakers, Sec. 19.7, p. 718 Electromagnets, Sec. 19.10, p. 726 Magnetic storage, Sec. 19.10, p. 727 Electric generators, Sec. 20.2, p. 745 DC generator, Sec. 20.2, p. 747 Ground fault interrupter, Sec. 20.3, p. 752 Moving coil microphone, Sec. 20.3, p. 752 Back emf in a motor, Sec. 20.5, p. 756 Transformers, Sec. 20.6, p. 756 Distribution of electricity, Sec. 20.6, p. 758 Eddy-current braking, Sec. 20.7, p. 759 Induction stove, Sec. 20.7, p. 759 Radio tuning circuit, Ex. 21.3, p. 790 Laptop power supply, Ex. 21.5, p. 793 Tuning circuits, Sec. 21.6, p. 795 Radio tuner, Ex. 21.6, p. 796 Rectifiers, Sec. 21.7, p. 796 Crossover networks, Sec. 21.7, p. 798 Electric dipole antenna, Ex. 21.1, p. 815 Microwave ovens, Sec. 22.3, p. 819 Liquid crystal displays, Sec. 22.7, p. 834 Radar guns, Ex. 22.9, p. 838 Periscope, Sec. 23.4, p. 861 Fiber optics, Sec. 23.4, p. 862 Zoom lens, Ex. 23.9, p. 879 Cameras, Sec. 24.2, p. 895 Microscopes, Sec. 24.5, p. 905 Reading a compact disk (CD), Sec. 25.1, p. 927 Michelson interferometer, Sec. 25.2, p. 927 Interference microscope, Sec. 25.2, p. 929 Antireflective coating, Sec. 25.3, p. 933 CD tracking, Sec. 25.5, p. 940 Spectroscopy, Sec. 25.5, p. 941 Diffraction and photolithography, Ex. 25.7, p. 943 Resolution of a laser printer, Ex. 25.9, p. 949 X-ray diffraction, Sec. 25.9, pp. 950-952 Holography, Sec. 25.10, p. 952 Photocells for sound tracks, burglar alarms, garage door openers, Sec. 27.3, p. 1005 Diagnostic x-rays in medicine, Ex. 27.4, p. 1006 Lasers, Sec. 28.9, p. 1034 Quantum corral, Sec. 28.5, p. 1041 Scanning tunneling microscope, Sec. 28.10, p. 1054 Atomic clock, Sec. 28.10, p. 1056 Nuclear fission reactors, Sec. 29.7, p. 1094 Fusion reactors, Sec. 29.8, p. 1098

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Technology/Machines (continued) High-energy particle accelerators, Sec. 30.4, p. 1113 Problems (4) P: 55. (5) P: 55, 68-69, 71, 85; R&S: 33. (6) P: 6, 25. (8) P: 6, 10, 12, 23, 25, 36, 48, 49, 56, 71, 84, 88, 96; R&S: 27. (10) CQ: 7; P: 28, 41, 78. (12) P: 13. (16) CQ: 6; P: 77, 87. (18) P: 4-5, 12, 105; R&S: 7, 16, 18. (19) CQ: 5, 13, 16; P: 15, 53-54, 90, 103, 104, 110. (20) CQ: 7, 10, 19; MC: 9; P: 28-30, 38, 44-49, 76, 79, 81, 91. (21) P: 3-10, 40, 51. (22) CQ: 1-2, 8-9, 16; MC: 1, 7-8; P: 5, 13, 43, 66, 67, 73. (23) CQ: 19; P: 69, 96. (24) CQ: 1, 6, 7, 16; MC: 1-2, 7, 10; P: 9-21, 29-38, 4-44, 54. (25) CQ: 6-11; MC: 6; P: 1-3, 18-19, 41, 42, 45, 56, 62, 77-78. (26) P: 24, 62. (27) CQ: 20; P: 14-20, 65, 69, 77. (28) CQ: 9-11, 13; P: 22, 48-53. (29) CQ: 13; P: 7. (30) P: 10, 12-13, 17, 26.

Transportation Acceleration of a sports car, Ex. 2.5, p. 35 Braking a car, Practice Prob. 2.5, p. 36 Relative velocities for pilots and sailors, Sec. 3.6, p. 74 Airplane flight in a wind, Ex. 3.10, p. 75 Length of runway for airplane takeoff, Ex. 4.17, p. 120 Angular speed of a motorcycle wheel, Ex. 5.3, p. 152 Banked roadways, Sec. 5.3, p. 157 Banked and unbanked curves, Ex. 5.7, p. 158 Banking angle of an airplane, Sec. 5.3, p. 160 Circular motion of stunt pilot, Ex. 5.14, p. 171 Damage in a high-speed collision, Ex. 6.3, p. 196 Power of a car climbing a hill, Ex. 6.13, p. 212 Momentum of a moving car, Ex. 7.1, p. 228 Force acting on a car passenger in a crash, Ex. 7.3, p. 231 Jet, rockets, and airplane wings, Sec. 7.4, p. 236 Collision at a highway entry ramp, Ex. 7.10, p. 245 Torque on a spinning bicycle wheel, Ex. 8.3, p. 268 How a ship can float, Sec. 9.6, p. 329 Airplane wings and lift, Sec. 9.8, p. 338 Shock absorbers in a car, Sec. 10.9, p. 380 Shock wave of a supersonic plane, Sec. 12.9, p. 443 Air temperature in car tires, Ex. 13.5, p. 469 Efficiency of an automobile engine, Ex. 15.7, p. 544 Starting a car using flashlight batteries, Ex. 18.5, p. 653 Regenerative braking, Sec. 20.6, p. 746 Bicycle generator, Ex. 20.2, p. 747 Problems (2) P: 26, 34, 37, 57, 72. (3) P: 68-72, 78, 81, 83, 86, 91, 95. (4) P: 152. (5) P: 9, 19-21, 24-28, 41, 43, 51, 81; R&S: 6-7, 26-27. (6) P: 4-5, 10, 18, 22, 32, 70-71, 80, 91. (7) P: 71, 86. (8) CQ: 6; P: 93; MCAT: 5. (9) CQ: 11, 16; P: 9-11, 28, 48, 96. (10) CQ: 16; P: 24, 39-40, 45; P: 70, 74. (12) P: 14. (13) P: 8-9, 24, 41-42, 91, 101. (14) CQ: 9, 10. (15) P: 18; R&S: 21. (18) P: 8, 10-11. (20) MC: 10; P: 80, 88-89. (23) CQ: 11, P: 51, 77. (26) P: 11.

Sports Velocity and acceleration of an inline skater, Ex. 3.5, p. 65 Rowing and current, Practice Problem 3.10, p. 75 Rowing across a river, Ex. 3.11, p. 75 The hammer throw, Ex. 5.5, p. 155 Bungee jumping, Ex. 6.4, p. 197 Rock climbers rappelling, Ex. 6.5, p. 199 Speed of a downhill skier, Ex. 6.6, p. 201 Work done in drawing a bow, Sec. 6.6, Ex. 6.9, p. 206 Energy in a dart gun, Ex. 6.11, p. 209 Elastic collision in a game of pool, Ex. 7.12, p. 248 Choking up on a baseball bat, Sec. 8.1, p. 263 Muscle forces for the iron cross (gymnastics), Sec. 8.5, p. 282 Rotational inertia of a figure skater, Sec. 8.8, p. 290 Pressure on a diver, Ex. 9.3, p. 323 Compressed air tanks for a scuba diver, Ex. 13.6, p. 470 Problems (1) P: 26. (2) P: 7, 8, 14, 56. (3) P: 70, 74, 77, 88. (4) P: 24, 111, 116. (5) P: 2, 5, 23, 67; R&S: 5, 8, 35, 38. (6) P: 12, 16, 31, 36, 47, 61,

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62, 68, 69, 75, 77-79, 86, 90, 94, 99. (7) CQ: 15, 17; P: 12, 16, 17, 24, 74, 75, 79, 81. (8) CQ: 7, 15, 19; MC: 9; P: 3, 8, 32-34, 53, 74, 75, 78, 79, 87, 109; R&S: 1, 7, 12, 18, 26. (9) CQ: 18; P: 75, 89. (10) CQ: 9, 10; P: 90. (11) P: 18. (12) P: 3. (14) P: 4, 6, 7.

Everyday Life Buying clothes, unit conversions, Ex. 1.5, p. 10 Snow shoveling, Ex. 4.3, p. 93 Hauling a crate up to a third-floor window, Ex. 4.16, p. 119 Circular motion of a CD, Ex. 5.4, p. 154 Speed of roller coaster car in vertical loop, Ex. 5.11, p. 166 Circular motion of a potter’s wheel, Ex. 5.13, p. 169 Antique chest delivery, Ex. 6.1, p. 192 Pulling a sled through snow, Ex. 6.2, p. 194 Getting down to nuts and bolts, Ex. 6.10, p. 207 Motion of a raft on a still lake, Ex. 7.5, p. 235 Automatic screen door closer, Ex. 8.4, p. 270 Work done on a potter’s wheel, Ex. 8.5, p. 272 Climbing a ladder on a slippery floor, Ex. 8.7, p. 276 Pushing a file cabinet so it doesn’t tip, Ex. 8.9, p. 279 Torque on a grinding wheel, Ex. 8.11, p. 285 Pressure exerted by high-heeled shoes, Ex. 9.1, p. 319 Cutting action of a pair of scissors, Ex. 10.4, p. 364 Difference between musical sound and noise, Sec. 11.4, p. 398 Sound of a horn in air and water, Ex. 11.5, p. 405 Sound from a guitar, Sec. 12.1, p. 421 Sound from a loudspeaker, Sec. 12.1, p. 421 Sound intensity of a jackhammer, Ex. 12.3, p. 427 Sound level of two lathes, Ex. 12.4, p. 428 Wind instruments, Sec. 12.4, p. 429 Tuning a piano, Sec. 12.7, Ex. 12.7, p. 438 Chill caused by perspiration, Sec. 14.5, p. 504 Double-paned windows and down jackets, Sec. 14.7, p. 510 Offshore and onshore breezes, Sec. 14.7, p. 510 Static charge from walking across a carpet, Ex. 16.1, p. 564 Grounding of fuel trucks, Sec. 16.2, p. 567 Electrostatic charge of adhesive tape, Sec. 16.2, p. 568 Resistance of an extension cord, Ex. 18.3, p. 650 Resistance heating, Sec. 21.1, p. 781 Polarized sunglasses, Sec. 22.7, p. 835 Colors from reflection and absorption of light, Sec. 23.1, p. 849 Mirages, Sec. 23.3, p. 857 Height needed for a full-length mirror, Ex. 23.5, p. 868 Cosmetic mirrors and automobile headlights, Sec. 23.8, p. 872 Side-view mirrors on cars, Ex. 23.7, p. 875 Colors in soap films, oil slicks, Sec. 25.3, p. 929 Neon signs and fluorescent lights, Sec. 27.6, p. 1009 Fluorescent dyes in laundry detergent, Sec. 27.7, p. 1018 Problems (1) P: 27. (4) P: 147. (5) P: 12, 65-66, 75; R&S: 3, 9, 10, 13, 15, 20, 22. (6) P: 7-9, 21, 26, 66, 67, 104, 107. (7) CQ: 1, 13: P: 1, 15, 31, 47, 78, 85. (8) CQ: 3, 12-14, 18; MC: 1, P: 11, 13-16, 18-19, 21, 26, 30, 32, 35, 37, 50, 54, 55, 68, 80, 92, 106, 110; R&S: 16. (9) CQ: 2, 13; MC: 2; P: 3, 5, 16, 21, 37, 41, 43-44, 49, 52, 56-58. (10) CQ: 2, 3; P: 1, 26, 37, 46, 73, 80. (11) CQ: 1-6; MC: 3-5, P: 2-4, 910, 15, 17, 36, 44, 46, 48, 50-59, 63-65, 67, 73, 77. (12) MC: 1-3, 910; P: 18, 20-28, 36-37, 40-45, 53, 55, 62-63, 69; R&S: 1-3, 6, 9, 1517. (13) CQ: 6, 8, 19, 20; P: 4, 6, 45-46, 78, 94, 107, 108. (14) CQ: 5, 11, 12, 17, 19, 22; MC: 5; P: 13, 24, 27-36, 43, 47, 56, 61-62, 65, 67, 69, 73, 81, 88, 100. (15) CQ: 1-2, 5-8, 11, 13; MC: 6; P: 24, 2931, 36, 39, 43-44, 56, 70, 77; R&S: 11, 17-19, 24. (16) CQ: 2, 12. (17) CQ: 3, 16; P: 67. (18) CQ: 1, 3, 9, 13, 18; P: 1, 29, 59, 59-62, 67, 70, 84, 87, 97-98, 110, 113-114, 117; R&S: 6, 21; MCAT: 2-13. (19) CQ: 9. (20) CQ: 10, 14, 17; MC: 5; P: 33, 71. (21) P: 1-2, 4, 68, 80, 82, 83. (22) P: 10, 13, 21, 24, 25, 55, 58-59. (23) CQ: 5, 8, 14, 22, 32-33, 38, 54, 68, 71, 80, 83, 85, 87. (25) CQ: 2; P: 7, 14-17. (27) P: 8, 61, 67.

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Preface Physics is intended for a two-semester college course in introductory physics using algebra and trigonometry. Our main goals in writing this book are • • •

to present the basic concepts of physics that students need to know for later courses and future careers, to emphasize that physics is a tool for understanding the real world, and to teach transferable problem-solving skills that students can use throughout their lives.

We have kept these goals in mind while developing the main themes of the book.

NEW TO THIS EDITION Although the fundamental philosophy of the book has not changed, detailed feedback from almost 60 reviewers (many of whom used the first edition in the classroom) has enabled us to fine-tune our approach to make the text even more user-friendly, conceptually based, and relevant for students. The second edition also has some added features to further facilitate student learning. A greater emphasis has been placed on fundamental physics concepts: •

•

•

•

Connections identify areas in each chapter where important concepts are revisited. A marginal Connections heading and summary adjacent to the coverage in the main text help students easily recognize that a previously introduced concept is being applied to the current discussion. Knowledge is being revisited and further developed—not newly introduced. Checkpoint questions have been added to applicable sections of the text to allow students to pause and test their understanding of the concept explored within the current section. The answers to the Checkpoints are found at the end of the chapter so that students can confirm their knowledge without jumping too quickly to the provided answer. The exercises in the Review & Synthesis sections have been revised to concentrate even more heavily on helping students to realize through practice problems how the concepts in the previously covered group of chapters are interrelated. The number of problems in the Review & Synthesis sections has also been increased in the new edition. (The MCAT review problems have been retained to also help premed students focus on the concepts covered in the upcoming exam.) Nonessential coverage and derivations have been moved to the text’s website. This will help students not only to focus further on the fundamental, core concepts in their reading of the text but also allow them to go online for additional information or explanation on topics of interest. identifiers in the text direct students to additional information online.

“G/R/R is as good as it gets as far as a college textbook in physics goes. One of the coauthors of this book has been teaching a course at this level for 30 years. This book is a direct result of her 30 years’ worth of personal experience, and there is no better substitute for that. It is, without any doubt, one of the best of its kind.” Dr. Abu Fasihuddin, University of Connecticut

In addition, the following general revisions occur in chapters of the text: •

• • •

The topical question from the chapter-opening vignette now appears in the margin (along with a reduced version of the chapter-opening image) to help students identify where in the main text the answer to the chapter-opening question is addressed. Applications have been clearly identified as such in the text with a complete listing in the front matter. Many helpful subheadings have been added to the text to help students quickly identify new subtopics. Portions of the text now caption images to establish a visual connection between the text’s concepts and terms and the art and photos. xiii

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Great care was taken by both the authors and the contributors to the second edition to revise the end-of-chapter and Review & Synthesis problems. Approximately 150 problems are new, and an emphasis has been placed on progressing difficulty level to help students gain confidence and reinforce new skills before tackling more challenging problems. The following lists major chapter-specific revisions to the text:

Chapter 2: Vector notation has been removed from Chapter 2. Discussion of vectors and components of vectors now begins in Chapter 3. Chapter 3: A discussion of Unit Vectors has been added to Section 3.2. A new example for finding average velocity has been added. Chapter 4: A more concise section on air resistance is provided with a more detailed discussion available online. A new Figure 4.20 emphasizes the normal and frictional forces as perpendicular components of a contact force. Chapter 7: Section 7.6 Motion of the Center of Mass has been simplified. Chapter 8: Example 8.1 has been replaced with a new problem on the rotational inertia of a barbell. Chapter 10: Section 10.8 The Pendulum has been made much more concise with a more detailed discussion of the physical pendulum available online. Chapter 11: A new “law box” highlights the physical properties that determine wave speed. The discussion on interference has been expanded for added clarity. Chapter 12: In Section 12.9, the discussion of shock waves has been shortened. A more detailed discussion is available online. Chapter 14: A detailed discussion of convection and Example 14.12 Roller Blading in Still Air have been moved online. Section 14.7 is now a brief, conceptual description of convection. Section 14.8 Thermal Radiation has been revised with a clearer description of solar radiation and global warming. Chapter 15: Section 15.5 Heat Engines has been revised to include a more accurate description of the development of the steam engine. The process of the internal combustion engine is now illustrated in Figure 15.12. Details of the Carnot cycle and discussion of the statistical interpretation of entropy are available online. Chapter 16: A new Example 16.7 Electric Field due to Three Point Charges has been added. Chapter 22: Section 22.1 has been simplified and is now titled Maxwell’s Equations and Electromagnetic Waves. A more detailed discussion appears online. The material on antennas has been made more concise. Chapter 27: The derivation of the radii of the Bohr orbits has been moved online. The section on atomic energy levels has been revised and made more concise. Chapter 28: Section 28.8 Electron Energy Levels in a Solid has been made much more concise with a more detailed discussion available online. Chapter 30: The discussions of quarks and leptons have been expanded and clarified. The discussion of the standard model is significantly more concise. Twenty-firstcentury particle physics has been updated, and the most recent information will be provided online. Please see your McGraw-Hill sales representative for a more detailed list of revisions.

ORGANIZATION OF CHAPTERS 2 THROUGH 4 In spite of the more traditional organization, Chapters 2–4 retain much of the flavor of the approach in College Physics. In particular, we use correct vector notation, diagrams, terminology, and methods from the very beginning. For example, we carefully distinguish components from magnitudes by writing “vx = −5 m/s” and never “v = −5 m/s,” even if the object moves only along the x-axis.

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COMPREHENSIVE COVERAGE Students should be able to get the whole story from the book. The text works well in our self-paced course, where students must rely on the textbook as their primary learning resource. Nonetheless, completeness and clarity are equally advantageous when the book is used in a more traditional classroom setting. Physics frees the instructor from having to try to “cover” everything. The instructor can then tailor class time to more important student needs—reinforcing difficult concepts, working through example problems, engaging the students in cooperative learning activities, describing applications, or presenting demonstrations.

INTEGRATING CONCEPTUAL PHYSICS INTO A QUANTITATIVE COURSE Some students approach introductory physics with the idea that physics is just the memorization of a long list of equations and the ability to plug numbers into those equations. We want to help students see that a relatively small number of basic physics concepts are applied to a wide variety of situations. Physics education research has shown that students do not automatically acquire conceptual understanding; the concepts must be explained and the students given a chance to grapple with them. Our presentation, based on years of teaching this course, blends conceptual understanding with analytical skills. The Conceptual Examples and Conceptual Practice Problems in the text and a variety of Conceptual and Multiple-Choice Questions at the end of each chapter give students a chance to check and to enhance their conceptual understanding.

“Conceptual ideas are important, ideas must be motivated, physics should be integrated, a coherent problem-solving approach should be developed. I’m not sure other books are as explicit in these goals, or achieve them as well as Giambattista, Richardson, and Richardson.” Dr. Michael G. Strauss, University of Oklahoma

INTRODUCING CONCEPTS INTUITIVELY We introduce key concepts and quantities in an informal way by establishing why the quantity is needed, why it is useful, and why it needs a precise definition. Then we make a transition from the informal, intuitive idea to a formal definition and name. Concepts motivated in this way are easier for students to grasp and remember than are concepts introduced by seemingly arbitrary, formal definitions. For example, in Chapter 8, the idea of rotational inertia emerges in a natural way from the concept of rotational kinetic energy. Students can understand that a rotating rigid body has kinetic energy due to the motion of its particles. We discuss why it is useful to be able to write this kinetic energy in terms of a single quantity common to all the particles (the angular speed), rather than as a sum involving particles with many different speeds. When students understand why rotational inertia is defined the way it is, they are better prepared to move on to the concepts of torque and angular momentum. We avoid presenting definitions or formulas without any motivation. When an equation is not derived in the text, we at least describe where the equation comes from or give a plausibility argument. For example, Section 9.9 introduces Poiseuille’s law with two identical pipes in series to show why the volume flow rate must be proportional to the pressure drop per unit length. Then we discuss why ΔV/Δt is proportional to the fourth power of the radius (rather than to r 2, as it would be for an ideal fluid).

“The authors are clearly very able to communicate in written English. The text is well written, not concise to the point of density, but not discursive to the point of longwindedness. A real pleasure to read.” Dr. Galen T. Pickett, California State University, Long Beach

WRITTEN IN CLEAR AND FRIENDLY STYLE We have kept the writing down-to-earth and conversational in tone—the kind of language an experienced teacher uses when sitting at a table working one-on-one with a student. We hope students will find the book pleasant to read, informative, and accurate without seeming threatening, and filled with analogies that make abstract concepts easier to grasp. We want students to feel confident that they can learn by studying the textbook.

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While learning correct physics terminology is essential, we avoid all unnecessary jargon—terminology that just gets in the way of the student’s understanding. For example, we never use the term centripetal force, since its use sometimes leads students to add a spurious “centripetal force” to their free-body diagrams. Likewise, we use radial component of acceleration because it is less likely to introduce or reinforce misconceptions than centripetal acceleration.

ACCURACY ASSURANCE The authors and the publisher acknowledge the fact that inaccuracies can be a source of frustration for both the instructor and students. Therefore, throughout the writing and production of this edition, we have worked diligently to eliminate errors and inaccuracies. Bill Fellers of Fellers Math & Science conducted an independent accuracy check and worked all end-of-chapter questions and problems in the final draft of the manuscript. He then coordinated the resolution of discrepancies between accuracy checks, ensuring the accuracy of the text, the end-of-book answers, and the solutions manuals. Corrections were then made to the manuscript before it was typeset. The page proofs of the text were double-proofread against the manuscript to ensure the correction of any errors introduced when the manuscript was typeset. The textual examples, practice problems and solutions, end-of-chapter questions and problems, and problem answers were accuracy checked by Fellers Math & Science again at the page proof stage after the manuscript was typeset. This last round of corrections was then cross-checked against the solutions manuals.

PROVIDING STUDENTS WITH THE TOOLS THEY NEED Problem-Solving Approach Problem-solving skills are central to an introductory physics course. We illustrate these skills in the example problems. Lists of problem-solving strategies are sometimes useful; we provide such strategies when appropriate. However, the most elusive skills— perhaps the most important ones—are subtle points that defy being put into a neat list. To develop real problem-solving expertise, students must learn how to think critically and analytically. Problem solving is a multidimensional, complex process; an algorithmic approach is not adequate to instill real problem-solving skills. “The major strength of this text is its approach, which makes students think out the problems, rather than always relying on a formula to get an answer. The way the authors encourage students to investigate whether the answer makes sense, and compare the magnitude of the answer with common sense is good also.” Dr. Jose D’Arruda, University of North Carolina, Pembroke

Strategy We begin each example with a discussion—in language that the students can understand—of the strategy to be used in solving the problem. The strategy illustrates the kind of analytical thinking students must do when attacking a problem: How do I decide what approach to use? What laws of physics apply to the problem and which of them are useful in this solution? What clues are given in the statement of the question? What information is implied rather than stated outright? If there are several valid approaches, how do I determine which is the most efficient? What assumptions can I make? What kind of sketch or graph might help me solve the problem? Is a simplification or approximation called for? If so, how can I tell if the simplification is valid? Can I make a preliminary estimate of the answer? Only after considering these questions can the student effectively solve the problem. Solution Next comes the detailed solution to the problem. Explanations are intermingled with equations and step-by-step calculations to help the student understand the approach used to solve the problem. We want the student to be able to follow the mathematics without wondering, “Where did that come from?” Discussion The numerical or algebraic answer is not the end of the problem; our examples end with a discussion. Students must learn how to determine whether their answer is consistent and reasonable by checking the order of magnitude of the answer,

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comparing the answer to a preliminary estimate, verifying the units, and doing an independent calculation when more than one approach is feasible. When there are several different approaches, the discussion looks at the advantages and disadvantages of each approach. We also discuss the implications of the answer—what can we learn from it? We look at special cases and look at “what if” scenarios. The discussion sometimes generalizes the problem-solving techniques used in the solution.

“I understood the math, mostly because it was worked out step-bystep, which I like.” Student, Bradley University

Practice Problem After each Example, a Practice Problem gives students a chance to gain experience using the same physics principles and problem-solving tools. By comparing their answers to those provided at the end of each chapter, they can gauge their understanding and decide whether to move on to the next section. Our many years of experience in teaching the college physics course in a one-onone setting has enabled us to anticipate where we can expect students to have difficulty. In addition to the consistent problem-solving approach, we offer several other means of assistance to the student throughout the text. A boxed problem-solving strategy gives detailed information on solving a particular type of problem, while an icon for problem-solving tips draws attention to techniques that can be used in a variety of contexts. A hint in a worked example or end-of-chapter problem provides a clue on what approach to use or what simplification to make. A warning icon emphasizes an explanation that clarifies a possible point of confusion or a common student misconception. An important problem-solving skill that many students lack is the ability to extract information from a graph or to sketch a graph without plotting individual data points. Graphs often help students visualize physical relationships more clearly than they can do with algebra alone. We emphasize the use of graphs and sketches in the text, in worked examples, and in the problems.

Review & Synthesis with MCAT Review® Eight Review & Synthesis sections appear throughout the text, following groups of related chapters. The MCAT ® Review includes actual reading passages and questions written for the Medical College Admission Test (MCAT). The Review Exercises are intended to serve as a bridge between textbook problems that are linked to a particular chapter and exam problems that are not. These exercises give students practice in formulating a problem-solving strategy without an external clue (section or chapter number) that indicates which concepts are involved. Many of the problems draw on material from more than one chapter to help the student integrate new concepts and skills with what has been learned previously.

“The warning signs about many of the misconceptions, traps, and common mistakes is a very helpful and novel idea. Those of us who have taught undergraduate students in service courses have spent considerable time on these. It is good to see them in a book.” Dr. H.R. Chandrasekhar, University of Missouri, Columbia

Using Approximation, Estimation, and Proportional Reasoning Physics is forthright about the constant use of simplified models and approximations in solving physics problems. One of the most difficult aspects of problem solving that students need to learn is that some kind of simplified model or approximation is usually required. We discuss how to know when it is reasonable to ignore friction, treat g as constant, ignore viscosity, treat a charged object as a point charge, or ignore diffraction. Some Examples and Problems require the student to make an estimate—a useful skill both in physics problem solving and in many other fields. Similarly, we teach proportional reasoning as not only an elegant shortcut but also as a means to understanding patterns. We frequently use percentages and ratios to give students practice in using and understanding them.

Showcasing an Innovative Art Program To help show that physics is more than a collection of principles that explain a set of contrived problems, in every chapter we have developed a system of illustration’s, ranging from simpler diagrams to ellaborate and beautiful illustrations, that brings to life the

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“I have tried a number of texts in this course over the past 30 years that I have taught Physics 116–117, and I can assure you that G/R/R is the one I (and the students . . .) like the best. The explanations are clear, and the graphics are excellent—the best I have seen anywhere. And the structure of the question and problem sets is very good. G/R/R is the best standard algebra-based text I have ever seen.” Dr. Carey E. Stronach, Virginia State University

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connections between physics concepts and the complex ways in which they are applied. We believe these illustrations, with subjects ranging from three-dimensional views of electric field lines to the biomechanics of the human body and from representations of waves to the distribution of electricity in the home, will help students see the power and beauty of physics.

Helping Students See the Relevance of Physics in Their Lives Students in an introductory college physics course have a wide range of backgrounds and interests. We stimulate interest in physics by relating the principles to applications relevant to students’ lives and in line with their interests. The text, examples, and end-of-chapter problems draw from the everyday world; from familiar technological applications; and from other fields such as biology, medicine, archaeology, astronomy, sports, environmental science, and geophysics. (Applications in the text are identified with a text heading or marginal note. An icon ( ) identifies applications in the biological or medical sciences.) The Physics at Home experiments give students an opportunity to explore and see physics principles operate in their everyday lives. These activities are chosen for their simplicity and for the effective demonstration of physics principles. Each Chapter Opener includes a photo and vignette, designed to capture student interest and maintain it throughout the chapter. The vignette describes the situation shown in the photo and asks the student to consider the relevant physics. A reduced version of the chapter opener photo and question marks where the topic from the vignette is addressed within the chapter.

Focusing on the Concepts To focus on the basic, core concepts of physics and reinforce for students that all of physics is based on a few, fundamental ideas, within chapters we have developed Connections to identify areas where important concepts are revisited. A marginal Connections heading and summary adjacent to the coverage in the main text help students easily recognize that a previously introduced concept is being applied to the current discussion. Knowledge is being built-up—not newly introduced. The exercises in the Review & Synthesis sections have been revised to increase the number of available exercises and to also concentrate even more heavily on helping students to realize through practice problems how the concepts in the previously covered group of chapters are interrelated. Checkpoint questions have been added to applicable sections of the text to allow students to pause and test their understanding of the concept explored within the current section. The answers to the Checkpoints are found at the end of the chapter so that students can confirm their knowledge without jumping too quickly to the provided answer. Applications are clearly identified as such in the text with a complete listing in the front matter. With Applications, students have the opportunity to see how physics concepts are experienced through their everyday lives. icons identify opportunities for students to access additional information or explanation of topics of interest online. This will help students to focus even further on just the very fundamental, core concepts in their reading of the text.

ADDITIONAL RESOURCES FOR INSTRUCTORS AND STUDENTS Online Homework and Resources McGraw-Hill’s Physics website offers online electronic homework along with a myriad of resources for both instructors and students. Instructors can create homework

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with easy-to-assign algorithmically generated problems from the text and the simplicity of automatic grading and reporting: • •

The end-of-chapter problems and Review & Synthesis exercises appear in the online homework system in diverse formats and with various tools. The online homework system incorporates new and exciting interactive tools and problem types: ranking problems, a graphing tool, a free-body diagram drawing tool, symbolic entry, a math palette, and multi-part problems.

Instructors also have access to PowerPoint lecture outlines, an Instructor’s Resource Guide with solutions, suggested demonstrations, electronic images from the text, clicker questions, quizzes, tutorials, interactive simulations, and many other resources directly tied to text-specific materials in Physics. Students have access to self-quizzing, interactive simulations, tutorials, selected solutions for the text’s problems, and more. See www.mhhe.com/grr to learn more and to register.

Electronic Media Integrated with the Text McGraw-Hill is proud to bring you an assortment of outstanding interactives and tutorials like no other. These activities offer a fresh and dynamic method to teach the physics basics by providing students with activities that work with real data. icons identify areas in the text where additional understanding can be gained through work with an interactive or tutorial on the text website. The interactives allow students to manipulate parameters and gain a better understanding of the more difficult physics concepts by watching the effect of these manipulations. Each interactive includes: • • •

Analysis tool (interactive model) Tutorial describing its function Content describing its principle themes

The text website contains accompanying interactive quizzes. An instructor’s guide for each interactive with a complete overview of the content and navigational tools, a quick demonstration description, further study with the textbook, and suggested end-of-chapter follow-up questions is also provided as an online instructor’s resource. The tutorials, developed and integrated by Raphael Littauer of Cornell University, provide the opportunity for students to approach a concept in steps. Detailed feedback is provided when students enter an incorrect response, which encourages students to further evaluate their responses and helps them progress through the problem.

Electronic Book Images and Assets for Instructors Build instructional materials wherever, whenever, and however you want! Accessed from the Physics website, an online digital library containing photos, artwork, interactives, and other media types can be used to create customized lectures, visually enhanced tests and quizzes, compelling course websites, or attractive printed support materials. Assets are copyrighted by McGraw-Hill Higher Education, but can be used by instructors for classroom purposes. The visual resources in this collection include •

Art Full-color digital files of all illustrations in the book can be readily incorporated into lecture presentations, exams, or custom-made classroom materials. In addition, all files are preinserted into PowerPoint slides for ease of lecture preparation.

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Active Art Library These key art pieces—formatted as PowerPoint slides—allow you to illustrate difficult concepts in a step-by-step manner. The artwork is broken into small, incremental pieces, so you can incorporate the illustrations into your lecture in whatever sequence or format you desire. Photos The photos collection contains digital files of photographs from the text, which can be reproduced for multiple classroom uses. Worked Example Library, Table Library, and Numbered Equations Library Access the worked examples, tables, and equations from the text in electronic format for inclusion in your classroom resources. Interactives Flash files of the physics interactives described earlier are included so that you can easily make use of the interactives in a lecture or classroom setting.

Also residing on your textbook’s website are • •

PowerPoint Lecture Outlines Ready-made presentations that combine art and lecture notes are provided for each chapter of the text. PowerPoint Slides For instructors who prefer to create their lectures from scratch, all illustrations and photos are preinserted by chapter into blank PowerPoint slides.

Computerized Test Bank Online A comprehensive bank of over 2000 test questions in multiple-choice format at a variety of difficulty levels is provided within a computerized test bank powered by McGrawHill’s flexible electronic testing program—EZ Test Online (www.eztestonline.com). EZ Test Online allows you to create paper and online tests or quizzes in this easy-to-use program! Imagine being able to create and access your test or quiz anywhere, at any time without installing the testing software. Now, with EZ Test Online, instructors can select questions from multiple McGraw-Hill test banks or create their own, and then either print the test for paper distribution or give it online. See www.mhhe.com/grr for more information.

Electronic Books If you or your students are ready for an alternative version of the traditional textbook, McGraw-Hill brings you innovative and inexpensive electronic textbooks. By purchasing E-books from McGraw-Hill, students can save as much as 50% on selected titles delivered on the most advanced E-book platforms available. E-books from McGraw-Hill are smart, interactive, searchable, and portable, with such powerful built-in tools as detailed searching, highlighting, note taking, and student-to-student or instructor-to-student note sharing. E-books from McGraw-Hill will help students to study smarter and quickly find the information they need. E-books also saves students money. Contact your McGraw-Hill sales representative to discuss E-book packaging options.

Personal Response Systems Personal response systems, or “clickers,” bring interactivity into the classroom or lecture hall. Wireless response systems give the instructor and students immediate feedback from the entire class. The wireless response pads are essentially remotes that are easy to use and engage students, allowing instructors to motivate student preparation, interactivity, and active learning. Instructors receive immediate feedback to gauge which concepts students understand. Questions covering the content of the Physics text (formatted in PowerPoint) are available on the website for Physics.

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Instructor’s Resource Guide The Instructor’s Resource Guide includes many unique assets for instructors, such as demonstrations, suggested reform ideas from physics education research, and ideas for incorporating just-in-time teaching techniques. It also includes answers to the end-ofchapter conceptual questions and complete, worked-out solutions for all the end-ofchapter problems from the text. The Instructors Resource Guide is available in the Instructor Resources on the text’s website.

ALEKS® Help students master the math skills needed to understand difficult physics problems. ALEKS® [Assessment and LEarning in Knowledge Spaces] is an artificial intelligence– based system for individualized math learning available via the World Wide Web. ALEKS® is • • • • • •

A robust course management system. It tells you exactly what your students know and don’t know. Focused and efficient. It enables students to quickly master the math needed for college physics. Artificial intelligence. It totally individualizes assessment and learning. Customizable. Click on or off each course topic. Web based. Use a standard browser for easy Internet access. Inexpensive. There are no setup fees or site license fees.

ALEKS® is a registered trademark of ALEKS Corporation.

Student Solutions Manual The Student Solutions Manual contains complete worked-out solutions to selected end-of-chapter problems and questions, selected Review & Synthesis problems, and the MCAT Review Exercises from the text. The solutions in this manual follow the problem-solving strategy outlined in the text’s examples and also guide students in creating diagrams for their own solutions. For more information, contact a McGraw-Hill customer service representative at (800) 338–3987, or by email at www.mhhe.com. To locate your sales representative, go to www.mhhe.com for Find My Sales Rep.

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To the Student HOW TO SUCCEED IN YOUR PHYSICS CLASS It’s true—how much you get out of your studies depends on how much you put in. Success in a physics class requires: • • • •

Commitment of time and perseverance Knowing and motivating yourself Getting organized Managing your time

This section will help you learn how to be effective in these areas, as well as offer guidance in: • • • •

Getting the most out of your lecture Finding extra help when you need it Getting the most out of your textbook How to study for an exam

Commitment of Time and Perseverance A good rule of thumb is to allow 2 hours of study time for every hour you spend in lecture. For instance, a 3-hour lecture deserves 6 hours of study time per week. If you commit to studying for this course daily, you’re investing a little less than one hour per day, including the weekend.

Learning and mastering takes time and patience. Nothing worthwhile comes easily. Be committed to your studies and you will reap the benefits in the long run. A regular, sustained effort is much more effective than sporadic bouts of cramming.

Begin each of the tasks assigned in your course with the goal of understanding the material. Simply completing the assignment does not mean that learning has taken place. Your fellow students, your instructor, and this textbook can all be important resources in broadening your knowledge.

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Knowing and Motivating Yourself What kind of learner are you? When are you most productive? Know yourself and your limits, and work within them. Know how to motivate yourself to give your all to your studies and achieve your goals. There are many types of learners, and no right or wrong way of learning. Which category do you fall into?

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Visual learner You respond best to “seeing” processes and information. Focus on text illustrations and graphs. Use course handouts and the animations on the course and text websites to help you. Draw diagrams in your notes to illustrate concepts. Auditory learner You work best by listening to—and possibly recording—the lecture and by talking information through with a study partner. Tactile/Kinesthetic Learner You learn best by being “hands on.” You’ll benefit by applying what you’ve learned during lab time. Writing and drawing are physical activities, so don’t neglect taking notes on your reading and the lecture to explain the content in your own words. Try pacing while you read the text. Stand up and write on a chalkboard during discussions in your study group.

Identify your own personal preferences for learning and seek out the resources that will best help you with your studies. Also remember, even though you have a preferred style of learning, most learners benefit when they engage in all styles of learning.

Getting Organized It’s simple, yet it’s fundamental. It seems the more organized you are, the easier things come. Take the time before your course begins to analyze your life and your study habits. Get organized now and you’ll find you have a little more time—and a lot less stress. •

Find a calendar system that works for you. The best kind is one that you can take with you everywhere. To be truly organized, you should integrate all aspects of your life into this one calendar—school, work, and leisure. Some people also find it helpful to have an additional monthly calendar posted by their desk for “at a

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glance” dates and to have a visual planner. If you do this, be sure you are consistently synchronizing both calendars so as not to miss anything. More tips for organizing your calendar can be found in the time management discussion below. By the same token, keep everything for your course or courses in one place—and at your fingertips. A three-ring binder works well because it allows you to add or organize handouts and notes from class in any order you prefer. Incorporating your own custom tabs helps you flip to exactly what you need at a moment’s notice. Find your space. Find a place that helps you be organized and focused. If it’s your desk in your dorm room or in your home, keep it clean. Clutter adds confusion and stress and wastes time. Perhaps your “space” is at the library. If that’s the case, keep a backpack or bag that’s fully stocked with what you might need—your text, binder or notes, pens, highlighters, Post-its, phone numbers of study partners. [Hint: A good place to keep phone numbers is in your “one place for everything calendar.”]

Add extra “padding” into your personal deadlines. If you have a report due on Friday, set a goal for yourself to have it done on Wednesday. Then, take time on Thursday to look over your project with a fresh eye. Make any corrections or enhancements and have it ready to turn in on Friday.

Managing Your Time Managing your time is the single most important thing you can do to help yourself, but it’s probably one of the most difficult tasks to successfully master. In college, you are expected to work much harder and to learn much more than you ever have before. To be successful you need to invest in your education with a commitment of time. We all lead busy lives, but we all make choices as to how we spend our time. Choose wisely. •

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Know yourself and when you’ll be able to study most efficiently. When are you most productive? Are you a night owl? Or an early bird? Plan to study when you are most alert and can have uninterrupted segments. This could include a quick 5minute review before class or a one-hour problem-solving study session with a friend. Create a set daily study time for yourself. Having a set schedule helps you commit to studying and helps you plan instead of cram. Find—and use—a planner that is small enough that you can take it with you everywhere. This may be a simple paper calendar or an electronic version. They all work on the same premise: organize all of your activities in one place. Schedule study time using shorter, focused blocks with small breaks. Doing this offers two benefits: (1) You will be less fatigued and gain more from your effort and (2) Studying will seem less overwhelming, and you will be less likely to procrastinate. Plan time for leisure, friends, exercise, and sleep. Studying should be your main focus, but you need to balance your time—and your life. Log your homework deadlines and exam dates in your personal calendar. Try to complete tasks ahead of schedule. This will give you a chance to carefully review your work before it is due. You’ll feel less stressed in the end. Know where help can be found. At the beginning of the semester, find your instructor’s office hours, your lab partner’s contact information, and the “Help Desk” or Learning Resource Center if your course offers one. Make use of all of the support systems that your college or university has to offer. Ask questions both in class and during your instructor’s office hours. Don’t be shy—your instructor is there to help you learn. Prioritize! In your calendar or planner, highlight or number key projects; do them first, and then cross them off when you’ve completed them. Give yourself a pat on the back for getting them done! Review your calendar and reprioritize daily. Resist distractions by setting and sticking to a designated study time. Multitask when possible. You may find a lot of extra time you didn’t think you had. Review material in your head or think about how to tackle a tough problem while walking to class or doing laundry.

Plan to study and plan for leisure. Being well balanced will help you focus when it is time to study.

Try combining social time with studying in a group, or social time with mealtime or exercise. Being a good student doesn’t mean you have to be a hermit. It does mean you need to know how to smartly budget your time.

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Your instructors want you to succeed. They put a lot of effort into preparing their lectures and other materials designed to help you learn. Attending class is one of the simplest, most valuable things you can do to help yourself. But it doesn’t end there—getting the most out of your lectures means being organized. Here’s how:

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Prepare Before You Go to Class Study the text on the lecture topic before attending class. Familiarizing yourself with the material gives you the ability to take notes selectively rather than scrambling to write everything down. You’ll be able to absorb more of the subtleties and difficult points from the lecture. You may also develop some good questions to ask your instructor. Don’t feel overwhelmed by this task. Spend time the night before class gaining a general overview of the topics for the next lecture using your syllabus. If your schedule does not allow this, plan to arrive at class 5–15 minutes before lecture. Bring your text with you and skim the chapter before lecture begins. Don’t try to read an entire chapter in one sitting; study one or two sections at a time. It’s difficult to maintain your concentration in a long session with so many new concepts and skills to learn. Be a Good Listener Most people think they are good listeners, but few really are. Are you? Important points to remember: • • •

You can’t listen if you are talking. You aren’t listening if you are daydreaming or constantly distracted by other concerns. Listening and comprehending are two different things. Listen carefully in class. The language of science is precise; be sure you understand your instructor. If you don’t understand something your instructor is saying, ask a question or jot a note and visit the instructor during office hours. You are likely doing others a favor when you ask questions because there are probably others in the class who have the same questions.

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Focus on main points and try to use an outline format to take notes to capture key ideas and organize sub-points. Take your text to lecture, and keep it open to the topics being discussed. You can also take brief notes in your textbook margin or reference textbook pages in your notebook to help you study later. Review and edit your notes shortly after class—within 24 hours—to make sure they make sense and that you’ve recorded core thoughts. You may also want to compare your notes with a study partner later to make sure neither of you have missed anything. This is a very IMPORTANT point: You can and should also add notes from your reading of the textbook.

Get a Study Partner Find a few study partners and get together regularly. Four or five study partners to a group is a good number. Too many students make the group unwieldy, but you want enough students to ensure the group can meet even if one or two people can’t make it. Having study partners has many benefits. First, they can help you keep your commitment to this class. By having set study dates, you can combine study and social time, and maybe even make it fun! In addition, you now have several minds to help digest the information from the lecture and the text: •

• •

• •

Talk through concepts and go over the difficulties you may be having. Take turns explaining things to each other. You learn a tremendous amount when you teach someone else. Compare your notes and solutions with the Practice Problems. Try a new approach to a problem or look at the problem from the perspective of your partner. There are often many ways to do the same problem. You can benefit from the insights of others—and they from you—but resist the temptation to simply copy solutions. You need to learn how to solve the problem yourself. Quiz each other and discuss some of the Conceptual Questions from the end of the chapter. Don’t take advantage of your study partner by skipping class or skipping study dates. You obviously won’t have a study partner—or a friend—much longer if it’s not a mutually beneficial arrangement!

Take Good Notes •

•

• •

Use a standard size notebook, or better yet, a three-ring binder with loose leaf notepaper. The binder will allow you to organize and integrate your notes and handouts, integrate easy-to-reference tabs, and the like. Color-code your notes. Use one color of ink pen to take your initial notes. You can annotate later using a pencil, which can be erased if need be. Start a new page with each lecture or note-taking session. Label each page with the date and a heading for each day.

Getting the Most Out of Your Textbook We hope that you enjoy your physics course using this text. While studying physics does require hard work, we have tried to remove the obstacles that sometimes make introductory physics unnecessarily difficult. We have also tried to reveal the beauty inherent in the principles of physics and how these principles are manifest all around you. In our years of teaching experience, we have found that studying physics is a skill that must be learned. It’s much more effective to study a physics textbook, which involves active participation on your part, than to read through

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passively. Even though active study takes more time initially, in the long run it will save you time; you learn more in one active study session than in three or four superficial readings. As you study, take particular note of the following elements: Consider the chapter opener. It will help you make the connection between the physics you are about to study and how it affects the world around you. Each chapter opener includes a photo and vignette designed to pique your interest in the chapter. The vignette describes the situation shown in the photo and asks you to consider the relevant physics. The question is then answered within the chapter. Look for the reduced opener photo and question on the referenced page.

Evaluate the Concepts & Skills to Review on the first page of each chapter. It lists important material from previous chapters that you should understand before you start reading. If you have problems recalling any of the concepts, you can revisit the sections referenced in the list.

CHAPTER

The Hubble Space Telescope, orbiting Earth at an altitude of about 600 km, was launched in 1990 by the crew of the Space Shuttle Discovery. What is the advantage of having a telescope in space when there are telescopes on Earth with larger lightgathering capabilities? What justifies the cost of $2 billion to place this 12.5-ton instrument into orbit? (See p. 910 for the answer.)

• • • • • •

Concepts & Skills to Review

24

Optical Instruments

distinction between real and virtual images (Section 23.6) magnification (Section 23.8) refraction (Section 23.3) thin lenses (Section 23.9) finding images with ray diagrams (Section 23.6) small-angle approximations (Appendix A.7)

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Force due to back muscles (Fb) Axis (sacrum)

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Force due to sacrum (Fs) 12°

Spine

Weight of upper body(mg)

38 cm

Pressure p variation p0

44 cm

t=0

x

F

(b)

Displacement of air elements

Rarefaction

Rarefaction

Study the figures and graphs carefully. Some elaborate illustrations and more straightforward diagrammatic illustrations are used in combination throughout the text to help you grasp concepts. Complex illustrations help you visualize the most difficult concepts. When looking at graphs, try to see the wealth of information displayed. Ask yourself about the physical meaning of the slope, the area under the curve, the overall shape of the graph, the vertical and horizontal intercepts, and any maxima and minima.

Compression

–p0

Compression

(a)

F

F

F

F

s t=0 s0

s

Right (+)

s

(c) Left (–)

Com pre ssi on Rar efa ctio n

–s0

s

s

s

x

Wavelength

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CONNECTION: Rotational and translational kinetic energies have the same form: _12 inertia × speed2.

Marginal Connections headings and summaries adjacent to the coverage in the main text identify areas where important concepts are revisited. Consider the notes carefully to help you recognize how a previously introduced concept is being applied to the current discussion. Checkpoint questions appear in applicable sections of the text to allow you to test your understanding of the concept explored within the current section. The answers to the Checkpoints are found at the end of the chapter so that you can confirm your knowledge without jumping too quickly to the provided answer.

CHECKPOINT 8.2 You are trying to loosen a nut, without success. Why might it help to switch to a wrench with a longer handle?

icons identify opportunities for you to access additional information or explanation of topics of interest online. Various Reinforcement Notes appear in the margin to emphasize the important points in the text.

Δr⃗ v⃗ = lim ___ Δt→0 Δt

(3-12)

(Δr⃗ is the displacement during a very short time interval Δt)

If an object moves along a curved path, the direction of the velocity vector at any point is tangent to the path at that point.

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Important Equations are numbered for easier reference. Equations that correspond to important laws are boxed for quick identification. Statements of important physics Rules and Laws are boxed to highlight the most important and central concepts.

The Law of Conservation of Energy The total energy in the universe is unchanged by any physical process: total energy before = total energy after.

Problem-Solving Strategy for Newton’s Second Law • • • •

Decide what objects will have Newton’s second law applied to them. Identify all the external forces acting on that object. Draw an FBD to show all the forces acting on the object. Choose a coordinate system. If the direction of the net force is known, choose axes so that the net force (and the acceleration) are along one of the axes. • Find the net force by adding the forces as vectors. • Use Newton’s second law to relate the net force to the acceleration. • Relate the acceleration to the change in the velocity vector during a time interval of interest.

A warning note describes possible points of confusion or any common misconceptions that may apply to a particular concept.

Boxed Problem-Solving Strategies give detailed information on solving a particular type of problem. These are supplied for the most fundamental physical rules and laws.

tor; the length of the arrow is proportional to the magnitude of the vector. By contrast, a scalar quantity can have magnitude, algebraic sign, and units, but not a direction in space. It wouldn’t make sense to draw an arrow to represent a scalar such as mass! In this book, an arrow over a boldface symbol indicates a vector quantity (r⃗). (Some books use boldface without the arrow or the arrow without boldface.) When writing by hand, always draw an arrow over a vector symbol to distinguish it from a scalar. When the symbol for a vector is written without the arrow and in italics rather than boldface (r), it stands for the magnitude of the vector (which is a scalar). Absolute value bars are also used to stand for the magnitude of a vector, so r = r⃗. The magnitude of a vector may have units and is never negative; it can be positive or zero. When scalars are added or subtracted, they do so in the usual way: 3 kg of water plus 2 kg of water is equal to 5 kg of water. Adding or subtracting vectors is different. Vectors follow rules of addition and subtraction that take into account the directions of the vectors as well as their magnitudes. Whenever you need to add or subtract quantities, check whether they are vectors. If so, be sure to add or subtract them correctly as vectors. Do not just add or subtract their magnitudes.

A problem-solving tip will guide you in applying problem-solving techniques.

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Example 6.4 Bungee Jumping A bungee jumper makes a jump in the Gorge du Verdon in southern France. The jumping platform is 182 m above the bottom of the gorge. The jumper weighs 780 N. If the jumper falls to within 68 m of the bottom of the gorge, how much work is done by the bungee cord on the jumper during his descent? Ignore air resistance. Strategy Ignoring air resistance, only two forces act on the jumper during the descent: gravity and the tension in the cord. Since the jumper has zero kinetic energy at both the highest and lowest points of the jump, the change in kinetic energy for the descent is zero. Therefore, the total work done by the two forces on the jumper must equal zero. Solution Let Wg and Wc represent the work done on the jumper by gravity and by the cord. Then Wtotal = Wg + Wc = ΔK = 0 The work done by gravity is

Then the work done by gravity is Wg = −(780 N) × (−114 m) = +89 kJ The work done by the cord is Wc = Wtotal − Wg = −89 kJ. Discussion The work done by gravity is positive, since the force and the displacement are in the same direction (downward). If not for the negative work done by the cord, the jumper would have a kinetic energy of 89 kJ after falling 114 m. The length of the bungee cord is not given, but it does not affect the answer. At first the jumper is in free fall as the cord plays out to its full length; only then does the cord begin to stretch and exert a force on the jumper, ultimately bringing him to rest again. Regardless of the length of the cord, the total work done by gravity and by the cord must be zero since the change in the jumper’s kinetic energy is zero.

Practice Problem 6.4 The Bungee Jumper’s Speed

Wg = Fy Δy = −mg Δy where the weight of the jumper is mg = 780 N. With y = 0 at the bottom of the gorge, the vertical component of the displacement is

Suppose that during the jumper’s descent, at a height of 111 m above the bottom of the gorge, the cord has done −21.7 kJ of work on the jumper. What is the jumper’s speed at that point?

When you come to an Example, pause after you’ve read the problem. Think about the strategy you would use to solve the problem. See if you can work through the problem on your own. Now study the Strategy, Solution, and Discussion in the textbook. Sometimes you will find that your own solution is right on the mark; if not, you can focus your attention on the areas of misunderstanding or any mistakes you may have made. Work the Practice Problem after each Example to practice applying the physics concepts and problem-solving skills you’ve just learned. Check your answer with the one given at the end of the chapter. If your answer isn’t correct, review the previous section in the textbook to try to find your mistake.

Δy = yf − yi = 68 m − 182 m = −114 m

CHECKPOINT 6.3 Kinetic energy and work are related. Can kinetic energy ever be negative? Can work ever be negative?

6.4

GRAVITATIONAL POTENTIAL ENERGY (1)

Gravitational Potential Energy When Gravitational Force Is Constant Toss a stone up with initial speed vi. Ignoring air resistance, how high does the stone go? We can solve this problem with Newton’s second law, but let’s use work and energy instead. The stone’s initial kinetic energy is Ki = _12 mv 2i . For an upward displacement Δy, gravity does negative work W grav = −mg Δy. No other forces act, so this is the total work done on the stone. The stone is momentarily at rest at the top, so K f = 0. Then

Application headings idenBanked Curves To help prevent cars from going into a skid or losing control, the tify places in the text where roadway is often banked (tilted at a slight angle) around curves so that the outer portion of the road—the part farthest from the center of curvature—is higher than the physics can be applied to ⃗ so inner portion. Banking changes the angle and magnitude of the normal force, N, other areas of your life. that it has a horizontal component Nx directed toward the center of curvature (in the Familiar topics and interests are discussed in the accompanying text, including examples from biology, archaeology, astronomy, sports, and the everyday world. The biology/life science examples have a special icon.

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Application of radial acceleration and contact forces: banked roadways

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Application of the 10/31/08 manometer: measuring blood pressure

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Try the Physics at Home experiments in your dorm room or at home. They reinforce key physics concepts and help you see how these concepts operate in the world around you.

PHYSICS AT HOME Drop a very tiny speck of dust or lint into a container of water and push the speck below the surface. The motion of the speck—called Brownian motion—is easily observed as it is pushed and bumped about randomly by collisions with water molecules. The water molecules themselves move about randomly, but at much higher speeds than the speck of dust due to their much smaller mass.

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Write your own chapter summary or outline, adding notes from class where appropriate, and Master the Concepts then compare it with the Master the Concepts equal in magnitude to the weight of the vol• Fluids are materials that flow and include both liquids ume of fluid displaced by the object: and gases. A liquid is nearly incompressible, whereas a provided at the end of the chapter. This will gas expands to fill its container. (9-7) F = rgV mg • Pressure is the perpendicular force per unit area that a help you identify the most important and funwhere V is the volume of the part of the F fluid exerts on any surface with which it comes in conobject that is submerged and r is the density damental concepts in each chapter. tact (P = F/A). The SI unit of pressure is the pascal of the fluid. (1 Pa = 1 N/m ). Along with working the problems assigned • In steady flow, the velocity of the fluid at any point is con• The average air pressure at sea level is 1 atm = 101.3 kPa. stant in time. In laminar flow, the fluid flows in neat layers • Pascal’s principle: A change in pressure at any point in by your instructor, try quizzing yourself on the so that each small portion of fluid that passes a particular a confined fluid is transmitted everywhere throughout point follows the same path as every other portion of fluid Multiple-Choice Questions. Check your the fluid. that passes the same point. The path that the fluid follows, • The average density of a substance is the ratio of its starting from any point, is called a streamline. Laminar answers against the answers at the end of the mass to its volume flow is steady. Turbulent flow is chaotic and unsteady. m book. Consider the Conceptual Questions to The viscous force opposes the flow of the fluid; it is the r = __ (9-2) V counterpart to the frictional force for solids. • The specific gravity of a material is the ratio of its dencheck your qualitative understanding of the key • An ideal fluid exhibits laminar flow, has no viscosity, sity to that of water at 4°C. and is incompressible. The flow of an ideal fluid is govideas from the chapter. Try writing some • Pressure variation with depth in a static fluid: erned by two principles: the continuity equation and P = P + rgd (9-3) Bernoulli’s equation. responses to practice your writing • The continuity equation states that the volume flow rate where point 2 is a depth d below point 1. skills and to help prepare for any for an ideal fluid is constant: • Instruments to measure pressure include the manometer 5.1 Description of Uniform Circular Motion and the barometer. The barometer measures the presΔV = A v = A v ___ essay problems on the exam. (9-12, 9-13) Δt sure of the atmosphere. The manometer measures a 1. A carnival swing is fixed on the end of an 8.0-m-long pressure difference. When working the Problems beam. If the swing and beam sweepx through an angle of and Comprehensive Problems 120°, what is the distance through which the riders assigned by your instructor, pay spemove? cial attention to the explanatory para2. A soccer ball of diameter 31 cm rolls without slipping graph below the Problem heading at a linear speed of 2.8 m/s. Through how many revolu- and the keys accompanying each tions has the soccer ball turned as it moves a linear dis- problem. tance of 18 m? • Paired Problems are connected 3. Find the average angular speed of the second hand of a with a bracket. Your instructor clock. may assign the even-numbered Problems problem, which has no answer at the end of the book. However, working the connected odd-numbered problem Combination conceptual/quantitative problem will allow you to check your answer at the back of the Biological or medical application book and apply what you have learned to working the ✦ Challenging problem even-numbered problem. Blue # Detailed solution in the Student Solutions Manual • Problem numbers highlighted in blue have a solution Problems paired by concept 1 2 available in the Student Solutions Manual if you need Text website interactive or tutorial additional help or would like to double-check your work. • The difficulty level for each problem is indicated. The least difficult problems and problems of intermediate difficulty 5.1 Description of Uniform Circular Motion have no diamond. The more challenging problems have one 1. A carnival swing is fixed on the end of an 8.0-m-l diamond ✦. beam. If the swing and beam sweep through an angle Read through all of the assigned problems and 120°, what is the distance through which the rid ✦114. A student’s head is bent over her physics book. The budget your time accordingly. move? head weighs 50.0 N and is supported by the muscle • indicates a combination Conceptual ⃗ m exerted by the neck extensor muscles and by force F and Quantitative problem. ⃗ c exerted at the atlantooccipital joint. • the contact force F indicates a problem with a biological or ⃗ m is 60.0 N and is directed Given that the magnitude of F medical application. 35° below the horizontal, find (a) the magnitude and • indicates a problem that has an ⃗ c. (b) the direction of F accompanying interactive or tutorial online. B

B

2

2

1

1 1

2 2

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While working your solutions to problems, try to keep your work in symbolic form until the very end. Symbolic solutions will allow you to view which factors affect the results and how the answer would change should any one of the variables in the problem change their value. In this fashion, your solution to any one problem becomes a solution to a whole series of similar problems.

Substituting values into your final symbolic solution will then enable you to judge if your answer is reasonable and provide greater ease in troubleshooting your error if it is not. Always perform a “reality check” at the end of each problem. Did you obtain a reasonable answer given the question being asked?

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Review & Synthesis: Chapters 1−5 Review Exercises 1. From your knowledge of Newton’s second law and dimensional analysis, find the units (in SI base units) of the spring constant k in the equation F = kx, where F is a force and x is a distance. 2. Harrison traveled 2.00 km west, then 5.00 km in a direction 53.0° south of west, then 1.00 km in a direction 60.0° north of west. (a) In what direction, and for how far, should Harrison travel to return to his starting point? (b) If Harrison returns directly to his starting point with a speed of 5.00 m/s, how long will the return trip take? 3. (a) How many center-stripe road reflectors, separated by 17.6 yd, are required along a 2.20-mile section of curving mountain roadway? (b) Solve the same problem for a road

his rapid descent and lost control? (It turns out that aircraft altitudes are given in feet throughout the world except in China, Mongolia, and the former Soviet states where meters are used.) 8. Paula swims across a river that is 10.2 m wide. She can swim at 0.833 m/s in still water, but the river flows with a speed of 1.43 m/s. If Paula swims in such a way that she crosses the river in as short a time as possible, how far downstream is she when she gets to the opposite shore? ✦ 9. Peter is collecting paving stones from a quarry. He harnesses two dogs, Sandy and Rufus, in tandem to the ⃗ at a 15° angle to the loaded cart. Sandy pulls with force F north of east; Rufus pulls with 1.5 times the force of Sandy and at an angle of 30.0° south of east. Use a ruler

After a group of related chapters, you will find a Review & Synthesis section. This section will provide Review Exercises that require you to combine two or more concepts learned in the previous chapters. Working these problems will help you to prepare for cumulative exams. This section also contains MCAT Review exercises. These problems were written for the actual MCAT exam and will provide additional practice if this exam is part of your future plans.

How to Study for an Exam •

• •

•

Be an active learner: • read • be an active participant in class; ask questions • apply what you’ve learned; think through scenarios rather than memorizing your notes Finish reading all material—text, notes, handouts—at least three days prior to the exam. Three days prior to the exam, set aside time each day to do self-testing, work practice problems, and review your notes. Useful tools to help: • end-of-chapter summaries • questions and practice problems • text website • your professor’s course website • the Student Solutions Manual • your study partner Analyze your weaknesses, and create an “I don’t know this yet” list. Focus on strengthening these areas and narrow your list as you study. If you find that you were unable to allow the full three days to study for the exam, the most important thing you can do is try some practice problems that are similar to those your instructor assigned for homework. Choose odd-numbered problems so that you can check your answer. The Review & Synthesis problems are designed to help you prepare for exams. Try to solve each problem under exam conditions—use a formula sheet, if your instructor provides one with the exam, but don’t look at the book or your notes. If you can’t solve the problem, then you have found an area of weakness. Study the material needed to solve that problem and closely related material. Then try another similar problem. VERY IMPORTANT—Be sure to sleep and eat well before the exam. Staying up late and memorizing the night before an exam doesn’t help much in physics. On a physics exam, you will be asked to demonstrate reasoning and analytical skills acquired by much practice. If you are fatigued or hungry, you won’t perform at your highest level.

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•

•

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We hope that these suggestions will help you get the most out of your physics course. After many years working with students, both in the classroom and one-on-one in a self-paced course, we wrote this book so you could benefit from our experience. In Physics, we have tried to address the points that have caused difficulties for our students in the past. We also wish to share with you some of the pleasure and excitement we have found in learning about the physical laws that govern our world. Alan Giambattista Betty Richardson Bob Richardson xxix

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Acknowledgments We are grateful to the faculty, staff, and students at Cornell University, who helped us in a myriad of ways. We especially thank our friend and colleague Bob Lieberman who shepherded us through the process as our literary agent and who inspired us as an exemplary physics teacher. Donald F. Holcomb, Persis Drell, Peter Lepage, and Phil Krasicky read portions of the manuscript and provided us with many helpful suggestions. Raphael Littauer contributed many innovative ideas and served as a model of a highly creative, energetic teacher. We are indebted to Jeevak Parpia, David G. Cassel, Edith Cassel, Richard Galik, Lou Hand, Chris Henley, and Tomás Arias for many helpful discussions while they taught Physics 101–102 using the second edition. We also appreciate the assistance of Leonard J. Freelove and Rosemary French. We thank our enthusiastic and capable graduate teaching assistants and, above all, the students in Physics 101–102, who patiently taught us how to teach physics. We are grateful for the guidance and enthusiasm of Debra Hash and Mary Hurley at McGraw-Hill, whose tireless efforts were invaluable in bringing this project to fruition. We would like to thank the entire team of talented professionals assembled by McGraw-Hill to publish this book, including Traci Andre, Tammy Ben, Carrie Burger, Linda Davoli, Laura Fuller, David Hash, Tammy Juran, Mary Jane Lampe, Lisa Nicks, Mary Reeg, Gloria Schiesl, Thomas Timp, Dan Wallace, and many others whose hard work has contributed to making the book a reality. We are grateful to Bill Fellers for accuracy-checking the manuscript and for many helpful suggestions. Our thanks to Janet Scheel, Warren Zipfel, Rebecca Williams, and Mike Nichols for contributing some of the medical and biological applications; to Nick Taylor and Mike Strauss for contributing to the end-of-chapter and Review & Synthesis problems; and to Nick Taylor for writing answers to the Conceptual Questions. From Alan: Above all, I am deeply grateful to my family. Marion, Katie, Charlotte, Julia, and Denisha, without your love, support, encouragement, and patience, this book could never have been written. From Bob and Betty: We thank our daughter Pamela’s classmates and friends at Cornell and in the Vanderbilt Master’s in Nursing program who were an early inspiration for the book, and we thank Dr. Philip Massey who was very special to Pamela and is dear to us. We thank our friends at blur, Alex, Damon, Dave, and Graham, who love physics and are inspiring young people of Europe to explore the wonders of physics through their work with the European Space Agency’s Mars mission. Finally we thank our daughter Jennifer, our grandsons Jasper, Dashiell, Oliver, and Quintin, and son-inlaw Jim who endured our protracted hours of distraction while this book was being written.

REVIEWERS, CLASS TESTERS, AND ADVISORS This text reflects an extensive effort to evaluate the needs of college physics instructors and students, to learn how well we met those needs, and to make improvements where we fell short. We gathered information from numerous reviews, class tests, and focus groups. The primary stage of our research began with commissioning reviews from instructors across the United States and Canada. We asked them to submit suggestions

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ACKNOWLEDGMENTS

xxxi

for improvement on areas such as content, organization, illustrations, and ancillaries. The detailed comments of these reviewers constituted the basis for the revision plan. We organized focus groups across the United States from 2006 through 2008. Participants reviewed our text in comparison to other books and suggested improvements to Physics and ways in which we as publishers could help to improve the content of the college physics course. Finally, we received extremely useful advice on the instructional design, quality, and content of the print and media ancillary packages from Pete Anderson, Gerry Feldman, Ajawad Haija, Hong Luo, David Mast, John Prineas, Michael Pravica, and Craig Wiegert. Considering the sum of these opinions, the Giambattista/Richardson/Richardson texts now embody the collective knowledge, insight, and experience of hundreds of college physics instructors. Their influence can be seen in everything from the content, accuracy, and organization of the text to the quality of the illustrations. We are grateful to the following instructors for their thoughtful comments and advice:

REVIEWERS AND FOCUS GROUP ATTENDEES David Aaron South Dakota State University Rhett Allain Southeastern Louisiana University Peter Anderson Oakland Community College Natalie Batalha San Jose State Thomas K. Bolland The Ohio State University Juan Burciaga Whitman College Peng Chen Dai University of Tennessee—Knoxville Carl Covatto Arizona State University Michael Crescimanno Youngstown State Steven Ellis University of Kentucky—Lexington Abbas Faridi Orange Coast College Gerald Feldman George Washington University David Gerdes University of Michigan Robert Hagood Washtenaw Community College Ajawad Haija Indiana University of Pennsylvania Grady Hendricks Blinn Community College Klaus Honschied The Ohio State University John Hopkins The Pennsylvania State University Brad Johnson Western Washington University Kyungseon Joo University of Connecticut Linda Jones College of Charleston Arya Karamjeet San Jose State Daniel Kennefick University of Arkansas Yuri Kholodenko Albany College of Pharmacy Dana Klinck Hillsborough Community College, Tampa Allen Landers Auburn University Paul Lee California State University Northridge Hong Luo University at Buffalo Stephanie Magelby Brigham Young University

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George Marion Texas State University, San Marcos David Mast University of Cincinnati Dan Mazilu Virginia Polytechnic Institute & State University Rahul Mehta University of Central Arkansas Meredith Newby Clemson University Miroslav Peric California State University Northridge Amy Pope Clemson University Michael Pravica University of Nevada, Las Vegas Kent Price Morehead State John Prineas University of Iowa Oren Quist South Dakota State University Larry Rowan University of North Carolina—Chapel Hill Ajit Rupaal Western Washington University Douglas Sherman San Jose State University Bjoern Siepel Portland State University Michael Sobel Brooklyn College Xiang-Ning Song Richland College Tim Stelzer University of Illinois James Taylor University of Central Missouri Marshall Thomsen Eastern Michigan University Ralf Widenhorn Portland State University Craig Wiegert University of Georgia Karen Williams East Central University Scott Wissink Indiana University Pei Xiong-Skiba Austin Peay State University Capp Yess Morehead State David Young Louisiana State University—Baton Rouge Michael Ziegler The Ohio State University

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ACKNOWLEDGMENTS

We are also grateful to our international reviewers for their comments and suggestions: Goh Hock Leong National Junior College—Singapore Mohammed Saber Musazay King Fahd University of Petroleum and Minerals

Contributors We are deeply indebted to: Professor Suzanne Willis of Northern Illinois University and Professor Susanne M. Lee, Visiting Scientist Rensselaer Polytechnic Institute, for creating the instructor resources and demonstrations in the Physics Instructors’ Resource Guide.

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Professor Jack Cuthbert of Holmes Community College, Ridgeland for the Test Bank to accompany Physics. Professor Lorin Swint Matthews of Baylor University for the clicker questions to accompany Physics. Professor Carl Covatto of Arizona State University for the PowerPoint Lectures to accompany Physics. Professor Allen Landers of Auburn University for his work on the Physics collection of Active Art on the text’s website.

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CHAPTER

Introduction

In 2004, the exploration rovers Spirit and Opportunity landed on sites on opposite sides of Mars. The primary goal of the mission was to examine a wide variety of rocks and soils that might provide evidence of the past presence of water on Mars and clues to where the water went. The mission sent back tens of thousands of photographs and a wealth of geologic data. By contrast, in a previous mission to Mars, a simple mistake caused the loss of the Mars Climate Orbiter as it entered orbit around Mars. In this chapter, you will learn how to avoid making this same mistake. (See p. 9.)

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The Mars Exploration Rover Opportunity looks back toward its lander in “Eagle Crater” on the surface of Mars.

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2

CHAPTER 1 Introduction

Concepts & Skills to Review

• algebra, geometry, and trigonometry (Appendix A) • To the Student: How to Succeed in Your Physics Class (p. xxii)

1.1

WHY STUDY PHYSICS?

Physics is the branch of science that describes matter, energy, space, and time at the most fundamental level. Whether you are planning to study biology, architecture, medicine, music, chemistry, or art, some principles of physics are relevant to your field. Physicists look for patterns in the physical phenomena that occur in the universe. They try to explain what is happening, and they perform experiments to see if the proposed explanation is valid. The goal is to find the most basic laws that govern the universe and to formulate those laws in the most precise way possible. The study of physics is valuable for several reasons:

A patient being prepared for magnetic resonance imaging (MRI). MRI provides a detailed image of the internal structures of the patient’s body.

• Since physics describes matter and its basic interactions, all natural sciences are built on a foundation of the laws of physics. A full understanding of chemistry requires a knowledge of the physics of atoms. A full understanding of biological processes in turn is based on the underlying principles of physics and chemistry. Centuries ago, the study of natural philosophy encompassed what later became the separate fields of biology, chemistry, geology, astronomy, and physics. Today there are scientists who call themselves biophysicists, chemical physicists, astrophysicists, and geophysicists, demonstrating how thoroughly the sciences are intertwined. • In today’s technological world, many important devices can be understood correctly only with a knowledge of the underlying physics. Just in the medical world, think of laser surgery, magnetic resonance imaging, instant-read thermometers, x-ray imaging, radioactive tracers, heart catheterizations, sonograms, pacemakers, microsurgery guided by optical fibers, ultrasonic dental drills, and radiation therapy. • By studying physics, you acquire skills that are useful in other disciplines. These include thinking logically and analytically; solving problems; making simplifying assumptions; constructing mathematical models; using valid approximations; and making precise definitions. • Society’s resources are limited, so it is important to use them in beneficial ways and not squander them on scientifically impossible projects. Political leaders and the voting public are too often led astray by a lack of understanding of scientific principles. Can a nuclear power plant supply energy safely to a community? What is the truth about the greenhouse effect, the ozone hole, and the danger of radon in the home? By studying physics, you learn some of the basic scientific principles and acquire some of the intellectual skills necessary to ask probing questions and to formulate informed opinions on these important matters. • Finally, by studying physics, we hope that you develop a sense of the beauty of the fundamental laws governing the universe.

1.2

TALKING PHYSICS

Some of the words used in physics are familiar from everyday speech. This familiarity can be misleading, since the scientific definition of a word may differ considerably from its common meaning. In physics, words must be precisely defined so that anyone reading a scientific paper or listening to a science lecture understands exactly what is meant. Some of the basic defined quantities, whose names are also words used in everyday speech, include time, length, force, velocity, acceleration, mass, energy, momentum, and temperature. In everyday language, speed and velocity are synonyms. In physics, there is an important distinction between the two. In physics, velocity includes the direction of motion as well as the distance traveled per unit time. When a moving object changes

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THE USE OF MATHEMATICS

3

direction, its velocity changes even though its speed may not have changed. Confusion of the scientific definition of velocity with its everyday meaning will prevent a correct understanding of some of the basic laws of physics and will lead to incorrect answers. Mass, as used in everyday language, has several different meanings. Sometimes mass and weight are used interchangeably. In physics, mass and weight are not interchangeable. Mass is a measure of inertia—the tendency of an object at rest to remain at rest or, if moving, to continue moving with the same velocity. Weight, on the other hand, is a measure of the gravitational pull on an object. (Mass and weight are discussed in more detail in Chapter 4.) There are two important reasons for the way in which we define physical quantities. First, physics is an experimental science. The results of an experiment must be stated unambiguously so that other scientists can perform similar experiments and compare their results. Quantities must be defined precisely to enable experimental measurements to be uniform no matter where they are made. Second, physics is a mathematical science. We use mathematics to quantify the relationships among physical quantities. These relationships can be expressed mathematically only if the quantities being investigated have precise definitions.

1.3

THE USE OF MATHEMATICS

A working knowledge of algebra, trigonometry, and geometry is essential to the study of introductory physics. Some of the more important mathematical tools are reviewed in Appendix A. If you know that your mathematics background is shaky, you might want to test your mastery by doing some problems from a math textbook. You may find it useful to visit www.mhhe.com to explore the Schaum’s Outline series, especially the Schaum’s Outlines of Precalculus, College Physics, or Physics for Pre-Med, Biology, and Allied Health Students. Mathematical equations are shortcuts for expressing concisely in symbols relationships that are cumbersome to describe in words. Algebraic symbols in the equations stand for quantities that consist of numbers and units. The number represents a measurement and the measurement is made in terms of some standard; the unit indicates what standard is used. In physics, a number to specify a quantity is useless unless we know the unit attached to the number. When buying silk to make a sari, do we need a length of 5 millimeters, 5 meters, or 5 kilometers? Is the term paper due in 3 minutes, 3 days, or 3 weeks? Systems of units are discussed in Section 1.5. There are not enough letters in the alphabet to assign a unique letter to each quantity. The same letter V can represent volume in one context and voltage in another. Avoid attempting to solve problems by picking equations that seem to have the correct letters. A skilled problem-solver understands specifically what quantity each symbol in a particular equation represents, can specify correct units for each quantity, and understands the situations to which the equation applies. Ratios and Proportions In the language of physics, the word factor is used frequently, often in a rather idiosyncratic way. If the power emitted by a radio transmitter has doubled, we might say that the power has “increased by a factor of two.” If the concentration of sodium ions in the bloodstream is half of what it was previously, we might say that the concentration has “decreased by a factor of two,” or, in a blatantly inconsistent way, someone else might say that it has “decreased by a factor of one-half.” The factor is the number by which a quantity is multiplied or divided when it is changed from one value to another. In other words, the factor is really a ratio. In the case of the radio transmitter, if P0 represents the initial power and P represents the power after new equipment is installed, we write P =2 ___ P0

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CHAPTER 1 Introduction

It is also common to talk about “increasing 5%” or “decreasing 20%.” If a quantity increases n%, that is the same as saying that it is multiplied by a factor of 1 + (n/100). If a quantity decreases n%, then it is multiplied by a factor of 1 − (n/100). For example, an increase of 5% means something is 1.05 times its original value and a decrease of 4% means it is 0.96 times the original value. Physicists talk about increasing “by some factor” because it often simplifies a problem to think in terms of proportions. When we say that A is proportional to B (written A ∝ B), we mean that if B increases by some factor, then A must increase by the same factor. For instance, the circumference of a circle equals 2p times the radius: C = 2p r. Therefore C ∝ r. If the radius doubles, the circumference also doubles. The area of a circle is proportional to the square of the radius (A = p r2, so A ∝ r2). The area must increase by the same factor as the radius squared, so if the radius doubles, the area increases by a factor of 22 = 4.

A ∝ B means A1/A2 = B1/B2

Example 1.1 Effect of Increasing Radius on the Volume of a Sphere The volume of a sphere is given by the equation V = _4 p r 3 3

where V is the volume and r is the radius of the sphere. If a basketball has a radius of 12.4 cm and a tennis ball has a radius of 3.20 cm, by what factor is the volume of the basketball larger than the volume of the tennis ball? Strategy The problem gives the values of the radii for the two balls. To keep track of which ball’s radius and volume we mean, we use subscripts “b” for basketball and “t” for tennis ball. The radius of the basketball is rb and the radius of the tennis ball is rt. Since _43 and p are constants, we can work in terms of proportions. Solution The ratio of the basketball radius to that of the tennis ball is rb ________ 12.4 [cm] __ rt = 3.20 [cm] = 3.875 The volume of a sphere is proportional to the cube of its radius:

Discussion A slight variation on the solution is to write out the proportionality in terms of ratios of the corresponding sides of the two equations: _4 p r 3 rb 3 Vb _____ b __ ___ = _43 3 = ( rt ) Vt p rt 3

Substituting the ratio of rb to rt yields Vb ___ = 3.8753 ≈ 58.2 Vb which says that Vb is approximately 58.2 times Vt.

Practice Problem 1.1 by a Lightbulb

Power Dissipated

The electric power P dissipated by a lightbulb of resistance R is P = V 2/R, where V represents the line voltage. During a brownout, the line voltage is 10.0% less than its normal value. How much power is drawn by a lightbulb during the brownout if it normally draws 100.0 W (watts)? Assume that the resistance does not change.

V ∝ r3 Since the basketball radius is larger by a factor of 3.875, and volume is proportional to the cube of the radius, the new volume should be bigger by a factor of 3.8753 ≈ 58.2.

CHECKPOINT 1.3 If the radius of the sphere is increased by a factor of 3, by what factor does the volume of the sphere change?

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1.4

1.4

SCIENTIFIC NOTATION AND SIGNIFICANT FIGURES

5

SCIENTIFIC NOTATION AND SIGNIFICANT FIGURES

In physics, we deal with some numbers that are very small and others that are very large. It can get cumbersome to write numbers in conventional decimal notation. In scientific notation, any number is written as a number between 1 and 10 times an integer power of ten. Thus the radius of Earth, approximately 6 380 000 m at the equator, can be written 6.38 × 106 m; the radius of a hydrogen atom, 0.000 000 000 053 m, can be written 5.3 × 10−11 m. Scientific notation eliminates the need to write zeros to locate the decimal point correctly. In science, a measurement or the result of a calculation must indicate the precision to which the number is known. The precision of a device used to make a measurement is limited by the finest division on the scale. Using a meterstick with millimeter divisions as the smallest separations, we can measure a length to a precise number of millimeters and we can estimate a fraction of a millimeter between two divisions. If the meterstick has centimeter divisions as the smallest separations, we measure a precise number of centimeters and estimate the fraction of a centimeter that remains.

Learn how to use the button on your calculator (usually labeled EE) to enter a number in scientific notation. To enter 1.2 × 108, press 1.2, EE, 8.

Significant Figures The most basic way to indicate the precision of a quantity is to write it with the correct number of significant figures. The significant figures are all the digits that are known accurately plus the one estimated digit. If we say that the distance from here to the state line is 12 km, that does not mean we know the distance to be exactly 12 kilometers. Rather, the distance is 12 km to the nearest kilometer. If instead we said that the distance is 12.0 km, that would indicate that we know the distance to the nearest tenth of a kilometer. More significant figures indicate a greater degree of precision.

Rules for Identifying Significant Figures 1. Nonzero digits are always significant. 2. Final or ending zeros written to the right of the decimal point are significant. 3. Zeros written to the right of the decimal point for the purpose of spacing the decimal point are not significant. 4. Zeros written to the left of the decimal point may be significant, or they may only be there to space the decimal point. For example, 200 cm could have one, two, or three significant figures; it’s not clear whether the distance was measured to the nearest 1 cm, to the nearest 10 cm, or to the nearest 100 cm. On the other hand, 200.0 cm has four significant figures (see rule 5). Rewriting the number in scientific notation is one way to remove the ambiguity. In this book, when a number has zeros to the left of the decimal point, you may assume a minimum of two significant figures. 5. Zeros written between significant figures are significant.

Example 1.2 Identifying the Number of Significant Figures For each of these values, identify the number of significant figures and rewrite it in standard scientific notation. (a) (b) (c) (d)

409.8 s 0.058700 cm 9500 g 950.0 × 101 mL

Strategy We follow the rules for identifying significant figures as given. To rewrite a number in scientific notation, we move the decimal point so that the number to the left of the decimal point is between 1 and 10 and compensate by multiplying by the appropriate power of ten.

continued on next page

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CHAPTER 1 Introduction

Example 1.2 continued

Solution (a) All four digits in 409.8 s are significant. The zero is between two significant figures, so it is significant. To write the number in scientific notation, we move the decimal point two places to the left and compensate by multiplying by 102: 4.098 × 102 s. (b) The first two zeros in 0.058700 cm are not significant; they are used to place the decimal point. The digits 5, 8, and 7 are significant, as are the two final zeros. The answer has five significant figures: 5.8700 × 10−2 cm. (c) The 9 and 5 in 9500 g are significant, but the zeros are ambiguous. This number could have two, three, or four significant figures. If we take the most cautious approach and assume the zeros are not significant, then the number in scientific notation is 9.5 × 103 g. (d) The final zero in 950.0 × 101 mL is significant since it comes after the decimal point. The zero to its left is also significant since it comes between two other significant digits.

The result has four significant figures. The number is not in standard scientific notation since 950.0 is not between 1 and 10; in scientific notation we write 9.500 × 103 mL. Discussion Scientific notation clearly indicates the number of significant figures since all zeros are significant; none are used only to place the decimal point. In (c), if we want to show that the zeros were significant, we would write 9.500 × 103 g.

Practice Problem 1.2 Identifying Significant Figures State the number of significant figures in each of these measurements and rewrite them in standard scientific notation. (a) 0.000 105 44 kg (b) 0.005 800 cm (c) 602 000 s

Significant Figures in Calculations 1. When two or more quantities are added or subtracted, the result is as precise as the least precise of the quantities (Example 1.3). If the quantities are written in scientific notation with different powers of ten, first rewrite them with the same power of ten. After adding or subtracting, round the result, keeping only as many decimal places as are significant in all of the quantities that were added or subtracted. 2. When quantities are multiplied or divided, the result has the same number of significant figures as the quantity with the smallest number of significant figures (Example 1.4). 3. In a series of calculations, rounding to the correct number of significant figures should be done only at the end, not at each step. Rounding at each step would increase the chance that roundoff error could snowball and have an adverse effect on the accuracy of the final answer. It’s a good idea to keep at least two extra significant figures in calculations, then round at the end.

Example 1.3 Significant Figures in Addition Calculate the sum 44.56005 s + 0.0698 s + 1103.2 s. Strategy The sum cannot be more precise than the least precise of the three quantities. The quantity 44.56005 s is known to the nearest 0.00001 s, 0.0698 s is known to the nearest 0.0001 s, and 1103.2 s is known to the nearest 0.1 s.

Therefore the least precise is 1103.2 s. The sum has the same precision; it is known to the nearest tenth of a second. Solution According to the calculator, 44.56005 + 0.0698 + 1103.2 = 1147.82985 continued on next page

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7

SCIENTIFIC NOTATION AND SIGNIFICANT FIGURES

Example 1.3 continued

We do not want to write all of those digits in the answer. That would imply greater precision than we actually have. Rounding to the nearest tenth of a second, the sum is written = 1147.8 s

or subtraction, we are concerned with the precision rather than the number of significant figures. The three quantities to be added have seven, three, and five significant figures, respectively, while the sum has five significant figures.

and there are five significant figures in the result. Discussion Note that the least precise measurement is not necessarily the one with the fewest number of significant figures. The least precise is the one whose rightmost significant figure represents the largest unit: the “2” in 1103.2 s represents 2 tenths of a second. In addition

Practice Problem 1.3 Significant Figures in Subtraction Calculate the difference 568.42 m − 3.924 m and write the result in scientific notation. How many significant figures are in the result?

Example 1.4 Significant Figures in Multiplication Find the product of 45.26 m/s and 2.41 s. How many significant figures does the product have? Strategy The product should have the same number of significant figures as the factor with the least number of significant figures. Solution A calculator gives

Discussion Writing the answer as 109.0766 m would give the false impression that we know the answer to a precision of about 0.0001 m, whereas we actually have a precision of only about 1 m. Note that although both factors were known to two decimal places, our solution is properly given with no decimal places. It is the number of significant figures that matters in multiplication or division. In scientific notation, we write 1.09 × 102 m.

45.26 × 2.41 = 109.0766 Since the answer should have only three significant figures, we round the answer to 45.26 m/s × 2.41 s = 109 m

Practice Problem 1.4 Significant Figures in Division Write the solution to 28.84 m divided by 6.2 s with the correct number of significant figures.

When an integer, or a fraction of integers, is used in an equation, the precision of the result is not affected by the integer or the fraction; the number of significant figures is limited only by the measured values in the problem. The fraction _12 in an equation is exact; it does not reduce the number of significant figures to one. In an equation such as C = 2p r for the circumference of a circle of radius r, the factors 2 and p are exact. We use as many digits for p as we need to maintain the precision of the other quantities. Order-of-Magnitude Estimates Sometimes a problem may be too complicated to solve precisely, or information may be missing that would be necessary for a precise calculation. In such a case, an order-of-magnitude solution is the best we can do. By order of magnitude, we mean “roughly what power of ten?” An order of magnitude calculation is done to at most one significant figure. Even when a more precise solution is feasible, it is often a good idea to start with a quick, “back-of-the-envelope estimate.” Why? Because we can often make a good guess about the correct order of magnitude of the answer to a problem, even before we start solving the problem. If the answer comes out with a different order of magnitude, we go back and search out an error. Suppose a problem concerns a vase that is knocked off a fourth-story window ledge. We can guess by experience the order of magnitude of the time it takes the vase to hit the ground. It might be 1 s, or 2 s, but we are certain that it is not 1000 s or 0.00001 s.

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Back-of-the-envelope estimate: a calculation so short that it could easily fit on the back of an envelope

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CHAPTER 1 Introduction

CHECK POINT 1.4 What are some of the reasons for making order-of-magnitude estimates?

1.5

kg⋅m/s2 can also be written kg⋅m⋅s−2

Silicon atoms (radius 10–10 m)

10–15

10–10

10–5

UNITS

A metric system of units has been used for many years in scientific work and in European countries. The metric system is based on powers of ten (Fig. 1.1). In 1960, the General Conference of Weights and Measures, an international authority on units, proposed a revised metric system called the Système International d’Unités in French (abbreviated SI), which uses the meter (m) for length, the kilogram (kg) for mass, the second (s) for time, and four more base units (Table 1.1). Derived units are constructed from combinations of the base units. For example, the SI unit of force is kg⋅m/s2; the combination of kg⋅m/s2 is given a special name, the newton (N), in honor of Isaac Newton. The newton is a derived unit because it is composed of a combination of base units. When units are named after famous scientists, the name of the unit is written with a lowercase letter, even though it is based on a proper name; the abbreviation for the unit is written with an uppercase letter. The inside front cover of the book has a complete listing of the derived SI units used in this book. As an alternative to explicitly writing powers of ten, SI uses prefixes for units to indicate power of ten factors. Table 1.2 shows some of the powers of ten and the SI prefixes used for them. These are also listed on the inside front cover of the book. Note that when an SI unit with a prefix is raised to a power, the prefix is also raised to that power. For example, 8 cm3 = 2 cm × 2 cm × 2 cm. SI units are preferred in physics and are emphasized in this book. Since other units are sometimes used, we must know how to convert units. Various scientific fields, even in physics, do use units other than SI units, whether for historical or practical reasons.

A child (height 100 m)

100

105

Earth (diameter 107 m)

1010

1015

A spiral galaxy (diameter 1019 m)

1020

1025

Distance to quasar observed by Hubble Telescope (1026 m)

Hydrogen nucleus (radius 10–15 m)

HIV (diameter 10–7 m) invading a T lymphocyte (a type of white blood cell)

The Duomo (cathedral) in Florence, Italy (height 102 m)

The Sun (diameter 109 m)

Figure 1.1 Scientific notation uses powers of ten to express quantities that have a wide range of values.

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1.5

Table 1.1

9

UNITS

SI Base Units

Quantity

Unit Name

Symbol

Length

meter

m

Mass Time

kilogram second

kg s

Electric current

ampere

A

Temperature

kelvin

K

Amount of substance

mole

mol

Luminous intensity

candela*

cd

Definition The distance traveled by light in vacuum during a time interval of 1/299 792 458 s. The mass of the international prototype of the kilogram. The duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom. The constant current in two long, thin, straight, parallel conductors placed 1 m apart in vacuum that would produce a force on the conductors of 2 × 10−7 N per meter of length. The fraction 1/273.16 of the thermodynamic temperature of the triple point of water. The amount of substance that contains as many elementary entities as there are atoms in 0.012 kg of carbon-12. The luminous intensity, in a given direction, of a source that emits radiation of frequency 540 × 1012 Hz and that has a radiant intensity in that direction of 1/683 watts per steradian.

*Not used in this book

For example, in atomic and nuclear physics, the SI unit of energy (the joule, J) is rarely used; instead the energy unit used is usually the electron-volt (eV). Biologists and chemists use units that are not familiar to physicists. One reason that SI is preferred is that it provides a common denominator—all scientists are familiar with the SI units. In most of the world, SI units are used in everyday life and in industry. In the United States, the U.S. customary units—sometimes called English units—are still used. The base units for this system are the foot, the second, and the pound. The pound is legally defined in the United States as a unit of mass, but it is also commonly used as a unit of force (in which case it is sometimes called pound-force). Since mass and force are entirely different concepts in physics, this inconsistency is one good reason to use SI units. In the autumn of 1999, to the chagrin of NASA, a $125 million spacecraft was destroyed as it was being maneuvered into orbit around Mars. The company building the booster rocket provided information about the rocket’s thrust in U.S. customary units, but the NASA scientists who were controlling the rocket thought the figures provided were in metric units. Arthur Stephenson, chairman of the Mars Climate Orbiter Mission Failure Investigation Board, stated that, “The ‘root cause’ of the loss of the spacecraft was the failed translation of English units into metric units in a segment of ground-based, navigation-related mission software.” After a journey of 122 million miles, the Climate Orbiter dipped about 15 miles too deep into the Martian atmosphere, causing the propulsion system to overheat. The discrepancy in units unfortunately caused a dramatic failure of the mission. Converting Units If the statement of a problem includes a mixture of different units, the units must be converted to a single, consistent set before the problem is solved. Quantities to be added or subtracted must be expressed in the same units. Usually the best way is to convert everything to SI units. Common conversion factors are listed on the inside front cover of this book. Examples 1.5 and 1.6 illustrate the technique for converting units. The quantity to be converted is multiplied by one or more conversion factors written as a fraction equal to 1. The units are multiplied or divided as algebraic quantities.

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What happened to the Mars Climate Orbiter?

Table 1.2 Prefix (abbreviation) peta- (P) tera- (T) giga- (G) mega- (M) kilo- (k) deci- (d) centi- (c) milli- (m) micro- (μ) nano- (n) pico- (p) femto- (f )

SI Prefixes Power of Ten 1015 1012 109 106 103 10−1 10−2 10−3 10−6 10−9 10−12 10−15

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CHAPTER 1 Introduction

Some conversions are exact by definition. One meter is defined to be exactly equal to 100 cm; all SI prefixes are exactly a power of ten. The use of an exact conversion factor such as 1 m = 100 cm, or 1 foot = 12 inches, does not affect the precision of the result; the number of significant figures is limited only by the other quantities in the problem.

Example 1.5 Buying Clothes in a Foreign Country Michel, an exchange student from France, is studying in the United States. He wishes to buy a new pair of jeans, but the sizes are all in inches. He does remember that 1 m = 3.28 ft and that 1 ft = 12 in. If his waist size is 82 cm, what is his waist size in inches? Strategy Each conversion factor can be written as a fraction. If 1 m = 3.28 ft, then 3.28 ft = 1 ______ 1m We can multiply any quantity by 1 without changing its value. We arrange each conversion factor in a fraction and multiply one at a time to get from centimeters to inches. Solution We first convert cm to meters. 1m 82 — cm × _______ 100 — cm Now, we convert meters to feet. 3.28 ft 1m – × ______ 82 — cm × _______ 100 — cm 1m –

Finally, we convert feet to inches. 3.28 ft × _____ 1m – × ______ 12 in = 32 in 82 — cm × _______ 100 — cm 1m – 1 ft – In each case, the fraction is written so that the unit we are converting from cancels out. As a check: ft – × __ m – __ in = in cm × ___ — – ft cm × m — – Discussion This problem could have been done in one step using a direct conversion factor from inches to cm (1 in = 2.54 cm). One of the great advantages of SI units is that all the conversion factors are powers of ten (see Table 1.2); there is no need to remember that there are 12 inches in a foot, 4 quarts in a gallon, 16 ounces in a pound, 5280 feet in a mile, and so on.

Practice Problem 1.5 Driving on the Autobahn A BMW convertible travels on the German autobahn at a speed of 128 km/h. What is the speed of the car (a) in meters per second? (b) in miles per hour?

Example 1.6 Conversion of Volume A beaker of water contains 255 mL of water. (1 mL = 1 milliliter; 1 L = 1000 cm3.) What is the volume of the water in (a) cubic centimeters? (b) cubic meters? Strategy First convert milliliters to liters; then convert liters to cubic centimeters. To convert cubic centimeters to cubic meters, use 100 cm = 1 m. Since there are three factors 1m of centimeters to convert, we have to multiply by _______ 100 cm three times.

(

)

Solution (a) The prefix milli- means 10−3, so 1 mL = 10−3 L. Then −3

3

10 L – × ________ 1000 cm = 255 cm3 mL × ______ 255 — 1m L 1L – — (b) 1 m = 100 cm. Since we need to convert cubic centimeters to cubic meters, we must raise the conversion factor to the third power:

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(1 m)3 1 m 3 = 255 cm3 × _________ 255 cm3 × _______ 100 cm (100 cm)3

(

)

3

1m 255 — = 2.55 × 10−4 m3 cm3 × ________ 1003 — cm3 Discussion Be careful when a unit is raised to a power other than one; the conversion factor must be raised to the same power. Writing out the units to make sure they cancel prevents mistakes. When a quantity is raised to a power, both the number and the unit must be raised to the same power. (100 cm)3 is equal to 1003 cm3 = 106 cm3; it is not equal to 100 cm3, nor is it equal to 106 cm.

Practice Problem 1.6 Surface Area of Earth The radius of Earth is 6.4 × 103 km. Find the surface area of Earth in square meters and in square miles. (Surface area of a sphere = 4p r 2.)

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1.6

11

DIMENSIONAL ANALYSIS

Whenever a calculation is performed, always write out the units with each quantity. Combine the units algebraically to find the units of the result. This small effort has three important benefits: 1. It shows what the units of the result are. A common mistake is to get the correct numerical result of a calculation but to write it with the wrong units, making the answer wrong. 2. It shows where unit conversions must be done. If units that should have canceled do not, we go back and perform the necessary conversion. When a distance is calculated and the result comes out with units of meter-seconds per hour (m⋅s/h), we should convert hours to seconds. 3. It helps locate mistakes. If a distance is calculated and the units come out as m/s, we know to look for an error.

CHECKPOINT 1.5 If 1 fluid ounce (fl oz) is approximately 30 mL, how many liters are in a half gallon (64 fl oz) of milk?

1.6

DIMENSIONAL ANALYSIS

Dimensions are basic types of units, such as time, length, and mass. (Warning: The word dimension has several other meanings, such as in “three-dimensional space” or “the dimensions of a soccer field.”) Many different units of length exist: meters, inches, miles, nautical miles, fathoms, leagues, astronomical units, angstroms, and cubits, just to name a few. All have dimensions of length; each can be converted into any other. We can add, subtract, or equate quantities only if they have the same dimensions (although they may not necessarily be given in the same units). It is possible to add 3 meters to 2 inches (after converting units), but it is not possible to add 3 meters to 2 kilograms. To analyze dimensions, treat them as algebraic quantities, just as we did

Example 1.7 Dimensional Analysis for a Distance Equation Analyze the dimensions of the equation d = vt, where d is distance traveled, v is speed, and t is elapsed time. Strategy Replace each quantity with its dimensions. Distance has dimensions [L]. Speed has dimensions of length per unit time [L/T]. The equation is dimensionally consistent if the dimensions are the same on both sides. Solution The right side has dimensions [L] ___ × [T] = [L] [T] Since both sides of the equation have dimensions of length, the equation is dimensionally consistent. Discussion If, by mistake, we wrote d = v/t for the relation between distance traveled and elapsed time, we could

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quickly catch the mistake by looking at the dimensions. On the right side, v/t would have dimensions [L/T2], which is not the same as the dimensions of d on the left side. A quick dimensional analysis of this sort is a good way to catch algebraic errors. Whenever we are unsure whether an equation is correct, we can check the dimensions.

Practice Problem 1.7 of Another Equation

Testing Dimensions

Test the dimensions of the following equation: 1 at d = __ 2 where d is distance traveled, a is acceleration (which has SI units m/s2), and t is the elapsed time. If incorrect, can you suggest what might have been omitted?

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CHAPTER 1 Introduction

with units in Section 1.5. Usually [M], [L], and [T] are used to stand for mass, length, and time dimensions, respectively. Equivalently, we can use the SI base units: kg for mass, m for length, and s for time. Applying Dimensional Analysis Dimensional analysis is good for more than just checking equations. In some cases, we can completely solve a problem—up to a dimen__ sionless factor like 1/(2p) or √ 3 —using dimensional analysis. To do this, first list all the relevant quantities on which the answer might depend. Then determine what combinations of them have the same dimensions as the answer for which we are looking. If only one such combination exists, then we have the answer, except for a possible dimensionless multiplicative constant.

Example 1.8 Violin String Frequency While it is being played, a violin string produces a tone with frequency f in s−1; the frequency is the number of vibrations per second of the string. The string has mass m, length L, and tension T. If the tension is increased 5.0%, how does the frequency change? Tension has SI unit kg⋅m/s2. Strategy We could make a study of violin strings, but let us see what we can find out by dimensional analysis. We want to find out how the frequency f can depend on m, L, and T. We won’t know if there is a dimensionless constant involved, but we can work by proportions so any such constant will divide out. Solution The unit of tension T is kg⋅m/s2. The units of f do not contain kg or m; we can get rid of them from T by dividing the tension by the length and the mass: T has SI unit s−2 ___ mL That is almost what we want; all we have to do is take the square root: ____

T has SI unit s− √ ___ mL Therefore,

1

where C is some dimensionless constant. To answer the question, let the original frequency and tension be f and T and the new frequency and tension be f ′ and T ′, where T ′ = 1.050T. Frequency is proportional to the square root of tension, so ___ _____ f′ T ′ = √ 1.050 = 1.025 __ = ___ T f

√

The frequency increases 2.5%. Discussion We’ll learn in Chapter 11 how to calculate the value of C, which is 1/2. That is the only thing we cannot get by dimensional analysis. There is no other way to combine T, m, and L to come up with a quantity that has the units of frequency.

Practice Problem 1.8 Increase in Kinetic Energy When a body of mass m is moving with a speed v, it has kinetic energy associated with its motion. Energy is measured in kg ⋅ m2⋅s−2. If the speed of a moving body is increased by 25% while its mass remains constant, by what percentage does the kinetic energy increase?

____

√

T f = C ___ mL

CHECKPOINT 1.6 If two quantities have different dimensions, is it possible to (a) multiply; (b) divide; (c) add; (d) subtract them?

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1.7

PROBLEM-SOLVING TECHNIQUES

13

PROBLEM-SOLVING TECHNIQUES

No single method can be used to solve every physics problem. We demonstrate useful problem-solving techniques in the examples in every chapter of this text. Even for a particular problem, there may be more than one correct way to approach the solution. Problem-solving techniques are skills that must be practiced to be learned. Think of the problem as a puzzle to be solved. Only in the easiest problems is the solution method immediately apparent. When you do not know the entire path to a solution, see where you can get by using the given information—find whatever you can. Exploration of this sort may lead to a solution by suggesting a path that had not been considered. Be willing to take chances. You may even find the challenge enjoyable! When having some difficulty, it helps to work with a classmate or two. One way to clarify your thoughts is to put them into words. After you have solved a problem, try to explain it to a friend. If you can explain the problem’s solution, you really do understand it. Both of you will benefit. But do not rely too much on help from others; the goal is for each of you to develop your own problem-solving skills.

General Guidelines for Problem Solving 1. Read the problem carefully and all the way through. 2. Reread the problem one sentence at a time and draw a sketch or diagram to help you visualize what is happening. 3. Write down and organize the given information. Some of the information can be written in labels on the diagram. Be sure that the labels are unambiguous. Identify in the diagram the object, the position, the instant of time, or the time interval to which the quantity applies. Sometimes information might be usefully written in a table beside the diagram. Look at the wording of the problem again for information that is implied or stated indirectly. 4. Identify the goal of the problem. What quantities need to be found? 5. If possible, make an estimate to determine the order of magnitude of the answer. This estimate is useful as a check on the final result to see if it is reasonable. 6. Think about how to get from the given information to the final desired information. Do not rush this step. Which principles of physics can be applied to the problem? Which will help get to the solution? How are the known and unknown quantities related? Are all of the known quantities relevant, or might some of them not affect the answer? Which equations are relevant and may lead to the solution to the problem? This step requires skills developed only with much practice in problem solving. 7. Frequently, the solution involves more than one step. Intermediate quantities might have to be found first and then used to find the final answer. Try to map out a path from the given information to the solution. Whenever possible, a good strategy is to divide a complex problem into several simpler subproblems. 8. Perform algebraic manipulations with algebraic symbols (letters) as far as possible. Substituting the numbers in too early has a way of hiding mistakes. 9. Finally, if the problem requires a numerical answer, substitute the known numerical quantities, with their units, into the appropriate equation. Leaving out the units is a common source of error. Writing the units shows when a unit conversion needs to be done—and also may help identify an algebra mistake. 10. Once the solution is found, don’t be in a hurry to move on. Check the answer—is it reasonable? Try to think of other ways to solve the same problem. Many problems can be solved in several different ways. Besides providing a check on the answer, finding more than one method of solution deepens our understanding of the principles of physics and develops problem-solving skills that will help solve other problems.

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CHAPTER 1 Introduction

1.8

APPROXIMATION

Physics is about building conceptual and mathematical models and comparing observations of the real world with the model. Simplified models help us to analyze complex situations. In various contexts we assume there is no friction, or no air resistance, no heat loss, or no wind blowing, and so forth. If we tried to take all these things into consideration with every problem, the problems would become vastly more complicated to solve. We never can take account of every possible influence. We freely make approximations whenever possible to turn a complex problem into an easier one, as long as the answer will be accurate enough for our purposes. A valuable skill to develop is the ability to know when an assumption or approximation is reasonable. It might be permissible to ignore air resistance when dropping a stone, but not when dropping a beach ball. Why? We must always be prepared to justify any approximation we make by showing the answer is not changed very much by its use. As well as making simplifying approximations in models, we also recognize that measurements are approximate. Every measured quantity has some uncertainty; it is impossible for a measurement to be exact to an arbitrarily large number of significant figures. Every measuring device has limits on the precision and accuracy of its measurements.

(a)

(b)

Figure 1.2 Approximation of human body by one or more cylinders.

Approximating the Surface Area of the Human Body Sometimes it is difficult or impossible to measure precisely a quantity that is needed for a problem. Then we have to make a reasonable estimate. Suppose we need to know the surface area of a human being to determine the heat loss by radiation in a cold room. We can estimate the height of an average person. We can also estimate the average distance around the waist or hips. Approximating the shape of a human body as a cylinder, we can estimate the surface area by calculating the surface area of a cylinder with the same height and circumference (Fig. 1.2a). If we need a better estimate, we use a slightly more refined model. For instance, we might approximate the arms, legs, trunk, and head and neck as cylinders of various sizes (Fig. 1.2b). How different is the sum of these areas from the original estimate? That gives an idea of how close the first estimate is.

Example 1.9 Number of Cells in the Human Body Average-sized cells in the human body are about 10 μm in length (Fig. 1.3). How many cells are in the human body? Make an order-of-magnitude estimate. Strategy We divide this problem into three subproblems: estimating the volume of a human, estimating the volume of the average cell, and finally estimating the number of cells. To find the volume of a human body, we approximate the body as a cylinder, as previously discussed. Next we assume the cells are cubical to find the volume of a cell. Third, the ratio of the two volumes (volume of the body to volume of the cell) shows how many cells are in the body. Solution Model the body as a cylinder. A typical height is about 2 m. A typical maximum circumference (think hip size) is about 1 m. The corresponding radius is 1/(2p) m, or about 1/6 m. The average radius is somewhat smaller; say

Figure 1.3 Scanning electron micrograph of a precursor T lymphocyte (a type of white blood cell in the human body). The cell is approximately 12 μm in diameter.

0.1 m. The volume of a cylinder is the height times the crosssectional area: V = Ah = p r2h ≈ 3 × (0.1 m)2 × (2 m) = 0.06 m3 continued on next page

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15

GRAPHS

Example 1.9 continued

The volume of a cube is V = s3. Then the volume of an average cell is about V cell ≈ (1 × 10−5 m)3 = 1 × 10−15 m3 The number of cells is the ratio of the two volumes: volume of body 6 × 10−2 m3 ≈ 6 × 1013 N = ___________________ ≈ ___________ average volume of cell 1 × 10−15 m3 Discussion Based on this rough estimate, we cannot rule out the possibility that a better estimate might be 3 × 1013. On the other hand, we can rule out the possibility that the number of cells is, say, 100 million (= 108).

1.9

Practice Problem 1.9 in the United States

Drinking Water Consumed

How many liters of water are swallowed by the people living in the United States in one year? This is a type of problem made famous by the physicist Enrico Fermi (1901–1954), who was a master at this sort of back-of-the-envelope calculation. Such problems are often called Fermi problems in his honor. (1 liter = 10−3 m3 ≈ 1 quart.)

GRAPHS

Graphs are used to help us see a pattern in the relationship between two quantities. It is much easier to see a pattern on a graph than to see it in a table of numerical values. When we do experiments in physics, we change one quantity (the independent variable) and see what happens to another (the dependent variable). We want to see how one variable depends on another. The value of the independent variable is usually plotted along the horizontal axis of the graph. In a plot of p versus q, which means p is plotted on the vertical axis and q on the horizontal axis, normally p is the dependent variable and q is the independent variable. Some general guidelines for recording data and making graphs are given next.

Recording Data and Making Data Tables 1. Label columns with the names of the data being measured and be sure to include the units for the measurements. Do not erase any data, but just draw a line through data that you think are erroneous. Sometimes you may decide later that the data were correct after all. 2. Try to make a realistic estimate of the precision of the data being taken when recording numbers. For example, if the timer says 2.3673 s, but you know your reaction time can vary by as much as 0.1 s, the time should be recorded as 2.4 s. When doing calculations using measured values, remember to round the final answer to the correct number of significant figures. 3. Do not wait until you have collected all of your data to start a graph. It is much better to graph each data point as it is measured. By doing so, you can often identify equipment malfunction or measurement errors that make your data unreliable. You can also spot where something interesting happens and take data points closer together there. Graphing as you go means that you need to find out the range of values for both the independent and dependent variables.

Graphing Data 1. Make large, neat graphs. A tiny graph is not very illuminating. Use at least half a page. A graph made carelessly obscures the pattern between the two variables. 2. Label axes with the name of the quantities graphed and their units. Write a meaningful title.

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CHAPTER 1 Introduction

The equation of a straight line on a graph of y versus x can be written y = mx + b, where m is the slope and b is the y-intercept (the value of y corresponding to x = 0).

3. When a linear relation is expected, use a ruler or straightedge to draw the best-fit straight line. Do not assume that the line must go through the origin—make a measurement to find out, if possible. Some of the data points will probably fall above the line and some will fall below the line. 4. Determine the slope of a best-fit line by measuring the ratio Δy/Δx using as large a range of the graph as possible. Do not choose two data points to calculate the slope; instead, read values from two points on the best-fit line. Show the calculations. Do not forget to write the units; slopes of graphs in physics have units, since the quantities graphed have units. 5. When a nonlinear relationship is expected between the two variables, the best way to test that relationship is to manipulate the data algebraically so that a linear graph is expected. The human eye is a good judge of whether a straight line fits a set of data points. It is not so good at deciding whether a curve is parabolic, cubic, or exponential. To test the relationship x = _12 at2, where x and t are the quantities measured, graph x versus t2 instead of x versus t. 6. If one data point does not lie near the line or smooth curve connecting the other data points, that data point should be investigated to see whether an error was made in the measurement or whether some interesting event is occurring at that point. If something unusual is happening there, obtain additional data points in the vicinity. 7. When the slope of a graph is used to calculate some quantity, pay attention to the equation of the line and the units along the axes. The quantity to be found may be the inverse of the slope or twice the slope or one half the slope. For example, if you wish to find the value of a in the relationship x = _12 at2, and you make a graph of x versus t 2, then the slope of the line is _12 a. The value of a you seek is twice the slope.

The symbol Δ, the Greek uppercase letter delta, stands for the difference between two measurements. The notation Δy is read aloud as “delta y” and represents a change in the value of y.

Example 1.10 Length of a Spring In an introductory physics laboratory experiment, students are investigating how the length of a spring varies with the weight hanging from it. Various weights (accurately calibrated to 0.01 N) ranging up to 6.00 N can be hung from the spring; then the length of the spring is measured with a meterstick (Fig. 1.4). The goal is to see if the weight F and length L are related by

5

5

F = kx

10

10

where x = (L − L0), L0 is the length of the spring when no weight is hanging from it, and k is called the spring constant of the spring. Graph the data in the table and calculate k for this spring.

15

15

20

20

25

25

30

30

F (N):

L (cm): 9.4

0.50

1.00

2.50

3.00

3.50

4.00

5.00

6.00

10.2

12.5

17.9

19.7

22.5

23.0

28.8

29.5

L0 x L

Figure 1.4 A weight causes an extension in the length of a spring.

Strategy Weight is the independent variable, so it is plotted on the horizontal axis. After plotting the data points, we draw the best-fit straight line. Then we calculate the slope of the line, using two points on the line that are widely separated

and that cross gridlines of the graph (so the values are easy to read). The slope of the graph is not k; we must solve the equation for L, since length is plotted on the vertical axis. continued on next page

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17

GRAPHS

Example 1.10 continued

Solution Figure 1.5 shows a graph with data points and a best-fit straight line. There is some scatter in the data, but a linear relationship is plausible. Two points where the line crosses gridlines of the graph are (0.80 N, 12.0 cm) and (4.40 N, 25.0 cm). From these, we calculate the slope:

Discussion As discussed in the graphing guidelines, the slope of the straight-line graph is calculated from two widely spaced values along the best-fit line. We do not subtract values of actual data points. We are looking for an average value from the data; using two data points to find the slope would defeat the purpose of plotting a graph or of taking more than two data measurements. The values read from the graph, including the units, are indicated in Fig. 1.5. The units for the slope are cm/N, since we plotted centimeters versus newtons. For this particular problem the inverse of the slope is the quantity we seek, the spring constant in N/cm.

25.0 cm − 12.0 cm = 3.61 ___ cm ΔL = ________________ slope = ___ N 4.40 N − 0.80 N ΔF By analyzing the units of the equation F = k(L − L0), it is clear that the slope cannot be the spring constant; k has the same units as weight divided by length (N/cm). Is the slope equal to 1/k? The units would be correct for that case. To be sure, we solve the equation of the line for L: F+L L = __ 0 k

Practice Problem 1.10 Another Weight on Spring

We recognize the equation of a line with a slope of 1/k. Therefore,

What is the length of the spring of Example 1.10 when a weight of 8.00 N is suspended? Assume that the relationship found in Example 1.10 still holds for this weight.

1 = 0.277 N/cm k = _________ 3.61 cm/N L (cm)

35.0 (4.40 N, 25.0 cm)

30.0 25.0 20.0

∆ L = 13.0 cm

15.0

∆ F = 3.60 N

10.0 (0.80 N, 12.0 cm)

5.0

Figure 1.5

0.0

1.00

Spring length versus weight hanging.

2.00

3.00

4.00

5.00

6.00 F (N)

Master the Concepts • Terms used in physics must be precisely defined. A term may have a different meaning in physics from the meaning of the same word in other contexts. • A working knowledge of algebra, geometry, and trigonometry is essential in the study of physics. • The factor by which a quantity is increased or decreased is the ratio of the new value to the original value. • When we say that A is proportional to B (written A ∝ B), we mean that if B increases by some factor, then A must increase by the same factor.

• In scientific notation, a number is written as the product of a number between 1 and 10 and a whole-number power of ten. • Significant figures are the basic grammar of precision. They enable us to communicate quantitative information and indicate the precision to which that information is known. • When two or more quantities are added or subtracted, the result is as precise as the least precise of the quantities. When quantities are multiplied or divided, the result has continued on next page

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CHAPTER 1 Introduction

Master the Concepts continued

the same number of significant figures as the quantity with the smallest number of significant figures. • Order-of-magnitude estimates and calculations are made to be sure that the more precise calculations are realistic. • The units used for scientific work are those from the Système International (SI). SI uses seven base units, which include the meter (m), the kilogram (kg), and the second (s) for length, mass, and time, respectively. Using combinations of the base units, we can construct other derived units. • If the statement of a problem includes a mixture of different units, the units should be converted to a single, consistent set before the problem is solved. Usually the best way is to convert everything to SI units.

Conceptual Questions 1. Give a few reasons for studying physics. 2. Why must words be carefully defined for scientific use? 3. Why are simplified models used in scientific study if they do not exactly match real conditions? 4. By what factor does tripling the radius of a circle increase (a) the circumference of the circle? (b) the area of the circle? 5. What are some of the advantages of scientific notation? 6. After which numeral is the decimal point usually placed in scientific notation? What determines the number of numerical digits written in scientific notation? 7. Are all the digits listed as “significant figures” precisely known? Might any of the significant digits be less precisely known than others? Explain. 8. Why is it important to write quantities with the correct number of significant figures? 9. List three of the base units used in SI. 10. What are some of the differences between the SI and the customary U.S. system of units? Why is SI preferred for scientific work? 11. Sort the following units into three groups of dimensions and identify the dimensions: fathoms, grams, years, kilometers, miles, months, kilograms, inches, seconds. 12. What are the first two steps to be followed in solving almost any physics problem? 13. Why do scientists plot graphs of their data instead of just listing values? 14. A student’s lab report concludes, “The speed of sound in air is 327.” What is wrong with that statement?

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• Dimensional analysis is used as a quick check on the validity of equations. Whenever quantities are added, subtracted, or equated, they must have the same dimensions (although they may not necessarily be given in the same units). • Mathematical approximations aid in simplifying complicated problems. • Problem-solving techniques are skills that must be practiced to be learned. • A graph is plotted to give a picture of the data and to show how one variable changes with respect to another. Graphs are used to help us see a pattern in the relationship between two variables. • Whenever possible, make a careful choice of the variables plotted so that the graph displays a linear relationship.

15. Once the solution of a problem has been found, what should be done before moving on to solve another problem?

Multiple-Choice Questions 1. One kilometer is approximately (a) 2 miles (b) 1/2 mile (c) 1/10 mile (d) 1/4 mile 2. 55 mi/h is approximately (a) 90 km/h (b) 30 km/h (c) 10 km/h (d) 2 km/h 3. By what factor does the volume of a cube increase if the length of the edges are doubled? __ (a) 16 (b) 8 (c) 4 (d) 2 (e) √ 2 4. If the length of a box is reduced to one third of its original value and the width and height are doubled, by what factor has the volume changed? (a) 2/3 (b) 1 (c) 4/3 (d) 3/2 (e) depends on relative proportion of length to height and width 5. If the area of a circle is found to be half of its original value after the radius is multiplied by a certain factor, what was the factor used? __ __ (b) 1/2 (c) √ 2 (d) 1/√ 2 (e) 1/4 (a) 1/(2p) 6. In terms of the original diameter d, what new diameter will result in a new spherical volume that is a factor of eight times the original volume? __ (a) 8d (b) 2d (c) d/2 (d) d × 3√ 2 (e) d/8 7. An equation for potential energy states U = mgh. If U is in kg⋅m2⋅s−2, m is in kg, and g is in m⋅s−2, what are the units of h? (a) s

(b) s2

(c) m−1

(d) m

(e) g−1

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PROBLEMS

8. The equation _______ for the speed of sound in a gas states that v = √ g k BT/m . Speed v is measured in m/s, g is a dimensionless constant, T is temperature in kelvins (K), and m is mass in kg. What are the units for the Boltzmann constant, kB? (a) kg⋅m2⋅s2⋅K (b) kg⋅m2⋅s−2⋅K−1 (c) kg−1⋅m−2⋅s2⋅K (d) kg⋅m/s (e) kg⋅m2⋅s−2 9. How many significant figures should be written in the sum 4.56 g + 9.032 g + 580.0078 g + 540.439 g? (a) 3 (b) 4 (c) 5 (d) 6 (e) 7 10. How many significant figures should be written in the product 0.007 840 6 m × 9.450 20 m? (a) 3 (b) 4 (c) 5 (d) 6 (e) 7

Problems Combination conceptual/quantitative problem Biological or medical application

✦ Blue # 1

2

Challenging problem Detailed solution in the Student Solutions Manual Problems paired by concept Text website interactive or tutorial

1.3 The Use of Mathematics 1. The gardener is told that he must increase the height of his fences 37% if he wants to keep the deer from jumping in to eat the foliage and blossoms. If the current fence is 1.8 m high, how high will the new fence be? 2. What is the ratio of the number of seconds in a day to the number of hours in a day? 3. A spherical balloon expands when it is taken from the cold outdoors to the inside of a warm house. If its surface area increases 16.0%, by what percentage does the radius of the balloon change? 4. A spherical balloon is partially blown up and its surface area is measured. More air is then added, increasing the volume of the balloon. If the surface area of the balloon expands by a factor of 2.0 during this procedure, by what factor does the radius of the balloon change? ( tutorial: car on curve) 5. For any cube with edges of length s, what is the ratio of the surface area to the volume? 6. Samantha is 1.50 m tall on her eleventh birthday and 1.65 m tall on her twelfth birthday. By what factor has her height increased? By what percentage? 7. The “scale” of a certain map is 1/10 000. This means the length of, say, a road as represented on the map is 1/10 000 the actual length of the road. What is the ratio of the area of a park as represented on the map to the actual area of the park? ( tutorial: scaling)

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8. On Monday, a stock market index goes up 5.00%. On Tuesday, the index goes down 5.00%. What is the net percentage change in the index for the two days? 9. According to Kepler’s third law, the orbital period T of a planet is related to the radius R of its orbit by T 2 ∝ R3. Jupiter’s orbit is larger than Earth’s by a factor of 5.19. What is Jupiter’s orbital period? (Earth’s orbital period is 1 yr.) 10. If the radius of a circular garden plot is increased by 25%, by what percentage does the area of the garden increase? 11. A poster advertising a student election candidate is too large according to the election rules. The candidate is told she must reduce the length and width of the poster by 20.0%. By what percentage will the area of the poster be reduced? 12. An architect is redesigning a rectangular room on the blueprints of the house. He decides to double the width of the room, increase the length by 50%, and increase the height by 20%. By what factor has the volume of the room increased?

1.4 Scientific Notation and Significant Figures 13. Perform these operations with the appropriate number of significant figures. (a) 3.783 × 106 kg + 1.25 × 108 kg (b) (3.783 × 106 m) ÷ (3.0 × 10−2 s) 14. Write these numbers in scientific notation: (a) the U.S. population, 290 000 000; (b) the diameter of a helium nucleus, 0.000 000 000 000 003 8 m. 15. In the following calculations, be sure to use an appropriate number of significant figures. (a) 3.68 × 107 g − 4.759 × 105 g 6.497 × 104 m2 (b) _____________ 5.1037 × 102 m 16. Write your answer to the following problems with the appropriate number of significant figures. (a) 6.85 × 10−5 m + 2.7 × 10−7 m (b) 702.35 km + 1897.648 km (c) 5.0 m × 4.3 m (d) (0.04/p) cm (e) (0.040/p) m 17. Solve the following problem and express the answer in scientific notation with the appropriate number of significant figures: (3.2 m) × (4.0 × 10−3 m) × (1.3 × 10−8 m). 18. How many significant figures are in each of these measurements? (a) 7.68 g (b) 0.420 kg (c) 0.073 m (d) 7.68 × 105 g 3 (e) 4.20 × 10 kg (f) 7.3 × 10−2 m 4 (g) 2.300 × 10 s

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CHAPTER 1 Introduction

19. Solve the following problem and express the answer in meters per second (m/s) with the appropriate number of significant figures. (3.21 m)/(7.00 ms) = ? [Hint: Note that ms stands for milliseconds.] 20. Solve the following problem and express the answer in meters with the appropriate number of significant figures and in scientific notation: 3.08 × 10−1 km + 2.00 × 103 cm

32. (a) How many square centimeters are in 1 square foot? (1 in. = 2.54 cm.) (b) How many square centimeters are in 1 square meter? (c) Using your answers to parts (a) and (b), but without using your calculator, roughly how many square feet are in one square meter? 33. A snail crawls at a pace of 5.0 cm/min. Express the snail’s speed in (a) ft/s and (b) mi/h. 34. An average-sized capillary in the human body has a cross-sectional area of about 150 μm2. What is this area in square millimeters (mm2)?

1.5 Units 21. A cell membrane is 7.0 nm thick. How thick is it in inches? 22. The label on a small soda bottle lists the volume of the drink as 355 mL. (a) How many fluid ounces are in the bottle? A competitor’s drink is labeled 16.0 fl oz. (b) How many milliliters are in that drink? 23. The length of the river span of the Brooklyn Bridge is 1595.5 ft. The total length of the bridge is 6016 ft. Find the length and the order of magnitude in meters of (a) the river span and (b) the total bridge length? 24. Convert 1.00 km/h to meters per second (m/s). 25. A sprinter can run at a top speed of 0.32 miles per minute. Express her speed in (a) m/s and (b) mi/h. 26. The first modern Olympics in 1896 had a marathon distance of 40 km. In 1908, for the Olympic marathon in London, the length was changed to 42.195 km to provide the British royal family with a better view of the race. This distance was adopted as the official marathon length in 1921 by the International Amateur Athletic Federation. What is the official length of the marathon in miles? 27. At the end of 2006 an expert economist from the Global Economic Institute in Kiel, Germany, predicted a drop in the value of the dollar against the euro of 10% over the next 5 years. If the exchange rate was $1.27 to 1 euro on November 5, 2006, and was $1.45 to 1 euro on November 5, 2007, what was the actual drop in the value of the dollar over the first year? 28. The intensity of the Sun’s radiation that reaches Earth’s atmosphere is 1.4 kW/m2 (kW = kilowatt; W = watt). Convert this to W/cm2. 29. Density is the ratio of mass to volume. Mercury has a density of 1.36 × 104 kg/m3. What is the density of mercury in units of g/cm3? 30. A molecule in air is moving at a speed of 459 m/s. How many meters would the molecule move during 7.00 ms (milliseconds) if it didn’t collide with any other molecules? 31. Express this product in units of km3 with the appropriate number of significant figures: (3.2 km) × (4.0 m) × (13 × 10−3 mm).

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1.6 Dimensional Analysis 35. An equation for potential energy states U = mgh. If U is in joules, with m in kg, h in m, and g in m/s2, find the combination of SI base units that are equivalent to joules. 36. One equation involving force states that Fnet = ma, where Fnet is in newtons, m is in kg, and a is in m⋅s−2. Another equation states that F = −kx, where F is in newtons, k is in kg⋅s−2, and x is in m. (a) Analyze the dimensions of ma and kx to show they are equivalent. (b) What are the dimensions of the force unit newton? 37. An equation for the period T of a planet (the time to make one orbit about the Sun) is 4p 2r3/(GM), where T is in s, r is in m, G is in m3/(kg⋅s2), and M is in kg. Show that the equation is dimensionally correct. 38. The relationship between kinetic energy K (SI unit kg⋅m2⋅s−2) and momentum p is K = p2/(2m), where m stands for mass. What is the SI unit of momentum? 39. An expression for buoyant force is FB = rgV, where FB has dimensions [MLT −2], r (density) has dimensions [ML−3], and g (gravitational field strength) has dimensions [LT−2]. (a) What must be the dimensions of V? (b) Which could be the correct interpretation of V: velocity or volume? 40. Use dimensional analysis to determine how the linear speed (v in m/s) of a particle traveling in a circle depends on some, or all, of the following properties: r is the radius of the circle; w is an angular frequency in s−1 with which the particle orbits about the circle, and m is the mass of the particle. There is no dimensionless constant involved in the relation.

1.8 Approximation 41. What is the approximate distance from your eyes to a book you are reading? 42. What is the approximate volume of your physics textbook in cubic centimeters (cm3)? 43. (a) Estimate the average mass of a person’s leg. (b) Estimate the length of a full-size school bus.

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PROBLEMS

44. Estimate the number of times a human heart beats during its lifetime. 45. Estimate the number of automobile repair shops in the city you live in by considering its population, how often an automobile needs repairs, and how many cars each shop can service per day. Then look in the yellow pages of your phone directory to see how accurate your estimate is. By what percentage was your estimate off? 46. What is the order of magnitude of the number of seconds in one year? 47. What is the order of magnitude of the height (in meters) of a 40-story building?

1.9 Graphs 48. You have just performed an experiment in which you measured many values of two quantities, A and B. According to theory, A = cB3 + A0. You want to verify that the values of c and A0 are correct by making a graph of your data that enables you to determine their values from a slope and a vertical axis intercept. What quantities do you put on the vertical and horizontal axes of the plot? 49. A nurse recorded the values shown in the temperature chart for a patient’s temperature. Plot a graph of temperature versus elapsed time and from the graph find (a) an estimate of the temperature at noon and (b) the slope of the graph. (c) Would you expect the graph to follow the same trend over the next 12 hours? Explain.

Weight (lb) 6.6 7.4 9.6 11.2 12.0 13.6 13.8 14.8 15.0 16.6 17.5 18.4

Temp (°F) 100.00 100.45 100.90 101.35 102.48

50. A graph of x versus t4, with x on the vertical axis and t4 on the horizontal axis, is linear. Its slope is 25 m/s4 and its vertical axis intercept is 3 m. Write an equation for x as a function of t. 51. A patient’s temperature was 97.0°F at 8:05 a.m. and 101.0°F at 12:05 p.m. If the temperature change with respect to elapsed time was linear throughout the day, what would the patient’s temperature be at 3:35 p.m.? 52. The weight of a baby measured over an 11-mon period is given in the weight chart for this problem. (a) Plot the baby’s weight versus age over the 11 mon. (b) What was the average monthly weight gain for this baby over the period from birth to 5 mon? How do you find this value from the graph? (c) What was the average monthly weight gain for the baby over the period from 5 mon to 10 mon? (d) If a baby continued to grow at the same rate as in the first five months of life, what would the child weigh at age 12 yr?

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Decays/s

15

30

45

60

75

90

405

237

140

90

55

32

19

(a) Plot the decays per second versus time. (b) Plot the natural logarithm of the decays per second versus the time. Why might the presentation of the data in this form be useful? 56. An object is moving in the x-direction. A graph of the distance it has moved as a function of time is shown. (a) What are the slope and vertical axis intercept? (Be sure to include units.) (b) What physical significance do the slope and intercept on the vertical axis have for this graph?

20 Distance (km)

Time

0 (birth) 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0

53. A physics student plots results of an experiment as v versus t. The equation that describes the line is given by at = v − v0. (a) What is the slope of this line? (b) What is the vertical axis intercept of this line? 54. A linear plot of speed versus elapsed time has a slope of 6.0 m/s2 and a vertical intercept of 3.0 m/s. (a) What is the change in speed in the time interval between 4.0 s and 6.0 s? (b) What is the speed when the elapsed time is equal to 5.0 s? 55. In a laboratory you measure the decay rate of a sample of radioactive carbon. You write down the following measurements: Time (min)

10:00 a.m. 10:30 a.m. 11:00 a.m. 11:30 a.m. 12:45 p.m.

Age (mon)

15 10 5 0

5.0 Time (h)

10.0

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CHAPTER 1 Introduction

Comprehensive Problems

If the volume of saliva coughed onto you by your friend with the flu is 0.010 cm3 and 10−9 of that volume consists of viral particles, how many influenza viruses have just landed on you? 67. The smallest “living” thing is probably a type of infectious agent known as a viroid. Viroids are plant pathogens that consist of a circular loop of single-stranded RNA, containing about 300 bases. (Think of the bases as beads strung on a circular RNA string.) The distance from one base to the next (measured along the circumference of the circular loop) is about 0.35 nm. What is the diameter of a viroid in (a) m, (b) μm, and (c) in.?

57. It is useful to know when a small number is negligible. Perform the following computations. (a) 186.300 + 0.0030 (b) 186.300 − 0.0030 (c) 186.300 × 0.0030 (d) 186.300/0.0030 (e) For cases (a) and (b), what percent error will result if you ignore the 0.0030? Explain why you can never ignore the smaller number, 0.0030, for case (c) and case (d)? (f) What rule can you make about ignoring small values? 58. The weight of an object at the surface of a planet is pro✦ portional to the planet’s mass and inversely proportional to the square of the radius of the planet. Jupiter’s radius 68. The largest living creature on Earth is the blue whale, is 11 times Earth’s and its mass is 320 times Earth’s. An which has an average length of 70 ft. The largest blue apple weighs 1.0 N on Earth. How much would it weigh whale on record (and therefore the largest animal ever on Jupiter? found) was 1.10 × 102 ft long. (a) Convert this length 59. In cleaning out the artery of a patient, a doctor to meters. (b) If a double-decker London bus is 8.0 m increases the radius of the opening by a factor of 2.0. long, how many double-decker-bus lengths is the By what factor does the cross-sectional area of the record whale? artery change? ✦60. A scanning electron micrograph of xylem vessels in a corn root shows the vessels magnified by a factor of 600. In the micrograph the xylem vessel is 3.0 cm in diameter. (a) What is the diameter of the vessel itself? (b) By what factor has the cross-sectional area of the vessel been increased in the micrograph? 61. The average speed of a nitrogen molecule in air is pro✦ 69. The record blue whale in Problem 68 had a mass of portional to the square root of the temperature in kel1.9 × 105 kg. Assuming that its average density was vins (K). If the average speed is 475 m/s on a warm 0.85 g/cm3, as has been measured for other blue whales, summer day (temperature = 300.0 K), what is the averwhat was the volume of the whale in cubic meters (m3)? age speed on a cold winter day (250.0 K)? (Average density is the ratio of mass to volume.) 62. A furlong is 220 yd; a fortnight is 14 d. How fast is 70. A sheet of paper has length 27.95 cm, width 8.5 in., and 1 furlong per fortnight (a) in μm/s? (b) in km/day? thickness 0.10 mm. What is the volume of a sheet of 63. Given these measurements, identify the number of sigpaper in m3? (Volume = length × width × thickness.) nificant figures and rewrite in scientific notation. ✦71. An object moving at constant speed v around a circle of (a) 0.00574 kg (b) 2 m (c) 0.450 × 10−2 m radius r has an acceleration a directed toward the center (d) 45.0 kg (e) 10.09 × 104 s (f) 0.09500 × 105 mL of the circle. The SI unit of acceleration is m/s2. (a) Use 64. A car has a gas tank that holds 12.5 U.S. gal. Using the dimensional analysis to find a as a function of v and r. conversion factors from the inside front cover, (a) deter(b) If the speed is increased 10.0%, by what percentage mine the size of the gas tank in cubic inches. (b) A cubit is does the radial acceleration increase? an ancient measurement of length that was defined as the ✦ 72. The speed of ocean waves depends on their wavelength distance from the elbow to the tip of the finger, about 18 in. l (measured in meters) and the gravitational field long. What is the size of the gas tank in cubic cubits? strength g (measured in m/s2) in this way: 65. You are given these approximate measurements: (a) the radius of Earth is 6 × 106 m, (b) the length of a human v = Kl pgq body is 6 ft, (c) a cell’s diameter is 10−6 m, (d) the width of the hemoglobin molecule is 3 × 10−9 m, and (e) the where K is a dimensionless constant. Find the values of distance between two atoms (carbon and nitrogen) is the exponents p and q. 3 × 10−10 m. Write these measurements in the simplest 73. In the United States, we often use miles per hour (mi/h) possible metric prefix forms (in either nm, Mm, μm, or when discussing speed, but the SI unit of speed is m/s. whatever works best). What is the conversion factor for changing m/s to 66. A typical virus is a packet of protein and DNA (or mi/h? If you want to make a quick approximation of the RNA) and can be spherical in shape. The influenza A speed in mi/h given the speed in m/s, what might be the virus is a spherical virus that has a diameter of 85 nm. easiest conversion factor to use?

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COMPREHENSIVE PROBLEMS

23

✦ 74. How many cups of water are required to fill a bathtub? ✦ 84. Use dimensional analysis to determine how ✦ 75. Without looking up any data, make an order-ofthe period T of a swingmagnitude estimate of the annual consumption of gasing pendulum (the oline (in gallons) by passenger cars in the United elapsed time for a comStates. Make reasonable order-of-magnitude estimates L plete cycle of motion) for any quantities you need. Think in terms of average depends on some, or quantities. (1 gal ≈ 4 L.) all, of these properties: 76. Some thieves, escaping after a bank robbery, drop a sack Pendulum m bob the length L of the penof money on the sidewalk. (a) Estimate the mass of the dulum, the mass m of sack if it contains $5000 in half-dollar coins. (b) Estimate the pendulum bob, and the mass if the sack contains $1 000 000 in $20 bills. the gravitational field 77. The weight W of an object is given by W = mg, where m strength g (in m/s2). Assume that the amplitude of the is the object’s mass and g is the gravitational field swing (the maximum angle that the string makes with the strength. The SI unit of field strength g, expressed in SI vertical) has no effect on the period. base units, is m/s2. What is the SI unit for weight, ✦ 85. The Space Shuttle astronauts use a massing chair to expressed in base units? measure their mass. The chair is attached to a spring 78. Kepler’s law of planetary motion says that the square of and is free to oscillate back and forth. The frequency of the period of a planet (T 2) is proportional to the cube of the oscillation is measured and that is used to calculate the distance of the planet from the Sun (r 3). Mars is the total mass m attached to the spring. If the spring about twice as far from the Sun as Venus. How does the constant of the spring k is measured in kg/s2 and the period of Mars compare with the period of Venus? chair’s frequency f is 0.50 s−1 for a 62-kg astronaut, 79. One morning you read in the New York Times that the what is the chair’s frequency for a 75-kg astronaut? net worth of the richest man in the world, Carlos Slim The chair itself has a mass of 10.0 kg. [Hint: Use Helu of Mexico, is $59 000 000 000. Later that day you dimensional analysis to find out how f depends on see him on the street, and he gives you a $100 bill. What m and k.] is his net worth now? (Think of significant figures.) 86. The average depth of the oceans is about 4 km, ✦80. Estimate the number of hairs on the average human and oceans cover about 70% of Earth’s surface. Make head. [Hint: Consider the number of hairs in an area of an order-of-magnitude estimate of the volume of water 1 in.2 and then consider the area covered by hair on the in the oceans. Do not look up any data in books. (Use head.] your ingenuity to estimate the radius or circumference of Earth.) 81. Suppose you have a pair of Seven League Boots. These are magic boots that enable you to stride along a distance ✦ 87. The population of a culture of yeast cells is studied in of 7.0 leagues with each step. (a) If you march along at a the laboratory to see the effects of limited resources military march pace of 120 paces/min, what will be your (food, space) on population growth. At 2-h intervals, speed in km/h? (b) Assuming you could march on top of the size of the population (measured as total mass of the oceans when you step off the continents, how long (in yeast cells) is recorded (see table on p. 24). (a) Make minutes) will it take you to march around the Earth at the a graph of the yeast population as a function of elapsed equator? (1 league = 3 mi = 4.8 km.) time. Draw a best-fit smooth curve. (b) Notice from the graph of part (a) that after a long time, the popula✦ 82. The electrical power P drawn from a generator by a tion asymptotically approaches a maximum known as lightbulb of resistance R is P = V2/R, where V is the line the carrying capacity. From the graph, estimate the voltage. The resistance of bulb B is 42% greater than carrying capacity for this population. (c) When the the resistance of bulb A. What is the ratio PB/PA of the population is much smaller than the carrying capacity, power drawn by bulb B to the power drawn by bulb A if the growth is expected to be exponential: m(t) = m0ert, the line voltages are the same? where m is the population at any time t, m0 is the ini✦ 83. Three of the fundamental constants of physics are the 8 tial population, r is the intrinsic growth rate (i.e., the speed of light, c = 3.0 × 10 m/s, the universal gravita−11 3 −1 −2 growth rate in the absence of limits), and e is the base tional constant, G = 6.7 × 10 m ⋅kg ⋅s , and Planck’s −34 2 −1 of natural logarithms (see Appendix A.3). To obtain a constant, h = 6.6 × 10 kg⋅m ⋅s . straight line graph from this exponential relationship, (a) Find a combination of these three constants that has we can plot the natural logarithm of m/m0: the dimensions of time. This time is called the Planck rt m time and represents the age of the universe before which ln ___ m 0 = ln e = rt the laws of physics as presently understood cannot be Make a graph of ln (m/m 0) versus t from t = 0 to applied. (b) Using the formula for the Planck time t = 6.0 h, and use it to estimate the intrinsic growth derived in part (a), what is the time in seconds?

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CHAPTER 1 Introduction

rate r for the yeast population. (The term ln stands for the natural logarithm; see Appendix A.3 if you need help with natural logs.) Time (h) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0

Mass (g) 3.2 5.9 10.8 19.1 31.2 46.5 62.0 74.9 83.7 89.3 92.5 94.0 95.1

Answers to Practice Problems 1.1 81.0 W 1.2 (a) five; 1.0544 × 10−4 kg; (b) four; 5.800 × 10−3 cm; (c) ambiguous, three to six; if three, 6.02 × 105 s 1.3 The least precise value is to the nearest hundredth of a meter, so we round the result to the nearest hundredth of a meter: 564.50 m or, in scientific notation, 5.6450 × 102 m; five significant figures.

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1.4 4.7 m/s 1.5 (a) 35.6 m/s; (b) 79.5 mi/h 1.6 5.1 × 1014 m2; 2.0 × 108 mi2 1.7 The equation is dimensionally inconsistent; the right side has dimensions [L/T]. To have matching dimensions we must multiply the right side by [T]; the equation must involve time squared: d = _12 at2. 1.8 kinetic energy = (constant) × mv2; kinetic energy increases by 56%. 1.9 1011 L (Make a rough estimate of the population to be about 3 × 108 people, each drinking about 1.5 L/day.) 1.10 38.3 cm

Answers to Checkpoints 1.3 The volume increases by a factor of 27. 1.4 Order-of-magnitude estimates provide a quick method for obtaining limited precision solutions to problems. Even if greater accuracy is required, order-of-magnitude calculations are still useful as they provide a check as to the accuracy of the higher precision calculation. 1.5 1.9 L 1.6 (a) and (b) It is possible to multiply or divide quantities with different dimensions. (c) and (d) To be added or subtracted, quantities must have the same dimensions.

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PART ONE

Mechanics

CHAPTER

Motion Along a Line

2

Despite its enormous mass (425 to 900 kg), the Cape buffalo is capable of running at a top speed of about 55 km/h (34 mi/h). Since the top speed of the African lion is about the same, how is it ever possible for a lion to catch the buffalo, especially since the lion typically makes its move from a distance of 20 to 30 m from the buffalo? (See p. 34 for the answer.)

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CHAPTER 2 Motion Along a Line

Concepts & Skills to Review

• • • •

scientific notation and significant figures (Section 1.4) converting units (Section 1.5) problem-solving techniques (Section 1.7) meaning of velocity in physics (Section 1.2)

2.1

POSITION AND DISPLACEMENT

Position

CONNECTION: The topic of Chapters 2 and 3 is kinematics: the mathematical description of motion. Beginning in Chapter 4, we will learn the principles of physics that predict and explain why objects move the way they do.

To describe motion unambiguously, we need a way to say where an object is located. Suppose that at 3:00 p.m. a train stops on an east-west track as a result of an engine problem. The engineer wants to call the railroad office to report the problem. How can he tell them where to find the train? He might say something like “three kilometers east of the old trestle bridge.” Notice that he uses a point of reference: the old trestle bridge. Then he states how far the train is from that point and in what direction. If he omits any of the three pieces (the reference point, the distance, and the direction), then his description of the train’s whereabouts is ambiguous. The same thing is done in physics. First, we choose a reference point, called the origin. Then, to describe the location of something, we give its distance from the origin and the direction. For motion along a line, we can choose the line of motion to be the x-axis of a coordinate system. The origin is the point x = 0. The position of an object can be described by its x-coordinate, which tells us both how far the object is from the origin and on which side. For the train in Fig. 2.1, we choose the origin at the center of the bridge and the +x-direction to the east. Then x = +3 km means the train is 3 km east of the bridge and x = −26 km means the train is 26 km west of the bridge.

Displacement Once the train’s engine is repaired and it goes on its way, we might want to describe its motion. At 3:14 p.m., it leaves its initial position, 3 km east of the origin (see Fig. 2.1). At 3:56 p.m., the train is 26 km west of the origin, which is 29 km to the west of its initial position. Displacement is defined as the change of the position—the final position minus the initial position. The displacement is written Δx where the symbol Δ (the uppercase Greek letter delta) means the change in the quantity that follows.

Displacement: final position minus initial position

Displacement: Δx = x f − x i

Final position 3:56 P.M.

Origin

–26 km

xf = –26 km

(2-1)

Initial position 3:14 P.M.

10 km

3 km

0 W

E

xi = +3 km

+x

Trestle bridge

Figure 2.1 Initial (xi) and final (xf) positions of a train. (Train not to scale.)

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2.1

27

POSITION AND DISPLACEMENT

Final position 3:56 P.M. xf = –26 km

Initial position 3:14 P.M. xi = +3 km

∆ x = xf – xi = –29 km (29 km west)

x

Figure 2.2 With the x-axis pointing east, Δx = xf − xi = −26 km − (+3 km) = −29 km. The train’s displacement is 29 km west.

We can subtract x-coordinates to find the displacement of the train. If we choose the x-axis to the east, then xi = +3 km and xf = −26 km. The displacement is Δx = x f − x i = (−26 km) − (+3 km) = −29 km The displacement is 29 km in the −x-direction (west) (Fig. 2.2). Displacement Versus Distance Notice that the magnitude of the displacement is not necessarily equal to the distance traveled. Suppose the train first travels 7 km to the east, putting it 10 km east of the origin, and then reverses direction and travels 36 km to the west. The total distance traveled in that case is (7 km + 36 km) = 43 km, but the magnitude of the displacement—which is the distance between the initial and final positions—is 29 km. The displacement depends only on the starting and ending positions, not on the path taken.

Example 2.1 A Mule Hauling Corn to Market A mule hauls the farmer’s wagon along a straight road for 4.3 km directly east to the neighboring farm where a few bushels of corn are loaded onto the wagon. Then the farmer drives the mule back along the same straight road, heading west for 7.2 km to the market. Find the displacement of the mule from the starting point to the market. (The train first travels 7 km to the east, then reverses direction and travels 36 km to the west.) Strategy The problem gives us two successive displacements along a straight line. Let’s choose the +x-axis to point east and an arbitrary point along the road to be the origin. Suppose the mule starts at position x1 (Fig. 2.3). It goes east until it reaches the neighbor’s farm at position x2. The displacement to the neighbor’s farm is x2 − x1 = 4.3 km east. North y x3 – x1 x3 – x2 x2 – x1 x3

x1

Origin

x2

x East

Figure 2.3 The total displacement is the sum of two successive displacements.

gia04535_ch02_025-054.indd 27

Then the mule goes 7.2 km west to reach the market at position x3. The displacement from the neighbor’s farm to the market is x3 − x2 = −7.2 km (negative because the displacement is in the −x-direction). The problem asks for the displacement of the mule from x1 to x3. Solution We can eliminate x2, the intermediate position, by adding the two displacements: (x 3 − x 2) + (x 2 − x 1) = −7.2 km + 4.3 km x 3 − x 1 = −2.9 km The displacement is 2.9 km west. Discussion When we added the two displacements, the intermediate position x2 dropped out, as it must since the displacement is independent of the path taken from the initial position to the final position. The result does not depend on the choice of origin.

Practice Problem 2.1 A Nervous Squirrel A nervous squirrel, trying to cross a road, first moves 3.0 m east, then 4.0 m west, then 1.2 m west, then 6.0 m east. What is the squirrel’s total displacement?

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CHAPTER 2 Motion Along a Line

Adding Displacements Generalizing the result of Example 2.1, the total displacement for a trip with several parts is the sum of the displacements for each part of the trip. Although x-coordinates depend on the choice of origin, displacements (changes in x-coordinates) do not depend on the choice of origin.

CHECKPOINT 2.1 In Example 2.1, is the magnitude of the displacement equal to the distance traveled? Explain.

2.2

VELOCITY: RATE OF CHANGE OF POSITION

We introduced velocity as a quantity with magnitude and direction in Section 1.2. The magnitude is the speed with which the object moves and the direction is the direction of motion. Now we develop a mathematical definition of velocity that fits that description. Note that displacement indicates by how much and in what direction the position has changed, but implies nothing about how long it took to move from one point to the other. Velocity depends on both the displacement and the time interval.

Average Velocity Reminder: the symbol Δ stands for the change in. If the initial value of a quantity Q is Qi and the final value is Qf, then ΔQ = Qf − Qi. ΔQ is read “delta Q.”

When a displacement Δx occurs during a time interval Δt, the average velocity during that time interval is Average velocity: Δx v av,x = ___ Δt

(2-2)

Since Δt is always positive, the direction of the average velocity is the same as the direction of the displacement. The symbol Δ does not stand alone and cannot be canceled in equations because it xf − xi Δx means ______ modifies the quantity that follows it; ___ , which is not the same as x/t. tf − ti Δt

Example 2.2 Average Velocity of a Train Find the average velocity in kilometers per hour of the train shown in Fig. 2.1 during the time interval between 3:14 p.m., when the train is 3 km east of the origin, and 3:56 p.m., when it is 26 km west of the origin. Strategy We choose the +x-axis to the east, as before. Then the displacement is Δx = −29 km, which means 29 km to the west. The average velocity is also to the west, so vav,x is negative. We convert Δt to hours to find the average velocity in kilometers per hour.

Solution The time interval is Δt = 56 min −14 min = 42 min. Converting to hours, 1 h = 0.70 h Δt = 42 min × ______ 60 min The average velocity is −29 km = − 41 km/h Δx = _______ v av,x = ___ 0.70 h Δt The negative sign means that the average velocity is directed along the negative x-axis, or to the west. continued on next page

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2.2 VELOCITY: RATE OF CHANGE OF POSITION

29

Example 2.2 continued

Discussion If the train had started at the same instant of time, 3:14 p.m., and had traveled directly west at a constant 41 km/h, it would have ended up in exactly the same place— 26 km west of the trestle bridge—at 3:56 p.m. Had we started measuring time from when we first spotted the motionless train at 3:00 p.m., instead of 3:14 p.m., we would have found the average velocity over a different time interval, changing the average velocity. The average velocity depends on the time interval considered. The magnitude of the train’s average velocity is not equal to the total distance traveled divided by the time interval for the complete trip. The latter quantity is called the average speed:

The distinction arises because the average velocity is the constant velocity that would result in the same displacement (during the given time interval), while the average speed is the constant speed that would result in the same distance traveled (during the same time interval).

Practice Problem 2.2 Average Velocity for a Different Time Interval What is the average velocity of the same train during the time interval from 3:28 p.m., when it is at x = 10 km, to 3:56 p.m., when it is at x = −26 km?

distance traveled = ______ 43 km = 61 km/h average speed = ______________ total time 0.70 h

Average Speed Versus Average Velocity The average velocity does not convey detailed information about the motion during the corresponding time interval Δt. The average velocity would be the same for any other motion that takes the object through the same displacement in the same amount of time. However, the average speed, defined as the total distance traveled divided by the time interval, depends on the path traveled.

CHECKPOINT 2.2 Can average speed ever be greater than the magnitude of the average velocity? Explain.

Instantaneous Velocity The speedometer of a car does not indicate the average speed for an entire trip. When a speedometer reads 55 mi/h, it does not necessarily mean that the car travels 55 miles in the next hour; the car could change its speed or direction or stop during that hour. The speedometer reading can be used to calculate how far the car travels during a very short time interval—short enough that the speed does not change appreciably. For instance, at 55 mi/h (= 25 m/s), we can calculate that in 0.010 s the car moves 25 m/s × 0.010 s = 0.25 m—as long as the speed does not change significantly during that 0.010-s interval. Similarly, the instantaneous velocity is a quantity whose magnitude is the speed and whose direction is the direction of motion. The instantaneous velocity can be used to calculate the displacement of the object during a very short time interval, as long as neither the speed nor the direction of motion change significantly during that time interval. Repeating the word instantaneous can get cumbersome. When we refer simply to the velocity, we always mean the instantaneous velocity.

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CHAPTER 2 Motion Along a Line

Thus, the velocity at some instant of time t is the average velocity during a very short time interval:

CONNECTION: Couldn’t we omit “x” subscripts in average (vav,x) and instantaneous (vx) velocity? If we wanted to understand only motion along a line, then we certainly would. However, in Chapter 3 we generalize the definitions of position, displacement, velocity, and acceleration as vector quantities in three dimensions. Using the “x” subscripts now lets us carry forward everything in Chapter 2 without requiring a change in notation. Then, when you look back to review Chapter 2, you won’t have to remember different definitions for the same symbol. For example, in Chapter 3 we’ll learn that v (without the subscript) stands for the magnitude of the velocity (the speed), which can never be negative.

Instantaneous velocity: Δx vx = lim ___ Δt→0 Δt (Δx is the displacement during a very short time interval Δt)

(2-3)

The notation lim is read “the limit, as Δt approaches zero, of . . . .” In other words, Δt→0

let the time interval get smaller and smaller, approaching—but never reaching—zero. This notation in Eq. (2-3) reminds you that Δt must be a very short time interval. How short a time interval is short enough? If you use a shorter time interval and the calculation of vx always gives the same value (to within the precision of your measurements), then Δt is short enough. In other words, Δt must be short enough that we can treat the velocity as constant during that time interval. When vx is constant, cutting Δt in half also cuts the displacement in half, giving the same value for Δx/Δt.

Graphical Relationships Between Position and Velocity For motion along the x-axis, the displacement is Δx. The average velocity can be represented on the graph of x(t) as the slope of a line connecting two points (called a chord). In Fig. 2.4a, the displacement Δx = x3 − x1 is the rise of the graph (the change along the vertical axis) and the time interval Δt = t3 − t1 is the run of the graph (the change along the horizontal axis). The slope of the chord is the rise over the run: Δx = v rise = ___ slope of chord = ____ av,x run Δt The slope of the chord is the average velocity for that time interval.

(2-4)

Slope of tangent gives instantaneous velocity

x2

x1

Slope of chord gives average velocity over time interval from t1 to t3 t1

t2 t3 Time (t) (a)

Position (x)

Position (x)

x3 x2

Slope of chord gives average velocity over a shorter time interval t2 Time (t) (b)

Figure 2.4 A graph of x(t) for an object moving along the x-axis. (a) The average velocity vx,av for the time interval t1 to t3 is the slope of the chord connecting those two points on the graph. (b) The average velocity measured over a shorter time interval. As the time interval gets shorter and shorter, the average velocity approaches the instantaneous velocity vx at the instant t2. The slope of the tangent line to the graph is vx at that instant.

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2.2 VELOCITY: RATE OF CHANGE OF POSITION

Finding vx on a Graph of x(t) To find the instantaneous velocity at some time t = t2, we draw lines showing the average velocity for shorter and shorter time intervals. As the time interval is reduced (Fig. 2.4b), the average velocity changes. As Δt gets shorter and shorter, the chord approaches a tangent line to the graph at t2. Thus, vx is the slope of the line tangent to the graph of x(t) at the chosen time. In Fig. 2.5, the position of the train considered in Example 2.2 is graphed as a function of time, where 3:00 p.m. is chosen as t = 0. The graph of position versus time shows a curving line, but that does not mean the train travels along a curved path. The motion of the train is along a straight line since the track runs in an east-west direction. The graph shows the train’s position as a function of time. A horizontal portion of the graph (as from t = 0 to t = 14 min and from t = 23 min to t = 28 min) indicates that the position is not changing during that time interval and, therefore, it is at rest (its velocity is zero). Sloping portions of the graph indicate that the train is moving. The steeper the graph, the larger the speed of the train. The sign of the slope indicates the direction of motion. A positive slope (t = 14 min to t = 23 min) indicates motion in the +x-direction, and a negative slope (t = 28 min to t = 56 min) indicates motion in the −x-direction.

The slope of the tangent line on a graph of x(t) is vx.

Example 2.3 Velocity of the Train Use Fig. 2.5 to estimate the velocity of the train in kilometers per hour at t = 40 min.

The velocity is approximately 89 km/h in the –x-direction (west).

Strategy Figure 2.5 is a graph of x(t). The slope of a line tangent to the graph at t = 40 min is vx at that instant. After sketching a tangent line on the graph, we find its slope from the rise divided by the run.

Discussion Since the slope of a line is constant, any two points on the tangent line would give the same value for the slope. Using widely spaced points tends to give a more accurate estimate for the slope.

Solution Figure 2.6 shows a tangent line drawn on the graph. Using the endpoints of the tangent line, the rise is (−25 km) − (15 km) = − 40 km. The run is approximately (57 min) − (30 min) = 27 min = 0.45 h. Then

Practice Problem 2.3 Maximum Eastward Velocity

vx ≈ −40 km/(0.45 h) ≈ −89 km/h

0 –10

0 –10 x = –25 km t = 57 min

–20

–20 –30

10 Position x (km)

Position x (km)

x = 15 km t = 30 min x (km) t (min) +3 0 +3 14 +10 23 +10 28 0 40 –26 56

10

Estimate the maximum velocity of the train in kilometers per hour during the time it moves east (t = 14 min to t = 23 min).

–30 0

10

20 30 40 Time t (min)

50

60

10

20 30 40 Time t (min)

50

60

Figure 2.5

Figure 2.6

Graph of position x versus time t for the train. The positions of the train at various times are marked with a dot. The position of the train would have to be measured at more frequent time intervals to accurately trace out the shape of the graph.

On the graph of x(t), the slope of a line tangent to the graph at t = 40 min is vx at t = 40 min.

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CHAPTER 2 Motion Along a Line

Finding ∆x with Constant Velocity What about the other way around? Given a graph of vx(t), how can we determine the displacement (change in position)? If vx is constant during a time interval, then the average velocity is equal to the instantaneous velocity: Δx (for constant v ) (2-5) vx = vav,x = ___ x Δt and therefore

vx v1

∆x

t1

t2

Δx = vx Δt

t

Figure 2.7 Displacement Δx between t1 and t2 is represented by the shaded area under the red vx(t) graph.

Δx is the area under the graph of vx(t). The area is negative when the graph is beneath the time axis (vx < 0).

(for constant vx)

(2-6)

The graph of Fig. 2.7 shows vx versus t for an object moving along the x-axis with constant velocity v1 from time t1 to t2. The displacement Δx during the time interval Δt = t2 − t1 is v1Δt. The shaded rectangle has “height” v1 and “width” Δt. Since the area of a rectangle is the product of the height and width, the displacement Δx is represented by the area of the rectangle between the graph of vx(t) and the time axis for the time interval considered. When we speak of the area under a graph, we are not talking about the literal number of square centimeters of paper or computer screen. The figurative area under a graph usually does not have dimensions of an ordinary area [L2]. In a graph of vx(t), vx has dimensions [L/T] and time has dimensions [T]; areas on such a graph have dimensions [L/T] × [T] = [L], which is correct for a displacement. The units of Δx are determined by the units used on the axes of the graph. If vx is in meters per second and t is in seconds, then the displacement is in meters. Finding ∆x with Changing Velocity What if the velocity is not constant? The displacement Δx during a very small time interval Δt can be found in the same way as for constant velocity since, during a short enough time interval, the velocity does not change appreciably. Then vx and Δt are the height and width of a narrow rectangle (Fig. 2.8a) and the displacement during that short time interval is the area of the rectangle. To find the total displacement during any time interval, the areas of all the narrow rectangles are added together (Fig. 2.8b). To improve the approximation, we let the time interval Δt approach zero and find that the displacement Δx during any time interval equals the area under the graph of vx(t) (Fig. 2.8c). When vx is negative, x is decreasing and the displacement is in the −x-direction, so we must count the area as negative when it is below the time axis. The magnitude of the train’s displacement is represented as the shaded areas in Fig. 2.9. The train’s displacement from t = 14 min to t = 23 min is +7 km (area above the t-axis means displacement in the +x-direction) and from t = 28 min to t = 56 min it is −36 km (area below the t-axis means displacement in the −x-direction). The total displacement from t = 0 to t = 56 min is Δx = (+7 km) + (−36 km) = −29 km.

vx

vx

vx

During a very small ∆t, ∆x = vx ∆t ∆t (a)

t

∆x = area

t1

t2 (b)

t

t2

t1

t

(c)

Figure 2.8 (a) Displacement Δx during a short time interval is approximately the area of a rectangle of height vx and width Δt. (b) During a longer time interval, the displacement is approximately the sum of the areas of the rectangles. (c) The area under the vx versus t graph for any time interval represents the displacement during that interval.

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33

ACCELERATION: RATE OF CHANGE OF VELOCITY

vx(t) 30 (m/s) 20 10 0

t (min)

–10 –20 –30 0

10

20

30

40

50

60

Figure 2.9 A graph of train velocity versus time. The train’s displacement from t = 14 min to t = 23 min is the shaded area under the graph during that time interval. To estimate the area, count the number of grid boxes under the curve, estimating the fraction of the boxes that are only partly below the curve. Each box is 2 m/s in height and 5 min (= 300 s) in width, so each box represents an “area” (displacement) of 2 m/s × 300 s = 600 m = 0.60 km. The total number of shaded boxes for this time interval is about 12, so the displacement is about Δx ≈ 12 × 0.60 km = +7.2 km, which is close to the actual value of 7 km (during this time interval the train went from +3 km to +10 km). The shaded area for the time interval t = 28 min to t = 56 min is below the time axis; this negative area represents displacement in the −x-direction (west). The number of shaded grid boxes in this interval is about 60, so the displacement during this time interval is Δx ≈ −(60) × 0.60 km = −36 km.

2.3

ACCELERATION: RATE OF CHANGE OF VELOCITY

The rate of change of the velocity is called the acceleration. The use of the word acceleration in everyday language is often imprecise and not in accord with its scientific definition. In everyday language, it usually means “an increase in speed” but sometimes is used almost as a synonym for speed itself. In physics, acceleration does not necessarily indicate an increase in speed. Acceleration can indicate any kind of change in velocity. The concept of acceleration is much less intuitive for most people than the concept of velocity. Keep reminding yourself that the acceleration tells you how the velocity is changing. The direction of the change in velocity is not necessarily the same as the direction of either the initial or final velocities.

Average Acceleration The average acceleration during a time interval Δt is: Δv aav,x = ___x (2-7) Δt Since average acceleration is the change in velocity divided by the corresponding time interval, the SI units of acceleration are (m/s)/s = m/s2, read as “meters per second squared.” Thinking of m/s2 as (m/s)/s can help you develop an understanding of what acceleration is. Suppose an object has a constant acceleration ax = +3.0 m/s2. Then vx increases 3.0 m/s during every second of elapsed time (the change in vx is +3.0 m/s per second). If ax = −2.0 m/s2, then vx would decrease 2.0 m/s during every second (the change in vx is −2.0 m/s per second). For example, suppose it takes 30 s for a truck to slow down from 25 m/s to 10 m/s while traveling east. With the x-axis pointing east, the truck’s average acceleration during that time interval is Δv −15 m/s aav,x = ___x = _______ = −0.50 m/s2 30 s Δt or 0.50 m/s2 to the west.

gia04535_ch02_025-054.indd 33

CONNECTION: Compare average acceleration [Eq. (2-7)] and average velocity [Eq. (2-2)]. Each is the change in a quantity divided by the time interval during which the change occurs. Each can have different values for different time intervals.

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CHAPTER 2 Motion Along a Line

Instantaneous Acceleration To find the instantaneous acceleration, we calculate the average acceleration during a very short time interval: Definition of instantaneous acceleration: Δv ax = lim ___x Δt→0 Δt (Δvx is the change in velocity during a very short time interval Δt)

Can the lion catch the buffalo?

(2-8)

The time interval Δt must be short enough that we can treat the acceleration as constant during that time interval. Just as with instantaneous velocity, the word instantaneous is not always repeated. Acceleration without the adjective means instantaneous acceleration. The chapter opener asked how an African lion can ever catch a Cape buffalo. Although Cape buffaloes and African lions have about the same top speed, lions are capable of much larger accelerations than are buffaloes. Starting from rest, it takes a buffalo much longer to get to its top speed. On the other hand, lions have much less stamina. Once the buffalo reaches its top speed, it can maintain that speed much longer than can the lion. Thus, a Cape buffalo is capable of outrunning a lion unless the stalking lion can get fairly close before charging.

Conceptual Example 2.4 Direction of Acceleration While Slowing Down Damon moves in the −x-direction on his motor scooter. He “decelerates” as he approaches a stop sign. While slowing down, is the scooter’s acceleration ax positive or negative? What is the direction of the acceleration? Strategy The acceleration has the same direction as the change in the velocity. Solution and Discussion The term decelerate is not a scientific term. In common usage it means the scooter is slowing: the scooter’s velocity is decreasing in magnitude. Damon is moving in the −x-direction, so vx is negative. He is slowing down, so the absolute value of vx, vx, is getting

smaller. To reduce the magnitude of a negative number, we have to add a positive number. Therefore, the change in vx is positive (Δvx > 0). In other words, vx is increasing. Since Δvx is positive, ax is positive. The acceleration is in the +x-direction.

Conceptual Practice Problem 2.4 Continuing on His Way As Damon pulls away from the stop sign, continuing in the −x-direction, his speed gradually increases. What is the sign of ax? What is the direction of the acceleration?

The Direction of the Acceleration Generalizing Example 2.4, suppose an object moves along the x-axis. When the acceleration is in the same direction as the velocity, the object is speeding up. If vx and ax are both positive, the object is moving in the +x-direction and is speeding up. If they are both negative, the object is moving in the −x-direction and is speeding up. When the acceleration and velocity are in opposite directions, the object is slowing down. When vx is positive and ax is negative, the object is moving in the positive x-direction and is slowing down. When vx is negative and ax is positive, the object is moving in the negative x-direction and is slowing down. In straight-line motion, the acceleration is always in the same direction as the velocity, in the direction opposite to the velocity, or zero.

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35

ACCELERATION: RATE OF CHANGE OF VELOCITY

Figure 2.10 In this graph of

vx (m/s)

2

4

t (s) 6

8

10

vx versus t, as Damon is stopping, vx is negative, but ax (the slope) is positive. The value of vx is increasing, but—since it is less than zero to begin with and is getting closer to zero as time goes on—the speed is decreasing. The slopes of the three tangent lines shown represent the instantaneous accelerations (ax) at three different times.

12

–2

–4

–6

Graphical Relationships Between Velocity and Acceleration

CONNECTION:

Both velocity and acceleration measure rates of change: velocity is the rate of change of position and acceleration is the rate of change of velocity. Therefore, the graphical relationship of acceleration to velocity is the same as the graphical relationship of velocity to position: ax is the slope on a graph of vx(t) and Δvx is the area under a graph of ax(t). Figure 2.10 shows a graph of vx versus t for Damon slowing down on his scooter. He is moving in the −x-direction, so vx < 0, and his speed is decreasing, so vx is decreasing. The slope of a tangent line to the graph is ax at that instant. Three tangent lines are drawn, showing that ax is positive (the slopes are positive) and is not constant (the slopes are not all the same).

On a graph of any quantity Q as a function of time, the slope of the graph represents the instantaneous rate of change of Q. On a graph of the rate of change of Q as a function of time, the area under the graph represents ΔQ.

Example 2.5 Acceleration of a Sports Car A sports car starting at rest can achieve 30.0 m/s in 4.7 s according to the advertisements. Figure 2.11 shows data for vx as a function of time as the sports car starts from rest and travels in a straight line in the +x-direction. (a) What is the average acceleration of the sports car from 0 to 30.0 m/s? (b) What is the maximum acceleration of the car? (c) What is the car’s displacement from t = 0 to t = 19.1 s (when it reaches 60.0 m/s)? (d) What is the car’s average velocity during the entire 19.1 s interval? Strategy (a) To find the average acceleration, the change in velocity for the time interval is divided by the time interval. (b) The instantaneous acceleration is the slope of the velocity graph, so it is maximum where the graph is steepest. At that point, the velocity is changing at a high rate. We expect the maximum acceleration to take place early on; the magnitude of acceleration must decrease as the velocity gets higher and higher—there is a maximum velocity for the car, after all. (c) The displacement Δx is the area under the vx(t) graph. The graph is not a simple shape such as a triangle or rectangle, so an estimate of the area is made. (d) Once we have a value for the displacement, we can apply the definition of average velocity.

vx (m/s) 60.0 Tangent at t = 0

50.0 40.0 30.0

55.0 m/s

20.0 10.0 6.0 s 0

2.0

vx (m/s) 0 t (s)

4.0

6.0

8.0 10.0 12.0 14.0 16.0 18.0 20.0 t (s)

15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 2.0 2.9 3.8 4.9 6.2 7.6 9.1 11.2 14.0 19.1

Figure 2.11 Data table and graph of vx(t) for a sports car. continued on next page

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CHAPTER 2 Motion Along a Line

Example 2.5 continued

(c) Δx is the area under the vx(t) graph shown shaded in Fig. 2.11. The area can be estimated by counting the number of grid boxes under the curve. Each box is 5.0 m/s in height and 2.0 s in width, so each represents an “area” (displacement) of 10 m. When counting the number of boxes under the curve, a best estimate is made for the fraction of the boxes that are only partly below the curve. Approximately 75 boxes lie below the curve, so the displacement is Δx = 75 × 10 m = 750 m. Since the car travels along a straight line and does not change direction, 750 m is also the distance traveled. (d) The average velocity during the 19.1-s interval is

vx (m/s) 60.0 Tangent at t = 0

50.0 40.0 30.0

55.0 m/s

20.0

750 m = 39 m/s Δx = ______ vav,x = ___ Δt 19.1 s

10.0 6.0 s 0

2.0

vx (m/s) 0 t (s)

4.0

6.0

8.0 10.0 12.0 14.0 16.0 18.0 20.0 t (s)

15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 2.0 2.9 3.8 4.9 6.2 7.6 9.1 11.2 14.0 19.1

Figure 2.11 Data table and graph of vx(t) for a sports car.

Given: Graph of vx(t) in Fig. 2.11. To find: (a) aav,x for vx = 0 to 30.0 m/s; (b) maximum value of ax; (c) Δx from vx = 0 to 60.0 m/s; (d) vav,x from t = 0 to 19.1 s Solution (a) The car starts from rest, so vxi = 0. It reaches vx = 30.0 m/s at t = 4.9 s, according to the data table. Then for this time interval, Δv 30.0 m/s − 0 m/s aav,x = ___x = ______________ = 6.1 m/s2 4.9 s − 0 s Δt The average acceleration for this time interval is 6.1 m/s2 in the +x-direction. (b) The acceleration ax, at any instant of time, is the slope of the tangent line to the vx(t) graph at that time. To find the maximum acceleration, we look for the steepest part of the graph. In this case, the largest slope occurs near t = 0, just as the car is starting out. In Fig. 2.11, a tangent line to the vx(t) graph at t = 0 passes through t = 0. Values for the rise and run to calculate the slope of the tangent line are read from the graph. The tangent line passes through the two points (t = 0, vx = 0) and (t = 6.0 s, vx = 55.0 m/s) on the graph, so the rise is 55.0 m/s for a run of 6.0 s. The slope of this line is 55.0 m/s − 0 m/s = +9.2 m/s2 rise = ______________ ax = ____ 6.0 s − 0 s run

Discussion The graph of velocity as a function of time is often the most helpful graph to have when solving a problem. If that graph is not given in the problem, it is useful to sketch one. The vx(t) graph shows displacement, velocity, and acceleration at once: the velocity vx is given by the points or the curve graphed, the displacement Δx is the area under the curve, and the acceleration ax is the slope of the curve. Why is the average velocity 39 m/s? Why is it not halfway between the initial velocity (0 m/s) and the final velocity (60 m/s)? If the acceleration were constant, the average velocity would indeed be _12 (0 + 60 m/s) = 30 m/s. The actual average velocity is somewhat higher than that—the acceleration is greater at the start, so less of the time interval is spent going (relatively) slow and more is spent going fast. The speed is less than 30 m/s for only 4.9 s, but is greater than 30 m/s for 14.2 s.

Practice Problem 2.5 Braking a Car An automobile is traveling along a straight road heading to the southeast at 24 m/s when the driver sees a deer begin to cross the road ahead of her. She steps on the brake and brings the car to a complete stop in an elapsed time of 8.0 s. A data recording device, triggered by the sudden braking action, records the following velocities and times as the car slows. Let the positive x-axis be directed to the southeast. Plot a graph of vx versus t and find (a) the average acceleration as the car comes to a stop and (b) the instantaneous acceleration at t = 2.0 s. vx (m/s)

24 17.3 12.0

8.7

6.0

3.5

2.0

0.75

t (s)

3.0

4.0

5.0

6.0

7.0

8.0

1.0

2.0

The maximum acceleration is 9.2 m/s in the +x-direction. 2

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37

MOTION ALONG A LINE WITH CONSTANT ACCELERATION

CHECKPOINT 2.3 What physical quantity does the slope of the tangent to a graph of vx versus time represent? vfx

2.4

MOTION ALONG A LINE WITH CONSTANT ACCELERATION

The graphical and mathematical relationships between position, velocity, and acceleration presented so far apply regardless of whether the acceleration is changing or is constant. In the important special case of an object whose acceleration is constant (both in magnitude and direction), we can write these relationships as algebraic equations. First, let us agree on a consistent notation:

vix

ti

tf (a)

• Choose an origin and a direction for the positive axis. (For vertical motion, it is conventional to use the y-axis instead of the x-axis, where the +y-direction is up.) • At an initial time ti, the initial position and velocity are xi and vix. • At a later time tf = ti + Δt, the final position and velocity are xf and vfx.

vfx 1

vav,x

2

From the following two essential relationships the others can be derived: 1. Since the acceleration ax is constant, the change in velocity over a given time interval Δt = tf − ti is the acceleration—the rate of change of velocity—times the elapsed time:

vix

ti

Δvx = vfx − vix = ax Δt

(2-9)

tf (b)

Figure 2.12 Finding the aver-

(if ax is constant during the entire time interval)

age velocity when the acceleration is constant.

Equation (2-9) is the definition of ax [Eq. (2-8)] with the assumption that ax is constant. 2. Since the velocity changes linearly with time, the average velocity is given by: vav,x = _12 (vfx + vix)

(constant ax)

(2-10)

Equation (2-10) is not true in general, but it is true for constant acceleration. To see why, refer to the vx(t) graph in Fig. 2.12a. The graph is linear because the acceleration—the slope of the graph—is constant. The displacement during any time interval is represented by the area under the graph. The average velocity is found by forming a rectangle with an area equal to the area under the curve in Fig. 2.12a, because the average velocity should give the same displacement in the same time interval. Figure 2.12b shows that, to make the excluded area above vav,x (triangle 1) equal to the extra area under vav,x (triangle 2), the average velocity must be exactly halfway between the initial and final velocities. Combining Eq. (2-10) with the definition of average velocity, Δx = x f − x i = vav,x Δt

(2-2)

gives our second essential relationship for constant acceleration: Δx = _12 (v fx + v ix) Δt

(2-11)

(if ax is constant during the entire time interval) If the acceleration is not constant, there is no reason why the average velocity has to be exactly halfway between the initial and the final velocity. As an illustration, imagine a trip where you drive along a straight highway at 80 km/h for 50 min and

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CHAPTER 2 Motion Along a Line

vx

∆t

vfx ∆vx = ax ∆t vix vix ti

tf

t

Figure 2.13 Graphical interpretation of Eq. (2-12).

then at 60 km/h for 30 min. Your acceleration is zero for the entire trip except during the few seconds while you slowed from 80 km/h to 60 km/h. The magnitude of your average velocity is not 70 km/h. You spent more time going 80 km/h than you did going 60 km/h, so the magnitude of your average velocity would be greater than 70 km/h. Other Useful Relationships for Constant Acceleration Two more useful relationships can be formed between the various quantities (displacement, initial and final velocities, acceleration, and time interval) by eliminating some quantity from Eqs. (2-9) and (2-11). For example, suppose we don’t know the final velocity vfx. Then we can solve Eq. (2-9) for vfx, substitute into Eq. (2-11), and simplify: Δx = _12 (v fx + v ix) Δt = _12 [(v ix + ax Δt) + v ix] Δt Δx = v ix Δt + _12 ax(Δt)2

(constant a x)

(2-12)

We can interpret Eq. (2-12) graphically. Figure 2.13 shows a vx(t) graph for motion with constant acceleration. The displacement that occurs between ti and a later time tf is the area under the graph for that time interval. Partition this area into a rectangle plus a triangle. The area of the rectangle is base × height = v ix Δt The height of the triangle is the change in velocity, which is equal to ax Δt. The area of the triangle is _1 base × height = _1 Δt × a Δt = _1 a (Δt)2 x 2 2 2 x

Adding these areas gives Eq. (2-12). Another useful relationship comes from eliminating the time interval Δt: 2

2

v fx − v ix _______ v − v ix Δx = _12 (v fx + v ix) Δt = _12 (v fx + v ix) _______ = fx 2ax ax Rearranging terms,

(

2

2

v fx − v ix = 2ax Δx

)

(constant ax)

(2-13)

CHECKPOINT 2.4 At 3:00 P.M., an airplane is moving due west at 460 km/h. At 3:05 P.M., it is moving due west at 480 km/h. Is its average velocity during the time interval necessarily 470 km/h west? Explain.

Example 2.6 A Sliding Brick Starting from rest, a brick slides along a straight line down an icy roof with a constant acceleration of magnitude 4.9 m/s2 (Fig. 2.14). How fast is the brick moving when it reaches the edge of the roof 0.90 s later?

Strategy What is the direction of the acceleration? It has to be downward along the roof, in the same direction as the brick’s velocity. An acceleration opposite the velocity would make the brick slow down, but since it starts from rest, a continued on next page

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39

MOTION ALONG A LINE WITH CONSTANT ACCELERATION

Example 2.6 continued

constant acceleration can only make it speed up. We choose the +x-axis in the direction of the acceleration. Then we use the acceleration to find how the velocity changes during the time interval. Solution With the x-axis in the direction of the acceleration, ax = +4.9 m/s2. The brick is initially at rest so vix = 0.

We want to know vfx at the end of the time interval Δt = 0.90 s. Since ax is constant, vx changes at a constant rate: Δv x = v fx − v ix = ax Δt = (+4.9 m/s2) × (0.90 s) = 4.4 m/s At the edge of the roof, the brick is moving at 4.4 m/s parallel to the roof. Discussion Conceptual check: ax = +4.9 m/s2 means that vx increases 4.9 m/s every second. The brick slides for a bit less than 1 s, so the increase in vx is a bit less than 4.9 m/s.

Figure 2.14 A brick sliding down an icy roof.

Practice Problem 2.6 Displacement of the Brick How far from the edge of the roof was the brick when it started sliding?

Example 2.7 Displacement of a Motorboat A motorboat starts from rest at a dock and heads due east with a constant acceleration of magnitude 2.8 m/s2. After traveling for 140 m, the motor is throttled down to slow down the boat at 1.2 m/s2 (while still moving east) until its speed is 16 m/s. Just as the boat attains the speed of 16 m/s, it passes a buoy due east of the dock. (a) Sketch a qualitative graph of vx(t) for the motorboat from the dock to the buoy. Let the +x-axis point east. (b) What is the distance between the dock and the buoy?

The boat is always headed to the east, so we choose east as the positive x-direction.

Strategy This problem involves two different values of acceleration, so it must be divided into two subproblems. The equations for constant acceleration cannot be applied to a time interval during which the acceleration changes. But for each of two time intervals, the acceleration of the boat is constant: from t1 to t2, a1x = +2.8 m/s2; from t2 to t3, a2x = −1.2 m/s2. The two subproblems are connected by the position and velocity of the boat at the instant the acceleration changes. This is reflected in the graph of vx(t): It consists of two different straightline segments with different slopes that connect with the same value of vx at time t2. For subproblem 1, the boat speeds up with a constant acceleration of 2.8 m/s2 to the east. We know the acceleration, the displacement (140 m east), and the initial velocity: the boat starts from rest, so the initial velocity v1x is zero. We need to calculate the final velocity v2x, which then becomes the initial velocity for the second subproblem.

For subproblem 2, we know acceleration, final velocity v3x, and we have just found the initial velocity v2x from subproblem 1. Because the boat is slowing down, its acceleration is in the direction opposite its velocity; therefore, a2x < 0. From these three quantities we can find the displacement of the boat during the second time interval.

Subproblem 1: Known: v1x = 0; a1x = +2.8 m/s2; Δx21 = x2 − x1 = 140 m. To find: v2x.

Subproblem 2: Known: v2x from subproblem 1; a2x = −1.2 m/s2; v3x = +16 m/s. To find: Δx32 = x3 − x2. Adding the displacements for the two time intervals gives the total displacement. The magnitude of the total displacement is the distance between the dock and the buoy. continued on next page

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CHAPTER 2 Motion Along a Line

Example 2.7 continued

The boat is moving east, in the +x-direction, so the correct sign here is positive: v2x = +28 m/s.

Solution (a) The graph starts with vx = 0 at t = t1. We choose t1 = 0 for simplicity. The graph is a straight line with slope +2.8 m/s2 until t = t2. Then, starting from where the graph left off, the graph continues as a straight line with slope −1.2 m/s2 until the graph reaches vx = 16 m/s at t = t3. Figure 2.15 shows the vx(t) graph. It is not quantitatively accurate because we have not calculated the values of t2 and t3.

(b2) The final velocity for the first interval (v2x) is the initial velocity for the second interval. The final velocity is v3x. Using the same equation just derived for this time interval, 2

v 2x − v 1x 1 (v + v ) Δt = __ 1 (v + v ) ________ Δx 21 = __ = 1x 1x 2 2x 2 2x a 1x

(

)

The total displacement is

2 2 v 2x − v 1x ________

x 3 − x 1 = (x 3 − x 2) + (x 2 − x 1) = 220 m + 140 m = +360 m

2a 1x

The buoy is 360 m from the dock.

Solving for v2x, ___________

√

v 2x = ±

2 v 1x

Discussion The natural division of the problem into two parts occurs because the boat has two different constant accelerations during two different time periods. In problems that can be subdivided in this way, the final velocity and position found in the first part becomes the initial velocity and position for the second part.

_____________________

+ 2a 1x Δx = ±√ 0 + 2 × 2.8 m/s2 × 140 m

= ± 28 m/s vx v2x 16 m/s (v3x) t1 = 0

Practice Problem 2.7 Time to Reach the Buoy

Figure 2.15 t2

t3 t

What is the time required by the boat in Example 2.7 to reach the buoy?

Graph of vx versus t for the motorboat.

2.5

2

v 3x − v 2x __________________ (16 m/s)2 − (28 m/s)2 = +220 m Δx 32 = ________ = 2 × (−1.2 m/s2) 2a 2x

(b1) To find v2x without knowing the time interval, we eliminate Δt from Eqs. (2-9) and (2-11) for constant acceleration:

VISUALIZING MOTION ALONG A LINE WITH CONSTANT ACCELERATION

Motion Diagrams In Fig. 2.16, three carts move in the same direction with three different values of constant acceleration. The position of each cart is depicted in a motion diagram as it would appear in a stroboscopic photograph with pictures taken at equal time intervals (here, the time interval is 1.0 s). The yellow cart has zero acceleration and, therefore, constant velocity. During each 1.0-s time interval its displacement is the same: 1.0 m/s × 1.0 s = 1.0 m to the right. Positions of the carts at 1.0-s intervals ax = 0, vix = 1.0 m/s ax = 0.2 m/s2, vix = 1.0 m/s

1.0 m/s 1.0 m/s 1.0 m/s 1.0 m/s 1.0 m/s 1.0 m/s 0s

1s

1.0 m/s

ax = –0.2 m/s2, vix = 2.0 m/s

0s

2s

1.2 m/s

3s

4s

1.4 m/s

1s

2s

3s

2.0 m/s

1.8 m/s

1.6 m/s

0s

1s 0

1

5s

1.6 m/s

3

2.0 m/s

4s 1.4 m/s

2s 2

1.8 m/s

3s 4

5s 1.2 m/s 4s

5

6

1.0 m/s 5s 7

8 x (m)

Figure 2.16 Each cart is shown as if photographs were taken at 1.0-s time intervals of 1.0 s. The arrows above each cart indicate the instantaneous velocities.

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2.5 VISUALIZING MOTION ALONG A LINE WITH CONSTANT ACCELERATION

41

Figure 2.17 Graphs of posi-

4

2

4 0

4

2

t (s)

1

4

2

t (s)

4

–0.2

t (s)

ax (m/s2)

ax

(m/s2)

Acceleration

4

2

0.2

2

4

–0.2 ax = 0 m/s2

4 t (s)

0.2

2

1

t (s)

0.2

4

2 vx (m/s)

vx (m/s)

vx (m/s)

Velocity

1

2 t (s)

2

2

4

t (s)

2

8

4

t (s)

ax (m/s2)

8

x (m)

8

x (m)

x (m)

Position

tion, velocity, and acceleration for the carts of Fig. 2.16.

2

4

t (s)

–0.2 ax = 0.2 m/s2

ax = – 0.2 m/s2

The red cart has a constant acceleration of 0.2 m/s2 to the right. Although m/s2 is normally read “meters per second squared,” it can be useful to think of it as “m/s per second”: the cart’s velocity changes by 0.2 m/s during each 1.0-s time interval. In this case, acceleration is in the same direction as the velocity, so the velocity increases. The displacement of the cart during successive 1.0-s time intervals gets larger and larger. The blue cart experiences a constant acceleration of 0.2 m/s2 in the −x-direction— the direction opposite to the velocity. The magnitude of the velocity then decreases; during each 1.0-s interval, the speed decreases by 0.2 m/s. Now the displacements during 1.0-s intervals get smaller and smaller. Graphs Figure 2.17 shows graphs of x(t), vx(t), and ax(t) for each of the carts. The acceleration graphs are horizontal since each of the carts has a constant acceleration. All three vx graphs are straight lines. Since ax is the rate of change of vx, the slope of the vx graph at any value of t is ax at that value of t. With constant acceleration, the slope is the same everywhere and the graph is linear. Remember that a positive ax does mean that vx is increasing, but not necessarily that the speed is increasing. If vx is negative, then a positive ax indicates a decreasing speed. (See Conceptual Example 2.4.) Speed is increasing when the acceleration and velocity are in the same direction (ax and vx both positive or both negative). Speed is decreasing when acceleration and velocity are in opposite directions—when ax and vx have opposite signs. The position graph is linear for the yellow cart because it has constant velocity. For the red cart, the x(t) graph curves with increasing slope, showing that vx is increasing. For the blue cart, the x(t) graph curves with decreasing slope, showing that vx is decreasing.

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CHAPTER 2 Motion Along a Line

Example 2.8 Two Spaceships Two spaceships are moving from the same starting point in the +x-direction with constant accelerations. The silver spaceship has an initial velocity of +2.00 km/s and an acceleration of +0.400 km/s2. The black spaceship has an initial velocity of +6.00 km/s and an acceleration of −0.400 km/s2. (a) Find the time at which the silver spaceship just overtakes the black spaceship. (b) Sketch graphs of vx(t) for the two spaceships. (c) Sketch a motion diagram (similar to Fig. 2.16) showing the positions of the two spaceships at 1.0-s intervals. Strategy We can find the positions of the spaceships at later times from the initial velocities and the accelerations. At first, the black spaceship is moving faster, so it pulls out ahead. Later, the silver ship overtakes the black ship at the instant their positions are equal. Solution (a) The position of either spaceship at a later time is given by Eq. (2-12): x f = x i + Δx = x i + v ix Δt + _12 ax(Δt)2 We set the final position of the silver spaceship equal to that of the black spaceship (xfs = xfb):

(b) Figure 2.18 shows the vx(t) graphs with ti = 0. Note that the area under the graphs from ti to tf is the same in the two graphs: the spaceships have the same displacement during that interval. (c) Equation (2-12) can be used to find the position of each spaceship as a function of time. Choosing xi = 0, ti = 0, and t = tf, the position at time t is x(t) = 0 + v ixt + _12 a xt2 Figure 2.19 shows the data table calculated this way and the corresponding motion diagram. Discussion Quick check: the two ships must have the same displacement at Δt = 10.0 s. Δ x s = v isx Δt + _12 a sx(Δt)2 = 2.00 km/s × 10.0 s + _12 × 0.400 km/s2 × (10.0 s)2 = 40.0 km Δ x b = v ibx Δt + _12 a bx(Δt)2 = 6.00 km/s × 10.0 s + _12 × (− 0.400 km/s2) × (10.0 s)2 = 40.0 km

x is + v isx Δt + _12 asx (Δt)2 = x ib + v ibx Δt + _12 a bx(Δt)2 Subscripts are useful for preventing you from mixing up similar quantities. The subscripts s and b stand for silver and black, respectively. The subscripts i and f stand for initial and final, respectively. A skilled problem-solver must be able to come up with algebraic symbols that are explicit and unambiguous. The initial positions are the same: xis = xib. Subtracting the initial positions from each side, moving all terms to one side, and factoring out one power of Δt yields

Silver vx (km/s) 6

2 0

10 t (s)

Δ t (v isx + _12 asx Δt − v ibx − _12 a bx Δt) = 0 Black

This equation has two solutions—there are two times at which the spaceships are at the same position. One solution is Δt = 0. We already knew that the two spaceships started at the same initial position. The other solution, which gives the time at which one spaceship overtakes the other, is found by setting the expression in parentheses equal to zero. Solving for Δt, 2(v isx − v ibx) _______________________ 2 × (2.00 km/s − 6.00 km/s) Δt = __________ a bx − a sx = − 0.400 km/s2 − 0.400 km/s2 = 10.0 s The silver spaceship overtakes the black spaceship 10.0 s after they leave the starting point.

vx (km/s) 6

2 0

10 t (s)

Figure 2.18 Graphs of vx versus t for the silver and black spaceships. The shaded area under each graph represents the displacement Δx during the time interval. continued on next page

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2.6

43

FREE FALL

Example 2.8 continued

Practice Problem 2.8 Velocity

Time to Reach the Same

When do the two spaceships have the same velocity? What is the value of the velocity then?

t (s) xs (km) xb (km) 0

0 0 0

1.0 2.2 5.8

1.0 s 2.0 s

2.0 4.8 11.2 3.0 s

1.0 s

3.0 7.8 16.2 4.0 s 2.0 s

4.0 11.2 20.8

5.0 15.0 25.0

5.0 s

6.0 19.2 28.8 6.0 s

3.0 s

10

7.0 23.8 32.2

4.0 s

8.0 28.8 35.2

7.0 s 5.0 s

20

9.0 34.2 37.8

10.0 40.0 40.0

8.0 s 6.0 s

9.0 s 7.0 s

30

10.0 s

8.0 s 9.0 s 10.0 s

40

x (km)

Figure 2.19 Calculated positions of the spaceships at 1.0-s time intervals and a motion diagram.

2.6

FREE FALL

Suppose you are standing on a bridge over a deep gorge. If you drop a stone into the gorge, how fast does it fall? You know from experience that it does not fall at a constant velocity; the longer it falls, the faster it goes. A better question is: What is the stone’s acceleration? First, let us simplify the problem. If the stone were moving very fast, air resistance would oppose its motion. When it is not falling so fast, the effect of air resistance is negligibly small. In free fall, no forces act on an object other than the gravitational force that makes the object fall. On Earth, free fall is an idealization since there is always some air resistance. We also assume that the stone’s change in altitude is small enough that Earth’s gravitational pull on it is constant.

CONNECTION: Free fall is an example of motion with constant acceleration.

Free-fall Acceleration An object in free fall has a constant downward acceleration, called the free-fall acceleration. The magnitude of this acceleration varies a little from one place to another near Earth’s surface, but at any given place, it has the same value for every object, regardless of the mass of the object. Unless another value is given in a particular problem, please assume that the magnitude of the free-fall acceleration near Earth’s surface is a free fall = g = 9.80 m/s2

(2-14)

The symbol g represents the magnitude of the free-fall acceleration. When dealing with vertical motion, the y-axis is usually chosen to be positive pointing upward. The direction of the free-fall acceleration is down, so ay = −g. The same techniques and equations used for other constant acceleration situations are used with free fall. Earth’s gravity always pulls downward, so the acceleration of an object in free fall is always downward and constant in magnitude, regardless of whether the object is moving up, down, or is at rest, and independent of its speed. If the object is moving downward, the downward acceleration makes it speed up; if it is moving upward, the downward acceleration makes it slow down.

gia04535_ch02_025-054.indd 43

In free fall, ay = −g (if the y-axis points up).

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CHAPTER 2 Motion Along a Line

Figure 2.20 Graph of vy versus t for an object thrown upward.

vy

vy > 0 Moving up

vy < 0 Moving down

Slope = –g 0

t Top of flight vy = 0

Acceleration at Highest Point If an object is thrown straight up, its velocity is zero at the highest point of its flight. Why? On the way up, its velocity vy is positive (if the positive y-axis is pointing up). On the way down, vy is negative. Since vy changes continuously, it must pass through zero to change sign (Fig. 2.20). At the highest point, the velocity is zero but the acceleration is not zero. If the acceleration were to suddenly become zero at the top of flight, the velocity would no longer change; the object would get stuck at the top rather than fall back down. The velocity is zero at the top but it does not stay zero; it keeps changing at the same rate.

CHECKPOINT 2.6 Is it possible for an object in free fall to be moving upward? Explain.

Example 2.9 Throwing Stones Standing on a bridge, you throw a stone straight upward. The stone hits a stream, 44.1 m below the point at which you release it, 4.00 s later. (a) What is the velocity of the stone just after it leaves your hand? (b) What is the velocity of the stone just before it hits the water? (c) Draw a motion diagram for the stone, showing its position at 0.1-s intervals during the first 0.9 s of its motion. (d) Sketch graphs of y(t) and vy(t). The positive y-axis points up. Strategy Ignoring air resistance, the stone is in free fall once your hand releases it and until it hits the water. For the time interval during which the stone is in free fall, the initial velocity is the velocity of the stone just after it leaves your hand and the final velocity is the velocity just before it hits the water. During free fall, the stone’s acceleration is constant and assumed to be 9.80 m/s2 downward. Known: ay = −9.80 m/s2; Δy = − 44.1 m at Δt = 4.00 s. To find: viy and vfy. Solution (a) Equation (2-12) can be used to solve for viy since all the other quantities in it (∆y, ∆t, and ay) are known and the acceleration is constant. Δy = v iy Δt + _12 ay (Δt)2

Solving for viy, Δy 1 v iy = ___ − __ a Δt Δt 2 y

(1)

− 44.1 m − __ 1 (−9.80 m/s2 × 4.00 s) = ________ 4.00 s 2 = −11.0 m/s + 19.6 m/s = 8.6 m/s The initial velocity is 8.6 m/s upward. (b) The change in vy is ay Δ t from Eq. (2-9): v fy = v iy + ay Δt Substituting the expression for viy in the preceding equation, v fy =

( __ΔyΔt − _12a Δt ) + a Δt = __ΔyΔt + _12 a Δt y

y

y

(2)

− 44.1 m + __ 1 (−9.80 m/s2 × 4.00 s) = ________ 4.00 s 2 = −11.0 m/s − 19.6 m/s = −30.6 m/s The final velocity is 30.6 m/s downward. continued on next page

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MASTER THE CONCEPTS

Example 2.9 continued

0.9 s 0.8 s 0.7 s

(c) Choosing yi = 0 and ti = 0, the position of the stone as a function of time is

y (m)

y (m)

4.0 t (s)

y(t) = v iyt + _12 ayt2

3.5

0.6 s

The motion diagram is shown in Fig. 2.21.

0.5 s

3.0

– 44.1

2.5

vy (m/s)

0.4 s

+8.6 0

0.3 s

(d) The graphs are shown in Fig. 2.22.

4.0 t (s)

2.0 –30.6 0.2 s

1.5

Figure 2.22 Graphs of y(t) and vy(t) for the stone.

Discussion The final speed is greater than the initial speed, as expected. Equations (1) and (2) have a direct interpretation, which is a good check on their validity. The first term, Δy/Δt, is the average velocity of the stone during the 4.00 s of free fall. The second term, _12 a y Δt, is half the change in vy since Δvy = ay Δt. Because the acceleration is constant, the average velocity is halfway between the initial and final velocities. Therefore, the initial velocity is the average velocity minus half of the change, while the final velocity is the average velocity plus half of the change.

1.0 0.1 s

Practice Problem 2.9 Height Attained by Stone 0.5

Figure 2.21 0.0 s

Motion diagram for a stone moving straight up.

(a) How high above the bridge does the stone go? [Hint: What is vy at the highest point?] (b) If you dropped the stone instead of throwing it, how long would it take to hit the water?

Master the Concepts • Displacement is the change in position: Δx = xf − xi. The displacement depends only on the starting and ending positions, not on details of the motion. The magnitude of the displacement is not necessarily equal to the total distance traveled; it is the straight-line distance from the initial position to the final position. • Average velocity is the constant velocity that would cause the same displacement in the same amount of time. Δx (for any time interval Δt) vav,x = ___ (2-2) Δt • Velocity is a measure of how fast and in what direction something moves. Its direction is the direction of the object’s motion and its magnitude is the instantaneous speed. It is the instantaneous rate of change of the position. Δx (for a very short time interval Δt) (2-3) vx = lim ___ Δt→0 Δt

• Average acceleration is the constant acceleration that would cause the same velocity change in the same amount of time. Δv aav,x = ____x (for any time interval Δt) (2-7) Δt • Acceleration is the instantaneous rate of change of the velocity. Δv ax = lim ____x (for a very short time interval Δt) (2-8) Δt→0 Δt Acceleration does not necessarily mean speeding up. A velocity can change by decreasing speed or by changing direction. • Interpreting graphs: On a graph of x(t ), the slope at any point is vx. On a graph of vx(t ), the slope at any point is ax, and the area under the graph during any time interval is the displacement Δ x during that time interval. If vx is negative, the displacement is also continued on next page

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CHAPTER 2 Motion Along a Line

Master the Concepts continued

negative, so we must count the area as negative when it is below the time axis. vx

Δx = v ix Δt + _21 ax(Δt)2

(2-12)

2 v fx

(2-13)

−

2 v ix

= 2ax Δx

These same relationships hold for position, velocity, and acceleration along the y-axis if ay is constant. vx

∆t

vfx ∆x = area

∆vx = ax ∆t vix

t1

t2

t

On a graph of ax (t ), the area under the curve is Δvx, the change in vx during that time interval. • Essential relationships for constant acceleration problems: if ax is constant during the entire time interval Δt from ti until a later time tf = ti + Δt, Δvx = v fx − v ix = ax Δt Δx = _1 ( v + v ) Δt 2

fx

ix

(2-9)

ti

tf

t

• An object in free fall has a constant downward acceleration. The magnitude of the acceleration g varies a little from place to place near Earth’s surface. A typical value is g = 9.80 m/s2.

(2-11)

Conceptual Questions 1. Explain the difference between distance traveled, displacement, and displacement magnitude. 2. Explain the difference between speed and velocity. 3. On a graph of vx versus time, what quantity does the area under the graph represent? 4. On a graph of vx versus time, what quantity does the slope of the graph represent? 5. On a graph of ax versus time, what quantity does the area under the graph represent? 6. On a graph of x versus time, what quantity does the slope of the graph represent? 7. What is the relationship between average velocity and instantaneous velocity? An object can have different instantaneous velocities at different times. Can the same object have different average velocities? Explain. 8. Can the velocity of an object be zero and the acceleration be nonzero at the same time? Explain. 9. You are bicycling along a straight north-south road. Let the x-axis point north. Describe your motion in each of the following cases. Example: ax > 0 and vx > 0 means you are moving north and speeding up. (a) ax > 0 and vx < 0. (b) ax = 0 and vx < 0. (c) ax < 0 and vx = 0. (d) ax < 0 and vx < 0. (e) Based on your answers, explain why it is not a good idea to use the expression “negative acceleration” to mean slowing down. 10. When a coin is tossed straight up, what can you say about its velocity and acceleration at the highest point of its motion?

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vix

Multiple-Choice Questions 1. A ball is thrown straight up into the air. Ignore air resistance. While the ball is in the air its acceleration (a) increases. (b) is zero. (c) remains constant. (d) decreases on the way up and increases on the way down. (e) changes direction. 2. Which car has a westward acceleration? (a) a car traveling westward at constant speed (b) a car traveling eastward and speeding up (c) a car traveling westward and slowing down (d) a car traveling eastward and slowing down (e) a car starting from rest and moving toward the east Questions 3 and 4. A toy rocket is propelled straight upward from the ground and reaches a height Δy. After an elapsed time Δ t, measured from the time the rocket was first fired off, the rocket has fallen back down to the ground, landing at the same spot from which it was launched. Answer choices: (a) zero

Δy (b) 2 ___ Δt Δy Δy 1 (c) ___ (d) __ ___ 2 Δt Δt 3. What is the magnitude of the average velocity of the rocket during this time? 4. What is the average speed of the rocket during this time?

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PROBLEMS

5. A leopard starts from rest at t = 0 and runs in a straight line with a constant acceleration until t = 3.0 s. The distance covered by the leopard between t = 1.0 s and t = 2.0 s is (a) the same as the distance covered during the first second. (b) twice the distance covered during the first second. (c) three times the distance covered during the first second. (d) four times the distance covered during the first second. Multiple-Choice Questions 6–15. A jogger is exercising along a long, straight road that runs north-south. She starts out heading north. A graph of vx (t) follows Question 10. 6. What is the displacement of the jogger from t = 18.0 min to t = 24.0 min? (a) 720 m, south (b) 720 m, north (c) 2160 m, south (d) 3600 m, north 7. What is the displacement of the jogger for the entire 30.0 min? (a) 3120 m, south (b) 2400 m, north (c) 2400 m, south (d) 3840 m, north 8. What is the total distance traveled by the jogger in 30.0 min? (a) 3840 m (b) 2340 m (c) 2400 m (d) 3600 m 9. What is the average velocity of the jogger during the 30.0 min? (a) 1.3 m/s, north (b) 1.7 m/s, north (c) 2.1 m/s, north (d) 2.9 m/s, north 10. What is the average speed of the jogger for the 30 min? (a) 1.4 m/s (b) 1.7 m/s (c) 2.1 m/s (d) 2.9 m/s vx (m/s) 5

C

D

11. In what direction is she running at time t = 20 min? (a) south (b) north (c) not enough information 12. In which region of the graph is ax positive? (a) A to B (b) C to D (c) E to F (d) G to H 13. In which region is ax negative? (a) A to B (b) C to D (c) E to F (d) G to H 14. In which region is the velocity directed to the south? (a) A to B (b) C to D (c) E to F (d) G to H ✦15. What distance does the jogger travel during the first 10.0 min (t = 0 to 10.0 min)? (a) 8.5 m (b) 510 m (c) 900 m (d) 1020 m 16. The figure shown here has four graphs of x versus time. Which graph shows a constant, positive, nonzero velocity?

(a)

(b)

(c)

(d)

Multiple-Choice Questions 16 and 17 17. The four graphs show vx versus time. (a) Which graph shows a constant velocity? (b) Which graph shows ax constant and positive? (c) Which graph shows ax constant and negative? (d) Which graph shows a changing ax that is always positive?

B G

2

Problems

A

H

t (min)

✦

–2

E 0

10.0

F 20.0

Multiple-Choice Questions 6–15

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Blue # 30.0

1

2

Combination conceptual/quantitative problem Biological or medical application Challenging problem Detailed solution in the Student Solutions Manual Problems paired by concept Text website interactive or tutorial

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CHAPTER 2 Motion Along a Line

2.1 Position and Displacement 1. A displacement of magnitude 32 cm toward the east is followed by displacements of magnitude 48 cm to the east and then 64 cm to the west. What is the total displacement? 2. A squirrel is trying to locate some nuts he buried for the winter. He moves 4.0 m to the right of a stone and digs unsuccessfully. Then he moves 1.0 m to the left of his hole, changes his mind, and moves 6.5 m to the right of that position and digs a second hole. No luck. Then he moves 8.3 m to the left and digs again. He finds a nut at last. What is the squirrel’s total displacement from its starting point? 3. A runner, jogging along a straight line path, starts at a position 60 m east of a milestone marker and heads west. After a short time interval he is 20 m west of the mile marker. Choose east to be the positive x-direction. (a) What is the runner’s displacement from his starting point? (b) What is his displacement from the milestone? (c) The runner then turns around and heads east. If at a later time the runner is 140 m east of the milestone, what is his displacement from the starting point at this time? (d) What is the total distance traveled from the starting point if the runner stops at the final position listed in part (c)? 4. Johannes bicycles from his dorm to the pizza shop that is 3.00 mi east. Darren’s apartment is located 1.50 mi west of Johannes’s dorm. If Darren is able to meet Johannes at the pizza shop by bicycling in a straight line, what is the distance and direction he must travel? 5. At 3 p.m. a car is located 20 km south of its starting point. One hour later it is 96 km farther south. After two more hours, it is 12 km south of the original starting point. (a) What is the displacement of the car between 3 p.m. and 6 p.m.? (b) What is the displacement of the car from the starting point to the location at 4 p.m.? (c) What is the displacement of the car between 4 p.m. and 6 p.m.?

2.2 Velocity: Rate of Change of Position 6. For the train of Example 2.2, find the average velocity between 3:14 p.m. when the train is at 3 km east of the origin and 3:28 p.m. when it is 10 km east of the origin. 7. A cyclist travels 10.0 km east in a time of 11 min 40 s. What is his average velocity in meters per second? 8. In a game against the White Sox, baseball pitcher Nolan Ryan threw a pitch measured at 45.1 m/s. If it was 18.4 m from Nolan’s position on the pitcher’s mound to home plate, how long did it take the ball

gia04535_ch02_025-054.indd 48

to get to the batter waiting at home plate? Treat the ball’s velocity as constant and ignore any gravitational effects. 9. Jason drives due west with a speed of 35.0 mi/h for 30.0 min, then continues in the same direction with a speed of 60.0 mi/h for 2.00 h, then drives farther west at 25.0 mi/h for 10.0 min. What is Jason’s average velocity for the entire trip? 10. Two cars, a Toyota Yaris and a Jeep, are traveling in the same direction, although the Yaris is 186 m behind the Jeep. The speed of the Yaris is 24.4 m/s and the speed of the Jeep is 18.6 m/s. How much time does it take for the Yaris to catch the Jeep? [Hint: What must be true about the displacement of the two cars when they meet?] ( tutorial: catchup) 11. Speedometer readings are obtained and graphed as a car comes to a stop along a straight-line path. How far does the car move between t = 0 and t = 16 s? ( tutorial: start/stop traffic) vx (m/s) 25 20 15 10 5 0

2

4

6

8 10 12 14 16 t (s)

Problems 11 and 29 12. A graph is plotted of the vertical velocity vy of an elevator versus time. The y-axis points up. (a) How high is the elevator above the starting point (t = 0) after 20 s has elapsed? (b) When is the elevator at its highest location above the starting point? vy (m/s) 2

t (s)

–2 0

4

8

12

16

20

13. A bicycle is moving along a straight line. The graph in the figure shows its position from the starting point as a function of time. (a) In which section(s) of the graph does the object have the highest speed? (b) At which time(s) does the object reverse its direction of

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PROBLEMS

motion? (c) How far does the object move from t = 0 to t = 3 s? x (m) 40 D

30 B

20 10 0

F

C A E 0

1

2

3

4

5

6 t (s)

14. A ball thrown by a pitcher on a women’s softball team is timed at 65.0 mph. The distance from the pitching rubber to home plate is 43.0 ft. In major league baseball the corresponding distance is 60.5 ft. If the batter in the softball game and the batter in the baseball game are to have equal times to react to the pitch, with what speed must the baseball be thrown? Assume the ball travels with a constant velocity. [Hint: There is no need to convert units; set up a ratio.] 15. A motor scooter travels east at a speed of 12 m/s. The driver then reverses direction and heads west at 15 m/s. What is the change in velocity of the scooter? Give magnitude and direction. 16. To pass a physical fitness test, Massimo must run 1000 m at an average rate of 4.0 m/s. He runs the first 900 m in 250 s. Is it possible for Massimo to pass the test? If so, how fast must he run the last 100 m to pass the test? Explain. 17. The graph shows speedometer readings, in meters per second (on the vertical axis), obtained as a skateboard travels along a straight-line path. How far does the board move between t = 3.00 s and t = 8.00 s? 18. The graph shows values of x(t) in meters, on the vertical axis, for a skater traveling in a straight line. (a) What is vav, x for the interval from t = 0 to t = 4.0 s? (b) from t = 0 to t = 5.0 s? 19. The graph shows values of x(t) in meters for a skater traveling in a straight line. What is vx at t = 2.0 s? 20. The graph shows values of x(t) in meters for an object traveling in a straight line. Plot vx as a function of time for this object from t = 0 to t = 8 s. 8

4

4

49

21. A chipmunk, trying to cross a road, first moves 80 cm to the right, then 30 cm to the left, then 90 cm to the right, and finally 310 cm to the left. (a) What is the chipmunk’s total displacement? (b) If the elapsed time was 18 s, what was the chipmunk’s average speed? (c) What was its average velocity? 22. Rita Jeptoo of Kenya was the first female finisher in the 110th Boston Marathon. She ran the first 10.0 km in a time of 0.5689 h. Assume the race course to be along a straight line. (a) What was her average speed during the first 10.0 km segment of the race? (b) She completed the entire race, a distance of 42.195 km, in a time of 2.3939 h. What was her average speed for the race? ✦23. A relay race is run along a straight-line track of length 300.0 m running south to north. The first runner starts at the south end of the track and passes the baton to a teammate at the north end of the track. The second runner races back to the start line and passes the baton to a third runner who races 100.0 m northward to the finish line. The magnitudes of the average velocities of the first, second, and third runners during their parts of the race are 7.30 m/s, 7.20 m/s, and 7.80 m/s, respectively. What is the average velocity of the baton for the entire race? [Hint: You will need to find the time spent by each runner in completing her portion of the race.]

2.3 Acceleration: Rate of Change of Velocity 24. If a pronghorn antelope accelerates from rest in a straight line with a constant acceleration of 1.7 m/s2, how long does it take for the antelope to reach a speed of 22 m/s? 25. If a car traveling at 28 m/s is brought to a full stop in 4.0 s after the brakes are applied, find the average acceleration during braking. 26. An 1100-kg airplane starts from rest; 8.0 s later it reaches its takeoff speed of 35 m/s. What is the average acceleration of the airplane during this time? 27. A rubber ball is attached to a paddle by a rubber band. The ball is initially moving away from the paddle with a speed of 4.0 m/s. After 0.25 s, the ball is moving toward the paddle with a speed of 3.0 m/s. What is the average acceleration of the ball during that 0.25 s? Give magnitude and direction. 28. (a) In Fig. 2.11, what is the instantaneous acceleration of the sports car of Example 2.5 at the time of 14 s from the start? (b) What is the displacement of the car from t = 12.0 s to t = 16.0 s? (c) What is the average velocity of the car in the 4.0-s time interval from 12.0 s to 16.0 s?

8 t (s)

Problems 17, 18, 19, and 20

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CHAPTER 2 Motion Along a Line

29. The graph with Problem 11 shows speedometer readings as a car comes to a stop. What is the magnitude of the acceleration at t = 7.0 s? ✦30. The figure shows a plot of vx(t) for a car traveling in a straight line. (a) What is aav,x between t = 6 s and t = 11 s? (b) What is vav,x for the same time interval? (c) What is vav,x for the interval t = 0 to t = 20 s? (d) What is the increase in the car’s speed between 10 s and 15 s? (e) How far does the car travel from time t = 10 s to time t = 15 s? vx (m/s) 20

15

10

5

5

10

15

20 t (s)

31. The graph shows vx versus t for a body moving along a straight line. (a) What is ax at t = 11 s? (b) What is ax at t = 3 s? (c) How far does the body travel from t = 12 s to t = 14 s? ( tutorial: x, v, a) vx (m/s) 40 20 0

2

4

6

8

10

12

14 t (s)

2.4 Motion Along a Line with Constant Acceleration; 2.5 Visualizing Motion Along a Line with Constant Acceleration 32. A toboggan is sliding in a straight line down a snowy slope. The table shows the speed of the toboggan at various times during its trip. (a) Make a graph of the speed as a function of time. (b) Judging by the graph, is it plausible that the toboggan’s acceleration is constant? If so, what is the acceleration? Time Elapsed, t (s) 0 1.14 1.62 2.29 2.80

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Speed of Toboggan, v (m/s) 0 2.8 3.9 5.6 6.8

33. The St. Charles streetcar in New Orleans starts from rest and has a constant acceleration of 1.20 m/s2 for 12.0 s. (a) Draw a graph of vx versus t. (b) How far has the train traveled at the end of the 12.0 s? (c) What is the speed of the train at the end of the 12.0 s? (d) Draw a motion diagram, showing the streetcar’s position at 2.0-s intervals. 34. An airplane lands and starts down the runway with a southwest velocity of 55 m/s. What constant acceleration allows it to come to a stop in 1.0 km? 35. A train is traveling south at 24.0 m/s when the brakes are applied. It slows down with a constant acceleration to a speed of 6.00 m/s in a time of 9.00 s. (a) Draw a graph of vx versus t for a 12-s interval (starting 2 s before the brakes are applied and ending 1 s after the brakes are released). Let the x-axis point to the north. (b) What is the acceleration of the train during the 9.00-s interval? (c) How far does the train travel during the 9.00 s? ✦36. A 1200-kg airplane starts from rest and moves forward with a constant acceleration of magnitude 5.00 m/s2 along a runway that is 250 m long. (a) How long does it take the plane to reach a speed of 46.0 m/s? (b) How far along the runway has the plane moved when it reaches 46.0 m/s? 37. A car is speeding up and has an instantaneous velocity of 1.0 m/s in the +x-direction when a stopwatch reads 10.0 s. It has a constant acceleration of 2.0 m/s2 in the +x-direction. (a) What change in speed occurs between t = 10.0 s and t = 12.0 s? (b) What is the speed when the stopwatch reads 12.0 s? 38. You are driving your car along a country road at a speed of 27.0 m/s. As you come over the crest of a hill, you notice a farm tractor 25.0 m ahead of you on the road, moving in the same direction as you at a speed of 10.0 m/s. You immediately slam on your brakes and slow down with a constant acceleration of magnitude 7.00 m/s2. Will you hit the tractor before you stop? How far will you travel before you stop or collide with the tractor? If you stop, how far is the tractor in front of you when you finally stop? 39. A train is traveling along a straight, level track at 26.8 m/s (60.0 mi/h). Suddenly the engineer sees a truck stalled on the tracks 184 m ahead. If the maximum possible braking acceleration has magnitude 1.52 m/s2, can the train be stopped in time? 40. In a cathode ray tube in an old TV, electrons are accelerated from rest with a constant acceleration of magnitude 7.03 × 1013 m/s2 during the first 2.0 cm of the tube’s length; then they move at essentially constant velocity another 45 cm before hitting the screen. (a) Find the speed of the electrons when they hit the

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COMPREHENSIVE PROBLEMS

screen. (b) How long does it take them to travel the length of the tube? 41. The graph is of vx versus t for an object moving along the x-axis. How far does the object move between t = 9.0 s and t = 13.0 s? Solve using two methods: a graphical analysis and an algebraic solution. vx (m/s) 40 20 0

2

4

6

8

10

12

14 t (s)

Problems 41–42 42. The graph is of vx versus t for an object moving along the x-axis. What is the average acceleration between t = 5.0 s and t = 9.0 s? Solve using two methods: a graphical analysis and an algebraic solution. 43. A train, traveling at a constant speed of 22 m/s, comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude 1.4 m/s2. (a) Draw a graph of vx versus t where the x-axis points up the incline. (b) What is the speed of the train after 8.0 s on the incline? (c) How far has the train traveled up the incline after 8.0 s? (d) Draw a motion diagram, showing the trains position at 2.0-s intervals.

2.6 Free Fall In the problems, please assume the free-fall acceleration g = 9.80 m/s2 unless a more precise value is given in the problem statement. Ignore air resistance. 44. A brick is thrown vertically upward with an initial speed of 3.00 m/s from the roof of a building. If the building is 78.4 m tall, how much time passes before the brick lands on the ground? 45. A penny is dropped from the observation deck of the Empire State building (369 m above ground). With what velocity does it strike the ground? 46. (a) How long does it take for a golf ball to fall from rest for a distance of 12.0 m? (b) How far would the ball fall in twice that time? 47. Grant Hill jumps 1.3 m straight up into the air to slamdunk a basketball into the net. With what speed did he leave the floor? 48. During a walk on the Moon, an astronaut accidentally drops his camera over a 20.0-m cliff. It leaves his hands with zero speed, and after 2.0 s it has attained a velocity of 3.3 m/s downward. How far has the camera fallen after 4.0 s?

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51

49. Glenda drops a coin from ear level down a wishing well. The coin falls a distance of 7.00 m before it strikes the water. If the speed of sound is 343 m/s, how long after Glenda releases the coin will she hear a splash? 50. A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s and the stone is 1.50 m above the ground when launched. (a) How high above the ground does the stone rise? (b) How much time elapses before the stone hits the ground? 51. A 55-kg lead ball is dropped from the leaning tower of Pisa. The tower is 55 m high. (a) How far does the ball fall in the first 3.0 s of flight? (b) What is the speed of the ball after it has traveled 2.5 m downward? (c) What is the speed of the ball 3.0 s after it is released? (d) If the ball is thrown vertically upward from the top of the tower with an initial speed of 4.80 m/s, where will it be after 2.42 s? ✦52. A balloonist, riding in the basket of a hot air balloon that is rising vertically with a constant velocity of 10.0 m/s, releases a sandbag when the balloon is 40.8 m above the ground. What is the bag’s speed when it hits the ground? ✦53. Superman is standing 120 m horizontally away from Lois Lane. A villain throws a rock vertically downward with a speed of 2.8 m/s from 14.0 m directly above Lois. (a) If Superman is to intervene and catch the rock just before it hits Lois, what should be his minimum constant acceleration? (b) How fast will Superman be traveling when he reaches Lois? 54. A student, looking toward his fourth-floor dormitory ✦ window, sees a flowerpot with nasturtiums (originally on a window sill above) pass his 2.0-m high window in 0.093 s. The distance between floors in the dormitory is 4.0 m. From a window on which floor did the flowerpot fall? ✦55. You drop a stone into a deep well and hear it hit the bottom 3.20 s later. This is the time it takes for the stone to fall to the bottom of the well, plus the time it takes for the sound of the stone hitting the bottom to reach you. Sound travels about 343 m/s in air. How deep is the well?

Comprehensive Problems In the problems, please assume the free-fall acceleration g = 9.80 m/s2 unless a more precise value is given in the problem statement. Ignore air resistance. 56. (a) If a freestyle swimmer traveled 1500 m in a time of 14 min 53 s, how fast was his average speed? (b) If the pool was rectangular and 50 m in length, how does the

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58.

59.

60.

61.

62.

63.

64.

speed you found compare with his sustained swimming should be her average speed over the last 500 m in speed of 1.54 m/s during one length of the pool after he order to finish with an overall average speed of had been swimming for 10 min? What might account 4.00 m/s? for the difference? 65. At 3:00 p.m., a bank robber is spotted driving north While passing a slower car on the highway, you acceleron I-15 at milepost 126. His speed is 112.0 mi/h. At ate uniformly from 17.4 m/s to 27.3 m/s in a time of 3:37 p.m., he is spotted at milepost 185 doing 105.0 mi/h. 10.0 s. (a) How far do you travel during this time? During this time interval, what are the bank robber’s (b) What is your acceleration magnitude? displacement, average velocity, and average acceleration? (Assume a straight highway.) A cheetah can accelerate from rest to 24 m/s in 2.0 s. Assuming the acceleration is constant over the time ✦ 66. Based on the information given in Problem 59, is it posinterval, (a) what is the magnitude of the acceleration of sible that the rocket moves with constant acceleration? the cheetah? (b) What is the distance traveled by the Explain. cheetah in these 2.0 s? (c) A runner can accelerate from ✦ 67. An elevator starts at rest on the ninth floor. At t = 0, a rest to 6.0 m/s in the same time, 2.0 s. What is the magpassenger pushes a button to go to another floor. The nitude of the acceleration of the runner? By what factor graph for this problem shows the acceleration ay of the is the cheetah’s average acceleration magnitude greater elevator as a function of time. Let the y-axis point than that of the runner? upward. (a) Has the passenger gone to a higher or lower floor? (b) Sketch a graph of the velocity vy of the elevaA rocket is launched from rest. After 8.0 min, it is 160 km tor versus time. (c) Sketch a graph of the position y of above the Earth’s surface and is moving at a speed of the elevator versus time. 7.6 km/s. Assuming the rocket moves up in a straight line, what are its (a) average velocity and (b) average acceleration? ay (m/s2) 0.25 A streetcar named Desire travels between two stations 0.60 km apart. Leaving the first station, it accelerates for 10.0 s at 1.0 m/s2 and then travels at a constant speed 0 t1 t3 t2 t (s) until it is near the second station, when it brakes at 2 2.0 m/s in order to stop at the station. How long did – 0.25 this trip take? [Hint: What’s the average velocity?] An unmarked police car starts from rest just as a speed– 0.50 ing car passes at a speed of v. If the police car speeds up with a constant acceleration of magnitude a, what is the ✦ 68. The graph for this problem shows the vertical velocity speed of the police car when it catches up to the speeder, vy of a bouncing ball as a function of time. The y-axis who does not realize she is being pursued and does not points up. Answer these questions based on the data in vary her speed? the graph. (a) At what time does the ball reach its maxiA stone is thrown vertically downward from the roof of mum height? (b) For how long is the ball in contact with a building. It passes a window 16.0 m below the roof the floor? (c) What is the maximum height of the ball? with a speed of 25.0 m/s. It lands on the ground 3.00 s (d) What is the acceleration of the ball while in the air? after it was thrown. What was (a) the initial velocity of (e) What is the average acceleration of the ball while in the stone and (b) how tall is the building? contact with the floor? A car traveling at 29 m/s (65 mi/h) runs into a bridge abutment after the driver falls asleep at the wheel. (a) If the driver is wearing a seat belt and comes to rest within a 4 1.0-m distance, what is his acceleration (assumed con3 stant)? (b) A passenger who isn’t wearing a seat belt is 2 thrown into the windshield and comes to a stop in a dis1 tance of 10.0 cm. What is the acceleration of the 0 passenger? 0 0.5 1.0 1.5 2.0 2.5 3.0 t (s) –1 To pass a physical fitness test, Marcella must run –2 1000 m at an average speed of 4.00 m/s. She runs the –3 first 500 m at an average of 4.20 m/s. (a) How much –4 time does she have to run the last 500 m? (b) What

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vy (m/s)

57.

CHAPTER 2 Motion Along a Line

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ANSWERS TO PRACTICE PROBLEMS

✦69. A rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of 20.0 m/s2. However, after 50.0 s of flight the engine fails. (a) What is the rocket’s altitude when the engine fails? (b) When does it reach its maximum height? (c) What is the maximum height reached? [Hint: A graphical solution may be easiest.] (d) What is the velocity of the rocket just before it hits the ground? ✦70. The graph shows the position x of a switch engine in a rail yard as a function of time t. At which of the labeled times t0 to t7 is (a) ax < 0, (b) ax = 0, (c) ax > 0, (d) vx = 0, (e) the speed decreasing?

influx of sodium ions through the membrane of a neuron.) The signal is passed from one neuron to another by the release of neurotransmitters in the synapse. Suppose someone steps on your toe. The pain signal travels along a 1.0-m-long sensory neuron to the spinal column, across a synapse to a second 1.0-m-long neuron, and across a second synapse to the brain. Suppose that the synapses are each 100 nm wide, that it takes 0.10 ms for the signal to cross each synapse, and that the action potentials travel at 100 m/s. (a) At what average speed does the signal cross a synapse? (b) How long does it take the signal to reach the brain? (c) What is the average speed of propagation of the signal?

x

t0

t1

t2

t6 t3 t4

Answers to Practice Problems

t7 t

t5

✦71. An airtrack glider, 8.0 cm long, blocks light as it goes through a photocell gate. The glider is released from rest on a frictionless inclined track and the gate is positioned so that the glider has traveled 96 cm when it is in the middle of the gate. The timer gives a reading of 333 ms for the glider to pass through this gate. Friction is negligible. What is the acceleration (assumed constant) of the glider along the track?

2.1 3.8 m east 2.2 77 km/h in the −x-direction (west) 2.3 About 100 to 110 km/h in the +x-direction (east) 2.4 The velocity is increasing in magnitude, so the acceleration is in the same direction as the velocity (the −x-direction). Thus, ax is negative; the acceleration is in the −x-direction. 2.5 vx (m/s)

Instantaneous Slope at acceleration = t = 2.0 s at t = 2.0 s 0 m/s – 20.5 m/s = –4.3 m/s2 ax = 4.8 s – 0 s – 24 m/s aav,x = = –3.0 m/s2 8.0 s

20 15

vix = 0 8.0 cm

Photogate

vfx

x 96 cm

10 5.0 0

✦72. Find the point of no return for an airport runway of 1.50 mi in length if a jet plane can accelerate at 10.0 ft/s2 and decelerate at 7.00 ft/s2. The point of no return occurs when the pilot can no longer abort the takeoff without running out of runway. What length of time is available from the start of the motion in which to decide on a course of action? ✦ 73. In the human nervous system, signals are transmitted along neurons as action potentials that travel at speeds of up to 100 m/s. (An action potential is a traveling

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1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0 t (s)

(a) aav,x = −3.0 m/s2 where the negative sign means the average acceleration is directed to the northwest; (b) ax = −4.3 m/s2 (northwest) 2.6 2.0 m 2.7 20 s 2.8 5.00 s after they leave the starting point; 4.00 km/s in the +x-direction 2.9 (a) 3.8 m; (b) 3.00 s

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CHAPTER 2 Motion Along a Line

Answers to Checkpoints 2.1 No. The magnitude of the displacement is the shortest distance between two points. The distance traveled can be greater than or equal to the displacement, depending on the path taken. In Example 2.1 the displacement is 2.9 km to the west, and the distance traveled is 11.5 km. 2.2 Yes. Average speed is the distance traveled divided by the time interval in moving from point A to point B. Average velocity is the displacement from point A to point B divided by the same time interval. The magnitude of the displacement

gia04535_ch02_025-054.indd 54

is the shortest possible distance from A to B. Thus the average velocity magnitude is less than or equal to the average speed. 2.3 The slope of the tangent to a graph of vx versus time is the instantaneous acceleration ax at the time. 2.4 Only if the plane’s acceleration is constant must its average velocity be 470 km/h west. If its acceleration is not constant, the average velocity is not necessarily 470 km/h west. To find the average velocity, we would divide the plane’s displacement by the time interval. 2.6 Yes. If you throw a ball upward, it is in free fall as soon as it loses contact with your hand.

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CHAPTER

Motion in a Plane

3

A gull scoops up a clam and takes it high above the ground. While flying parallel to the ground, the gull lets go of the clam. The clam lands on a rock below and cracks open. Then the gull alights and enjoys lunch. A beachcomber on the beach sees the clam fall along a parabolic path, just as a projectile would. Why does the clam not drop straight down? What does the path of the falling clam look like to the gull? (See pp. 73 and 76–77 for the answers.)

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CHAPTER 3 Motion in a Plane

Concepts & Skills to Review

• • • • •

trigonometric functions: sine, cosine, and tangent (Appendix A.7) Pythagorean theorem (Appendix A.6) position, displacement, velocity, and acceleration (Sections 2.1–2.3) average and instantaneous quantities (Sections 2.2–2.3) motion along a line with constant acceleration (Sections 2.4–2.6)

3.1

CONNECTION: Vector quantities must be added and subtracted according to special rules that take their directions into account. All vector quantities follow the same rules of addition and subtraction. Vector quantities have both magnitude and direction.

GRAPHICAL ADDITION AND SUBTRACTION OF VECTORS

Chapter 2 introduced the quantities position, displacement, velocity, and acceleration to describe motion along a line—that is, motion in one dimension of space. To describe motion in more than one dimension, we need a full treatment of vector addition and subtraction because position, displacement, velocity, and acceleration are vectors. (Other vectors you will study in this book include force, momentum, angular momentum, torque, and the electric and magnetic fields.) Vectors and Scalars All vectors have a direction as well as a magnitude. The direction of any vector is always a physical direction in space such as up, down, north, or 35° south of west. Vector quantities are usually drawn as arrows pointing in the direction of the vector; the length of the arrow is proportional to the magnitude of the vector. By contrast, a scalar quantity can have magnitude, algebraic sign, and units, but not a direction in space. It wouldn’t make sense to draw an arrow to represent a scalar such as mass! In this book, an arrow over a boldface symbol indicates a vector quantity (r⃗). (Some books use boldface without the arrow or the arrow without boldface.) When writing by hand, always draw an arrow over a vector symbol to distinguish it from a scalar. When the symbol for a vector is written without the arrow and in italics rather than boldface (r), it stands for the magnitude of the vector (which is a scalar). Absolute value bars are also used to stand for the magnitude of a vector, so r = r⃗. The magnitude of a vector may have units and is never negative; it can be positive or zero.

Conceptual Example 3.1 Vector or Scalar? Is temperature a vector quantity? Strategy If a quantity is a vector, it must have both a magnitude and a physical direction in space. Solution and Discussion Does temperature have a direction? A temperature in Fahrenheit or Celsius can be above or below zero—is that a direction? No. A vector must have a physical direction in space. It does not make sense to say that the temperature of your coffee is “85 degrees Celsius in the

southwest direction.” “The temperature is up 5 degrees today,” means that it has increased, not that it is pointing vertically upward. Temperature is a scalar, not a vector.

Conceptual Practice Problem 3.1 Bank Balance When you deposit a paycheck, the balance of your checking account “goes up.” When you pay a bill, it “goes down.” Is the balance of your account a vector quantity?

When scalars are added or subtracted, they do so in the usual way: 3 kg of water plus 2 kg of water is equal to 5 kg of water. Adding or subtracting vectors is different. Vectors follow rules of addition and subtraction that take into account the directions of the vectors as well as their magnitudes. Whenever you need to add or subtract quantities, check whether they are vectors. If so, be sure to add or subtract them correctly as vectors. Do not just add or subtract their magnitudes.

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3.1 GRAPHICAL ADDITION AND SUBTRACTION OF VECTORS

A

A

B

A

B

B

Thiss is not n A + B; the direc ection is wrong.

A+ B

A

N (a)

(b)

(c)

(d)

Figure 3.1 Adding two vectors graphically. (a) Draw one vector arrow. (b) Draw the second, starting where the first arrow ended. (c) The sum of the two. (d) A common mistake. Graphical Vector Addition We start with a graphical method to help develop your intuition. To add two vectors graphically, first draw an arrow to represent one of them ⃗ + B ⃗ = B ⃗ + A. ⃗ ) The (Fig. 3.1a). (It does not matter in what order vectors are added; A arrow points in the direction of the vector and its length is proportional to the magnitude of the vector. It doesn’t matter where you start drawing the arrow. The value of a vector is not changed by moving it as long as its direction and magnitude are not changed. Now draw the second vector arrow starting where the first ends. In other words, place the “tail” of the second arrow at the “tip” of the first (Fig 3.1b). Finally, draw an arrow starting from the tail of the first and ending at the tip of the second. This arrow represents the sum of the two vectors (Fig. 3.1c). A common error is to draw the sum from the tip of the second to the tail of the first (Fig. 3.1d). If the lengths and directions of the vectors are drawn accurately to scale, using a ruler and a protractor, then the length and direction of the sum can be determined with the ruler and protractor. To add more than two vectors, continue drawing them tip to tail. Vector Subtraction To subtract a vector is to add its opposite (that is, a vector with the same magnitude but opposite direction): r⃗ f − r⃗ i = r⃗ f + (−r⃗ i). Multiplying a vector by the scalar −1 reverses the vector’s direction while leaving its magnitude unchanged, so −r⃗ i = −1 × r⃗ i is a vector equal in magnitude and opposite in direction to r⃗ i. Using Compass Headings It is common to use compass headings to specify vector directions in a horizontal plane. For example, the direction of the vector in Fig. 3.2 is “20° north of east,” which means that the vector makes a 20° angle with the east direction and is on the north (rather than the south) side of east. The same direction could be described as “70° east of north,” although it is customary to use the smaller angle. Northeast means “45° north of east” or, equivalently, “45° east of north.”

Position and Displacement The position r⃗ of an object can be represented as a vector arrow drawn from the origin to the location of the object (Fig. 3.3). Its magnitude is the distance from the origin. The displacement is literally the change in position (the final position vector minus the initial position vector): Δr⃗ = r⃗ f − r⃗ i (3-1) Figure 3.4 shows the graphical subtraction of two position vectors to illustrate the displacement for a trip from Killarney to Kenmare. This same procedure is used to subtract any kind of vector quantity (velocity, acceleration, etc.).

W

E

20°

S

Figure 3.2 Measuring angles with respect to compass headings. The direction of this vector is 20° north of east (20° N of E).

A plus sign (+) between vector quantities indicates vector addition, not ordinary addition. An equals sign (=) between vector quantities means that the vectors are identical in magnitude and direction, not simply that their magnitudes are equal.

⃗ − B ⃗ = Vector Subtraction: A ⃗ + (− B), ⃗ where −B ⃗ has the same A ⃗ but is opposite in magnitude as B direction. Note that the order ⃗ − A ⃗ = −(A ⃗ − B). ⃗ matters: B y r

Origin

x

Figure 3.3 A position vector r⃗.

Addition of Displacement Vectors As in Example 2.1, the total displacement for a trip with several parts is the vector sum of the displacements for each part of the trip because r⃗ 3 − r⃗ 1 = (r⃗ 3 − r⃗ 2) + (r⃗ 2 − r⃗ 1)

(3-2)

Example 3.2 explores this idea further.

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CHAPTER 3 Motion in a Plane

Figure 3.4 (a) Two position vectors, r⃗ i and r⃗ f, drawn from an arbitrary origin to the starting point (Killarney) and to the ending point (Cork) of a trip. (b) The final position vector minus the initial position vector is the displacement Δr⃗, found by adding −r⃗ i + r⃗ f.

Killarney

Killarney

ri

∆r = –ri + rf

–ri

Cork

Cork

rf Origin

rf Origin

(a)

(b)

Example 3.2 An Irish Adventure (1) In a trip from Killarney to Cork, Charlotte and Shona drive at a compass heading of 27° west of south for 18 km to Kenmare, then directly south for 17 km to Glengariff, then at a compass heading of 13° north of east for 48 km to Cork. Find the displacement vector for the entire trip by adding the three displacements graphically.

Killarney

N 30°

1 2

18 km A 27°

3

A+ 4

W

B+

5 cm

Kenmare

E

C

S

6 7

Cork

8 9

17 km B

48 km

10

C

13°

Blarney castle.

Strategy To add the displacement vectors, place the tail of each successive vector at the tip of the preceding vector. The value of a vector is not changed by moving it as long as its direction and magnitude are not changed, so a vector can be drawn starting at any point. The sum of the three displacements is then drawn from the tail of the first vector to the tip of the last vector. To add vectors graphically and get an accurate result, we use a ruler and a protractor. The protractor is used to draw the vector arrows in the correct directions and the ruler is used to draw them with the correct lengths. Then the length and direction of the sum can be determined with the ruler and protractor. Solution Let’s call the four positions r⃗ 1 (Killarney), r⃗ 2 (Kenmare), r⃗ 3 (Glengariff), and r⃗ 4 (Cork). The displacement for the whole trip is r⃗ 4 − r⃗ 1. The problem gives the displacements for the three parts of the trip; let’s call them ⃗ = r⃗ 2 − r⃗ 1 = 18 km, 27° west of south; B ⃗ = r⃗ 3 − r⃗ 2 = 17 km, A ⃗ = r⃗ 4 − r⃗ 3 = 48 km, 13° north of east. The sum of south; and C these three displacements is the total displacement because ⃗ + B ⃗ + C ⃗ = (r⃗ 2 − r⃗ 1) + (r⃗ 3 − r⃗ 2) + (r⃗ 4 − r⃗ 3) = r⃗ 4 − r⃗ 1 A Next we choose a convenient scale for the lengths of the vector arrows. Here we choose to represent 1 km as an arrow ⃗ length of 0.2 cm, so the length of the vector arrow for A should be

Glengariff

Figure 3.5 Graphical addition of the displacement vectors for the trip from Killarney to Cork via Kenmare and Glengariff.

0.2 cm = 3.6 cm 18 km × ______ 1 km ⃗ ⃗ should be 3.4 cm and Similarly, the arrows for B and C 9.6 cm long, respectively. After drawing the three vector arrows tip to tail, the arrow from the tail of the first vector to the tip of the last vector represents the sum (Fig. 3.5). This arrow is measured to have length 8.9 cm and its direction is 30° south of east. The total displacement has magnitude 1 km = 44.5 km 8.9 cm × ______ 0.2 cm Rounding to two significant figures, the total displacement ⃗ + B ⃗ + C ⃗ has magnitude 45 km and is directed 30° south A of east. Discussion Note that the answer includes both the magnitude and direction of the displacement. If a homework or exam question has you calculate a vector quantity such as position or velocity, don’t forget to specify the direction as well as the magnitude in your answer. One without the other is incomplete. Although the magnitude and direction of a position vector depends on the choice of origin, the magnitude and continued on next page

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3.2 VECTOR ADDITION AND SUBTRACTION USING COMPONENTS

Example 3.2 continued

direction of a displacement (change of position) does not depend on the choice of origin. The total distance traveled by Charlotte and Shona is 18 km + 17 km + 48 km = 83 km, which is not equal to the magnitude of the total displacement. Finding the total distance involves adding three scalars, while finding the total displacement involves adding three vectors. The magnitude of the total displacement is the straight-line distance from Killarney to Cork.

Practice Problem 3.2 A Traveling Executive An executive flies from Kansas City to Chicago (displacement = 400 mi in the direction 30° north of east) and then from Chicago to Tulsa (600 mi, 45° south of west). Add the two displacements graphically to find the total displacement from Kansas City to Tulsa.

y vx

3.2

x

58°

VECTOR ADDITION AND SUBTRACTION USING COMPONENTS

vy v

Components of a Vector Any vector can be expressed as the sum of vectors parallel to the x-, y-, and (if needed) z-axes. The x-, y-, and z-components of a vector indicate the magnitude and direction of the three vectors along the three perpendicular axes. The sign of a component indi⃗ are writcates the direction along that axis. The x-, y-, and z-components of vector A ten with subscripts as follows: Ax, Ay, and Az. One exception to this otherwise consistent notation is that the x-, y-, and z-components of a position vector r⃗ are usually written x, y, and z (instead of rx, ry, and rz). For now we will deal only with vectors in the xy-plane. The x-component of a position vector r⃗ is x, the x-coordinate. For all other vectors, the x-component is designated by a subscript x. For example, the x-component of a velocity vector v⃗ is written vx. Components of vectors have magnitude, units, and an algebraic sign. The sign indicates the direction: a positive x-component indicates the direction of the positive x-axis, while a negative x-component indicates the opposite direction (the negative x-axis). Finding Components The process of finding the components of a vector is called resolving the vector into its components. Consider the velocity vector v⃗ in Fig. 3.6. We can think of v⃗ as the sum of two vectors, one parallel to the x-axis and the other parallel to the y-axis. The magnitudes of these two vectors are the magnitudes (absolute values) of the x- and y-components of v⃗. We can find the magnitudes of the components using the right triangle in Fig. 3.6 and the trigonometric functions in Fig. 3.7. The length of the arrow represents the magnitude of the vector (v = 9.4 m/s), so adjacent | vx | cos 58° = __________ = ___ v hypotenuse

and

opposite | vy | sin 58° = __________ = ___ v hypotenuse

Figure 3.6 Resolving a velocity vector v⃗ into x- and y-components. f

Right triangle c

b 90°

q a f = 90° – q

side opposite ∠q sin q = ______________ = hypotenuse side adjacent ∠q cos q = ______________ = hypotenuse side opposite ∠q tan q = ______________ = side adjacent ∠q

b_ c a_ c b_ a

Figure 3.7 Trigonometric functions (see Appendix A.7 for more information). y

x

(3-3) 32°

Now we must determine the correct algebraic sign for each of the components. From Fig. 3.6, the vector along the x-axis points in the positive x-direction and the vector along the y-axis points in the negative y-direction, so in this case, v x = +v cos 58° = 5.0 m/s

and

v y = −v sin 58° = −8.0 m/s

(3-4)

Using the right triangle in Fig. 3.8 gives the same values for the x- and y-components of v⃗ since cos 32° = sin 58° and sin 32° = cos 58°.

gia04535_ch03_055-086.indd 59

v vy vx

Figure 3.8 Resolving the velocity vector into components using a different right triangle.

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CHAPTER 3 Motion in a Plane

Problem-Solving Strategy: Finding the x- and y-Components of a Vector from Its Magnitude and Direction 1. Draw a right triangle with the vector as the hypotenuse and the other two sides parallel to the x- and y-axes. 2. Determine one of the unknown angles in the triangle. 3. Use trigonometric functions to find the magnitudes of the components. Make sure your calculator is in “degree mode” to evaluate trigonometric functions of angles in degrees and “radian mode” for angles in radians. 4. Determine the correct algebraic sign for each component.

Finding Magnitude and Direction We must also know how to reverse the process to find a vector’s magnitude and direction from its component.

Problem-Solving Strategy: Finding the Magnitude and Direction of a ⃗ from Its x- and y-Components Vector A 1. Sketch the vector on a set of x- and y-axes in the correct quadrant, according to the signs of the components. 2. Draw a right triangle with the vector as the hypotenuse and the other two sides parallel to the x- and y-axes. 3. In the right triangle, choose which of the unknown angles you want to determine. 4. Use the inverse tangent function to find the angle. The lengths of the sides of the triangle represent Ax and Ay. If q is opposite the side parallel to the x-axis, then tan q = opposite/adjacent = Ax/Ay. If q is opposite the side parallel to the y-axis, then tan q = opposite/adjacent = Ay/Ax. If your calculator is in “degree mode,” then the result of the inverse tangent operation will be in degrees. [In general, the inverse tangent has two possible values between 0 and 360° because tan a = tan (a + 180°). However, when the inverse tangent is used to find one of the angles in a right triangle, the result can never be greater than 90°, so the value the calculator returns is the one you want.] 5. Interpret the angle: specify whether it is the angle below the horizontal, or the angle west of south, or the angle clockwise from the negative y-axis, etc. 6. Use the Pythagorean theorem to find the magnitude of the vector.

√

_______ 2

2

A = Ax + Ay

(3-5)

Suppose we knew the components of the velocity vector in Fig. 3.6, but not the magnitude and direction. Let us find the angle q between v⃗ and the +x-axis: vy opposite 8.0 m/s = 58° q = tan−1 _______ = tan−1 ___ = tan−1 _______ adjacent 5.0 m/s vx

(3-6)

The magnitude of v⃗ is _______

√

_____________________

v = v x + v y = √ (+5.0 m/s)2 + (−8.0 m/s)2 = 9.4 m/s 2

2

Adding Vectors Using Components It is generally easier and more accurate to add vectors algebraically rather than graphically. The algebraic method relies on adding the components of the vectors. Remember that each vector is thought of as the sum of vectors parallel to the axes (Fig. 3.9a). When

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3.2 VECTOR ADDITION AND SUBTRACTION USING COMPONENTS

Bx By

Ax

Bx

A

⃗ = A ⃗ + B, ⃗ Figure 3.9 (a) C shown graphically with the x- and y-components of each vector illustrated. (b) Cx = Ax + Bx; (c) Cy = Ay + By.

Ax Cx

B

By

Ay

Ay

C=A+B Cy

61

Cy Cx (a)

(b)

(c)

adding vectors, we can add them in any order and group them as we please. So we can sum the x-components to find the x-component of the sum (Fig. 3.9b) and then do the same with the y-components (Fig. 3.9c): ⃗ = A ⃗ + B ⃗ C

if and only if

Cx = Ax + Bx

and

Cy = Ay + By

(3-7)

In Eq. (3-7), remember that Ax + Bx represents ordinary addition since the signs of the components carry the direction information.

Problem-Solving Strategy: Adding Vectors Using Components 1. Find the x- and y-components of each vector to be added. 2. Add the x-components (with their algebraic signs) of the vectors to find the x-component of the sum. (If the signs are not correct, the sum will not be correct.) 3. Add the y-components (with their algebraic signs) of the vectors to find the y-component of the sum. 4. If necessary, use the x- and y-components of the sum to find the magnitude and direction of the sum.

Estimation Using Graphical Addition Even when using the component method to add vectors, the graphical method is an important first step. A rough sketch of vector addition, even one made without carefully measuring the lengths or the angles, has important benefits. Sketching the vectors makes it much easier to get the signs of the components correct. The graphical addition also serves as a check on the answer—it provides an estimate of the magnitude and direction of the sum, which can be used to check the algebraic answer. Graphical addition gives you a mental picture of what is going on and an intuitive feel for the algebraic calculations.

CHECKPOINT 3.2 ⃗ and B ⃗ have x- and y-components as follows: Ax = +3.0 km, Two displacements A ⃗ = A ⃗ + B. ⃗ Ay = − 6.0 km, Bx = − 8.5 km, By = −1.2 km. The total displacement is C ⃗ What are the x- and y-components of C?

Choosing x- and y-Axes A problem can be made easier to solve with a good choice of axes. We can choose any direction we want for the x- and y-axes, as long as they are perpendicular to one another. Three common choices are • x-axis horizontal and y-axis vertical, when the vectors all lie in a vertical plane; • x-axis east and y-axis north, when the vectors all lie in a horizontal plane; and • x-axis parallel to an inclined surface and y-axis perpendicular to it.

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CHAPTER 3 Motion in a Plane

Example 3.3 An Irish Adventure (2) In the trip of Example 3.2, Charlotte and Shona drive at a compass heading of 27° west of south for 18 km to Kenmare, then directly south for 17 km to Glengariff, then at a compass heading of 13° north of east for 48 km to Cork. Use the component method to find the magnitude and direction of the displacement vector for the entire trip. Strategy As before, let’s call the three successive dis⃗ B, ⃗ and C, ⃗ respectively. To add the vectors placements A, using components, we first choose directions for the x- and y-axes. Then we find the x- and y-components of the three displacements. Adding the x- or y-components of the three displacements gives the x- or y-component of the total displacement. Finally, from the components we find the magnitude and direction of the total displacement. Solution A good choice is the conventional one: x-axis to the east and the y-axis to the north. The first displacement ⃗ is directed 27° west of south. Both of its components are (A) negative since west is the −x-direction and south is the −y-direction. Using the right triangle in Fig. 3.10, the side of the triangle opposite the 27° angle is parallel to the x-axis. The sine function relates the opposite side to the hypotenuse: Ax = −A sin 27° = −18 km × 0.454 = −8.17 km ⃗ The cosine relates the adjawhere A is the magnitude of A. cent side to the hypotenuse: Ay = −A cos 27° = −18 km × 0.891 = −16.0 km ⃗ has no x-component since its direction is Displacement B south. Therefore, Bx = 0

and

By = −17 km

⃗ is 13° north of east. Both its compoThe direction of C nents are positive. From Fig. 3.10, the side of the right triangle opposite the 13° angle is parallel to the y-axis, so

A = 18 km

y

A x

27° B = 17 km Ay

Cx = +C cos 13° = +48 km × 0.974 = +46.8 km Cy = +C sin 13° = +48 km × 0.225 = +10.8 km Now we sum the x- and y-components separately to find the x- and y-components of the total displacement: Δx = Ax + Bx + Cx = (−8.17 km) + 0 + 46.8 km = +38.63 km Δy = Ay + By + Cy = (−16.0 km) + (−17 km) + 10.8 km = −22.2 km The magnitude and direction of Δr⃗ can be found from the right triangle in Fig. 3.11. The magnitude is represented by the hypotenuse: ___________

______________________

Δr = √ (Δx)2 + (Δy)2 = √ (38.63 km)2 + (−22.2 km)2 = 45 km The angle q is opposite 22.2 km = 30° q = tan−1 _______ = tan−1 ________ adjacent 38.63 km Since +x is east and −y is south, the direction of the displacement is 30° south of east. The magnitude and direction of the displacement found using components agree with the displacement found graphically in Fig. 3.5.

y 38.63 km q –22.2 km

x

∆r

Figure 3.11 Finding the magnitude and direction of Δr⃗.

Discussion Note that the x-component of one displacement was found using the sine function while another was found using the cosine. The x-component (or the y-component) of the vector can be related to either the sine or the cosine, depending on which angle in the right triangle is used.

C = 48 km C

B

Cy

13°

Ax

Cx

Figure 3.10 ⃗ B, ⃗ and C ⃗ into x- and y-components. Resolving A,

Practice Problem 3.3 Axes

Changing the Coordinate

Find the x- and y-components of the displacements for the three legs of the trip if the x-axis points south and the y-axis points east.

Unit Vectors The connection between a vector and its components may be expressed using the unit vectors xˆ (read aloud as “x hat”), yˆ, and zˆ, which are defined as vectors of magnitude 1 that point in the +x-, +y-, and +z-directions, respectively. (In some books, you may see them

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3.3 VELOCITY

63

written as ˆi, jˆ, and kˆ.) They are called unit vectors because the magnitude of each is the pure number 1—they do not have physical units such as kilograms or meters. Any vector ⃗ can be written as the sum of three vectors along the coordinate axes: A ⃗ = Axxˆ + Ayyˆ + Azzˆ A

(3-8)

⃗ which has physical units and can be positive or negative. Here Ax is the x-component of A, Axxˆ is a vector of magnitude |Ax| directed in the +x-direction if Ax > 0 and in the −x-direction if Ax < 0. For example, consider the velocity vector v⃗ of Fig. 3.8. v⃗ has x-component vx = +5.0 m/s and y-component vy = −8.0 m/s, so v⃗ = (+5.0 m/s)xˆ + (−8.0 m/s)yˆ. Using unit vector notation is one way to keep track of vector components in vector addition and subtraction without writing separate equations for each component. Adding two vectors in the xy-plane looks like this: ⃗ 1 + A ⃗ 2 = ( A 1xxˆ + A 1yyˆ ) + ( A 2xxˆ + A 2yyˆ ) A

(3-9)

Regrouping the terms shows that the x-component of the sum is the sum of the x-components and likewise for the y-components: ⃗ 1 + A ⃗ 2 = ( A 1x + A 2x ) xˆ + ( A 1y + A 2y ) yˆ A

3.3

(3-10)

VELOCITY

The definitions of average velocity, instantaneous velocity, average acceleration, and instantaneous acceleration from Chapter 2 still apply when the motion is not in a straight line as long as we add and subtract them as vectors. Suppose we want to know the instantaneous velocity of a race car at point P as it goes around a curved section of a racetrack (Fig. 3.12a). At a slightly later time the race car is at point Q. Let r⃗ i be the position of the car at P and r⃗ f be the position at point Q. Average Velocity The displacement Δr⃗ = r⃗ f − r⃗ i is represented as an arrow from P to Q. Alternatively, to subtract r⃗ i from r⃗f, the two vectors can be drawn with their tails at the same point. After reversing the direction of r⃗ i to represent −r⃗ i, the arrows are tip to tail and ready to add r⃗f + (−r⃗ i)—see Fig. 3.12b. The average velocity during this time interval is the displacement Δr⃗ divided by the time interval: r⃗ f − r⃗ i ___ Δr⃗ v⃗ av = ______ tf − ti = Δt

(3-11)

The direction of the average velocity is the direction of the displacement Δr⃗. Instantaneous Velocity The instantaneous velocity at P is the limit of the average velocity as Δt approaches zero. As we shorten the time interval between the initial and final positions by moving point Q closer and closer to P, the direction of the displacement y Q

rf

Q1

rf ∆r

∆r

∆r1 Q2

P ri

–ri x

(a)

P

∆r2

v

P vx Tangent at P

rf – ri = ∆r (b)

vy

(c)

(d)

Figure 3.12 (a) Position vectors for two points on the curve. (b) The displacement Δr⃗ from point P to point Q. (c) As the time interval is decreased, the final point moves closer and closer to P; the direction of the displacement Δr⃗ approaches the tangent to the curve at P. (d) Instantaneous velocity can be resolved into components along perpendicular axes.

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vector Δr⃗ gradually changes, approaching the tangent to the curved path at P (Fig. 3.12c). Expressed in mathematical terminology, the instantaneous velocity is the limit of Δr⃗/Δt as the time interval approaches zero: Δr⃗ v⃗ = lim ___ Δt→0 Δt

(3-12)

(Δr⃗ is the displacement during a very short time interval Δt)

If an object moves along a curved path, the direction of the velocity vector at any point is tangent to the path at that point.

With this definition, the instantaneous velocity at P becomes tangent to the curve at P (Fig. 3.12d). Here we are talking about a tangent to the actual path through space, not a tangent line on a graph of position versus time. The magnitude of the velocity vector is the speed at which the object moves and the direction of the velocity vector is the direction of motion. Component Equations A vector equation is always equivalent to a set of equations, one for each component. The x- and y-components of the average velocity are Δy Δx and v = ___ (3-13) vav,x = ___ av,y Δt Δt The x- and y-components of the instantaneous velocity are Δy Δx and v = lim ___ vx = lim ___ y Δt→0 Δt Δt→0 Δt

(3-14)

To put Eq. (3-14) into words, the x-component of an object’s velocity is the rate of change of its x-coordinate and the y-component of its velocity is the rate of change of its y-coordinate.

Example 3.4 An Irish Adventure (3) In their trip from Kenmare to Cork via Glengariff, Charlotte and Shona travel a total distance of 83 km in 1.4 h. The total displacement for the trip is 45 km, 30° south of east. What is their average velocity? Contrast it with their average speed, defined as the total distance divided by the time interval. Strategy The average velocity is calculated from the displacement—not from the distance traveled. Solution The magnitude of the average velocity is Δr⃗ 45 km v⃗av = ____ = ______ = 32 km/h 1.4 h Δt

83 km = 59 km/h average speed = ______ 1.4 h Therefore, v⃗av is not equal to the average speed. Furthermore, average velocity is a vector quantity with a direction in space, and average speed is a scalar.

Practice Problem 3.4 Average Speed

Average Velocity Versus

In Example 3.4, v⃗av was less than the average speed. Can v⃗av ever be greater than the average speed? Can v⃗av ever be equal to the average speed? Explain.

The average velocity has the same direction as the displacement, so v⃗av = 32 km/h, 30° south of east. The average speed is

3.4

ACCELERATION

The average acceleration a⃗ av is the change in velocity divided by the elapsed time: v⃗ f − v⃗ i ___ Δv⃗ a⃗ av = ______ tf − ti = Δt

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(3-15)

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illustrate that the average acceleration is always in the same direction as the change in velocity Δv⃗ during the same time interval.

vi

vi

∆v ∆v

65

Figure 3.13 Two examples to

Turning while increasing speed

Turning while keeping speed constant

vf

ACCELERATION

a av

vf a av

For motion in a plane, this vector equation is equivalent to two component equations: Δv aav,x = ____x Δt

and

Δv y aav,y = ____ Δt

(3-16)

The direction of a⃗ av is the same as the direction of Δv⃗ (Fig. 3.13). Instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero: Δv⃗ a⃗ = lim ___ Δt→0 Δt

(3-17)

(Δv⃗ is the change in velocity during a very short time interval Δt) In component form, Δv ax = lim ____x Δt→0 Δt

and

Δvy ay = lim ___ Δt→0 Δt

(3-18)

In straight-line motion the acceleration is always along the same line as the velocity. For motion in two dimensions, the acceleration vector can make any angle with the velocity vector because the velocity vector can change in magnitude, in direction, or both. The direction of the acceleration is the direction of the change in velocity Δv⃗ during a very short time interval.

CHECKPOINT 3.4 An airplane is initially moving due north at 400 km/h. After making a slight course correction, it is moving at the same speed but in a direction 2.0° east of north. Is the plane’s average acceleration during this time interval zero? Explain.

Example 3.5 Skating Uphill An inline skater is traveling on a level road with a speed of 8.94 m/s; 120.0 s later she is climbing a hill with a 15.0° angle of incline at a speed of 7.15 m/s. (a) What is the change in her velocity? (b) What is her average acceleration during the 120.0-s time interval?

Strategy The change in velocity is not 1.79 m/s (= 8.94 m/s −7.15 m/s). That is the change in speed. The change in velocity is found by subtracting the initial velocity vector from the final velocity vector. After first making a graphical sketch, we use the component method. The average acceleration is the change in velocity divided by the elapsed time.

continued on next page

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Example 3.5 continued

To find the magnitude of Δv⃗, we apply the Pythagorean theorem (Fig. 3.16):

vf = 7.15 m/s vi = 8.94 m/s

Δv⃗2 = (Δvx)2 + (Δvy)2 = (−2.03 m/s)2 + (1.85 m/s)2

15.0°

= 7.54 (m/s)2 (a)

Figure 3.14 vf

∆v

(a) Change in velocity as the skater slows going uphill and (b) graphical subtraction of velocity vectors.

vi vf – vi = ∆v (b)

Solution (a) Figure 3.14a shows the initial and final velocity vectors and the slope of the hill. The initial velocity is horizontal as the skater skates on level ground. The final velocity is 15.0° above the horizontal. To subtract the two velocity vectors graphically, we place the tails of the vectors together. The change in velocity Δv⃗ is found by drawing a vector arrow from the tip of v⃗ i to the tip of v⃗ f. Judging by the graphical subtraction in Fig. 3.14b, the change in velocity is roughly at a 45° angle above the −x-axis. Its magnitude is smaller than the magnitudes of the initial and final velocity vectors—something like 2 to 3 m/s. The components vfx and vfy can be found from a right triangle (Fig. 3.15): vfx = v f cos q = 7.15 m/s × 0.9659 = 6.91 m/s v fy = v f sin q = 7.15 m/s × 0.2588 = 1.85 m/s Since v i has only an x-component, v iy = 0

and

v ix = v i = 8.94 m/s

Now we subtract the components to find the components of Δv⃗: Δvx = v fx − v ix = (6.91 − 8.94) m/s = −2.03 m/s

Δv⃗ = 2.75 m/s The angle is found from opposite Δvy 1.85 m/s tan f = _______ = ___ = ________ = 0.9113 adjacent Δvx 2.03 m/s

| |

f = tan

−1

0.9113 = 42.3°

The direction of the change in velocity Δv⃗ is 42.3° above the negative x-axis. (b) The magnitude of the average acceleration is Δv⃗ 2.75 m/s a⃗ av = ____ = ________ = 0.0229 m/s2 120.0 s Δt The direction of the average acceleration is the same as the direction of Δv⃗: 42.3° above the negative x-axis. Discussion Checking back with the y graphical subtraction in Fig. 3.14b, the magnitude of Δv⃗ appears to be x roughly _14 to _13 the magnitude of v⃗ i. 1 1 _ _ Since 4 × 8.94 m/s = 2.24 m/s and 3 × 8.94 m/s = 2.98 m/s, the answer of ∆v ∆vy 2.75 m/s is reasonable. f Figure 3.14b also shows the direc∆vx tion of Δv⃗ to be roughly midway between the +y- and −x-axes. We found Figure 3.16 the direction of Δv⃗ to be 42.3° above Reconstruction of Δv⃗ the −x-axis and, therefore, 47.7° from from its components the +y-axis. So the direction we calcu- (not to scale). lated is also reasonable based on the graphical subtraction.

and Δvy = v fy − v iy = (1.85 − 0) m/s = +1.85 m/s y

x vi vix vf

Figure 3.15 q

vfx

gia04535_ch03_055-086.indd 66

vfy

Initial and final velocity vectors resolved into components.

Practice Problem 3.5 Change in Sailboat Velocity A C&C 30 sailboat is sailing at 12.0 knots (6.17 m/s) heading directly east across the harbor. When a gust of wind comes up, the boat changes its heading to 11.0° north of east and its speed increases to 14.0 knots (7.20 m/s). [A boat’s speed is customarily expressed in knots, which means nautical miles per hour. A nautical mile (6076 ft) is a little longer than a statute mile (5280 ft).] (a) What is the magnitude and direction of the change in velocity of the sailboat in m/s? (b) If this velocity change occurs during a 2.0-s time interval, what is the average acceleration of the sailboat during that interval?

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MOTION IN A PLANE WITH CONSTANT ACCELERATION

MOTION IN A PLANE WITH CONSTANT ACCELERATION

If an object moves in the xy-plane with constant acceleration, then both ax and ay are constant. By looking separately at the motion along two perpendicular axes, the y-direction and the x-direction, each component becomes a one-dimensional problem, which we studied in Chapter 2. We can apply any of the constant acceleration relationships from Section 2.4 separately to the x-components and to the y-components. It is generally easiest to choose the axes so that the acceleration has only one nonzero component. Suppose we choose the axes so that the acceleration is in the positive or negative y-direction. Then ax = 0 and vx is constant. With this choice, the constant acceleration relationships [Eqs. (2-9) through (2-13)] become x-axis: ax = 0

y-axis: constant ay

Δvx = 0 (vx is constant)

Δvy = ay Δt

(3-19)

Δ x = vx Δt

Δy = _12 (v fy + v iy) Δt

(3-20)

Δy = v iy Δt + _12 ay (Δt)2

(3-21)

2 v fy

(3-22)

−

2 v iy

= 2ay Δy

Why are only two equations shown in the column for the x-axis? The other two are redundant when ax = 0. Note that there is no mixing of components in Eqs. (3-19) through (3-22). Each equation pertains either to the x-components or to the y-components; none contains the x-component of one vector quantity and the y-component of another. The only quantity that appears in both x- and y-component equations is the time interval—a scalar.

Motion of Projectiles An object in free fall near the Earth’s surface has a constant acceleration. As long as air resistance is negligible, the constant downward pull of gravity gives the object a constant downward acceleration with magnitude g. In Section 2.6 we considered objects in free fall, but only when they had no horizontal velocity component, so they moved straight up or straight down. Now we consider objects (called projectiles) in free fall that have a nonzero horizontal velocity component. The motion of a projectile takes place in a vertical plane. Suppose some medieval marauders are attacking a castle. They have a catapult that propels large stones into the air to bombard the walls of the castle (Fig. 3.17). Picture a stone leaving the catapult with initial velocity v⃗ i. (v⃗ i is the initial velocity for the time interval during which it moves as a projectile. It is also the final velocity for the time interval during which it is in contact with the catapult.) The angle of elevation is the angle of the initial velocity above the horizontal. Once the stone is in the air, the only force acting on it is the downward gravitational force, provided that the air resistance has a negligible effect on the motion. The trajectory (path) of the stone is shown in Fig. 3.18. The positive x-axis is chosen in the horizontal direction (to the right) and the positive y-axis is upward. If the initial velocity v⃗ i is at an angle q above the horizontal, then resolving it into components gives v ix = v i cos q

and

v iy = v i sin q

(3-23)

(+y-axis up, q measured from the horizontal x-axis)

CONNECTION: Projectile motion is free fall for objects with a horizontal velocity component.

vi

Figure 3.17 A medieval catapult.

With the y-axis pointing up, ay = −g because the acceleration is downward (in the −y-direction). The acceleration has no x-component (ax = 0), so the stone’s horizontal

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Figure 3.18 Motion diagram

CHAPTER 3 Motion in a Plane

y

showing the trajectory of a projectile. The position is drawn at equal time intervals. Superimposed are the velocity vectors along with their x- and y-components.

vy = 0

vy vy vix viy

vix

vix

vix

vy

vix vy

q

vix x

vix vfy

The horizontal and vertical motions of a projectile can be treated separately; they are independent of each other.

velocity component vx is constant. The vertical velocity component vy changes at a constant rate, exactly as if the stone were propelled straight up with an initial speed of viy. The initially positive vy decreases until, at the top of flight, vy = 0. Then the pull of gravity makes the projectile fall back downward. During the downward trip, vy is still changing at the same constant rate with which it changed on the way up and at the top of the path. The acceleration has the same constant value—magnitude and direction—for the entire path. The motion of a projectile when air resistance is negligible is the superposition of horizontal motion with constant velocity and vertical motion with constant acceleration. The vertical and horizontal motions each proceed independently, as if the other motion were not present. In the experiment of Fig. 3.19, one ball was dropped and, at the same instant, another was projected horizontally. The strobe photo shows snapshots of the two balls at equally spaced time intervals. The vertical motion of the two is identical; at every instant, the two are at the same height. The fact that they have different horizontal motion does not affect their vertical motion. (This statement would not be true if air resistance were significant.)

PHYSICS AT HOME Take a nickel and a penny to a room with a high table or countertop. Place the penny at the edge of the table and then slide the nickel so it collides with the penny. Listen for the sound of the two coins hitting the floor. The two coins will slide off the table with different horizontal velocities but will land at the same time.

Figure 3.19 Independence of horizontal and vertical motion of a projectile in the absence of air resistance. The vertical motion of the projectile (white) is the same as that of an object (red) that falls straight down.

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Conceptual Example 3.6 Trajectory of a Projectile Discussion The same conclusion can be drawn algebraically. With the +y-axis upward and the origin and t = 0 at the top of flight, xi, yi, and viy are all zero. Then x = vixt and

The graph of an equation of the form y = kx , k = a nonzero constant 2

is a parabola. Show that the trajectory of a projectile is a parabola. [Hint: Choose the origin at the highest point of the trajectory and let ti = 0 at that instant.]

( )

( )

g 2 1 gt2 = − __ x 2 = − ____ 1 a t2 = − __ 1 g ___ y = viyt + __ x 2 2 2 y 2 v ix 2v ix 2

Strategy and Solution We start at the high point of the path and look at displacements from there. The horizontal displacement is proportional to the elapsed time t since the horizontal velocity is constant. The vertical displacement is the average vertical velocity component times the elapsed time t. The average vertical velocity component is itself proportional to t since it changes at a constant rate. Therefore, the vertical displacement is proportional to t2. Thus, the vertical displacement y is proportional to the square of the horizontal displacement x and y = kx2, where k is a constant of proportionality. The path followed by a projectile in free fall is a parabola.

So y is proportional to x and the constant of proportionality 2 is −g/(2v ix).

Conceptual Practice Problem 3.6 Throwing Stones You stand at the edge of a cliff and throw stones horizontally into the river below. To double the horizontal displacement of a stone from the cliff to where it lands, by what factor must you increase the stone’s initial speed? Ignore air resistance.

gia04535_ch03_055-086.indd 69

1– 2 tf

tf Time

Horizontal position x

tf Time

Horizontal velocity vx

Vertical velocity vy

Vertical position y

Graphing Projectile Motion Figure 3.20 shows graphs of the x- and y-components of the velocity and position of a projectile as functions of time. In this case, the projectile is launched above flat ground at t = 0 and returns to the same elevation at a later time tf. Note that the y-component graphs are symmetrical about the vertical line through the highest point in the trajectory. The y-component of velocity decreases linearly from its initial value; the slope of the line is ay = −g. When vy = 0, the projectile is at the apex of its trajectory. Then vy continues to decrease at the same rate and is now negative with its magnitude getting larger and larger. At tf, when the projectile has returned to its original altitude, the y-component of the velocity has the same magnitude as at t = 0 but with the opposite sign (vy = −viy). The graph of y(t) indicates that the projectile moves upward, quickly at first and then gradually slowing, until it reaches the maximum height. The slope of the tangent to the y(t) graph at any particular moment of time is vy at that instant. At the highest point

viy 0 0 vfy

1– 2 tf

0 0

1– 2 tf

tf Time

1– 2 tf

tf Time

vix 0

Figure 3.20 Projectile motion: separate vertical and horizontal quantities versus time.

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CHAPTER 3 Motion in a Plane

of the y(t) graph, the tangent is horizontal and vy = 0. After that, gravity makes the projectile start to fall downward. The horizontal velocity is constant, so the graph of vx(t) is a horizontal line. The horizontal position x increases uniformly in time because the object is moving with a constant vx.

CHECKPOINT 3.5 When a basketball is thrown in an arc toward the net, what can you say about its velocity and acceleration at the highest point of the arc?

Example 3.7 Attacking the Castle Walls The catapult used by the marauders hurls a stone with a velocity of 50.0 m/s at a 30.0° angle of elevation (Fig. 3.21). (a) What is the maximum height reached by the stone? (b) What is its range (defined as the horizontal distance traveled when the stone returns to its original height)? (c) How long has the stone been in the air when it returns to its original height?

Solution (a) First we find the x- and y-components of the initial velocity for an angle of elevation q = 30.0°.

Strategy The problem gives both the magnitude and direction of the initial velocity of the stone. Ignoring air resistance, the stone has a constant downward acceleration once it has been launched—until it hits the ground or some obstacle. We choose the positive y-axis upward and the positive x-axis in the direction of horizontal motion of the stone (toward the castle). When the stone reaches its maximum height, the velocity component in the y-direction is zero since the stone goes no higher. When the stone returns to its original height, Δy = 0 and vy = −viy. The range can be found once the time of flight tf is known—time is the quantity that connects the x-component equations to the y-component equations. Therefore, we solve (c) before (b). One way to find tf is to find the time to reach maximum height and then double it (see Fig. 3.20). (Other methods include setting Δy = 0 or setting vy = −viy.)

Eliminating the time interval using vfy − viy = ayΔt yields

viy

vi

v iy = v i sin q

and

v ix = v i cos q

The maximum height is the vertical displacement Δy when vfy = 0. Δy = _12 (v fy + v iy) Δt = _12 (0 + v i sin q ) Δt

(

)

0 − v i sin q (v i sin q)2 1 (v sin q ) __________ _________ Δy = __ = − ay 2 i 2a y −(50.0 m/s × sin 30.0°) = 31.9 m = ____________________ 2 × (−9.80 m/s2) 2

The maximum height of the projectile is 31.9 m above its launch height. (c) The initial and final heights are the same. Due to this symmetry, the time of flight (tf) is twice the time it takes the projectile to reach its maximum height. The time to reach the maximum height can be found from v fy = 0 = v iy + ay Δt

Maximum height

30.0° vix

Initial launch height

Range

Figure 3.21 A catapult projects a stone into the air in an attack on a castle wall. continued on next page

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MOTION IN A PLANE WITH CONSTANT ACCELERATION

71

Example 3.7 continued

Solving for Δt,

Since we analyze the horizontal motion independently from the vertical motion, we start by resolving the given initial velocity into x- and y-components. Time is what connects the horizontal and vertical motions.

−v iy Δt = ____ ay The time of flight is −50.0 m/s × sin 30.0° = 5.10 s t f = 2 Δt = 2 × __________________ −9.80 m/s2

Practice Problem 3.7 Maximum Height for Arrows

(b) The range is

Archers have joined in the attack on the castle and are shooting arrows over the walls. If the angle of elevation for an arrow is 45°, find an expression for the maximum height of the arrow in terms of vi and g.__[Hint: Simplify the expression using sin 45° = cos 45° = 1/√ 2 .]

Δx = v ix t f = (50.0 m/s × cos 30.0°) × 5.10 s = 221 m Discussion Quick check: using y f − y i = v iy Δt + _12 ay (Δt)2 we can check that Δy = 31.9 m when Δt = _12 × 5.10 s and that Δy = 0 when Δt = 5.10 s. Here we check the first of these:

Δy = (50.0 m/s × sin 30.0°) × 2.55 s + _21 × (−9.80 m/s2) × (2.55 s)2 = 63.8 m + (−31.9 m) = 31.9 m which is correct. This is not an independent check, since this equation can be derived from the others, but it can reveal algebra or calculation errors.

PHYSICS AT HOME On a warm day, take a garden hose and aim the nozzle so that the water streams upward at an angle above the horizontal. Set the nozzle for a fast, narrow stream for best effect. Once the water leaves the nozzle, it becomes a projectile with a constant downward acceleration (ignoring the small effect of air resistance). The continuous stream of water lets us see the parabolic path easily. Stand in one place and try aiming the nozzle at different angles of elevation to find an angle that gives the maximum range. Aim for a particular spot on the ground (at a distance less than the maximum range) and see if you can find two different angles of elevated nozzle position that allow the stream to hit the target spot (see Fig. 3.22).

100 75°

y (m)

60°

Figure 3.22 Parabolic trajec-

50

45° 30° 15°

gia04535_ch03_055-086.indd 71

50

100 x (m)

150

200

tories of projectiles launched with the same initial speed (vi = 44.3 m/s) at five different angles. The ranges of projectiles launched at angles q and 90° − q are the same. The maximum range occurs for q = 45°.

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Conceptual Example 3.8 Monkey and Hunter An inexperienced hunter aims and shoots an arrow straight at a coconut that is being held by a monkey in a tree (Fig. 3.23). At the same instant that the arrow leaves the bow, the monkey drops the coconut. Ignoring air resistance, does the arrow hit the coconut, the monkey, or neither? Strategy and Solution If there were no gravity, the arrow would fly straight to the coconut (along the dashed blue line in Fig. 3.23). Since gravity gives the dropped coconut and the released arrow the same constant acceleration downward, they each fall the same vertical distance below the positions they would have had with no gravity. The coconut falls along the dashed red line; the distance fallen at 0.25-s intervals is marked. The arrow falls below the blue dashed line by the same distances, marked along its trajectory at 0.25-s intervals. The arrow ends up hitting the coconut no matter what the initial speed of the arrow (as long as the arrow’s range is at least as large as the horizontal distance to the coconut). The

higher the speed of the arrow, the sooner they meet and the shorter the vertical distance that the coconut falls before being hit. Discussion An experienced hunter would have aimed above the initial position of the coconut to compensate for gravity; he would have missed the coconut but might have hit the monkey unless the monkey jumped down to retrieve the coconut.

Conceptual Practice Problem 3.8 Changes in Position and Velocity for Consecutive Arrows An arrow is shot into the air. One second later, a second arrow is shot with the same initial velocity. While the two are both in the air, does the difference in their positions (r⃗ 2 − r⃗ 1) stay constant or does it change with time? Does the difference in their velocities (v⃗ 2 − v⃗ 1) stay constant or does it change with time?

t=0s t = 0.25 s

0.3 m 1.2 m

t = 0.50 s

2.8 m

t = 0.75 s

2.8 m 1.2 m 0.3 m t = 0.25 s

t = 0.50 s

t = 0.75 s 4.9 m t = 1.00 s

Figure 3.23 A monkey drops a coconut at the very instant an arrow is shot toward the coconut. In each quarter second, the coconut and arrow have fallen the same distance below where their positions would be if there were no gravity.

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3.6 VELOCITY IS RELATIVE; REFERENCE FRAMES

Example 3.9 A Bullet Fired Horizontally A bullet is fired horizontally from the top of a cliff that is 20.0 m above a long lake. If the muzzle speed of the bullet is 500.0 m/s, how far from the bottom of the cliff does the bullet strike the surface of the lake? Ignore air resistance. Strategy We need to find the total time of flight so that we can find the horizontal displacement. The bullet is starting from the high point of the parabolic path because viy = 0. As usual in projectile problems, we choose the y-axis to be the positive vertical direction. Known: Δy = −20.0 m; v iy = 0; v ix = 500.0 m/s. To find: Δx. Solution The vertical displacement through which the bullet falls is 20.0 m. The relationship between Δy and Δt is Δy = _12 (v fy + v iy) Δt Substituting viy = 0 and vfy = viy + ayΔt = ayΔt yields

√

____

2 Δy 1 a (Δt)2 ⇒ Δt = ____ Δy = __ ay 2 y

The horizontal displacement of the bullet is

√

____

2 Δy Δx = v ix Δt = v ix ____ ay

2 × (−20.0 m) = 1.01 km = 500.0 m/s × ____________ −9.8 m/s2 Discussion How did we know to start with the y-component equation when the question asks about the horizontal displacement? The question gives vix and asks for Δx. The missing information needed is the time during which the bullet is in the air; the time can be found from analysis of the vertical motion. We ignored air resistance in this problem, which is not very realistic. The actual distance would be less than 1.01 km.

Practice Problem 3.9 Bullet Velocity Find the horizontal and vertical components of the bullet’s velocity just before it hits the surface of the lake. At what angle does it strike the surface?

At the beginning of the chapter, we asked why the clam does not fall straight down when the gull lets go. The gull is flying horizontally with the clam, so the clam has the same horizontal velocity as the gull. When the gull lets go, the clam falls toward Earth, but since ax = 0 the clam retains the same horizontal component of velocity as the gull. Therefore, the clam is a projectile starting at the top of its parabolic trajectory.

3.6

√

____________

Why does the clam not drop straight down?

VELOCITY IS RELATIVE; REFERENCE FRAMES

The idea of relativity arose in physics centuries before Einstein’s theory. Nicole Oresme (1323–1382) wrote that motion of one object can only be perceived relative to some other object. Until now, we have tacitly assumed in most situations that displacements, velocities, and accelerations should be measured in a reference frame attached to Earth’s surface—that is, by choosing an origin fixed in position relative to Earth’s surface and a set of axes whose directions are fixed relative to Earth’s surface. After learning about relative velocities, we will take another look at this assumption.

Relative Velocity Suppose Wanda is walking down the aisle of a train moving along the track at a constant velocity (Fig. 3.24). Imagine asking, “How fast is Wanda moving?” This question is not well defined. Do we mean her speed as measured by Tim, a passenger on the train, or her speed as measured by Greg, who is standing on the ground and looking into the train as it passes by? The answer to the question “How fast?” depends on the observer. Figure 3.25 shows Wanda walking from one end of the car to the other during a time interval Δt. The displacement of Wanda as measured by Tim—her displacement relative

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Figure 3.24 Tim and Greg watch Wanda walk down the aisle of a train. Wanda’s velocity with respect to Tim (or with respect to the train) is v⃗ WT; Tim’s velocity with respect to Greg (or with respect to the ground) is v⃗ TG.

Wanda vWT

Greg

Tim

vTG

to the train—is Δr⃗ WT = v⃗ WT Δt. During the same time interval, the train’s displacement relative to Greg is Δr⃗ TG = v⃗ TG Δt. As measured by Greg, Wanda’s displacement is partly due to her motion relative to the train and partly due to the motion of the train relative to the ground. Figure 3.25 shows that Δr⃗ WT + Δr⃗ TG = Δr⃗ WG. Dividing by the time interval Δt gives the relationship between the three velocities: v⃗ WT + v⃗ TG = v⃗ WG

(3-24)

To be sure that you are adding the velocity vectors correctly, think of the subscripts as if they were fractions that get multiplied when the velocity vectors are added. In Eq. (3-24), W × __ W so the equation is correct. T = ___ ___ T G G Applications of Relative Velocities for Pilots and Sailors Relative velocities are of enormous practical interest to pilots of aircraft, sailors, and captains of ocean freighters. The pilot of an airplane is ultimately concerned with the motion of the plane with respect to the ground—the takeoff and landing points are fixed points on the ground. However, the controls of the plane (engines, rudder, ailerons, and spoilers) affect the motion of the plane with respect to the air. A sailor has to consider three different velocities of the boat: with respect to shore (for launching and landing), with respect to the air (for the behavior of the sails), and with respect to the water (for the behavior of the rudder).

CHECKPOINT 3.6 In Fig. 3.24, if the train is moving at 18.0 m/s with respect to the ground and Wanda walks at 1.5 m/s with respect to the train, how fast is Wanda moving (a) with respect to Greg and (b) with respect to Tim?

tf = ti + ∆t

ti

∆rTG = vTG ∆t

∆rWT = vWT ∆t

∆rWG = vWG ∆t

Figure 3.25 Wanda’s displacement relative to the ground is the sum of her displacement relative to the train and the displacement of the train relative to the ground.

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3.6 VELOCITY IS RELATIVE; REFERENCE FRAMES

Example 3.10 Flight from Denver to Chicago An airplane flies from Denver to Chicago (1770 km) in 4.4 h when no wind blows. On a day with a tailwind, the plane makes the trip in 4.0 h. (a) What is the wind speed? (b) If a headwind blows with the same speed, how long does the trip take? Strategy We assume the plane has the same airspeed— the same speed relative to the air—in both cases. Once the plane is up in the air, the behavior of the wings, control surfaces, etc., depends on how fast the air is rushing by; the ground speed is irrelevant. But it is not irrelevant for the passengers, who are interested in a displacement relative to the ground. Solution Let v⃗ PG and v⃗ PA represent the velocity of the plane relative to the ground and the velocity of the _plane rel_ _ ative to the air, respectively. The wind velocity—the velocity of the air relative to the ground—can be written _ v⃗ AG. Then v⃗ PA + v⃗ AG = v⃗ PG. The equation is correct since A = __ P . With no wind, P × __ __ A G G 1770 km = 400 km/h v PA = v PG = ________ 4.4 h (a) On the day with the tailwind, 1770 km = 440 km/h v PG = ________ 4.0 h We expect v PA to be the same regardless of whether there is a wind or not. Since we are dealing with a tailwind, v⃗ PA and

x vPA (400 km/h)

v⃗ AG are in the same direction, which we label as the +x-direction in Fig. 3.26. Then, v PAx + v AGx = v PGx v AGx = v PGx − v PAx = 440 km/h − 400 km/h = 40 km/h vAGy = 0, so the wind speed is vAG = 40 km/h. (b) With a 40 km/h headwind, v⃗ PA and v⃗ AG are in opposite directions (Fig. 3.27). The velocity of the plane with respect to the ground is v PGx = v PAx + v AGx = 400 km/h + (−40 km/h) = 360 km/h The ground speed of the plane is 360 km/h and the trip takes 1770 km = 4.9 h ________ 360 km/h Discussion Quick check: the trip takes longer with a headwind (4.9 h) than with no wind (4.4 h), as we expect.

Practice Problem 3.10 Rowing Across the Bay Jamil, practicing to get on the crew team at school, rows a one-person racing shell to the north shore of the bay for a distance of 3.6 km to his friend’s dock. On a day when the water is still (no current flowing), it takes him 20 min (1200 s) to reach his friend. On another day when a current flows southward, it takes him 30 min (1800 s) to row the same course. Ignore air resistance. (a) What is the speed of the current in m/s? (b) How long does it take Jamil to return home with that same current flowing? x

Figure 3.26 vAG (40 km/h)

vPG (440 km/h)

Addition of velocity vectors in the case of a tailwind. Lengths of vectors are not to scale.

vPA (400 km/h)

vPG (360 km/h)

vAG (40 km/h)

Figure 3.27 Addition of velocity vectors in the case of a headwind. Lengths of vectors are not to scale.

The vector equation (3-24) applies to situations where the velocities are not all along the same line, as illustrated in Example 3.11.

Example 3.11 Rowing Across a River Jack wants to row directly across a river from the east shore to a point on the west shore. The width of the river is 250 m and the current flows from north to south at 0.61 m/s. The

trip takes Jack 4.2 min. In what direction did he head his rowboat to follow a course due west across the river? At what speed with respect to still water is Jack able to row? continued on next page

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CHAPTER 3 Motion in a Plane

Example 3.11 continued

Strategy We start with a sketch of the situation (Fig. 3.28). To keep the various velocities straight, we choose subscripts as follows: R = rowboat; W = water; S = shore. The velocity of the current given is the velocity of the water relative to the shore: v⃗ WS = 0.61 m/s, south. The velocity of the rowboat relative to shore (v⃗ RS) is due west. The magnitude of v⃗ RS can be found from the displacement relative to shore and the time interval, both of which are given. The question asks for the magnitude and direction of the velocity of the rowboat relative to the water (v⃗ RW). The three velocities are related by v⃗ RW + v⃗ WS = v⃗ RS To compensate for the current carrying the rowboat south with respect to shore, Jack heads (points) the rowboat upstream (against the current) at some angle to the north of west. Solution In a sketch of the vector addition (Fig. 3.29), the velocity of the rowboat with respect to the water is at an angle q north of west. With respect to shore, Jack travels 250 m in 4.2 min, so his speed with respect to shore is 250 m = 0.992 m/s v RS = ________________ 4.2 min × 60 s/min We can find the angle at which the rowboat should be headed by finding the tangent of the angle between v⃗ RW and v⃗ RS: v WS _________ 0.61 m/s tan q = ____ v RS = 0.992 m/s q = 32° N of W

Water current

Shore Path of rowboat relative to shore 250 m

Shore

N W

E

_________

theorem in this way. Rather, we would use the component method to add the two vectors. If Jack had headed the rowboat directly west, the current would have carried him south, so he would have traveled in a direction south of west relative to shore. He has to compensate by heading upstream at just such an angle that his velocity relative to shore is directed west.

Practice Problem 3.11 Heading Straight Across If Jack were to head straight across the river, in what direction with respect to shore would he travel? How long would it take him to cross? How far downstream would he be carried? Assume that he rows at the same speed with respect to the water as in Example 3.11.

vWS is velocity of water with respect to shore

______________________

v⃗ RW = v WS + v RS = √ (0.61 m/s)2 + (0.992 m/s)2 2

vWS

Discussion If v⃗ RS and v⃗ WS had not been perpendicular, we could not have used the Pythagorean

vRW q

Jack rows at a speed of 1.16 m/s with respect to the water.

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vRS is velocity of rowboat with respect to shore

2

= 1.16 m/s

What does the path of the falling clam look like to the gull?

Rowing across a river.

S

Not to scale

The speed at which Jack is able to row with respect to still water is the magnitude of v⃗ RW. Since v⃗ RS and v⃗ WS are perpendicular, the Pythagorean theorem yields

√

Figure 3.28

vRW is velocity of rowboat with respect to water

vRS

Figure 3.29 Graphical addition of the velocity vectors.

At the beginning of this chapter, we asked what the path followed by the falling clam looks like as seen by the gull flying through the air. With respect to a beachcomber on the ground and ignoring air resistance, the clam has a constant horizontal velocity component given to it by the gull and a changing vertical component of velocity due to gravity (Fig. 3.30a); the clam moves in a parabolic path. If the gull continues to fly at the same horizontal velocity after dropping the clam, it is directly overhead when the clam hits the rock because they both have the same constant horizontal component of velocity with respect to Earth.

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MASTER THE CONCEPTS

Figure 3.30 (a) Beachcomber vGR

view: The gull flies along a horizontal line while the clam follows a parabolic path. (b) Bird’s eye view: The gull sees the rocks moving while the clam drops straight down, landing on the rocks just as the rocks move under the clam.

vGG = 0 vCR

vCG

vRR = 0

vRG

G = gull C = clam R = rocks

(a)

(b)

In its own reference frame—that is, using its own position as the origin of the coordinate axes—the gull sees the clam drop straight down toward the ground while rocks and other objects on the beach are moving horizontally (Fig. 3.30b). The bird sees a collision between the horizontally moving rocks and the vertically falling clam. At any instant, if the velocity of the clam with respect to the gull is v⃗ CG, the velocity of the gull with respect to the rocks is v⃗ GR, and the velocity of the clam with respect to the rocks is v⃗ CR, then v⃗ CG + v⃗ GR = v⃗ CR.

Master the Concepts • Vectors are added graphically by drawing each vector so that its tail is placed at the tip of the previous vector. The sum is drawn as a vector arrow from the tail of the first vector to the tip of the last. Addition of vectors is ⃗ + B ⃗ = B ⃗ + A. ⃗ commutative: A

find the magnitude and direction of a vector if its components are known. y vx x

58° A vy v

B A+ B

• Vectors are subtracted by adding the opposite of the ⃗ − B ⃗ = A ⃗ + (−B). ⃗ second vector: A • Addition and subtraction of vectors algebraically using components is generally easier and more accurate than the graphical method. The graphical method is still a useful first step to get an approximate answer. • To find the components of a vector, first draw a right triangle with the vector as the hypotenuse and the other two sides parallel to the x- and y-axes. Then use the trigonometric functions to find the magnitudes of the components. The correct algebraic sign must be determined for each component. The same triangle can be used to

• To add vectors algebraically, add their components to find the components of the sum: ⃗ + B ⃗ = C ⃗ if and only if A Ax + Bx = Cx and A y + By = Cy • The x- and y-axes are chosen to make the problem easiest to solve. Any choice is valid as long as the two are perpendicular. If the direction of the acceleration is known, choose x- and y-axes so that the acceleration vector is parallel to one of the axes. • Position, displacement, velocity, and acceleration are vector quantities with both magnitude and direction. They must be added and subtracted as vectors. continued on next page

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CHAPTER 3 Motion in a Plane

Master the Concepts continued

• The equations for position, displacement, average velocity, instantaneous velocity, average acceleration, and instantaneous acceleration in Chapter 2 apply to each perpendicular component of the corresponding vector quantities for motion in two or three dimensions. • The instantaneous velocity vector is tangent to the path of motion.

vy

y-directions can be treated separately. Since ax = 0, vx is constant. Thus, the motion is a superposition of constant velocity motion in the x-direction and constant acceleration motion in the y-direction. • The kinematic equations for an object moving in two dimensions with constant acceleration along the y-axis are x-axis: ax = 0

v

• The instantaneous acceleration vector does not have to be tangent to the path of motion, since velocities can change both in direction and in magnitude. • For a projectile or any object moving with constant acceleration in the ± y-direction, the motion in the x- and

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(3-19)

Δx = vx Δt

(3-20)

Δy = _12 (vfy + viy) Δt

2

2

v fy = v iy = 2ay Δy

(3-22)

• To relate the velocities of objects measured in different reference frames, use the vector equation v⃗AC = v⃗AB + v⃗BC

(3-24)

where v⃗AC represents the velocity of A relative to C, and so forth.

Conceptual Questions 1. If two vectors have the same magnitude, are they necessarily equal? If not, why not? Can two vectors with different magnitudes ever be equal? 2. (a) Is it possible for the sum of two vectors to be smaller in magnitude than the magnitude of either vector? (b) Is it possible for the magnitude of the sum of two vectors to be larger than the sum of the magnitudes of the two vectors? 3. What is the distinction between a vector and a scalar quantity? Give two examples of each. 4. Is it possible for two identical projectiles with identical initial speeds, but with two different angles of elevation, to land in the same spot? Explain. Ignore air resistance and sketch the trajectories. 5. If the trajectory is parabolic in one reference frame, is it always, never, or sometimes parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else can it be? 6. You are standing on a balcony overlooking the beach. You throw a ball straight up into the air with speed vi and throw an identical ball straight down with speed vi. Ignoring air resistance, how do the speeds of the balls compare just before they hit the ground? 7. You throw a ball up with initial speed vi and when it reaches its high point at height h, you throw another ball into the air with the same initial speed vi. Will the two

Δvx = 0 (vx is constant) Δvy = ay Δt

Δy = v iy Δt + _12 ay (Δt)2 (3-21)

P vx Tangent at P

y-axis: constant a y

8. 9. 10.

11. 12.

13.

14. 15. 16.

17.

balls cross at half the height h, or more than half, or less than half? Explain. If an object is traveling at a constant velocity, is it necessarily traveling in a straight line? Explain. Can the average speed and the magnitude of the average velocity ever be equal? If so, under what circumstances? Give an example of an object whose acceleration is (1) in the same direction as its velocity, (2) opposite its velocity, and (3) perpendicular to its velocity. Name a situation where the speed of an object is constant while the velocity is not. Tell whether or not each of the following objects has a constant velocity and explain your reasoning. (a) A car driving around a curve at constant speed on a flat road. (b) A car driving straight up a 6° incline at constant speed. (c) The Moon. Explain how to add two displacement vectors of magnitudes 3L and 4L so that the vector sum has magnitude (a) L; (b) 7L; (c) 5L. Compare the advantages and disadvantages of the two methods of vector addition (graphical and algebraic). Can the x-component of a vector ever be greater than the magnitude of the vector? Explain. Why is the muzzle of a rifle not aimed directly at the center of the target? Why is this more important at longer ranges? Does the monkey, coconut, and hunter demonstration still work if the hunter is in a higher tree and the arrow is pointed downward at the monkey and coconut? Explain.

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MULTIPLE-CHOICE QUESTIONS

Multiple-Choice Questions ⃗ in the drawing is equal to 1. Vector A ⃗ + D ⃗ ⃗ + D ⃗ + E ⃗ ⃗ + F ⃗ (a) C (b) C (c) C ⃗ + C ⃗ ⃗ + F ⃗ (d) B (e) B B E

8.

D A

F

C

Multiple-Choice Questions 1 and 2 2. Which vector sum is not equal to zero? ⃗ + D ⃗ + E ⃗ ⃗ + C ⃗ + F ⃗ (a) C (b) B ⃗ + F ⃗ ⃗ + B ⃗ + F ⃗ (c) D (d) A 3. A hunter spots a pheasant flying along horizontally. If he shoots the pheasant, the time interval between the bird being shot and the dead bird hitting the ground depends on (a) the speed with which the bird was flying. (b) the height of the bird above the ground. (c) the speed of the bird and its height above the ground. 4. A runner moves along a circular track at a constant speed. (a) Her acceleration is zero. (b) Her velocity is constant. (c) Both (a) and (b) are true. (d) Both her acceleration and her velocity are changing. 5. A boy plans to cross a river in a rubber raft. The current flows from north to south at 1 m/s. In what direction should he head to get across the river to the east bank in the least amount of time if he is able to paddle the raft at 1.5 m/s in still water? (a) directly to the east (b) south of east (c) north of east (d) The three directions require the same time to cross the river. 6. A boy plans to paddle a rubber raft across a river to the east bank while the current flows downriver from north to south at 1 m/s. He is able to paddle the raft at 1.5 m/s in still water. In what direction should he head the raft to go straight east across the river to the opposite bank? (a) directly to the east (b) south of east (c) north of east (d) north (e) south 7. A kicker kicks a football from the 5-yard line to the 45-yard line (both on the same half of the field). Ignoring air resistance, where along the trajectory is the speed of the football a minimum? (a) at the 5-yard line, just after the football leaves the kicker’s foot (b) at the 45-yard line, just before the football hits the ground

gia04535_ch03_055-086.indd 79

9.

10.

11.

79

(c) at the 15-yard line, while the ball is still going higher (d) at the 35-yard line, while the ball is coming down (e) at the 25-yard line, when the ball is at the top of its trajectory Two balls, identical except for color, are projected horizontally from the roof of a tall building at the same instant. The initial speed of the red ball is twice the initial speed of the blue ball. Ignoring air resistance, (a) the red ball reaches the ground first. (b) the blue ball reaches the ground first. (c) both balls land at the same instant with different speeds. (d) both balls land at the same instant with the same speed. A person stands on the roof garden of a tall building with one ball in each hand. If the red ball is thrown horizontally off the roof and the blue ball is simultaneously dropped over the edge, which statement is true? (a) Both balls hit the ground at the same time, but the red ball has a higher speed just before it strikes the ground. (b) The blue ball strikes the ground first, but with a lower speed than the red ball. (c) The red ball strikes the ground first with a higher speed than the blue ball. (d) Both balls hit the ground at the same time with the same speed. A ball is thrown into the air and follows a parabolic trajectory. At the highest point in the trajectory, (a) the velocity is zero, but the acceleration is not zero. (b) both the velocity and the acceleration are zero. (c) the acceleration is zero, but the velocity is not zero. (d) neither the acceleration nor the velocity are zero. A ball is thrown into the air and follows a parabolic trajectory. Point A is the highest point in the trajectory and point B is a point as the ball is falling back to the ground. Choose the correct relationship between the speeds and the magnitudes of the acceleration at the two points. (b) vA < vB and aA > aB (a) vA > vB and aA = aB (c) vA = vB and aA ≠ aB (d) vA < vB and aA = aB

Questions 12–14. Two projectiles launched with the same initial speed but at different launch angles 30° and 60° land at the same spot (see Fig. 3.22). Ignore air resistance. Answer choices: (a) projectile launched at 30° (b) projectile launched at 60° (c) They are equal. 12. Which has the larger horizontal velocity component vx? 13. Which has a longer time of flight Δ t (time interval between launch and hitting the ground)? 14. For which is the product vx Δ t larger?

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CHAPTER 3 Motion in a Plane

Problems Combination conceptual/quantitative problem

✦ Blue # 1

2

Biological or medical application Challenging problem Detailed solution in the Student Solutions Manual Problems paired by concept Text website interactive or tutorial

8. Two vectors, each of magnitude 4.0 cm, are directed at a small angle a below the horizontal as shown. (The ⃗ = A ⃗ + B. ⃗ Sketch grid is 1 cm on a side.) (a) Let C ⃗ and estimate its magnitude. (b) Let the direction of C ⃗ = A ⃗ − B ⃗ . Sketch the direction of D ⃗ and estimate its D magnitude. ( tutorial: vectors) 4.0 cm

a A

3.1 Graphical Addition and Subtraction of Vectors ⃗ is directed to the west and has 1. Displacement vector A magnitude 2.56 km. A second displacement vector is also directed to the west and has magnitude 7.44 km. ⃗ + B? ⃗ (a) What are the magnitude and direction of A ⃗ − B? ⃗ (b) What are the magnitude and direction of A ⃗ − A? ⃗ (c) What are the magnitude and direction of B ⃗ is directed along the positive x-axis and has 2. Vector A ⃗ is directed along the magnitude 1.73 units. Vector B negative x-axis and has magnitude 1.00 unit. (a) What ⃗ + B ⃗ ? (b) What are are the magnitude and direction of A ⃗ − B ⃗ ? (c) What are the the magnitude and direction of A ⃗ − A? ⃗ magnitude and direction of B 3. Two vectors have magnitudes 3.0 and 4.0. How are the directions of the two vectors related if (a) the sum has magnitude 7.0, or (b) if the sum has magnitude 5.0? (c) What relationship between the directions gives the smallest magnitude sum and what is this magnitude? 4. A runner is practicing on a circular track that is 300 m in circumference. From the point farthest to the west on the track, he starts off running due north and follows the track as it curves around toward the east. (a) If he runs halfway around the track and stops at the farthest eastern point of the track, what is the distance he traveled? (b) What is his displacement? 5. Two displacement vectors each have magnitude 20 km. One is directed 60° above the +x-axis; the other is directed 60° below the +x-axis. What is the vector sum of these two displacements? Use graph paper to find your answer. 6. Orville walks 320 m due east. He then continues walking along a straight line, but in a different direction, and stops 200 m northeast of his starting point. How far did he walk during the second portion of the trip and in what direction? ⃗ B, ⃗ and C ⃗ are shown in the figure. (a) Draw 7. Vectors A, ⃗ ⃗ ⃗ = A ⃗ + B ⃗ and E ⃗ = A ⃗ + C. ⃗ vectors D and E, where D ⃗ ⃗ ⃗ ⃗ (b) Show that A + B = B + A by graphical means. A C

B

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a

4.0 cm

B

9. Michaela is planning a trip in Ireland from Killarney to Cork to visit Blarney Castle. (See Example 3.2.) She also wants to visit Mallow, which is located 39 km due east of Killarney and 22 km due north of Cork. Draw the displacement vectors for the trip when she travels from Killarney to Mallow to Cork. (a) What is the magnitude of her displacement once she reaches Cork? (b) How much additional distance does Michaela travel in going to Cork by way of Mallow instead of going directly from Killarney to Cork? 10. A scout troop is practicing its orienteering skills with map and compass. First they walk due east for 1.2 km. Next, they walk 45° west of north for 2.7 km. In what direction must they walk to go directly back to their starting point? How far will they have to walk? Use graph paper, ruler, and protractor to find a geometrical solution. 11. Prove that the displacement for a trip is equal to the vector sum of the displacements for each leg of the trip. [Hint: Imagine a trip that consists of n segments. The trip starts at position r⃗1, proceeds to r⃗2, then to r⃗3, . . . , then to r⃗n−1, then finally to r⃗n. Write an expression for each displacement as the difference of two position vectors and then add them.] 12. A sailboat sails from Marblehead Harbor directly east for 45 nautical miles, then 60° south of east for 20.0 nautical miles, returns to an easterly heading for 30.0 nautical miles, and sails 30° east of north for 10.0 nautical miles, then west for 62 nautical miles. At that time the boat becomes becalmed and the auxiliary engine fails to start. The crew decides to notify the Coast Guard of their position. Using graph paper, ruler, and protractor, sketch a graphical addition of the displacement vectors and estimate their position.

3.2 Vector Addition and Subtraction Using Components 13. A vector is 20.0 m long and makes an angle of 60.0° counterclockwise from the y-axis (on the side of the −x-axis). What are the x- and y-components of this vector? ⃗ has magnitude 4.0 units; vector B ⃗ has magni14. Vector A ⃗ and B ⃗ is 60.0°. tude 6.0 units. The angle between A ⃗ + B? ⃗ What is the magnitude of A ⃗ is directed 15. Vector A along the positive y-axis and has ___ ⃗ is directed along the magnitude √ 3.0 units. Vector B negative x-axis and has magnitude 1.0 unit. (a) What are

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PROBLEMS

⃗ + B? ⃗ (b) What are the the magnitude and direction of A ⃗ − B? ⃗ (c) What are the magnitude and direction of A ⃗ − A? ⃗ x- and y-components of B 16. Vector a⃗ has components ax = −3.0 m/s2 and ay = +4.0 m/s2. (a) What is the magnitude of a⃗? (b) What is the direction of a⃗? Give an angle with respect to one of the coordinate axes. 17. In Problem 8, let a = 10° and find the magnitude of ⃗ using the component method. vector C 18. In Problem 8, let a = 10° and find the magnitude of ⃗ using the component method. vector D 19. Find the x- and y-components of the four vectors shown in the drawing.

Cindy is able to meet Jerry at the fitness center by bicycling in a straight line, what is the length and direction she must travel? 26. Repeat Problem 10 using the component (algebraic) method. 27. Use the component method to obtain a more accurate description of the sailboat’s location in Problem 12. 28. You will be hiking to a lake with some of your friends by following the trails indicated on a map at the trailhead. The map says that you will travel 1.6 mi directly north, then 2.2 mi in a direction 35° east of north, then finally 1.1 mi in a direction 15° north of east. At the end of this hike, how far will you be from where you started, and what direction will you be from your starting point?

y

y

7.0 m

3.3 Velocity

A

20.0° x

20.0°

7.0 m/s

x y C

B

y 20.0°

x 7.0 m/s

7.0 m x

D

20.0°

20. The velocity vector of a sprinting cheetah has x- and y-components vx = + 16.4 m/s and vy = −26.3 m/s. (a) What is the magnitude of the velocity vector? (b) What angle does the velocity vector make with the +x- and −y-axes? 21. In each of these, the x- and y-components of a vector are given. Find the magnitude and direction of the vector. (a) Ax = −5.0 m/s, Ay = +8.0 m/s. (b) Bx = +120 m, By = −60.0 m. (c) Cx = −13.7 m/s, Cy = −8.8 m/s. (d) Dx = 2.3 m/s2, Dy = 6.5 cm/s2. ⃗ has a magnitude of 22.2 cm and makes an 22. A vector A angle of 130.0° with the positive x-axis. What are the x- and y-components of this vector? ⃗ has magnitude 7.1 and direction 14° below 23. Vector B ⃗ has x-component Cx = −1.8 and the +x-axis. Vector C y-component Cy = −6.7. Compute (a) the x- and ⃗ (b) the magnitude and direction of y-components of B; ⃗ ⃗ + B; ⃗ (d) the C; (c) the magnitude and direction of C ⃗ ⃗ magnitude and direction of C − B; (e) the x- and ⃗ − B. ⃗ y-components of C 24. Margaret walks to the store using the following path: 0.500 miles west, 0.200 miles north, 0.300 miles east. What is her total displacement? That is, what is the length and direction of the vector that points from her house directly to the store? Use vector components to find the answer. 25. Jerry bicycles from his dorm to the local fitness center: 3.00 miles east and 2.00 miles north. Cindy’s apartment is located 1.50 miles west of Jerry’s dorm. If

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29. A runner times his speed around a circular track with a circumference of 0.478 mi. At the start he is running toward the east and the track starts bending toward the north. If he goes halfway around, he will be running toward the west. He finds that he has run a distance of 0.750 mi in 4.00 min. What is his (a) average speed and (b) average velocity in m/s? 30. A runner times his speed around a track with a circumference of 0.50 mi. He finds that he has run a distance of 1.00 mi in 4.0 min. What is his (a) average speed and (b) average velocity magnitude in m/s? 31. Peggy drives from Cornwall to Atkins Glen in 45 min. Cornwall is 73.6 km from Illium in a direction 25° west of south. Atkins Glen is 27.2 km from Illium in a direction 15° south of west. Using Illium as your origin, (a) draw the initial and final position vectors, (b) find the displacement during the trip, and (c) find Peggy’s average velocity for the trip. 32. To get to a concert in time, a harpsichordist has to drive 122 mi in 2.00 h. (a) If he drove at an average speed of 55.0 mi/h in a due west direction for the first 1.20 h, what must be his average speed if he is heading 30.0° south of west for the remaining 48.0 min? (b) What is his average velocity for the entire trip? 33. A bicycle travels 3.2 km due east in 0.10 h, then 4.8 km at 15.0° east of north in 0.15 h, and finally another 3.2 km due east in 0.10 h to reach its destination. The time lost in turning is negligible. What is the average velocity for the entire trip? 34. A car travels east at 96 km/h for 1.0 h. It then travels 30.0° east of north at 128 km/h for 1.0 h. (a) What is the average speed for the trip? (b) What is the average velocity for the trip? 35. A speedboat heads west at 108 km/h for 20.0 min. It then travels at 60.0° south of west at 90.0 km/h for 10.0 min. (a) What is the average speed for the trip? (b) What is the average velocity for the trip? 36. See Problem 9. During Michaela’s travel from Killarney to Cork via Mallow, her actual travel time in the car is

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48 min. (a) What is her average speed in m/s? (b) What is the magnitude of her average velocity in m/s? ✦37. Geoffrey drives from his home town due east at 90.0 km/h for 80.0 min. After visiting a friend for 15.0 min, he drives in a direction 30.0° south of west at 76.0 km/h for 45.0 min to visit another friend. (a) How far is it to his home from the second town? (b) If it takes him 45.0 min to drive directly home, what is his average velocity on the third leg of the trip? (c) What is his average velocity during the first two legs of his trip? (d) What is his average velocity over the entire trip? (e) What is his average speed during the entire trip if he spent 55.0 min visiting the second friend?

3.4 Acceleration 38. A hawk is flying north at 2.0 m/s with respect to the ground; 10.0 s later, it is flying south at 5.0 m/s. What is its average acceleration during this time interval? 39. A skydiver is falling straight down at 55 m/s when he opens his parachute and slows to 8.3 m/s in 3.5 s. What is the average acceleration of the skydiver during those 3.5 s? 40. A car travels three quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. The initial velocity is west and the final velocity is south. (a) Find its average velocity for this trip. (b) What is the car’s average acceleration during these 3.0 s? (c) Explain how a car moving at constant speed has a nonzero average acceleration. 41. At t = 0, an automobile traveling north begins to make a turn. It follows one-quarter of the arc of a circle with a radius of 10.0 m until, at t = 1.60 s, it is traveling east. The car does not alter its speed during the turn. Find (a) the car’s speed, (b) the change in its velocity during the turn, and (c) its average acceleration during the turn. 42. At the beginning of a 3.0-h plane trip, you are traveling due north at 192 km/h. At the end, you are traveling 240 km/h in the northwest direction (45° west of north). (a) Draw your initial and final velocity vectors. (b) Find the change in your velocity. (c) What is your average acceleration during the trip? 43. John drives 16 km directly west from Orion to Chester at a speed of 90 km/h, then directly south for 8.0 km to Seiling at a speed of 80 km/h, then finally 34 km southeast to Oakwood at a speed of 100 km/h. Assume he travels at constant velocity during each of the three segments. (a) What was the change in velocity during this trip? [Hint: Do not assume he starts from rest and stops at the end.] (b) What was the average acceleration during this trip? 44. A particle’s constant acceleration is south at 2.50 m/s2. At t = 0, its velocity is 40.0 m/s east. What is its velocity at t = 8.00 s? 45. A particle’s constant acceleration is north at 100 m/s2. At t = 0, its velocity vector is 60 m/s east. At what time will the magnitude of the velocity be 100 m/s?

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3.5 Motion in a Plane with Constant Acceleration 46. A baseball is thrown horizontally from a height of 9.60 m above the ground with a speed of 30.0 m/s. Where is the ball after 1.40 s has elapsed? 47. A clump of soft clay is thrown horizontally from 8.50 m above the ground with a speed of 20.0 m/s. Where is the clay after 1.50 s? Assume it sticks in place when it hits the ground. 48. A tennis ball is thrown horizontally from an elevation of 14.0 m above the ground with a speed of 20.0 m/s. (a) Where is the ball after 1.60 s? (b) If the ball is still in the air, how long before it hits the ground and where will it be with respect to the starting point once it lands? 49. A ball is thrown from a point 1.0 m above the ground. The initial velocity is 19.6 m/s at an angle of 30.0° above the horizontal. (a) Find the maximum height of the ball above the ground. (b) Calculate the speed of the ball at the highest point in the trajectory. 50. An arrow is shot into the air at an angle of 60.0° above the horizontal with a speed of 20.0 m/s. (a) What are the x- and y-components of the velocity of the arrow 3.0 s after it leaves the bowstring? (b) What are the x- and y-components of the displacement of the arrow during the 3.0-s interval? 51. You are working as a consultant on a video game designing a bomb site for a World War I airplane. In this game, the plane you are flying is traveling horizontally at 40.0 m/s at an altitude of 125 m when it drops a bomb. (a) Determine how far horizontally from the target you should release the bomb. (b) What direction is the bomb moving just before it hits the target? 52. You have been employed by the local circus to plan their human cannonball performance. For this act, a spring-loaded cannon will shoot a human projectile, the Great Flyinski, across the big top to a net below. The net is located 5.0 m lower than the muzzle of the cannon from which the Great Flyinski is launched. The cannon will shoot the Great Flyinski at an angle of 35.0° above the horizontal and at a speed of 18.0 m/s. The ringmaster has asked that you decide how far from the cannon to place the net so that the Great Flyinski will land in the net and not be splattered on the floor, which would greatly disturb the audience. What do you tell the ringmaster? ( interactive: projectile motion) 53. A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 40 m/s at an angle of 37° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time. (a) What is the maximum height above the ground reached by the cannonball? (b) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land? (c) What are the x- and y-components of the cannonball’s velocity just before it lands? The y-axis points up.

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PROBLEMS

54. After being assaulted by flying cannonballs, the knights on the castle walls (12 m above the ground) respond by propelling flaming pitch balls at their assailants. One ball lands on the ground at a distance of 50 m from the castle walls. If it was launched at an angle of 53° above the horizontal, what was its initial speed? 55. From the edge of the rooftop of a building, a boy throws a stone at an angle 25.0° above the horizontal. The stone hits the ground 4.20 s later, 105 m away from the base of the building. (Ignore air resistance.) (a) For the stone’s path through the air, sketch graphs of x, y, vx , and vy as functions of time. These need to be only qualitatively correct—you need not put numbers on the axes. (b) Find the initial velocity of the stone. (c) Find the initial height h from which the stone was thrown. (d) Find the maximum height H reached by the stone. 56. Two angles are complementary when their sum is 90.0°. Find the ranges for two projectiles launched with identical initial speeds of 36.2 m/s at angles of elevation above the horizontal that are complementary pairs. (a) For one trial, the angles of elevation are 36.0° and 54.0°. (b) For the second trial, the angles of elevation are 23.0° and 67.0°. (c) Finally, the angles of elevation are both set to 45.0°. (d) What do you notice about the range values for each complementary pair of angles? At which of these angles was the range greatest? 57. The range R of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at t = 0 with initial speed vi at an angle q above the horizontal. (a) Find the time t at which the projectile returns to its original altitude. (b) Show that the range is

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time the ball leaves the bat until it reaches the fielder? (c) At what distance from home plate will the fielder be when he catches the ball? You are planning a stunt to be used in an ice skating show. For this stunt a skater will skate down a frictionless ice ramp that is inclined at an angle of 15.0° above the horizontal. At the bottom of the ramp, there is a short horizontal section that ends in an abrupt drop off. The skater is supposed to start from rest somewhere on the ramp, then skate off the horizontal section and fly through the air a horizontal distance of 7.00 m while falling vertically for 3.00 m, before landing smoothly on the ice. How far up the ramp should the skater start this stunt? A suspension bridge is 60.0 m above the level base of a gorge. A stone is thrown or dropped from the bridge. Ignore air resistance. At the location of the bridge g has been measured to be 9.83 m/s2. (a) If you drop the stone, how long does it take for it to fall to the base of the gorge? (b) If you throw the stone straight down with a speed of 20.0 m/s, how long before it hits the ground? (c) If you throw the stone with a velocity of 20.0 m/s at 30.0° above the horizontal, how far from the point directly below the bridge will it hit the level ground? A circus performer is shot out of a cannon and flies over a net that is placed horizontally 6.0 m from the cannon. When the cannon is aimed at an angle of 40° above the horizontal, the performer is moving in the horizontal direction and just barely clears the net as he passes over it. What is the muzzle speed of the cannon and how high is the net? Show that for a projectile launched at an angle of 45° the maximum height of the projectile is one quarter of the range (the distance traveled on flat ground).

2

v i sin 2q R = ________ g [Hint: Use the trigonometric identity sin 2q = 2 sin q cos q.] (c) What value of q gives the maximum range? What is this maximum range? 58. Use the expression in Problem 57 to find (a) the maximum range of a projectile with launch speed vi and (b) the launch angle q at which the maximum range occurs. 59. A projectile is launched at t = 0 with initial speed vi at an angle q above the horizontal. (a) What are vx and vy at the projectile’s highest point? (b) Find the time t at which the projectile reaches its maximum height. (c) Show that the maximum height H of the projectile is (v i sin q )2 H = _________ 2g ✦60. A ballplayer standing at home plate hits a baseball that is caught by another player at the same height above the ground from which it was hit. The ball is hit with an initial velocity of 22.0 m/s at an angle of 60.0° above the horizontal. ( tutorial: projectile) (a) How high will the ball rise? (b) How much time will elapse from the

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3.6 Velocity Is Relative; Reference Frames 65. Two cars are driving toward each other on a straight, flat Kansas road. The Jeep Wrangler is traveling at 82 km/h north and the Ford Taurus is traveling at 48 km/h south, both measured relative to the road. What is the velocity of the Jeep relative to an observer in the Ford? 66. Two cars are driving toward each other on a straight and level road in Alaska. The BMW is traveling at 100.0 km/h north and the VW is traveling at 42 km/h south, both velocities measured relative to the road. At a certain instant, the distance between the cars is 10.0 km. Approximately how long will it take from that instant for the two cars to meet? [Hint: Consider a reference frame in which one of the cars is at rest.] 67. A car is driving directly north on the freeway at a speed of 110 km/h and a truck is leaving the freeway driving 85 km/h in a direction that is 35° west of north. What is the velocity of the truck relative to the car? 68. A Nile cruise ship takes 20.8 h to go upstream from Luxor to Aswan, a distance of 208 km, and 19.2 h to

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make the return trip downstream. Assuming the ship’s speed relative to the water is the same in both cases, calculate the speed of the current in the Nile. An airplane has a velocity relative to the ground of 210 m/s toward the east. The pilot measures his airspeed (the speed of the plane relative to the air) to be 160 m/s. What is the minimum wind velocity possible? A small plane is flying directly west with an airspeed of 30.0 m/s. The plane flies into a region where the wind is blowing at 10.0 m/s at an angle of 30° to the south of west. (a) If the pilot does not change the heading of the plane, what will be the ground speed of the airplane? (b) What will be the new directional heading, relative to the ground, of the airplane? ( tutorial: flight of crow) A small plane is flying directly west with an airspeed of 30.0 m/s. The plane flies into a region where the wind is blowing at 10.0 m/s at an angle of 30° to the south of west. In that region, the pilot changes the directional heading to maintain her due west heading. (a) What is the change she makes in the directional heading to compensate for the wind? (b) After the heading change, what is the ground speed of the airplane? A boat that can travel at 4.0 km/h in still water crosses a river with a current of 1.8 km/h. At what angle must the boat be pointed upstream to travel straight across the river? In other words, in what direction is the velocity of the boat relative to the water? At an antique car rally, a Stanley Steamer automobile travels north at 40 km/h and a Pierce Arrow automobile travels east at 50 km/h. Relative to an observer riding in the Stanley Steamer, what are the x- and y-components of the velocity of the Pierce Arrow car? The x-axis is to the east and the y-axis is to the north. Sheena can row a boat at 3.00 mi/h in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at 1.60 mi/h. Not having her calculator ready, she guesses that to go straight across, she should head 60.0° upstream. (a) What is her speed with respect to the starting point on the bank? (b) How long does it take her to cross the river? (c) How far upstream or downstream from her starting point will she reach the opposite bank? (d) In order to go straight across, what angle upstream should she have headed? A dolphin wants to swim directly back to its home bay, which is 0.80 km due west. It can swim at a speed of 4.00 m/s relative to the water, but a uniform water current flows with speed 2.83 m/s in the southeast direction. (a) What direction should the dolphin head? (b) How long does it take the dolphin to swim the 0.80-km distance home? Demonstrate with a vector diagram that a displacement is the same when measured in two different reference frames that are at rest with respect to each other.

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✦77. A boy is attempting to swim directly across a river; he is able to swim at a speed of 0.500 m/s relative to the water. The river is 25.0 m wide and the boy ends up at 50.0 m downstream from his starting point. (a) How fast is the current flowing in the river? (b) What is the speed of the boy relative to a friend standing on the riverbank? ✦78. An aircraft has to fly between two cities, one of which is 600.0 km north of the other. The pilot starts from the southern city and encounters a steady 100.0 km/h wind that blows from the northeast. The plane has a cruising speed of 300.0 km/h in still air. (a) In what direction (relative to east) must the pilot head her plane? (b) How long does the flight take?

Comprehensive Problems 79. Jason is practicing his tennis stroke by hitting balls against a wall. The ball leaves his racquet at a height of 60 cm above the ground at an angle of 80° with respect to the vertical. (a) The speed of the ball as it leaves the racquet is 20 m/s and it must travel a distance of 10 m before it reaches the wall. How far above the ground does the ball strike the wall? (b) Is the ball on its way up or down when it hits the wall? 80. Imagine a trip where you drive along an east-west highway at 80.0 km/h for 45.0 min and then you turn onto a highway that runs 38.0° north of east and travel at 60.0 km/h for 30.0 min. (a) What is your average velocity for the trip? (b) What is your average velocity on the return trip when you head the opposite way and drive 38.0° south of west at 60.0 km/h for the first 30.0 min and then west at 80.0 km/h for the last 45.0 min? 81. A jetliner flies east for 600.0 km, then turns 30.0° toward the south and flies another 300.0 km. (a) How far is the plane from its starting point? (b) In what direction could the jetliner have flown directly to the same destination (in a straight-line path)? (c) If the jetliner flew at a constant speed of 400.0 km/h, how long did the trip take? (d) Moving at the same speed, how long would the direct flight have taken? 82. An African swallow carrying a very small coconut is flying horizontally with a speed of 18 m/s. (a) If it drops the coconut from a height of 100 m above the Earth, how long will it take before the coconut strikes the ground? (b) At what horizontal distance from the release point will the coconut strike the ground? 83. A pilot starting from Athens, New York, wishes to fly to Sparta, New York, which is 320 km from Athens in the direction 20.0° N of E. The pilot heads directly for Sparta and flies at an airspeed of 160 km/h. After flying for 2.0 h, the pilot expects to be at Sparta, but instead he finds himself 20 km due west of Sparta. He has forgotten to correct for the wind. (a) What is the velocity of

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the plane relative to the air? (b) Find the velocity (magnitude and direction) of the plane relative to the ground. (c) Find the wind speed and direction. The citizens of Paris were terrified during World War I when they were suddenly bombarded with shells fired from a long-range gun known as Big Bertha. The barrel of the gun was 36.6 m long and it had a muzzle speed of 1.46 km/s. When the gun’s angle of elevation was set to 55°, what would be the range? For the purposes of solving this problem, neglect air resistance. (The actual range at this elevation was 121 km; air resistance cannot be ignored for the high muzzle speed of the shells.) You are serving as a consultant for the newest James Bond film. In one scene, Bond must fire a projectile from a cannon and hit the enemy headquarters located on the top of a cliff 75.0 m above and 350 m from the cannon. The cannon will shoot the projectile at an angle of 40.0° above the horizontal. The director wants to know what the speed of the projectile must be when it is fired from the cannon so that it will hit the enemy headquarters. What do you tell him? [Hint: Don’t assume the projectile will hit the headquarters at the highest point of its flight.] The pilot of a small plane finds that the airport where he intended to land is fogged in. He flies 55 mi west to another airport to find that conditions there are too icy for him to land. He flies 25 mi at 15° east of south and is finally able to land at the third airport. (a) How far and in what direction must he fly the next day to go directly to his original destination? (b) How many extra miles beyond his original flight plan has he flown? A particle has a constant acceleration of 5.0 m/s2 to the east. At time t = 0, it is 2.0 m east of the origin and its velocity is 20 m/s north. What are the components of its position vector at t = 2.0 s? A baseball batter hits a long fly ball that rises to a height of 44 m. An outfielder on the opposing team can run at 7.6 m/s. What is the farthest the fielder can be from where the ball will land so that it is possible for him to catch the ball? A locust jumps at an angle of 55.0° and lands 0.800 m from where it jumped. (a) What is the maximum height of the locust during its jump? Ignore air resistance. (b) If it jumps with the same initial speed at an angle of 45.0°, would the maximum height be larger or smaller? (c) What about the range? (d) Calculate the maximum height and range for this angle. A helicopter is flying horizontally at 8.0 m/s and an altitude of 18 m when a package of emergency medical supplies is ejected horizontally backward with a speed of 12 m/s relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?

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91. An airplane is traveling from New York to Paris, a distance of 5.80 × 103 km. Ignore the curvature of the Earth. (a) If the cruising speed of the airplane is 350.0 km/h, how much time will it take for the airplane to make the round-trip on a calm day? (b) If a steady wind blows from New York to Paris at 60.0 km/h, how much time will the round-trip take? (c) How much time will it take if there is a crosswind of 60.0 km/h? 92. A gull is flying horizontally 8.00 m above the ground at 6.00 m/s. The bird is carrying a clam in its beak and plans to crack the clamshell by dropping it on some rocks below. Ignoring air resistance, (a) what is the horizontal distance to the rocks at the moment that the gull should let go of the clam? (b) With what speed relative to the rocks does the clam smash into the rocks? (c) With what speed relative to the gull does the clam smash into the rocks? 93. A beanbag is thrown horizontally from a dorm room window a height h above the ground. It hits the ground a horizontal distance h (the same distance h) from the dorm directly below the window from which it was thrown. Ignoring air resistance, find the direction of the beanbag’s velocity just before impact. ✦94. In a plate glass factory, sheets of glass move along a conveyor belt at a speed of 15.0 cm/s. An automatic cutting tool descends at preset intervals to cut the glass to size. Since the assembly belt must keep moving at constant speed, the cutter is set to cut at an angle to compensate for the motion of the glass. If the glass is 72.0 cm wide and the cutter moves across the width at a speed of 24.0 cm/s, at what angle should the cutter be set? ✦95. A pilot wants to fly from Dallas to Oklahoma City, a distance of 330 km at an angle of 10.0° west of north. The pilot heads directly toward Oklahoma City with an air speed of 200 km/h. After flying for 1.0 h, the pilot finds that he is 15 km off course to the west of where he expected to be after one hour assuming there was no wind. (a) What is the velocity and direction of the wind? (b) In what direction should the pilot have headed his plane to fly directly to Oklahoma City without being blown off course? 96. A ball is thrown horizontally off the edge of a cliff with an initial speed of 20.0 m/s. (a) How long does it take for the ball to fall to the ground 20.0 m below? (b) How long would it take for the ball to reach the ground if it were dropped from rest off the cliff edge? (c) How long would it take the ball to fall to the ground if it were thrown at an initial velocity of 20.0 m/s but 18° below the horizontal? 97. A marble is rolled so that it is projected horizontally off ✦ the top landing of a staircase. The initial speed of the marble is 3.0 m/s. Each step is 0.18 m high and 0.30 m wide. Which step does the marble strike first? ✦98. A motor scooter rounds a curve on the highway at a constant speed of 20.0 m/s. The original direction of the

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scooter was due east; after rounding the curve the scooter is heading 36° north of east. The radius of curvature of the road at the location of the curve is 150 m. What is the average acceleration of the scooter as it rounds the curve? ✦ 99. You want to make a plot of the trajectory of a projectile. That is, you want to make a plot of the height y of the projectile as a function of horizontal distance x. The projectile is launched from the origin with an initial speed vi at an angle q above the horizontal. Show that the equation of the trajectory followed by the projectile is

( ) ( )

v iy −g 2 ____ y = ___ v ix x + 2v 2 x ix ✦100. A person climbs from a Paris metro station to the street level by walking up a stalled escalator in 94 s. It takes 66 s to ride the same distance when standing on the escalator when it is operating normally. How long would it take for him to climb from the station to the street by walking up the moving escalator?

Answers to Practice Problems 3.1 No; the checkbook balance may increase or decrease, but there is no spatial direction associated with it. When we say it “goes down,” we do not mean that it moves in a direction toward the center of Earth! Rather, we really mean that it decreases. The balance is a scalar. 3.2 240 mi 20° W of S 3.3 Ax = +16 km; A y = −8.2 km; B x = +17 km; B y = 0 km; Cx = −11 km; Cy = +47 km 3.4 v⃗av can never be greater than the average speed because the magnitude of the displacement cannot be greater

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than the distance traveled. v⃗av can be equal to the average speed if the magnitude of the displacement is equal to the distance traveled, which is true when the motion is along a straight line with no change in direction. 3.5 (a) 1.64 m/s directed 33° east of north; (b) 0.82 m/s2 directed 33° east of north 3.6 2 2 3.7 v i /(4g) 3.8 Ignoring air resistance, the two arrows have the same constant horizontal velocity component: v2x − v1x = 0 (choosing the x-axis horizontal and the y-axis up). Their vertical velocity components are different, but they change at the same rate, so v2y − v1y stays constant. The difference in their velocities (v⃗2 − v⃗1) stays constant. This constant difference in their velocities makes the difference in their positions (r⃗2 − r⃗1) change with time 3.9 vfx = 500.0 m/s; v fy = −19.8 m/s; bullet enters the water at an angle of 2.27° below the horizontal 3.10 (a) 1.0 m/s; (b) 15 min 3.11 28° south of west; 3.6 min; 130 m

Answers to Checkpoints 3.2 Cx = −5.5 km and Cy = −7.2 km 3.4 Velocity is a vector quantity. The plane’s speed does not change, but its velocity does. Therefore, Δv⃗ ≠ 0 and a⃗av = Δv⃗/Δt ≠ 0. 3.5 The horizontal velocity component does not change. The vertical component is zero at the highest point, so the velocity vector is directed horizontally. The acceleration is constant and directed vertically downward throughout the flight, including at the highest point. 3.6 (a) 19.5 m/s (b) 1.5 m/s

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CHAPTER

Force and Newton’s Laws of Motion

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A sailplane (or “glider”) is a small, unpowered, high-performance aircraft. A sailplane must be initially towed a few thousand feet into the air by a small airplane, after which it relies on regions of upwardmoving air such as thermals and ridge currents to ascend further. Suppose a small plane requires about 120 m of runway to take off by itself. When it is towing a sailplane, how much runway does it need? (See p. 120 for the answer.)

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Concepts & Skills to Review

CHAPTER 4 Force and Newton’s Laws of Motion

• • • • •

addition of vectors (Sections 3.1 and 3.2) vector components (Section 3.2) acceleration (Sections 2.3 and 3.4) motion with constant acceleration (Sections 2.4 and 3.5) motion diagrams (Section 2.5)

4.1

Force: a push or pull that one object exerts on another

The weight of an object near a planet or moon is the magnitude of the gravitational force exerted on it by that planet or moon.

FORCE

Just as human life would be dull without social interactions, the physical universe would be dull without physical interactions. Social interactions with friends and family change our behavior; physical interactions change the “behavior” (motion, temperature, etc.) of matter. An interaction between two objects can be described and measured in terms of two forces, one exerted on each of the two interacting objects. A force is a push or a pull. When you play soccer, your foot exerts a force on the ball while the two are in contact, thereby changing the speed and direction of the ball’s motion. At the same time, the ball exerts a force on your foot, the effect of which you can feel. To understand the motion of an object, whether it be a soccer ball or the International Space Station, we need to analyze the forces acting on the object. Long-Range Forces Forces exerted on macroscopic objects—objects that are large enough for us to observe without instrumentation—can be either long-range forces or contact forces. Long-range forces do not require the two objects to be touching. These forces can exist even if the two objects are far apart and even if there are other objects between the two. For example, gravity is a long-range force. The gravitational force exerted on the Earth by the Sun keeps the Earth in orbit around the Sun, despite the great distance between them and despite other planets that occasionally come between them. The Earth also exerts a long-range gravitational force on objects on or near its surface. We call the magnitude of the gravitational force that a planet or moon exerts on a nearby object the object’s weight.

PHYSICS AT HOME Besides gravity, other long-range forces are electric or magnetic in nature. On a dry day, run a comb vigorously through your hair until you hear some crackling. Now hold the comb a few centimeters from small pieces of a torn paper napkin. Observe the long-range electrical interaction between the paper and the comb. Now take a refrigerator magnet. Hold it near but not touching the refrigerator door. You can feel the effect of a long-range magnetic interaction. Part 3 of this book treats electromagnetic forces in detail. Until then, you can safely assume that gravity is the only significant long-range interaction unless the statement of a problem indicates otherwise.

Contact forces exist only as long as the objects are touching one another.

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Contact Forces All forces exerted on macroscopic objects, other than long-range gravitational and electromagnetic forces, involve contact. Contact forces exist only as long as the objects are touching one another. Your foot has no noticeable effect on a soccer ball’s motion until the two come into contact, and the force lasts only as long as they are in contact. Once the ball moves away from your foot, your foot has no further influence over the ball’s motion. The idea of contact is a useful simplification for macroscopic objects. What we call a single contact force is really the net effect of enormous numbers of electromagnetic

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forces between atoms on the surfaces of the two objects. On an atomic scale, the idea of “contact” breaks down. There is no way to define “contact” between two atoms—in other words, there is no unique distance between the atoms at which the forces they exert on one another suddenly become zero.

Measuring Forces If the concept of force is to be useful in physics, there must be a way to measure forces. Consider a simple spring scale (Fig. 4.1). As the scale’s pan is pulled down, a spring is stretched. The harder you pull, the more the spring stretches. As the spring stretches, an attached pointer moves. Then all we have to do to measure the applied force is to calibrate the scale so the amount of stretch measures the magnitude of the force. For many springs, the extension is approximately proportional to the force, which makes calibration easy. In the United States, supermarket scales are generally calibrated to measure forces in pounds (lb). In the SI system, the unit of force is the newton (N). To convert pounds to newtons, use the approximate conversion factors 1 lb = 4.448 N

or

1 N = 0.2248 lb

(4-1)

Ceiling pulls up on scale Newtons 1 3

There are more sophisticated means for measuring forces than a supermarket scale. Even so, many operate on the same principle as the supermarket scale: a force is measured by the deformation—change of size or shape—it produces in some object.

5 7 9

0 2

x

4

Extension of spring

6 8 10

Force Is a Vector Quantity The magnitude of a force is not a complete description of the force. The direction of the force is equally important. The direction of the brief contact force exerted by a soccer player’s foot on the ball can make the difference between scoring a goal or not (Fig. 4.2). Force is a vector quantity that must be added (or subtracted) using the same methods used for other vector quantities such as position, velocity, and acceleration.

Hand pulls down on scale

Figure 4.1 As the bottom of a spring scale is pulled downward, the spring stretches. We can measure the force by measuring the extension of the spring. For many springs, the extension is approximately proportional to the force, which makes calibration easy. Note that there is a pull on both ends of the scale. The ceiling pulls up on the scale and supports the scale from above.

Figure 4.2 A soccer player’s foot exerts a force on the ball only when they are touching.

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Example 4.1 Traction on a Foot In a traction apparatus, three cords pull on the central pulley, each with magnitude 22.0 N, in the directions shown in Fig. 4.3. What is the sum of the forces exerted on the central pulley by the three cords? Give the magnitude and direction of the sum.

The y-components of the forces are F1y = F2y = (22.0 N) sin 45.0° F3y = (−22.0 N) sin 30.0° The sum of the x-components is

Strategy First, we sketch the graphical addition of the three forces to get an estimate of the magnitude and direction of the sum. Then, to get an accurate answer, we resolve the three forces into their x- and y-components, sum the components, and then calculate the magnitude and direction of the sum.

Fx = F1x + F2x + F3x = 2 × (22.0 N) cos 45.0° + (22.0 N) cos 30.0° = 31.11 N + 19.05 N = 50.16 N We keep an extra decimal place for now to minimize roundoff error. The sum of the y-components is

Solution Figure 4.4 shows the graphical addition of the three forces exerted on the central pulley by the cords. From this sketch, we can tell that the sum of the three forces is at a relatively small angle above the horizontal (roughly half of 45°) and has a magnitude a bit larger than 44 N. To find an algebraic solution, we find the components along the x- and y-axes and add them (Fig. 4.5). The x-components of the forces are

Fy = F 1y + F 2y + F 3y = 2 × (22.0 N) sin 45.0° + (−22.0 N) sin 30.0° = 31.11 N − 11.00 N = 20.11 N The magnitude of the sum is (Fig. 4.6): _______

√

F1x = F2x = (22.0 N) cos 45.0°

F2 F1 45.0° 30.0°

30.0°

2

2

and the direction of the sum is opposite 20.11 N = 21.8° q = tan−1 _______ = tan−1 _______ adjacent 50.16 N The sum of the forces exerted on the pulley by the three cords is 54.0 N at an angle 21.8° above the +x-axis.

F3x = (22.0 N) cos 30.0°

45.0°

____________________

F = F x + F y = √ (50.16 N)2 + (20.11 N)2 = 54.0 N

Discussion To check the answer, look back at the graphical estimate. The magnitude of the sum (54 N) is somewhat larger than 44 N and the direction is at an angle very nearly half of 45° above the horizontal.

F3

Practice Problem 4.1 Changing the Pulley Angles

22.0 N

(a)

⃗1 and F ⃗2 are at an angle The pulleys are moved, after which F ⃗3 is 60.0° below the x-axis. of 30.0° above the x-axis and F (a) What is the sum of these three forces in component form? (b) What is the magnitude of the sum? (c) At what angle with the horizontal is the sum?

(b)

Figure 4.3 (a) A foot in traction; (b) the three forces exerted on the central pulley by the cords. y

y

y F2

F3 F1

F1

q

x

30.0° F1y = F1 sin 45.0°

45.0° x

F3x = F3 cos 30.0°

F1x = F1 cos 45.0° (a)

x

F3y = –F3 sin 30.0° F3

q (b)

20.11 N

50.16 N

Figure 4.4

Figure 4.5

Figure 4.6

Graphical sum of the forces on the pulley due to the cords.

⃗3. For clarity, the vector ⃗1 and (b) F Finding the components of (a) F arrows are drawn twice as long as they were in Fig. 4.4.

Finding the sum from its components.

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Net Force When more than one force acts on an object, the subsequent motion of the object is determined by the net force acting on the object. The net force is the vector sum of all the forces acting on an object.

Definition of net force: ⃗1, F ⃗2, . . . , F ⃗n are all the forces acting on an object, then the net force F ⃗net actIf F ing on that object is the vector sum of those forces: ⃗net = ∑F ⃗ = F ⃗1 + F ⃗2 + ⋅ ⋅ ⋅ + F ⃗n F

(4-2)

The symbol ∑ is a capital Greek letter sigma that stands for “sum.”

Free-Body Diagrams An essential tool used to find the net force acting on an object is a free-body diagram (FBD): a simplified sketch of a single object with force vectors drawn to represent every force acting on that object. (For example, the sum of three forces calculated in Example 4.1 is not the net force on the central pulley because the forces on the pulley due to the patient’s leg and due to gravity are not included.) The net force must not include any forces that act on other objects. To draw an FBD: • Draw the object in a simplified way—you don’t have to be Michelangelo to solve physics problems! Almost any object can be represented as a box or a circle, or even a dot. • Identify all the forces that are exerted on the object. Take care not to omit any forces that are exerted on the object. Consider that everything touching the object may exert one or more contact forces. Then identify long-range forces (for now, just gravity unless electric or magnetic forces are specified in the problem). • Check your list of forces to make sure that each force is exerted on the object of interest by some other object. Make sure you have not included any forces that are exerted on other objects. • Draw vector arrows representing all the forces acting on the object. We usually draw the vectors as arrows that start on the object and point away from it. Draw the arrows so they correctly illustrate the directions of the forces. If you have enough information to do so, draw the lengths of the arrows so they are proportional to the magnitudes of the forces.

Example 4.2 Net Force on an Airplane The forces on an airplane in flight heading eastward are as follows: gravity = 16.0 kN (kilonewtons), downward; lift = 16.0 kN, upward; thrust = 1.8 kN, east; and drag = 0.8 kN, west. (Lift, thrust, and drag are three forces that the air exerts on the plane.) What is the net force on the plane? Strategy All the forces acting on the plane are given in the statement of the problem. After drawing these forces in

the FBD for the plane, we add the forces to find the net force. To resolve the force vectors into components, we choose x- and y-axes pointing east and north, respectively. All four forces are then lined up with the axes, so each will have only one nonzero component, with a sign that indicates the direction along that axis. For example, the drag force points in the −x-direction, so its x-component is negative and its y-component is zero. continued on next page

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Example 4.2 continued

⃗ T, ⃗ Solution Figure 4.7a is the FBD for the plane, using L, ⃗ ⃗ and D for the lift, thrust, and drag, respectively. W stands for the gravitational force on the plane; its magnitude is the plane’s weight W. The sum of the x-components of the forces is

∑Fx = Lx + Tx + Wx + Dx

= (16 kN) + 0 + (−16 kN) + 0 = 0

Discussion A graphical check of the vector addition is a good idea. Figure 4.7b shows that the sum of the four forces is indeed in the +x-direction (east).

y (Up) x (East)

L

Practice Problem 4.2 New Forces on the Airplane

T

(Down) D

∑ Fy = Ly + Ty + Wy + Dy The net force is 1.0 kN east.

= 0 + (1.8 kN) + 0 + (− 0.8 kN) = 1.0 kN

(West)

The sum of the y-components of the forces is

T L

W

Find the net force on the airplane if the forces are gravity = 16.0 kN, downward; lift = 15.5 kN, upward; thrust = 1.2 kN, north; drag = 1.2 kN, south.

W

ΣF

D (b)

(a)

4.2

Figure 4.7 (a) FBD for the airplane. (b) Graphical addition of the four force vectors.

INERTIA AND EQUILIBRIUM: NEWTON’S FIRST LAW OF MOTION

In 1687, Isaac Newton (1643–1727) published one of the greatest scientific works of all time, his Philosophiae Naturalis Principia Mathematica (or Principia for short). The Latin title translates as The Mathematical Principles of Natural Philosophy. In the Principia, Newton stated three laws of motion that form the basis of classical physics. To pre-Newtonian thinkers, it seemed that there must be two different sets of physical laws: one set to describe the motion of the heavenly bodies, thought to be perfect and enduring, and another to describe the motion of earthly bodies that always come to rest. Together with his law of universal gravitation, Newton’s laws of motion showed for the first time that the motion of the heavenly bodies (the Sun, the planets, and their satellites) and the motion of earthly bodies can be understood using the same physical principles.

Newton’s First Law of Motion Newton’s first law says that an object acted on by zero net force moves in a straight line with constant speed, or, if it is at rest, remains at rest. Using the concept of the velocity vector, which is a measure of both the speed and the direction of motion of an object, we can state the first law:

Newton’s First Law of Motion An object’s velocity vector v⃗ remains constant if and only if the net force acting on the object is zero.

This concise statement of Newton’s first law includes both the case of an object at rest (zero velocity) and a moving object (nonzero velocity). Certainly it makes sense that an object at rest remains at rest unless some force acts on it to make it start to move. On the

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INERTIA AND EQUILIBRIUM: NEWTON’S FIRST LAW OF MOTION

Start

(a)

Start

(b)

Figure 4.8 (a) Galileo found

Stop

h1

h1

h2 Stop

h2

Start Rolls on and on (c)

93

h1

that a ball rolled down an incline stops when it reaches almost the same height on the second incline. He decided that it would reach the same height if resistive forces could be eliminated. (b) As the second incline is made less and less steep, the ball rolls farther and farther before stopping. (c) If the second incline is horizontal, and there are no resistive forces, the ball would never stop.

other hand, it may not be obvious that an object can continue to move with constant speed in a straight line without forces acting to keep it moving. In our experience, most moving objects come to rest because of forces that oppose motion, like friction and air resistance. A hockey puck can slide the entire length of a rink with very little change in speed or direction because the ice is slippery (frictional forces are small). If we could remove all the resistive forces, including friction and air resistance, the puck would slide without changing its speed or direction at all. No force is required to keep an object in motion if there are no forces opposing its motion. When a hockey player strikes the puck with his stick, the brief contact force exerted on the puck by the stick changes the puck’s velocity, but once the puck loses contact with the stick, it slides along the ice even though the stick no longer exerts a force on it. Inertia Newton’s first law is also called the law of inertia. In physics, inertia means resistance to changes in velocity. It does not mean resistance to the continuation of motion (or the tendency to come to rest). Newton based the law of inertia on the ideas of some of his predecessors, including Galileo Galilei (1564–1642) and René Descartes (1596–1650). In a series of clever experiments in which he rolled a ball up inclines of different angles, Galileo postulated that, if he could eliminate all resistive forces, a ball rolling on a horizontal surface would never stop (Fig. 4.8). Galileo made a brilliant conceptual leap from the real world with friction to an imagined, ideal world, free of friction. The law of inertia contradicted the view of the Greek philosopher Aristotle (384–322 b.c.e.). Almost 2000 years before Galileo, Aristotle had formulated his view that the natural state of an object is to be at rest; and, for an object to remain in motion, a force would have to act on it continuously. Galileo conjectured that, in the absence of friction and other resistive forces, no continued force is needed to keep an object moving. However, Galileo thought that the sustained motion of an object would be in a great circle around the Earth. Shortly after Galileo’s death, Descartes argued that the motion of an object free of any forces should be along a straight line rather than a circle. Newton acknowledged his debt to Galileo, Descartes, and others when he wrote: “If I have seen farther, it is because I was standing on the shoulders of giants.”

Conceptual Example 4.3 Snow Shoveling The task of shoveling newly fallen snow from the driveway can be thought of as a struggle against the inertia of the snow. Without the application of a net force, the snow

remains at rest on the ground. However, there is an important way that the inertia of the snow makes it easier to shovel. Explain. continued on next page

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Conceptual Example 4.3 continued

Strategy Think about the physical motions used when shoveling snow. (If you live where there is no snow, think about shoveling gravel from a wheelbarrow to line a garden path.) In order for the shoveling to be facilitated by the snow’s inertia, there must be a time when the snow is moving on its own, without the shovel pushing it. Solution and Discussion Imagine scooping up a shovelful of snow and swinging the shovel forward toward the side

of the driveway. The snow and the shovel are both in motion. Then suddenly the forward motion of the shovel stops, but the snow continues to move forward because of its inertia; it slides forward off the shovel, to be pulled down to the ground by gravity. The snow does not stop moving forward when the forward force due to the shovel is removed. This procedure works best with fairly dry snow. Wet sticky snow tends to cling to the shovel. The frictional force on the snow due to the shovel keeps it from moving forward and makes the job far more difficult. In this case, it might help to give the shovel a thin coating of cooking oil to reduce the frictional force the shovel exerts on the snow.

Conceptual Practice Problem 4.3 Inertia on the Subway Negar, a college student, stands on a subway car, holding on to an overhead strap. As the train starts to pull out of the station, she feels thrust toward the rear of the car; as the train comes to a stop at the next station, she feels thrust forward. Explain the role played by inertia in this situation.

PHYSICS AT HOME For an easy demonstration of inertia, place a quarter on top of an index card, or a credit card, balanced on top of a drinking glass (Fig. 4.9a). With your thumb and forefinger, flick the card so it flies out horizontally from under the quarter. What happens to the quarter? The horizontal force on the coin due to friction is small. With a negligibly small horizontal force, the coin tends to remain motionless while the card slides out from under it (Fig. 4.9b). Once the card is gone, gravity pulls the coin down into the glass (Fig. 4.9c).

Equilibrium An object in translational equilibrium has a net force of zero acting on it.

When the net force acting on an object is zero, the object is said to be in translational equilibrium. Equilibrium conveys the idea that the forces are in balance; there is as much force upward as there is downward, as much to the right as to the left, and so forth. Any object moving with a constant velocity, whether at rest or moving in a

Figure 4.9 A demonstration of inertia.

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(a)

(b)

(c)

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INERTIA AND EQUILIBRIUM: NEWTON’S FIRST LAW OF MOTION

straight line at constant speed, is in translational equilibrium. A vector can only have zero magnitude if all of its components are zero, so For an object in equilibrium,

∑ Fx = 0 and ∑ Fy = 0 (and ∑ Fz = 0)

(4-3)

In an equilibrium problem, choose x- and y-axes so the fewest number of force vectors have both x- and y-components. It is always good practice to make a conscious choice of axes and then to draw them in the FBDs and any other sketches that you make in solving the problem.

Example 4.4 Sliding a Chest In order to slide a chest that weighs 750 N across the floor at constant velocity, you must push it horizontally with a force of 450 N (Fig. 4.10). Find the contact force that the floor exerts on the chest. Strategy The chest moves with constant velocity, so it is in equilibrium. The net force acting on it is zero. We will identify all the forces acting on the chest, draw an FBD, do a graphical addition of the forces, choose x- and y-axes, resolve the forces into their x- and y-components, and then set ΣFx = 0 and ΣFy = 0. Solution There are three forces acting on the chest. The ⃗ has magnitude 750 N and is directed gravitational force W ⃗ has magnitude 450 N and its direcdownward. Your push F ⃗ has tion is horizontal. The contact force due to the floor C unknown magnitude and direction. However, remembering that the chest is in equilibrium, upward and downward force components must balance, as must the horizontal force com⃗ must be roughly in the direction shown ponents. Therefore, C in the FBD (Fig. 4.11a), as is confirmed by adding the three forces graphically (Fig. 4.11b). The sum is zero because the tip of the last vector ends up at the tail of the first one. Figure 4.10 Sliding a chest across the floor. C

Choosing the x-axis to the right and the y-axis up means that two of the ⃗ and F, ⃗ have one three force vectors, W component that is zero:

y Cx

q C

Cy

Wx = 0

q x

and

Fx = 450 N

W y = −750 N and Fy = 0

Now we set the x- and y-components of

Figure 4.12

Finding the magni- the net force each equal to zero because the chest is in equilibrium. tude and direction of the contact force.

∑Fx = Wx + Fx + Cx = 0 + 450 N + Cx = 0 ∑Fy = Wy + Fy + Cy = −750 N + 0 + Cy = 0 ⃗ Cx = − 450 N These equations tell us the components of C: and Cy = +750 N. Then the magnitude of the contact force is (Fig. 4.12) _______

√

__________________

C = C x + C y = √ (−450 N)2 + (750 N)2 = 870 N 2

2

opposite 750 N = 59° q = tan−1 _______ = tan−1 ______ adjacent 450 N The contact force due to the floor is 870 N, directed 59° above the leftward horizontal (−x-axis). Discussion The x- and y-components of the contact force and its magnitude and direction are all reasonable based on the graphical addition, so we can be confident that we did not make an error such as a sign error with one of the components.

F F

Practice Problem 4.4 The Chest at Rest

C

W

W (a)

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(b)

Figure 4.11 (a) An FBD for the chest; (b) graphical addition of the three forces showing that the sum is zero.

Suppose the same chest is at rest. You push it horizontally with a force of 110 N but it does not budge. What is the contact force on the chest due to the floor during the time you are pushing?

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Using Newton’s first law, we can understand how a spring scale can be used to measure weight (the magnitude of the gravitational force exerted on an object). If a melon remains at rest in the pan of the scale, the net force on the melon must be zero. There are only two forces acting on the melon: gravity pulls down and the scale pulls up. Then these two forces must be equal in magnitude and opposite in direction. The scale measures the magnitude of the force it exerts on the melon, which is equal to the weight of the melon.

4.3

NET FORCE, MASS, AND ACCELERATION: NEWTON’S SECOND LAW OF MOTION

When a nonzero net force acts on an object, the object’s velocity changes. Newton’s second law says that the rate of change of the object’s velocity—that is, the object’s acceleration—is proportional to the net force acting on it and inversely proportional to its mass:

Newton’s Second Law 1 ⃗ a⃗ = __ m ∑F

or

⃗ = ma⃗ ∑F

(4-4)

If the net force is zero, then the acceleration is zero, in accordance with Newton’s first law. If the net force is not zero, then the acceleration has the same direction as the net force. When the net force is constant, the acceleration is also constant. In component form, Newton’s second law is

∑Fx = max and ∑Fy = may ΣF

(4-5)

If all the forces acting on an object are known, then Eq. (4-4) can be used to calculate its acceleration. Alternatively, sometimes we know the object’s acceleration but we have incomplete information about the forces acting on it; then Eq. (4-4) provides information about the unknown forces.

a

SI Unit of Force The SI unit of force, the newton, is defined so that a net force of 1 N gives a 1-kg mass an acceleration of 1 m/s2: 1 N = 1 kg⋅m/s2

(4-6)

Defining the unit of force in this way makes it possible to write Eqs. (4-4) and (4-5) without needing a constant of proportionality to convert between the force unit and kg·m/s2. ΣF

What Is Mass?

a

The acceleration of an object is proportional to the net force on it and is in the same direction (Fig. 4.13). A larger net force causes a more rapid change in the velocity vector. Newton’s second law also says that the acceleration is inversely proportional to the object’s mass. The same net force acting on two different objects causes a smaller acceleration on the object with greater mass (Fig. 4.14). Mass is a measure of an object’s inertia—the amount of resistance to changes in velocity. Newton’s second law serves as our definition of mass. In everyday language mass and weight are sometimes used as synonyms, but in physics, mass and weight are different physical properties. The mass of an object is a measure of its inertia, while weight is the magnitude of the gravitational force acting on it. Imagine taking a shuffleboard puck to the Moon. Since the Moon’s surface gravity is

Figure 4.13 The acceleration of a baseball is proportional to the net force acting on it.

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4.4 INTERACTION PAIRS: NEWTON’S THIRD LAW OF MOTION

Figure 4.14 The same net force acting on two different objects produces accelerations in inverse proportion to the masses.

ΣF

ΣF a

a

weaker than the Earth’s, the puck’s weight would be smaller on the Moon, but the puck’s mass would be the same as on Earth. Ignoring the effects of friction, an astronaut playing shuffleboard on the Moon would have to exert the same horizontal force on the puck as on Earth to give it the same acceleration (Fig. 4.15).

4.4

INTERACTION PAIRS: NEWTON’S THIRD LAW OF MOTION

Forces always exist in pairs. Every force is part of an interaction between two objects and each of the interacting objects exerts a force on the other. We call the two forces an interaction pair; each force is the interaction partner of the other. When you push open a door, the door pushes you. When two cars collide, each exerts a force on the other. Note that interaction partners act on different objects—the two objects that are interacting.

Fcourt

Fcourt

Fstick

Fstick

W a

a W ΣF

Earth (a)

(b)

ΣF

Moon (c)

(d)

Figure 4.15 An astronaut playing shuffleboard (a) on Earth and (c) on the Moon. FBDs for a puck of mass m being given the same acceleration a⃗ on a frictionless court on (b) Earth and (d) on the Moon. The contact force on the puck due to the ⃗stick) must be the same since the mass of the puck is the same: ∑F ⃗ = F ⃗stick = ma⃗. pushing stick (F

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Newton’s third law of motion says that interaction partners always have the same magnitude and are in opposite directions.

Newton’s Third Law of Motion In an interaction between two objects, each object exerts a force on the other. These two forces are equal in magnitude and opposite in direction.

Conceptual Example 4.5 An Orbiting Satellite Earth exerts a gravitational force on an orbiting communications satellite. What is the interaction partner of this force? Strategy The question concerns a gravitational interaction between two objects: Earth and the satellite. In this interaction, each object exerts a gravitational force on the other. Solution The interaction partner is the gravitational force exerted on the Earth by the satellite. Discussion Does the satellite really exert a force on the Earth with the same magnitude as the force Earth exerts on the satellite? If so, why does the satellite orbit Earth rather than Earth orbiting the satellite? Newton’s third law says that the interaction partners are equal in magnitude, but does not say that these two forces have equal effects. The effect of a net force on an object’s motion depends on the object’s mass. These two forces of equal magnitude have

vastly different effects due to the great discrepancy between the masses of the Earth and the satellite. On the other hand, if a massive planet orbits a star in a relatively small orbit, the gravitational force that the planet exerts on the star can make the star wobble enough to be observed. The wobble enables astronomers to discover planets orbiting stars other than the Sun. The planets do not reflect enough light toward Earth to be seen, but their presence can be inferred from the effect they have on the star’s motion.

Conceptual Practice Problem 4.5 Interaction Partner of a Surface Contact Force In Example 4.4, the contact force exerted on the chest by the floor was 870 N, directed 59° above the leftward horizontal (−x-axis). Describe the interaction partner of this force—in other words, what object exerts it on what other object? What are the magnitude and direction of the interaction partner?

Do not assume that Newton’s third law is involved every time two forces happen to be equal and opposite—it ain’t necessarily so! You will encounter many situations in which two equal and opposite forces act on a single object. These forces cannot be interaction partners because they act on the same object. Interaction partners act on different objects, one on each of the two objects that are interacting.

CHECKPOINT 4.4 In the photo, two children are pulling on a toy. If they are exerting equal and opposite forces on the toy, are these two forces interaction partners? The forces exerted by these two children on a toy cannot be interaction partners because they act on the same object (the toy). The interaction of the force exerted by a child on the toy is the force that the toy exerts on that child.

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PHYSICS AT HOME The next time you go swimming, notice that you use Newton’s third law to get the water to push you forward. When you push down and backward on the water with your arms and legs, the water pushes up and forward on you. The various swimming strokes are devised so that you exert as large a force as possible backward on the water during the power part of the stroke, and then as small a force as possible forward on the water during the return part of the stroke.

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Internal and External Forces When we say that a baseball has interactions with the Earth (gravity), with a baseball bat, and with the air, we are treating the baseball as a single entity. But the ball really consists of an enormous number of protons, neutrons, and electrons, all interacting with each other. The protons and neutrons interact with each other to form atomic nuclei; the nuclei interact with electrons to form atoms; interactions between atoms form molecules; and the molecules interact to form the structure of the thing we call a baseball. It would be difficult to have to deal with all of these interactions to predict the motion of a baseball. Defining a System Let us call the set of particles comprising the baseball a system. Once we have defined a system, we can classify all the interactions that affect the system as either internal or external to the system. For an internal interaction, both interacting objects are part of the system. When we add up all the forces acting on the system to find the net force, every internal interaction contributes two forces—an interaction pair—that always add to zero. For an external interaction, only one of the two interaction partners is exerted on the system. The other partner is exerted on an object outside the system and does not contribute to the net force on the system. Therefore, to find the net force on the system, we can ignore all the internal forces and just add the external forces. The insight that internal forces always add to zero is particularly powerful because the choice of what constitutes a system is completely arbitrary. We can choose any set of objects and define it to be a system. In one problem, it may be convenient to think of the baseball as a system; in another, we may choose a system consisting of both the baseball and the bat. The second choice might be useful if we do not have detailed information about the interaction between the bat and the ball.

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GRAVITATIONAL FORCES

Newton’s Law of Universal Gravitation Now we turn our attention to learning about some forces in more detail, beginning with gravity. According to Newton’s law of universal gravitation, any two objects exert gravitational forces on each other that are proportional to the masses (m1 and m2) of the two objects and inversely proportional to the square of the distance (r) between their centers. Strictly speaking, the law of gravitation as presented here only applies to point particles and symmetrical spheres. (The point particle is a common model in physics used when the size of an object is negligibly small and the internal structure is irrelevant.) Nevertheless, the law of gravitation is approximately true for any two objects if the distance between their centers is large compared with their sizes. In mathematical language, the magnitude of the gravitational force is written: Gm 1m 2 F = _______ r2

(4-7)

where the constant of proportionality (G = 6.674 × 10−11 N·m2/kg2) is called the universal gravitational constant. Equation (4-7) is only part of the law of universal gravitation because it gives only the magnitudes of the gravitational forces that each object exerts on the other. The directions are equally important: each object is pulled toward the other’s center (Fig. 4.16). In other words, gravity is an attractive force. The forces on the two objects are equal in magnitude and the directions are opposite, as they must be since they form an interaction pair. Gravitational forces exerted by ordinary objects on each other are so small as to be negligible in most cases (see Practice Problem 4.6). Gravitational forces exerted by Earth, on the other hand, are much larger due to Earth’s large mass.

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Figure 4.16 Gravity is always an attractive force. The force that each body exerts on the other is equal in magnitude, even though the masses may be very different. The force exerted on the Moon by the Earth is of the same magnitude as the force exerted on the Earth by the Moon. The directions are opposite.

Earth

Gravitational force exerted on the Earth by the Moon

Gravitational force exerted on the Moon by the Earth

Moon

Example 4.6 Weight at High Altitude When you are in a commercial airliner cruising at an altitude of 6.4 km, by what percentage has your weight (as well as the weight of the airplane) changed compared with your weight on the ground? Strategy Your weight is the magnitude of Earth’s gravitational force exerted on you. Newton’s law of universal gravitation gives the magnitude of the gravitational force at a distance r from the center of the Earth. For your weight on the ground W1, we can use the mean radius of the Earth RE as the distance between the Earth’s center and you: r1 = RE = 6.37 × 106 m (Fig. 4.17). At an altitude of h = 6.4 × 103 m above the surface, your weight is W2 and your distance from Earth’s center is r2 = RE + h. Your mass m, the mass of the Earth ME (= 5.97 × 1024 kg), and G are the same in the two cases, so it is efficient to write a ratio of the weights and let those factors cancel out. h

Solution The ratio of your weight in the airplane to your weight on the ground is GM Em ______ 2 2 W 2 ______ r r2 R E2 ___ = = __12 = ________ GM Em r W 1 ______ (R E + h)2 2 2 r1

(

)

2 6.37 × 106 m = _______________________ = 0.998 6 3 6.37 × 10 m + 6.4 × 10 m

Since 0.998 = 1 − 0.002 and 0.002 = 0.2/100, your weight decreases by 0.2%. Discussion Although 6400 m may seem like a significant altitude to us, it’s a small fraction of the Earth’s radius (0.10%), so the weight change is a small percentage. When judging whether a quantity is small or large, always ask: “Small (or large) compared to what?”

r

Practice Problem 4.6 A Creative Defense

RE

Figure 4.17 Earth

The gravitational force depends on the distance r to the center of the Earth.

After an automobile collision, one driver claims that the gravitational force between the two cars caused the collision. Estimate the magnitude of the gravitational force exerted by one car on another when they are driving side-byside in parallel lanes and comment on the driver’s claim.

Gravitational Field Strength For an object near Earth’s surface, the distance between the object and the Earth’s center is very nearly equal to the Earth’s mean radius, RE = 6.37 × 106 m. The mass of the Earth is ME = 5.97 × 1024 kg, so the weight of an object of mass m near Earth’s surface is

( )

GM Em GM E W = ______ = m _____ 2 2 RE RE

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(4-8)

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Notice that for objects near Earth’s surface, the constants in the parentheses are always the same and the weight of the object is proportional to its mass. Rather than recalculate that combination of constants over and over, we call the combination the gravitational field strength g near Earth’s surface: GM E ___________________________________ 6.674 × 10−11 N⋅m2⋅kg−2 × (5.97 × 1024 kg) = g = _____ ≈ 9.8 N/kg (4-9) 2 (6.37 × 106 m)2 RE The units newtons per kilogram reinforce the conclusion that weight is proportional to mass: g tells us how many newtons of gravitational force are exerted on an object for every kilogram of the object’s mass. The weight of a 1.0-kg object near Earth’s surface is 9.8 N (2.2 lb). Using g, the weight of an object of mass m near Earth’s surface is usually written Relationship between mass and weight: W = mg

(4-10)

Variations in Earth’s Gravitational Field The Earth is not a perfect sphere; it is slightly flattened at the poles. Since the distance from the surface to the center of the Earth is smaller there, the field strength at sea level is greatest at the poles (9.832 N/kg) and smallest at the equator (9.814 N/kg). Altitude also matters; as you climb above sea level, your distance from Earth’s center increases and the field strength decreases. Tiny local variations in the field strength are also caused by geologic formations. On top of dense bedrock, g is a little greater than above less dense rock. Geologists and geophysicists measure these variations to study Earth’s structure and also to locate deposits of various minerals, water, and oil. The device they use, a gravimeter, is essentially a mass hanging on a spring. As the gravimeter is carried from place to place, the extension of the spring increases where g is larger and decreases where g is smaller. The mass hanging from the spring does not change, but its weight does (W = mg). Furthermore, due to Earth’s rotation, the effective value of g that we measure in a coordinate system attached to Earth’s surface is slightly less than the true value of the field strength. This effect is greatest at the equator, where the effective value of g is 9.784 N/kg, about 0.3% smaller than the true value of g. The effect gradually decreases with latitude to zero at the poles. We learn more about this effect in Chapter 5. The most important thing to remember from this discussion is that, unlike G, g is not a universal constant. The value of g is a function of position. Near Earth’s surface, the variations are small, so we can adopt an average value g = 9.80 N/kg as a default.

Gravitational Field and Free-Fall Acceleration An object in free fall is assumed to have only one force acting on it: gravity. Other forces, such as air resistance, must be negligibly small for this approximation to be ⃗ = m⃗ valid. We can write the gravitational force on the object as W g, where the gravitational field vector g⃗ has magnitude g and is directed downward (in the direction of the gravitational force). From Newton’s second law, ⃗net = mg⃗ = m a⃗ F Dividing by the mass yields a⃗ = g⃗

(4-11)

Therefore, the acceleration of an object in free fall is g, ⃗ regardless of the object’s mass. Since 1 N = 1 kg·m/s2, 9.80 N/kg = 9.80 m/s2—the magnitude of the free-fall acceleration near Earth’s surface has average value 9.80 m/s2.

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More massive objects have the same free-fall acceleration as less massive objects. True, a more massive object is harder to accelerate: the acceleration of an object subjected to a given force is inversely proportional to its mass. However, the stronger gravitational force on a more massive object compensates for its greater inertia, giving it the same free-fall acceleration as a less massive object.

Gravitational Field Strength on Other Planets Equation (4-10) can be used to find the weight of an object at or above the surface of any planet or moon, but the value of g will be different due to the different mass M of the planet or moon and the different distance r from the planet’s center: GM g = ____ r2

(4-12)

CHECKPOINT 4.5 If you climb Mt. McKinley, what happens to the weight of your gear? What happens to its mass?

Example 4.7 “Weighing” Figs in Kilograms In most countries other than the United States, produce is sold in mass units (grams or kilograms) rather than in force units (pounds or newtons). The scale still measures a force, but the scale is calibrated to show the mass of the produce instead of its weight. What is the weight of 350 g of fresh figs, in newtons and in pounds? Strategy Weight is mass times the gravitational field strength. We will assume g = 9.80 N/kg. The weight in newtons can be converted to pounds using the conversion factor 1 N = 0.2248 lb.

Converting to pounds, W = 3.43 N × 0.2248 lb/N = 0.771 lb The figs weigh 3.4 N or 0.77 lb. Discussion This is the weight of the figs at a location where g has its average value of 9.80 N/kg. The figs would weigh a little more in the northern city of St. Petersburg, Russia, and a little less in Quito, Ecuador, which is near the equator.

Practice Problem 4.7 Figs on the Moon Solution The weight of the figs in newtons is W = mg = 0.35 kg × 9.80 N/kg = 3.43 N

What would those figs weigh on the surface of the Moon, where g = 1.62 N/kg?

CONNECTION: In Example 4.4, we resolved the contact force on a sliding chest into components perpendicular to and parallel to the contact surface. It is often convenient to think of these components as two separate but related contact forces: the normal force and the frictional force. Normal force: a contact force between two solid objects that is perpendicular to the contact surfaces. Each object pushes the other one away.

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We have already solved some problems involving forces exerted between two solid objects in contact. Now we look at contact forces in more detail.

Normal Force A contact force perpendicular to the contact surface that prevents two objects from passing through one another is called the normal force. (In geometry, the word normal means perpendicular.) Consider a book resting on a horizontal table surface. The normal force due to the table must have just the right magnitude to keep the book from falling through the table. If no other vertical forces act, the normal force on the book is equal in magnitude to the book’s weight because the book is in equilibrium (Fig. 4.18a).

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N N

N

Book

W

W

(a)

(b)

F W (c)

103

Figure 4.18 (a) The normal force is equal in magnitude to the weight of the book; the two forces sum to zero. (b) On an incline, the normal force is smaller than the weight of the book and is not vertical. (c) If ⃗ you push down on the book ( F), the normal force on the book due to the table is larger than the book’s weight.

According to Newton’s third law, there is also a normal force exerted on the table by the book; this normal force acts downward and is of equal magnitude. In everyday language, we might say that the table “feels the book’s weight.” That is not an accurate statement in the language of physics. The table cannot “feel” the gravitational force on the book; the table can only feel forces exerted on the table. What the table does “feel” is the normal force—a contact force—exerted on the table by the book. If the table’s surface is horizontal, the normal force on the book will be vertical and equal in magnitude to the book’s weight. If the surface of the table is not horizontal, the normal force is not vertical and is not equal in magnitude to the weight of the book. Remember that the normal force is perpendicular to the contact surface (Fig. 4.18b). Even on a horizontal surface, if there are other vertical forces acting on the book, then the normal force is not equal in magnitude to the book’s weight (Fig. 4.18c). Never assume anything about the magnitude of the normal force. In general, we can figure out what the magnitude of the normal force must be in various situations if we have enough information about other forces. What Causes Normal Forces? How does the table “know” how hard to push on the book? First imagine putting the book on a bathroom scale instead of the table. A spring inside the scale provides the upward force. The spring “knows” how hard to push because, as it is compressed, the force it exerts increases. When the book reaches equilibrium, the spring is exerting just the right amount of force, so there is no tendency to compress it further. The spring is compressed until it pushes up with a force equal to the book’s weight. If the spring were stiffer, it would exert the same upward force but with less compression. The forces that bind atoms together in a rigid solid, like the table, act like extremely stiff springs that can provide large forces with little compression—so little that it’s usually not noticed. The book makes a tiny indentation in the surface of the table (Fig. 4.19); a heavier book would make a slightly larger indentation. If the book were to be placed on a soft foam surface, the indentation would be much more noticeable.

CHECKPOINT 4.6 Your laptop is resting on the surface of your desk, which stands on four legs on the floor. Identify the normal forces acting on the desk and give their directions.

Friction

Figure 4.19 The book com-

A contact force parallel to the contact surface is called friction. We distinguish two types: static friction and kinetic (or sliding) friction. When the two objects are slipping or sliding across one another, as when a loose shingle slides down a roof, the friction is kinetic. When no slipping or sliding occurs, such as between the tires of a car parked on a hill and the road surface, the friction is called static. Static friction acts to prevent objects from starting to slide; kinetic friction acts to try to make sliding objects

presses the “atomic springs” in the table until they push up on the book to hold it up. The slight decrease in the distance between atoms is greatly exaggerated here.

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stop sliding. Note that two objects in contact with one another that move with the same velocity exert static frictional forces on one another, because there is no relative motion between the two. For example, if a conveyor belt carries an air freight package up an incline and the package is not sliding, the two move with the same velocity and the friction is static. Static Friction Frictional forces are complicated on the microscopic level and are an active field of current research. Despite the complexities, we can make some approximate statements about the frictional forces between dry, solid surfaces. In a simplified model, the maximum magnitude of the force of static friction fs,max that can occur in a particular situation is proportional to the magnitude of the normal force N acting between the two surfaces. fs,max ∝ N If you want better traction between the tires of a rear-wheel-drive car and the road, it helps to put something heavy in the trunk to increase the normal force between the tires and the road. The constant of proportionality is called the coefficient of static friction (symbol m s): Maximum force of static friction: fs,max = m s N

(4-13)

Since fs,max and N are both magnitudes of forces, ms is a dimensionless number. Its value depends on the condition and nature of the surfaces. Equation (4-13) provides only an upper limit on the force of static friction in a particular situation. The actual force of friction in a given situation is not necessarily the maximum possible. It tells us only that, if sliding does not occur, the magnitude of the static frictional force is less than or equal to this upper limit: fs ≤ msN

(4-14)

Kinetic (Sliding) Friction For sliding or kinetic friction, the force of friction is only weakly dependent on the speed and is roughly proportional to the normal force. In the simplified model we will use, the force of kinetic friction is assumed to be proportional to the normal force and independent of speed: Force of kinetic (sliding) friction: f k = m kN

(4-15)

where fk is the magnitude of the force of kinetic friction and mk is called the coefficient of kinetic friction. The coefficient of static friction is always larger than the coefficient of kinetic friction for an object on a given surface. On a horizontal surface, a larger force is required to start the object moving than is required to keep it moving at a constant velocity. Direction of Frictional Forces Equations (4-13) through (4-15) relate only the magnitudes of the frictional and normal forces on an object. Remember that the frictional force is perpendicular to the normal force between the same two surfaces. Friction is always parallel to the contact surface, but there are many directions parallel to a given contact surface. Here are some rules of thumb for determining the direction of a frictional force. • The static frictional force acts in whatever direction necessary to prevent the objects from beginning to slide or slip.

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• Kinetic friction acts in a direction that tends to make the sliding stop. If a book slides to the left along a table, the table exerts a kinetic frictional force on the book to the right, in the direction opposite to the motion of the book. • From Newton’s third law, frictional forces come in interaction pairs. If the table exerts a frictional force on the sliding book to the right, the book exerts a frictional force on the table to the left with the same magnitude.

Example 4.8 Coefficient of Kinetic Friction for the Sliding Chest Example 4.4 involved sliding a 750-N chest to the right at constant velocity by pushing it with a horizontal force of 450 N. We found that the contact force on the chest due to the floor had components Cx = − 450 N and Cy = +750 N, where the x-axis points to the right and the y-axis points up (see Fig. 4.20). What is the coefficient of kinetic friction for the chest-floor surface? Strategy To find the coefficient of friction, we need to know what the normal and frictional forces are. They are the

fk C F

fk

F

q

Solution The magnitude of the force due to sliding friction is fk = Cx = 450 N. The magnitude of the normal force is N = Cy = 750 N. Now we can calculate the coefficient of kinetic friction from fk = mkN: f N

450 N = 0.60 m k = __k = ______ 750 N

Discussion If we had written fk = Cx = − 450 N, we would have ended up with a negative coefficient of friction. The coefficient of friction is a relationship between the magnitudes of two forces, so it cannot be negative.

y

N

C

components of the contact force that are perpendicular and parallel to the contact surface. Since the surface is horizontal (in the x-direction), the x-component of the contact force is friction and the y-component is the normal force.

N

q x W (a)

W (b)

Practice Problem 4.8 Chest at Rest (c)

⃗ is the contact force Figure 4.20 (a) FBD for the chest. C due to the floor. (b) FBD in which the contact force is ⃗ replaced by two perpendicular forces, the normal force N ⃗ ⃗ and the kinetic frictional force f k. (c) Resolving C into normal and frictional components.

Suppose the same chest is at rest. You push to the right with a force of 110 N but the chest does not budge. What are the normal and frictional forces on the chest due to the floor while you are pushing? Explain why you do not need to know the coefficient of static friction to answer this question.

Conceptual Example 4.9 Horse and Sleigh A horse pulls a sleigh to the right at constant velocity on level ground. The horse exerts a horizontal force ⃗ Fsh on the sleigh. (The subscripts indicate the force on the sleigh due to the horse.) (a) Draw three FBDs, one for the horse, one for the sleigh, and one for the system horse + sleigh. (b) To make the sleigh increase its velocity, there must be a nonzero net force to the right acting on the sleigh. Suppose the horse pulls harder (Fsh increases in magnitude).

According to Newton’s third law, the sleigh always pulls back on the horse with a force of the same magnitude as the force with which the horse pulls the sleigh. Does this mean that no matter how hard it pulls, the horse can never make the net force on the sleigh nonzero? Explain. (c) Identify the interaction partner of each force acting on the sleigh.

continued on next page

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Conceptual Example 4.9 continued s = sleigh g = ground h = horse E = Earth Nsg

v (constant)

Fsh

fsg

in magnitude and opposite in direc⃗hs tion. From the FBDs, f⃗hg = −F ⃗sh. Because F ⃗hs and F ⃗sh and f⃗sg = −F are interaction partners, they are equal and opposite. Therefore, f⃗hg and f⃗sg are equal and opposite. The system is in equilibrium.

(b) The FBD for the sleigh (see Fig. 4.21) shows that if the horse pulls the sleigh with a force greater in Figure 4.21 magnitude than the force of friction FBD for the sleigh. on the sleigh (Fsh > fsg), then the net force on the sleigh is nonzero and to the right. From Fig. 4.22, we need fhg > Fhs to have a nonzero net force to the right on the horse. So the frictional force on the horse would have to increase to enable it to pull the sleigh with a greater force. Then in Fig. 4.23, the two frictional forces are no longer equal in magnitude. The forward frictional force on the horse is greater than the backward frictional force on the sleigh, so the net force on the system horse + sleigh is to the right. FsE

Strategy (a) In each FBD, we include only the external forces acting on that system. All three systems move with constant velocity, so the net force on each is zero. (b) Looking at the FBD for the sleigh, we can determine the conditions under which the net force on the sleigh can be nonzero. (c) For a force exerted on the sleigh by X, its interaction partner must be the force exerted on X by the sleigh. Solution and Discussion (a) If we think of the normal and frictional forces as separate forces, then there are four forces ⃗sh, the acting on the sleigh: the force exerted by the horse F ⃗ gravitational force due to Earth FsE, the normal force on the ⃗ sg, and kinetic (sliding) friction sleigh due to the ground N ⃗ due to the ground f sg. Figure 4.21 shows the FBD for the sleigh. The net force is zero, so its horizontal and vertical com⃗sh+ f⃗sg = 0 and N ⃗ sg + F ⃗sE = 0. ponents must each be zero: F Similarly, four forces are acting on the horse: the force ⃗hs, the gravitational force F ⃗hE, the norexerted by the sleigh F ⃗ mal force due to the ground Nhg, and friction due to the ⃗hs = −F ⃗sh; the sleigh ground f⃗hg. Newton’s third law says that F pulls back on the horse with a force equal in magnitude to ⃗hs is the forward pull of the horse on the sleigh. Therefore, F ⃗ to the left and has the same magnitude as Fsh. The horse is in ⃗hs + f⃗hg = 0 and N ⃗ hg + F ⃗hE = 0. The first of equilibrium, so F these equations means that the frictional force has to be to the right. How does the horse get friction to push it forward? By pushing backward on the ground with its feet. We all do the same thing when taking a step; by pushing backward on the ground, we get the ground to push forward on us. This is static friction because the horse’s hoof is not sliding along the ground. If there were no friction (imagine the ground to be icy), the hoof might slide backward. Static friction acts to prevent sliding, so the frictional force on the hoof is forward. Figure 4.22 shows the FBD for the horse. Of the eight forces acting either on the horse or on the sleigh, two are internal forces for the horse + sleigh system: ⃗sh and F ⃗hs. They add to zero since they are interaction partF ners, so we can omit them from the FBD for the system (Fig. 4.23). The two frictional forces on the system horse + sleigh are not interaction partners, but they are equal

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Nhg

Nhg Nsg

Figure 4.23 FBD for the system,

fhg

Fhs

fsg

Figure 4.22 FhE

FBD for the horse.

fhg horse and sleigh.

FsE FhE

The internal forces ⃗sh and F ⃗hs are F omitted—they form an interaction pair, so they add to zero.

(c) Force Exerted on Sleigh

Interaction Partner

Force on the sleigh due to the horse Force on the horse due to the sleigh ⃗sh ⃗hs F F Gravitational force on the sleigh due ⃗sE to Earth F

Gravitational force on Earth due to ⃗Es the sleigh F

Normal force on the sleigh due to ⃗ sg the ground N

Normal force on the ground due to ⃗ gs the sleigh N

Friction on the sleigh due to the Friction on the ground due to the ground f⃗sg sleigh f⃗gs

Practice Problem 4.9 Passing a Truck A car is moving north and speeding up to pass a truck on a level road. The combined contact force exerted on the road by all four tires has vertical component 11.0 kN downward and horizontal component 3.3 kN southward. The drag force exerted on the car by the air is 1.2 kN southward. (a) Draw the FBD for the car. (b) What is the weight of the car? (c) What is the net force acting on the car?

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Microscopic Origin of Friction What looks like the smooth surface of a solid to the unaided eye is generally quite rough on a microscopic scale (Fig. 4.24). Friction is caused by atomic or molecular bonds between the “high points” on the surfaces of the two objects. These bonds are formed by microscopic electromagnetic forces that hold the atoms or molecules together. If the two objects are pushed together harder, the surfaces deform a little more, enabling more “high points” to bond. That is why the force of kinetic friction and the maximum force of static friction are proportional to the normal force. A bit of lubricant drastically decreases the frictional forces, because the two surfaces can float past one another without many of the “high points” coming into contact. In static friction, when these molecular bonds are stretched, they pull back harder. The bonds have to be broken before sliding can begin. Once sliding begins, molecular bonds are continually made and broken as “high points” come together in a hit-or-miss fashion. These bonds are generally not as strong as those formed in the absence of sliding, which is why m s > m k. For dry, solid surfaces, the amount of friction depends on how smooth the surfaces are and how many contaminants are present on the surface. Does polishing two steel surfaces decrease the frictional forces when they slide across each other? Not necessarily. In an extreme case, if the surfaces are extremely smooth and all surface contaminants are removed, the steel surfaces form a “cold weld”—essentially, they become one piece of steel. The atoms bond as strongly with their new neighbors as they do with the old.

Equilibrium on an Inclined Plane Suppose we wish to pull a large box up a frictionless incline to a loading dock platform. ⃗a represents the applied force Figure 4.25 shows the three forces acting on the box. F with which we pull. The force is parallel to the incline. If we choose the x- and y-axes to be horizontal and vertical, respectively, then two of the three forces have both x- and ycomponents. On the other hand, if we choose the x-axis parallel to the incline and the y-axis perpendicular to it, then only one of the three forces has both x- and ycomponents (the gravitational force). With axes chosen, the weight of the box is then resolved into two perpendicular ⃗, components (Fig. 4.26a). To find the x- and y-components of the gravitational force W ⃗ makes with one of the axes. The right triangle of we must determine the angle that W Fig. 4.26b shows that a + f = 90°, since the interior angles of a triangle add up to 180°. The x- and y-axes are perpendicular, so a + b = 90°. Therefore, b = f . ⃗ is perpendicular to the surface of the incline. From The y-component of W Fig. 4.26a, the side parallel to the y-axis is adjacent to angle b , so

Figure 4.24 Friction is caused by bonds between atoms that form between the “high points” of the two surfaces that come into contact. y x

a Wy b

f W

Wx (a)

a f (b)

Wy adjacent cos b = __________ = ____ hypotenuse W

+y N

Since Wy is negative and W = mg, Wy = −mg cos b = −mg cos f −Fa

d

N

Wx = mg sin f +x

Wy = –mg cos f (c)

Fa

Figure 4.26 (a) Resolving the h

f W

Figure 4.25 A box of mass m pulled up an incline.

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weight into components parallel to and perpendicular to the incline. (b) A right triangle shows that a + f = 90°. (c) FBD for the box on the incline.

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The x-component of the weight tends to make the box slide down the incline (in the positive x-direction). Using the same triangle, W x = +mg sin f When the box is pulled with a force equal in magnitude to Wx up the incline (in the negative x-direction), it will slide up with constant velocity. The component of the box’s ⃗ that pushes the weight perpendicular to the incline is supported by the normal force N box away from the incline. Figure 4.26c is an FBD in which the gravitational force is separated into its x- and y-components. If the box is in equilibrium, whether at rest or moving along the incline at constant velocity, the force components along each axis sum to zero:

∑ Fx = (−Fa) + mg sin f = 0 and

∑ Fy = N + (−mg cos f ) = 0 On an incline, the normal force is not equal in magnitude to the weight and it does not point straight up. If the applied force has magnitude mg sin f , we can pull the box up the incline at constant velocity. If friction acts on the box, we must pull with a force greater than mg sin f to slide the box up the incline at constant velocity.

Example 4.10 Pushing a Safe up an Incline A new safe is being delivered to the Corner Book Store. It is to be placed in the wall at a height of 1.5 m above the floor. The delivery people have a portable ramp, which they plan to use to help them push the safe up and into position. The mass of the safe is 510 kg, the coefficient of static friction along the incline is m s = 0.42, and the coefficient of kinetic friction along the incline is m k = 0.33. The ramp forms an angle q = 15° above the horizontal. (a) How hard do the movers have to push to start the safe moving up the incline? Assume that they push in a direction parallel to the incline. (b) To slide the safe up at a constant speed, with what magnitude force must the movers push? Strategy (a) When the safe starts to move, its velocity is changing, so the safe is not in equilibrium. Nevertheless, to find the minimum applied force to start the safe moving, we can find the maximum applied force for which the safe remains at rest—an equilibrium situation. (b) The safe is in equilibrium

Solution First we draw a diagram to show forces acting (Fig. 4.27). Before resolving the forces into components, we must choose x- and y-axes. To use the coefficient of friction, we have to resolve the contact force on the safe due to the incline into components parallel and perpendicular to the incline—friction and the normal force, respectively—rather than into horizontal and vertical components. Therefore, we choose x- and y-axes parallel and perpendicular to the incline so friction is along the x-axis and the normal force is along the y-axis. ⃗ can be resolved into its comThe gravitational force W ponents: Wx = −mg sin q and Wy = −mg cos q (Fig. 4.28a). +y

q

q f 1.5 m

Figure 4.27 Forces acting on the safe as it is moved up the incline.

–mg sin q

Fa

–f

+x

–mg cos q (a)

W

N

–mg sin q

q q = 15°

+x

–mg cos q

mg

N

Fa

as it slides with a constant velocity. Both parts of the problem can be solved by drawing the FBD, choosing axes, and setting the x- and y-components of the net force equal to zero.

(b)

Figure 4.28 (a) Resolving the weight into x- and y-components, and (b) an FBD in which the weight is replaced with its x- and y-components. continued on next page

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4.7 TENSION

Example 4.10 continued

⃗ replaced by its components Now we draw the FBD with W (Fig. 4.28b). (a) Suppose that the safe is initially at rest. As the movers start to push, Fa gets larger and the force of static friction gets larger to “try” to keep the safe from sliding. Eventually, at some value of Fa, static friction reaches its maximum possible value m sN. If the movers continue to push harder, increasing Fa further, the force of static friction cannot increase past its maximum value m sN, so the safe starts to slide. The direction of the frictional force is along the incline and downward since friction is “trying” to keep the safe from sliding up the incline. The normal force is not equal in magnitude to the weight of the safe. To find the normal force, sum the y-components of the forces: ∑ Fy = N + (−mg cos q ) = 0 Then N = mg cos q. The normal force is less than the weight since cos q < 1. When the movers push with the largest force for which the safe does not slide,

∑ Fx = Fax + fx + Wx = 0 The applied force is in the +x-direction, so Fax = +Fa. The frictional force has its maximum magnitude and is in the −x-direction, so fx = −fs,max = −m sN = −m smg cos q. From the FBD, Wx = −mg sin q. Then,

∑ Fx = Fa − m smg cos q − mg sin q = 0 Solving for Fa, Fa = mg ( m s cos q + sin q ) = 510 kg × 9.80 m/s2 × (0.42 × cos 15° + sin 15°) = 3300 N

An applied force that exceeds 3300 N starts the box moving up the incline. (b) Once the safe is sliding, the movers need only push hard enough to make the net force on the safe equal to zero if they want the safe to slide at constant velocity. We are now dealing with sliding friction, so the frictional force is now fx = −m kN = −m kmg cos q.

∑ Fx = Fax + fx + Wx = Fa − m kmg cos q − mg sin q =0 Fa = mg ( m k cos q + sin q ) = 510 kg × 9.80 m/s2 × (0.33 × cos 15° + sin 15°) = 2900 N ⃗a of magnitude 2900 N The movers push with a force F directed up the incline. Discussion In (b), the expression Fa = mg ( mk cos q + sin q ) shows that the applied force up the incline has to balance the sum of two forces down the incline: the frictional force ( m kmg cos q ) and the component of the gravitational force down the incline (mg sin q ). This balance of forces is shown graphically in the FBD (Fig. 4.28b).

Practice Problem 4.10 Smoothing the Infield Dirt During the seventh-inning stretch of a baseball game, groundskeepers drag mats across the infield dirt to smooth it. A groundskeeper is pulling a mat at a constant velocity by applying a force of 120 N at an angle of 22° above the horizontal. The coefficient of kinetic friction between the mat and the ground is 0.60. Find (a) the magnitude of the frictional force between the dirt and the mat and (b) the weight of the mat.

PHYSICS AT HOME To estimate the coefficient of static friction between a penny and the cover of your physics book, place the penny on the book and slowly lift the cover. Note the angle of the cover when the penny starts to slide. Explain how you can use this angle to find the coefficient of static friction. Can you devise an experiment to find the coefficient of kinetic friction?

4.7

TENSION

Consider a heavy chandelier hanging by a chain from the ceiling (Fig. 4.29a). The chandelier is in equilibrium, so the upward force on it due to the chain is equal in magnitude to the chandelier’s weight. With what force does the chain pull downward on the ceiling? The ceiling has to pull up with a force equal to the total weight of the chain and the chandelier. The interaction partner of this force—the force the chain exerts on the ceiling—is

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Figure 4.29 (a) The chain pulls up on the chandelier and pulls down on the ceiling. (b) The chain is under tension. Each link is pulled in opposite directions by its neighbors.

Force on ceiling due to chain Force on chandelier due to chain

Force pulling up on top of link Force pulling down on bottom of link

(a)

(b)

equal in magnitude and opposite in direction. Therefore, if the weight of the chain is negligibly small compared with the weight of the chandelier, then the chain exerts forces of equal magnitude at its two ends. The forces at the ends would not be equal, however, if you grabbed the chain in the middle and pulled it up or down or if we could not neglect the weight of the chain. We can generalize this observation: An ideal cord (or rope, string, tendon, cable, or chain) pulls in the direction of the cord with forces of equal magnitude on the objects attached to its ends as long as no external force is exerted on it anywhere between the ends. An ideal cord has zero mass and zero weight. A single link of the chain (Fig. 4.29b) is pulled at both ends by the neighboring links. The magnitude of these forces is called the tension in the chain. Similarly, a little segment of a cord is pulled at both its ends by the tension in the neighboring pieces of the cord. If the segment is in equilibrium, then the net force acting on it is zero. As long as there are no other forces exerted on the segment, the forces exerted by its neighbors must be equal in magnitude and opposite in direction. Therefore, the tension has the same value everywhere and is equal to the force that the cord exerts on the objects attached to its ends.

Example 4.11 Archery Practice Figure 4.30 shows the bowstring of a bow and arrow just before it is released. The archer is pulling back on the midpoint of the bowstring with a horizontal force of 162 N. What is the 72 cm tension in the bowstring? 162 N

35 cm

Figure 4.30 The force applied to the bowstring by an archer.

Strategy Consider a small segment of the bowstring that touches the archer’s finger. That piece of the string is in equilibrium, so the net force acting on it is zero. We draw the FBD, choose coordinate axes, and apply the equilibrium condition: ΣFx = 0 and ΣFy = 0. We know the force exerted on the

segment of string by the archer’s fingers. That segment is also pulled on each end by the tension in the string. Can we assume the tension in the string is the same everywhere? The weight of the string is small compared with the other forces acting on it. The archer pulls sideways on the bowstring, exerting little or no tangential force, so we can assume the tension is the same everywhere. Solution Figure 4.31a is an FBD for the segment of bowstring being considered. The forces are labeled with their magnitudes: Fa for the force applied by the archer’s finger and T for each of the tension forces. Figure 4.31b shows these three forces adding to zero. From this sketch, we expect the tension T to be roughly the same as Fa. We choose the x-axis to the right and the y-axis upward. To find the continued on next page

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4.7 TENSION

Example 4.11 continued

components of the forces due to tension in the string, we draw a triangle (Fig. 4.31c). From the measurements given, we can find the angle q. opposite 35 cm = 0.486 sin q = __________ = ______ hypotenuse 72 cm q = sin−1 0.486 = 29.1° The x-component of the tension force exerted on the upper end of the segment is Tx = −T sin q The x-component of the force exerted on the lower end of the string is the same. Therefore,

Solving for T, Fa 162 N = 170 N T = ______ = ________ 2 × 0.486 2 sin q y T q

Fa x

T

q

T

72 cm

Practice Problem 4.11 Tightrope Practice

y T

(a)

Fa 1F ________ = __ 2 sin 90° 2 a That is correct because the archer would be pulling to the right with a force Fa, while each side of the bowstring would pull to the left with a force of magnitude T. For equilibrium, Fa = 2T or T = _12 Fa. As q gets smaller, sin q decreases and the tension increases (for a fixed value of Fa ). That agrees with our intuition. The larger the tension, the smaller the angle the string needs to make in order to supply the necessary horizontal force.

∑Fx = −2T sin q + Fa = 0

T

Discussion The tension is only slightly larger than Fa, a reasonable result given the picture of graphical vector addition in Fig. 4.31b. In this problem, only the x-components of the forces had to be used. The y-components must also add to zero. At the upper end of the string, the y-component of the force exerted by the bow is +T cos q, while at the lower end it is −T cos q. Therefore, ΣFy = 0. The expression T = Fa/(2 sin q ) can be evaluated for limiting values of q to make sure that the expression is correct. As q approaches 90°, the tension approaches

Fa

35 cm

(b)

x (c)

Figure 4.31 (a) FBD for a point on the bowstring with the magnitudes of the forces labeled. (b) Graphical addition of the three forces showing that the sum is zero. (c) The angle q is used to find the x- and y-components of the forces exerted at each end of the bowstring.

Jorge decides to rig up a tightrope in the backyard so his children can develop a good sense of balance (Fig. 4.32). For safety reasons, he positions a horizontal cable only 0.60 m above the ground. If the 6.00-m-long cable sags by 0.12 m from its taut horizontal position when Denisha (weight 250 N) is standing on the middle of it, what is the tension in the cable? Ignore the weight of the cable.

6.00 m Eyebolt

0.12 m 250 N

Figure 4.32 Tightrope for balancing practice.

Application: Tensile Forces in the Body Tensile forces are central in the study of animal motion, or biomechanics. Muscles are usually connected by tendons, one at each end of the muscle, to two different bones, which in turn are linked at a joint (Fig. 4.33). Usually one of the bones is more easily moved than the other. When the muscle contracts, the tension in the tendons increases, pulling on both of the bones.

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Figure 4.33 A muscle contracts, increasing the tension in the attached tendons. The tendons exert forces on two different bones. tendon

muscle

tendon joint

PHYSICS AT HOME Sit with your arm bent at the elbow with a heavy object on the palm of your hand. You can feel the contraction of the biceps muscle. With your other hand, feel the tendon that connects the biceps muscle to your forearm. Now place your hand palm down on the desktop and push down. Now it is the triceps muscle that contracts, pulling up on the bone on the other side of the elbow joint. Muscles and tendons cannot push; they can only pull. The biceps muscle cannot push the forearm downward, but the triceps muscle can pull on the other side of the joint. In both cases, the arm acts as a lever. F

Figure 4.34 Using a pulley to lift an object by pulling down⃗ ward on a rope with force F.

Application: Ideal Pulleys A pulley can change the direction of the force exerted by a cord under tension. To lift something heavy, it is easier to stand on the ground and pull down on the rope than to get above the weight on a platform and pull up on the rope (Fig. 4.34). An ideal pulley has no mass and no friction. An ideal pulley exerts no forces on the cord that are tangent to the cord—it is not pulling in either direction along the cord. As a result, the tension of an ideal cord that runs through an ideal pulley is the same on both sides of the pulley. An ideal pulley changes the direction of the force exerted by a cord without changing its magnitude. As long as a real pulley has a small mass and negligible amount of friction, we can approximate it as an ideal pulley.

Example 4.12 A Two-Pulley System A 1804-N engine is hauled upward at constant speed (Fig. 4.35). What are the tensions in the three ropes labeled A, B, and C? Assume the ropes and the pulleys labeled L and R are ideal. Strategy The engine and pulley L move up at constant speed, so the net force on each of them is zero. Pulley R is at

rest, so the net force on it is also zero. We can draw the FBD for any or all of these objects and then apply the equilibrium condition. If the pulleys are ideal, the tension in the rope is the same on both sides of the pulley. Therefore, rope C—which is attached to the ceiling, passes around both pulleys, and is pulled downward at the other end—has the same tension continued on next page

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APPLYING NEWTON’S SECOND LAW

Example 4.12 continued

B Pulley R C Pulley L

throughout. Call the tensions in the three ropes TA, TB, and TC. To analyze the forces exerted on a pulley, we define our system so the part of the rope wrapped around the pulley is considered part of the pulley. Then there are two cords pulling on the pulley, each with the same tension.

A

1804 N

Figure 4.35 A system of pulleys used to raise a heavy weight.

Solution There are two forces acting on the engine: the gravitational force (1804 N, downward) and the upward pull of rope A. These must be equal and opposite (Fig. 4.36a), since the net force is zero. Therefore TA = 1804 N.

Discussion The engine is raised by pulling down on a rope—the pulleys change the direction of the applied force needed to lift the engine. In this case they also change the magnitude of the required force. They do that by making the rope pull up on the engine twice, so the person pulling the rope only needs to exert a force equal to half the engine’s weight.

Practice Problem 4.12 and Engine

Consider the entire collection of ropes, pulleys, and the engine to be a single system. Draw the FBD for this system and show that the net force on the system is zero. [Hint: Remember that only forces exerted by objects external to the system are included in the FBD.]

2T C = T A

T B = 2T C = 1804 N

TB

Pulley L

W

T C = _12 T A = 902.0 N Figure 4.36c is the FBD for pulley R. Rope B pulls upward on it with a force of magnitude TB. On each side of the pulley, rope C pulls downward. For the net force to equal zero,

TC

TC

TA

The FBD for pulley L (Fig. 4.36b) shows rope A pulling down with a force of magnitude TA and rope C pulling upward on each side. The rope has the same tension throughout, so all forces labeled TC in Fig. 4.36b,c have the same magnitude. For the net force to equal zero,

4.8

System of Ropes, Pulleys,

(a)

Pulley R

TA

(b)

TC

TC

(c)

Figure 4.36 (a) FBD for the engine. (b) FBD for pulley L and (c) FBD for pulley R.

APPLYING NEWTON’S SECOND LAW

We can now apply Newton’s second law to a great variety of situations involving the forces we have encountered so far—gravity, contact forces, and tension. The following steps are helpful in most problems that involve Newton’s second law.

Problem-Solving Strategy for Newton’s Second Law • • • •

Decide what object will have Newton’s second law applied to it. Identify all the external forces acting on that object. Draw an FBD to show all the forces acting on the object. Choose a coordinate system. If the direction of the net force is known, choose axes so that the net force (and the acceleration) are along one of the axes. • Find the net force by adding the forces as vectors. • Use Newton’s second law to relate the net force to the acceleration. • Relate the acceleration to the change in the velocity vector during a time interval of interest.

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Example 4.13 The Broken Suitcase The wheels fall off Beatrice’s suitcase, so she ties a rope to it and drags it along the floor of the airport terminal (Fig. 4.37). The rope makes a 40.0° angle with the horizontal. The suitcase has a mass of 36.0 kg and Beatrice pulls on the rope with a force of 65.0 N. (a) What is the magnitude of the normal force acting on the suitcase due to the floor? (b) If the coefficient of kinetic friction between the suitcase and the marble floor is m k = 0.13, find the frictional force acting on the suitcase. (c) What is the acceleration of the suitcase while Beatrice pulls with a 65.0 N force at 40.0°? (d) Starting from rest, for how long a time must she pull with this force until the suitcase reaches a comfortable walking speed of 0.5 m/s?

Solution (a) Figure 4.38 shows the forces acting on the ⃗ is the force exerted by Beatrice. All the suitcase, where F other forces are either parallel or perpendicular to the floor, ⃗ needs to be resolved into x- and y-components. so only F Fx = F cos 40.0° = 65.0 N × 0.766 = 49.8 N Fy = F sin 40.0° = 65.0 N × 0.643 = 41.8 N ⃗ is replaced by its comFigure 4.39 is an FBD in which F ponents. The vertical force components add to zero since ay = 0.

∑Fy = may = 0 N + F sin 40.0° − W = 0 We can solve this equation for the magnitude of the normal force. The magnitude of the gravitational force is W = mg, so N = mg − F sin 40.0° = (36.0 kg × 9.80 N/kg) − (65.0 N × sin 40.0°) = 352.8 N − 41.8 N = 311 N

40.0°

(b) The magnitude of the kinetic frictional force is f k = mkN = 0.13 × 311 N = 40.43 N Rounding to two significant figures, the frictional force is 40 N in the −x-direction (opposite the motion of the suitcase).

Figure 4.37 Beatrice dragging her suitcase.

Strategy Since the suitcase is dragged horizontally along the floor, the vertical component of its velocity is always zero. The vertical acceleration component of the suitcase is zero because the vertical velocity component does not change. (If it did have a vertical acceleration component, the suitcase would begin to move either down through the floor or up into the air.) If we choose the +y-axis up and the +x-axis to be horizontal, then ay = 0. We resolve the forces acting on the suitcase into their components, draw a free-body diagram for the suitcase, and apply Newton’s second law.

(c) The y-component of the acceleration is zero. To find the x-component, we apply Newton’s second law to the x-components of the forces acting on the suitcase:

∑Fx = +F cos 40.0° + (−fk) = 49.79 N − 40.43 N = 9.36 N

∑F 9.36 N ax = ____x = _______ = 0.260 m/s2 m 36.0 kg

N

N F sin 40.0°

F 40.0°

fk

−fk

F cos 40.0°

Figure 4.38 y

W

x

Forces acting on a suitcase dragged along the floor. The lengths of the vector arrows are not to scale.

y

Figure 4.39 x

−mg

FBD for the suitcase, with the forces represented by their x- and y-components. continued on next page

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Example 4.13 continued

Here we have replaced newtons per kilogram with the equivalent meters per second squared, the usual way to write the SI units of acceleration. The acceleration is 0.3 m/s2 in the +x-direction. (d) With constant ax,

first accelerate the suitcase from rest. Once the suitcase is moving at the desired velocity, she pulls a little less hard, so the net force is zero and the suitcase slides at constant speed. She would do so without thinking much about it, of course!

Δvx = ax Δt The suitcase starts from rest so vix = 0 and Δvx = vfx − vix = vfx. Then, vfx _________ Δt = ___ = 0.5 m/s = 2 s ax 0.260 m/s2 Discussion What Beatrice probably wants to do is to drag the suitcase along at constant velocity. To do that, she must

Practice Problem 4.13 The Continuing Story . . . (a) How hard does Beatrice pull at a 40.0° angle while the suitcase slides along the floor at constant velocity? [Hint: Do not assume that the normal force is the same as in the previous discussion.] (b) The suitcase is moving at 0.50 m/s. Beatrice changes the force to 42 N at 40.0°. How long does it take the suitcase to come to rest?

Sometimes two or more objects are constrained to have the same acceleration by the way they are connected. In Example 4.14, we look at a train engine pulling five freight cars. The couplings maintain a fixed distance between the cars, so at any instant the cars move with the same velocity; if they didn’t, the distance between them would change. The velocities don’t have to be constant, they just have to change in exactly the same way, which implies that the accelerations must also be the same at any instant.

Example 4.14 Coupling Force on First and Last Freight Cars A train engine pulls out of a station along a straight horizontal track with five identical freight cars behind it, each of which weighs 90.0 kN. The train reaches a speed of 15.0 m/s within 5.00 min of starting out. Assuming the engine pulls with a constant force during this interval, with what magnitude of force does the coupling between cars pull forward on the first and last of the freight cars? Ignore air resistance and friction on the freight cars. Strategy A sketch of the situation is shown in Fig. 4.40. To find the force exerted by the first coupling, we consider all five cars to be one system so we do not have to worry about the force exerted on the first car by the second car. The only external forces on the group of five cars are the normal force, T4

T5 5

4

gravity, and the pull of the first coupling. To find the force exerted by the fifth coupling, we consider car five by itself to be a system. In each case, once we identify a system, we draw a free-body diagram, choose a coordinate system, and then apply Newton’s second law. As discussed previously, the engine and the cars must all have the same acceleration at any instant. We expect the acceleration to be constant because the engine pulls with a constant force. We can calculate the acceleration of the train from the initial and final velocities and the elapsed time.

T3 3

T2 2

T1

Engine

1

a

Figure 4.40 An engine pulling five identical freight cars. The entire train has a constant acceleration a⃗ to the right. continued on next page

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Example 4.14 continued

N1–5

y x

Cars 1–5

T1

W1–5

Figure 4.41 FBD for the system consisting of cars 1–5 (but not the engine).

Solution For the tension T1 in the first coupling, we consider the five cars as one system of mass M. Figure 4.41 shows the FBD in which cars 1 to 5 are treated as a single object. We choose the x-axis in the direction of motion of the train and the y-axis up. Since the train moves along the x-axis, the acceleration vector is along the x-axis. Therefore, ay = 0. Using the y-component of Newton’s second law, the vertical forces add to zero:

⃗5; the FBD is ⃗ 5, and the gravitational force W normal force N ⃗ 5 + W ⃗5 = 0, the net force is equal shown in Fig. 4.42. Since N ⃗5. From Newton’s second law, to T Wa ∑Fx = T 5 = max = __ g x 4 Δvx ___________ W × ___ 15.0 m/s = 459 N T 5 = __ = 9.00 × 10 2N × ________ g Δt 9.80 m/s 300 s

y

5

∑Fy = May = N1−5 − W1−5 = 0

The only external horizontal force is ⃗1 due to the tension in the first coupling. This the force T force is constant according to the problem statement, so we know that the acceleration ax is constant:

∑ Fx = T1 = Max The mass of the system M is five times the mass of one car m. We are given the weight of one car (W = 90.0 kN = 9.00 × 104 N). From the relation between mass and weight, W = mg, the mass of one car is m = W/g and the mass of five cars is M = 5W/g. The constant acceleration of the train is Δv v fx − v ix ___________ 15.0 m/s − 0 2 ax = ___x = _______ t f − t i = 300 s − 0 = 0.0500 m/s Δt Therefore, 4 Δvx ______________ 5W × ___ 15.0 m/s T 1 = Max = ___ = 5 × 9.00 × 102 N × ________ g 300 s 9.80 m/s Δt

= 2.30 kN Now consider the last freight car (car 5). If we ignore friction and air resistance, the only external forces acting ⃗5 due to the tension in the fifth coupling, the are the force T

x

N5

W5

T5

Figure 4.42 FBD for car 5. (Vector lengths are not to the same scale as those in Fig. 4.41.)

Discussion We considered two systems (cars 1 to 5 and car 5) that have the same acceleration and different masses. As expected, the net force is proportional to the mass: the net force on five cars is five times the net force on one car. The solution to this problem is much simpler when Newton’s second law is applied to a system comprised of all five cars, rather than to each car individually. Although the problem can be solved by looking at individual cars, to find the tension in the first coupling you would have to draw five FBDs (one for each car) and apply Newton’s second law five times. That’s because each car, except the fifth, is acted on by the unequal tensions in the couplings on either side. You’d have to first find the tension in the fifth coupling, then the fourth, then the third, and so on.

Practice Problem 4.14 Coupling Force Between First and Second Freight Cars With what force does the coupling between the first and second cars pull forward on the second car? [Hint: Try two methods. One of them is to draw the FBD for the first car and apply Newton’s third law as well as the second.]

Example 4.15 deals with two objects connected by an ideal cord. Although it may have a nonzero acceleration, the net force on an ideal cord is still zero because it has ⃗ = ma⃗ = 0. As a result, the tension is the same at the two zero mass: if m = 0, then ∑F ends as long as no external force acts on the cord between the ends (Fig. 4.43a). An ideal cord that passes over an ideal pulley has the same tension at its ends. The pulley exerts an external force on part of the cord, but this force is everywhere perpendicular to the cord. As Fig. 4.43b shows, an external force that has no component tangent to the cord does not affect the tension in the cord.

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N

q T2

q T1 x

x

T2

T1 a

Pulley Cord

(b)

(a)

Figure 4.43 (a) FBD for an ideal cord with acceleration a⃗. Applying Newton’s second law along the x-axis: ΣFx = T1 − T2 = max. The ideal cord has mass m = 0, so T1 = T2: the tensions at the ends are equal. (b) An ideal cord passing around an ideal pulley and the FBD for a short segment of the cord at the top of the pulley. Choosing the x-axis to be horizontal, the normal force has no x-component. Applying Newton’s second law along the x-axis: ΣFx = T1 cos q − T2 cos q = max. With m = 0, T1 = T2. The same reasoning can be applied to any segment of cord in contact with the pulley to show that the tensions are the same on either side of the pulley.

Example 4.15 Two Blocks Hanging on a Pulley In Fig. 4.44, two blocks are connected by an ideal cord that does not stretch; the cord passes over an ideal pulley. If the masses are m1 = 26.0 kg and m2 = 42.0 kg, what are the accelerations of each block and the tension in the cord? Strategy Since m2 is greater than m1, the downward force of gravity is stronger on the right side than on the left. We expect block 2’s acceleration to be downward and block 1’s to be upward. The cord does not stretch, so blocks 1 and 2 move at the same speed at any instant (in opposite directions). Therefore, the accelerations of the two blocks are equal in magnitude and opposite in direction. If the accelerations had different magnitudes, then soon the two blocks would be moving with different speeds. That could only happen if the cord either stretches or contracts. The tension in the cord must be the same everywhere along the cord since the masses of the cord and pulley are negligible and the pulley turns without friction. We treat each block as a separate system, draw FBDs for each, and then apply Newton’s second law to each. It is convenient to choose the positive y-direction differently for the two blocks since we know their accelerations are in opposite directions. For each block, we choose the +y-axis in the direction of the acceleration of that block: upward for block m1

and downward for m2. Doing so means that ay has the same magnitude and sign (both positive) for the two blocks. Solution Figure 4.45 shows FBDs for the two blocks. Two forces act on each: gravity and the pull of the cord. The T T +y 2

a

1

a +y

m1g m2 m1

Figure 4.44 Two hanging blocks connected on either side of a frictionless pulley by a massless, flexible cord that does not stretch.

m2g

Figure 4.45 FBDs for the hanging blocks. We draw the acceleration vector next to each FBD as a guide—the net force has to be in the direction of the acceleration. However, the acceleration vector is not part of the FBD (it is not a force to be added to the others). continued on next page

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Example 4.15 continued

Solving for T yields acceleration vectors are drawn next to the FBDs. Thus, we know the direction of the net force: it is always the same as the direction of the acceleration. Then we know that the tension must be greater than m1g to give block 1 an upward acceleration and less than m2g to give block 2 a downward acceleration. The +y-axes are drawn for each block to be in the direction of the acceleration. From the FBD of block 1, the pull of the cord is in the +y-direction and the gravitational force is in the −y-direction. Then Newton’s second law for block 1 is

∑F 1y = T − m 1g = m 1a 1y For block 2, the pull of the cord is in the −y-direction and the gravitational force is in the +y-direction. Newton’s second law for block 2 is

∑ F 2y = m 2g − T = m 2a 2y The tension T in the cord is the same in the two equations. Also a1y and a2y are identical, so we write them simply as ay. We then have a system of two equations with two unknowns. We can add the equations to obtain m 2g − m 1g = m 2ay + m 1ay Solving for ay, we find (m 2 − m 1)g ay = __________ m +m 2

1

Substituting numerical values, (42.0 kg − 26.0 kg) × 9.80 N/kg ay = __________________________ 42.0 kg + 26.0 kg = 2.31 m/s2 since kg ⋅ m/s2 N = 1 _______ 1 ___ = 1 m/s2 kg kg The blocks have the same magnitude acceleration. For block 1 the acceleration points upward and for block 2 it points downward. To find T we can substitute the expression for ay into either of the two original equations. Using the first equation,

2m 1m 2 T = _______ m1 + m2 g Substituting, 2 × 26.0 kg × 42.0 kg T = __________________ × 9.80 N/kg = 315 N 68.0 kg Discussion A few quick checks: • ay is positive, which means that the accelerations are in the directions we expect. • The tension (315 N) is between m1g (255 N) and m2g (412 N), as it must be for the accelerations to be in opposite directions. • The units and dimensions are correct for all equations. • We can check algebraic expressions in special cases for which we have some intuition. For example, if the masses had been equal, we expect the blocks to hang in equilibrium (either at rest or moving at constant velocity) due to the equal pull of gravity on the two blocks. Substituting m1 = m2 into the expressions for ay and T gives ay = 0 and T = m1g = m2g, which is just what we expect. Note that we did not find out which way the blocks move. We found the directions of their accelerations. If the blocks start out at rest, then the block of mass m2 moves downward and the block of mass m1 moves upward. However, if initially m2 is moving up and m1 down, they continue to move in those directions, slowing down since their accelerations are opposite to their velocities. Eventually they come to rest and then reverse directions.

Practice Problem 4.15 Another Check Using the numerical values of the tension and the acceleration calculated in Example 4.15, verify Newton’s second law directly for each of the two blocks.

(m 2 − m 1)g T − m 1g = m 1 __________ m +m 2

1

Examples 4.16, 4.17, and 4.18 illustrate how different concepts and problemsolving techniques from Chapters 2– 4 can be brought together to find the solution to a physics problem.

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Example 4.16 Hauling a Crate up to a Third-Floor Window A student is moving into a dorm room on the third floor and he decides to use a block and tackle arrangement (Fig. 4.46) to move a crate of mass 91 kg from the ground up to his window. If the breaking strength of the available rope is 550 N, what is the minimum time required to haul the crate to the level of the window, 30.0 m above the ground, without breaking the rope? Strategy The tension in the rope is T and is the same at both ends or anywhere along the rope, assuming the rope and pulleys are ideal. Two pieces of rope support the lower pulley, each pulling upward with a force of magnitude T. The gravitational force acts downward. We draw an FBD for the system consisting of the crate and the lower pulley and set the tension equal to the breaking force of the rope to find the maximum possible acceleration of the crate. Then we use the maximum acceleration to find the minimum time to move the required distance to the third-floor window. We choose the y-axis to be upward. Known: m = 91 kg; Δy = 30.0 m; Tmax = 550 N; viy = 0. To find: Δt, the time to raise the crate 30.0 m with the maximum tension in the cable.

y

T

T

Figure 4.47 mg

FBD for the crate and lower pulley. (This system is outlined by dashed lines in Fig. 4.46.)

Setting T = 550 N, the maximum possible value before the cable breaks, and substituting the other known values: 550 N + 550 N − 91 kg × 9.80 m/s2 ay = ____________________________ = 2.288 m/s2 91 kg The time to move the crate up a distance Δy starting from rest can be found from Δy = v iy Δt + _12 ay(Δt)2

(3-21)

Setting viy = 0 and solving for Δt, we find

√

____

2 Δy Δt = ± ____ ay

Our equation applies only for Δt ≥ 0 (the crate reaches the window after it leaves the ground). Taking the positive root and substituting numerical values,

4th-floor window T

√

2 × 30.0 m = 5.1 s Δt = _________ 2.288 m/s2

T

3rd-floor window

This is the minimum possible to haul the crate up without breaking the rope. T

2nd-floor window

mg

Figure 4.46 Block and tackle setup.

Solution From the FBD (Fig. 4.47), if the forces acting up are greater than the force acting down, the net force is upward and the crate’s acceleration is upward. In terms of components, with the +y-direction chosen to be upward,

∑Fy = T + T − mg = may Solving for the acceleration, T + T − mg ay = __________ m

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__________

Discussion In reality, the student is not likely to achieve this minimum possible time. To do so would mean pulling the rope at an unrealistic speed. At the end of the 5.1-s interval, vfy = 2.288 m/s2 × 5.1 s = 12 m/s! More likely, the student would hoist the crate at a roughly constant velocity (except at the beginning, to get it moving, and at the end, to let it come to rest). For motion with a constant velocity, the tension in the rope would be equal to half the weight of the crate (450 N).

Practice Problem 4.16 Single Pulley

Hauling the Crate with a

If only a single pulley, attached to the beam above the fourth floor, were available and if the student had a few friends to help him pull on the cable, could they haul the crate up to the third-floor window using the same rope? If so, what is the minimum time required to do so?

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Example 4.17 Towing a Glider What length runway does the plane need?

A small plane of mass 760 kg requires 120 m of runway to take off by itself. (120 m is the horizontal displacement of the plane just before it lifts off the runway, not the entire length of the runway.) As a simplified model, ignore friction and drag forces and assume the plane’s engine exerts a constant forward force on the plane. (a) When the plane is towing a 330-kg glider, how much runway does it need? (b) If the final speed of the plane just before it lifts off the runway is 28 m/s, what is the tension in the tow cable while the plane and glider are moving along the runway? Strategy We draw FBDs for the two cases: plane alone, then plane + glider. The motion in both cases is horizontal (along the runway), because we are told the displacement before it lifts off the runway. Until the plane begins to lift off the runway, its vertical acceleration component is zero. We need not be concerned with the vertical forces (gravity, the normal force, and lift—the upward force on the plane’s wings due to the air) since they cancel one another to produce zero vertical acceleration. We use Newton’s second law to compare the accelerations in the two cases and then use the accelerations to compare the displacements. Solution (a) When the plane takes off by itself, four forces act on it (see Fig. 4.48). Three are vertical and the third—the thrust due to the engine—is horizontal. Choosing the x-axis to be horizontal, Newton’s second law says

∑F1x = F = m 1a 1x where F is the thrust, m1 is the plane’s mass, and a1x is its horizontal acceleration component. When the glider is towed, we can consider the plane, glider, and cable to be a single system (see Fig. 4.49). There Lift Normal force

∑F 2x = F = (m 1 + m 2)ax where m1 + m2 is the total mass of the system (plane mass m1 plus glider mass m2) and ax is the horizontal acceleration component of plane and glider. We ignore the mass of the cable. The problem statement gives neither the thrust nor either of the accelerations. We can continue by setting the thrusts equal and finding the ratio of the accelerations: m1 ax _______ m 1a 1x = (m 1 + m 2)ax ⇒ ___ a1x = m 1 + m 2 The magnitude of the acceleration is inversely proportional to the mass of the system for the same net force. How is the acceleration related to the runway distance? The plane must get to the same final speed in order to lift off the runway. From our two basic constant acceleration equations Δvx = vfx − vix = ax Δt Δx = _12 (v fx + v ix) Δt

Normal force Thrust

(2-9) (2-11)

we can substitute vix = 0 and eliminate Δt to find 2

( )

v fx vfx _____ 1 (v + 0) ___ Δ x = __ ax = 2ax 2 fx In both cases, the displacement is inversely proportional to the acceleration and the acceleration is inversely proportional to the mass of the system. Therefore, the displacement is directly proportional to the mass. Letting Δ x1 = 120 m be the displacement of the plane without the glider, we can set up a proportion: a 1x _______ m + m 2 _______ 1090 kg Δ x = ___ ____ = 1 = = 1.434 m1 760 kg Δ x 1 ax Δ x = 1.434 × 120 m = 172.08 m → 170 m (b) The final speed given enables us to find the acceleration: 2

v fx Δ x = ___ 2ax

Lift

2

or

v fx ax = ____ 2 Δx

With vfx = 28 m/s, vix = 0, and Δx = 172.08 m, (28 m/s)2 ax = ____________ = 2.278 m/s2 2 × 172.08 m The tension in the cable is the only horizontal force acting on the glider. Therefore,

Thrust

Gravity

is still only one horizontal external force and it is the same thrust as before. The tension in the cable is an internal force. Therefore,

Gravity

Figure 4.48

Figure 4.49

FBD for the plane.

FBD for the system plane + glider.

∑Fx = T = m 2ax = 330 kg × 2.278 m/s2 = 751.7 N → 750 N continued on next page

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Example 4.17 continued

Discussion This solution is based on a simplified model, so we can only regard the answers as approximate. Nevertheless, it illustrates Newton’s second law. The same net force produces an acceleration inversely proportional to the mass of the object upon which it acts. Here we have the same net force acting on two different objects: first the plane alone, then the plane and glider together. Alternatively, we can look at forces acting only on the plane. When towing the glider, the cable pulls backward on

the plane. The net force on the plane is smaller, so its acceleration is smaller. The smaller acceleration means that it takes more time to reach takeoff speed and travels a longer distance before lifting off the runway.

Practice Problem 4.17 Engine Thrust What is the thrust provided by the airplane’s engines in Example 4.17?

Example 4.18 A Pulley, an Incline, and Two Blocks A block of mass m1 = 2.60 kg rests on an incline that is angled at 30.0° above the horizontal (Fig. 4.50). An ideal cord is connected from block 1 over an ideal, frictionless pulley to another block of mass m2 = 2.20 kg that is hanging 2.00 m above the ground. The coefficient of kinetic friction between the incline and block 1 is 0.180. The blocks are initially at rest. (a) How long does it take for block 2 to reach the ground? (b) Sketch a motion diagram for block 2 with a time interval of 0.5 s.

T +y

a1

m1g sin 30.0°

m2g

+x

Figure 4.52 FBD for block 2 with the downward direction chosen as +x.

Figure 4.50 m2

30.0°

a2

fk m1g cos 30.0°

FBD for block 1.

+y

2

1

Figure 4.51 m1

+x

T

N

Block on an incline connected to a hanging block by a cord passing over a pulley.

Strategy The problem says that the blocks start from rest and that block 2 hits the floor, so block 2’s acceleration is downward and block 1’s is up the incline. For block 1, we choose axes parallel and perpendicular to the incline so that its acceleration has only one nonzero component. The magnitudes of the accelerations of the two blocks are equal since they are connected by an ideal cord that does not stretch. Since the cord and pulley are ideal, the tension is the same at the two ends. Solution (a) We start by drawing separate FBDs for each block (Figs. 4.51 and 4.52). Since block 1 slides up the incline, the frictional force f⃗k acts down the incline to oppose the sliding. The gravitational force on block 1 is resolved into two components, one along the incline and one perpendicular to the incline. Using the FBDs, we write Newton’s second law in component form for each block. Block 1 has no acceleration

component perpendicular to the incline. It does not sink into the incline or rise above it; it can only slide along the incline. Thus, the net force on block 1 in the direction perpendicular to the incline—the direction we have chosen as the y-axis for block 1—is zero.

∑Fy = N − m 1g cos q = 0 or N = m1g cos q Here q = 30.0°. Along the incline, in the x-direction for block 1, the acceleration is nonzero:

∑Fx = T − m 1g sin q − f k = m 1ax The kinetic frictional force is related to the normal force: f k = m kN = mkm 1g cos q By substitution, T − m1g sin q − m km1g cos q = m1ax

(1)

continued on next page

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Example 4.18 continued

For block 2, we choose an x-axis pointing downward. Doing so simplifies the solution, since then the two blocks have the same ax. Applying Newton’s second law,

∑Fx = m 2g − T = m 2ax

(2)

The tension in the cord T and the x-component of acceleration ax are both unknown in Eqs. (1) and (2). We solve for T in Eq. (2) and substitute into Eq. (1): T = m 2g − m 2ax = m 2(g − ax) m 2(g − ax) − m 1g sin q − m km 1g cos q = m 1ax Rearranging and solving for ax yields m 2 − m 1(sin q + mk cos q ) g ax = _____________________ m +m 1

(3)

2

Substituting the known and given values, 2.20 kg − 2.60 kg × (0.50 + 0.180 × 0.866) ax = ___________________________________ × 9.80 m/s2 2.60 kg + 2.20 kg = 1.01 m/s2 Block 2 has a distance of 2.00 m to travel starting from rest with a constant downward acceleration of 1.01 m/s2. From Eq. (2-12) with vix = 0, Δ x = _12 ax(Δt)2

dimensional analysis can easily be used 0 2 0 to check for errors. In Eq. (3), the quan2 0.5 s tity in parentheses is dimensionless—the values of trigonometric functions are pure numbers as are coefficients of friction. 0.5 2 1.0 s Therefore, the numerator is the sum of two quantities with dimensions of force, the denominator is the sum of two masses, and force divided by mass gives an 1.0 acceleration. 2 1.5 s What if the problem did not tell us the directions of the blocks’ accelerations? We could figure it out by comparing the force with which gravity pulls down on 1.5 block 2 (m2g) with the component of the gravitational force pulling block 1 down the incline (m1g sin q ). Whichever is 2.0 s greater “wins the tug-of-war,” assuming 2.0 2 that static friction doesn’t prevent the x blocks from starting to slide. Once we (m) know the direction of block 1’s acceleration, we can determine the direction of Figure 4.53 the kinetic frictional force. If block 1 is Motion diagram not initially at rest, the kinetic frictional for block 2. t (s) x (m) force opposes the direction of sliding, 0 0 even though that may be opposite to the 0.5 0.125 direction of the acceleration. 1.0 0.50 1.5 2.0

The time to travel that distance is

√

____

√

__________

Practice Problem 4.18 and an Incline

2 × 2.00 m = 2.0 s 2 Δx = _________ Δt = ____ ax 1.01 m/s2 (b) Figure 4.53 shows the motion diagram for block 2. Choosing xi = 0 and ti = 0, the position as a function of time is x = _12 axt2. Discussion One advantage to solving for ax algebraically in Eq. (3) before substituting numerical values is that

1.125 2.0

More Fun with a Pulley

Suppose that m1 = 3.8 kg and m2 = 1.2 kg and the coefficient of kinetic friction is 0.18. The blocks are released from rest and block 1 starts to slide. (a) Does block 1 slide up or down the incline? (b) In which direction does the kinetic frictional force act? (c) Find the acceleration of block 1.

CHECKPOINT 4.8 Is it ever useful to choose the x- and y-axes so the x-axis is not horizontal? If yes, give an example.

4.9

REFERENCE FRAMES

Imagine a train moving at constant velocity with respect to the ground (Fig. 4.54). Suppose Tim does some experiments using the train’s reference frame for his measurements. Greg does similar experiments using the reference frame of the ground. Tim and Greg disagree about the numerical value of an object’s velocity, but since their velocity

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Figure 4.54 Greg’s frame of Tim

Greg

vTG

reference is that of the ground; Tim’s is that of the train, which moves at constant velocity v⃗TG with respect to the ground.

measurements differ by a constant, they will always agree about changes in velocity and about accelerations. Both observers can use Newton’s second law to relate the net force to the acceleration. The basic laws of physics, such as Newton’s laws of motion, work equally well in any two reference frames if they move with a constant relative velocity. Newton’s First Law Defines an Inertial Reference Frame You might wonder why ⃗ = 0? we need Newton’s first law—isn’t it just a special case of the second law when ∑F No, the first law defines what kind of reference frame we can use when applying the second law. For the second law to be valid, we must use an inertial reference frame—a reference frame in which the law of inertia holds—to observe the motion of objects. The law of inertia is a postulate of classical mechanics—an assumption that is used as a starting point. It is not something we can prove experimentally. Is a reference frame attached to Earth’s surface truly inertial? No, but it is close enough in many circumstances. When analyzing the motion of a soccer ball, the fact that Earth rotates about its axis does not have much effect. But if we want to analyze the motion of a meteor falling from a great distance toward Earth, Earth’s rotation must be considered. We will take a closer look at the effect of Earth’s rotation in Chapter 5.

4.10

APPARENT WEIGHT

Imagine being in an elevator when the cable snaps. Assume that some safety mechanism brings you to rest after you have been in free fall for a while. While you are in free fall, you seem to be “weightless,” but your weight has not changed; the Earth still pulls downward with the same gravitational force. In free fall, gravity gives the elevator and everything in it a downward acceleration equal to g. ⃗ If you jump up from the elevator floor, you seem to “float” up to the ceiling of the elevator. Your weight hasn’t changed, but your apparent weight is zero while you are in free fall. Similarly, astronauts in a space station in orbit around the Earth are in free fall (their acceleration is equal to the local value of g⃗ ). Earth exerts a gravitational force on them so they are not weightless; their apparent weight is zero. Imagine an object that appears to be resting on a bathroom scale. The scale measures the object’s apparent weight W′, which is equal to the true weight only if the object and the scale have zero acceleration. Newton’s second law requires that ⃗ = N ⃗ + m⃗ g = ma⃗ ∑F

⃗ is the normal force of the scale pushing up. The apparent weight W′ is the where N ⃗ reading of the scale—that is, the magnitude of N: ⃗ = N W′ = N In Fig. 4.55a, the acceleration of the elevator is upward. The normal force must be larger than the weight for the net force to be upward (Fig. 4.55b). Writing the forces in component form where the +y-direction is upward

∑Fy = N − mg = may or N = mg + may

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Figure 4.55 (a) Apparent weight in an elevator with acceleration upward. (b) FBD for the passenger. (c) The normal force must be greater than the weight to have an upward net force.

y N

Vector sum of forces

a N

mg

mg

ΣF

Free-body diagram

Σ F = N + mg = ma Σ F is upward so

(b)

(c)

(a)

N > mg

Figure 4.56 (a) Apparent weight in an elevator with acceleration downward. (b) FBD for the passenger. (c) The normal force must be less than the weight to have a downward net force.

y N

Vector sum of forces N

a mg

(a)

mg

ΣF

Free-body diagram

Σ F = N + mg = ma Σ F is downward so

(b)

(c)

N < mg

Therefore, W′ = N = m(g + ay)

(4-16)

Since the elevator’s acceleration is upward, ay > 0; the apparent weight is greater than the true weight (Fig. 4.55c). In Fig. 4.56a, the acceleration is downward. Then the net force must also point downward. The normal force is still upward, but it must be smaller than the weight in order to produce a downward net force (Fig. 4.56b). It is still true that W′ = m(g + ay), but now the acceleration is downward (ay < 0). The apparent weight is less than the true weight (Fig. 4.56c). If the elevator is in free fall, then ay = −g and the apparent weight of the unfortunate passenger is zero.

Example 4.19 Apparent Weight in an Elevator A passenger weighing 598 N rides in an elevator. What is the apparent weight of the passenger in each of the following situations? In each case, the magnitude of the elevator’s acceleration is 0.500 m/s2. (a) The passenger is on the first

floor and has pushed the button for the fifteenth floor; the elevator is beginning to move upward. (b) The elevator is slowing down as it nears the fifteenth floor.

continued on next page

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APPARENT WEIGHT

Example 4.19 continued

Strategy In each case, we sketch the FBD for the passenger. The apparent weight is equal to the magnitude of the normal force exerted by the floor on the passenger. The only other force acting is gravity. Newton’s second law lets us find the normal force from the weight and the acceleration. Known: W = 598 N; magnitude of the acceleration is a = 0.500 m/s2. To find: W′. Solution (a) Let the +y-axis be upward. When the elevator starts up from the first floor it has acceleration in the upward direction as its speed increases. Since the elevator’s acceleration is upward, ay > 0 (as in Fig. 4.55). We expect the apparent weight W′ = N to be greater than the true weight—the floor must push up with a force greater than W to cause an upward acceleration. Figure 4.57 is the FBD. Newton’s second law says

Since W = mg, we can substitute m = W/g.

(

ay W a = W 1 + __ W′ = N = W + may = W + __ g y g 2

)

(

ay N = W 1 + __ g

(

)

)

−0.500 m/s = 567 N = 598 N × 1 + __________ 9.80 m/s2 2

Discussion The apparent weight is greater when the direction of the elevator’s acceleration is upward. That can happen in two cases: either the elevator is moving up with increasing speed, or it is moving down with decreasing speed.

Practice Problem 4.19 Elevator Descending

∑Fy = N − W = may

(

(b) When the elevator approaches the fifteenth floor, it is slowing down while still moving upward; its acceleration is downward (ay < 0) as in Fig. 4.56. The apparent weight is less than the true weight. Figure 4.58 is the FBD. Again, ∑ Fy = N − W = may, but this time ay = −0.500 m/s2.

)

0.500 m/s = 629 N = 598 N × 1 + _________ 9.80 m/s2

What is the apparent weight of a passenger of mass 42.0 kg traveling in an elevator in each of the following situations? In each case, the magnitude of the elevator’s acceleration is 0.460 m/s2. (a) The passenger is on the fifteenth floor and has pushed the button for the first floor; the elevator is beginning to move downward. (b) The elevator is slowing down as it nears the first floor.

N a

N

Figure 4.57 W

FBD for the passenger in an elevator with upward acceleration.

a W

Figure 4.58 FBD for the passenger in an elevator with downward acceleration.

PHYSICS AT HOME Take a bathroom scale to an elevator. Stand on the scale inside the elevator and push a button for a higher floor. When the elevator’s acceleration is upward, you can feel the increase in your apparent weight and can see the increase by the reading on the scale. When the elevator slows down to stop, the elevator’s acceleration is downward and your apparent weight is less than your true weight. What is happening in your body while the elevator accelerates? The inertia principle means that your blood and internal organs cannot have the same acceleration as the elevator until the correct net force acts on them. Blood tends to collect in the lower extremities during acceleration upward and in the upper body during acceleration downward until the forces exerted on the blood by the body readjust to give the blood the same acceleration as the elevator. Likewise, the internal organs shift position within the body cavity, resulting in a funny feeling in the gut as the elevator starts and stops. To avoid this problem, high-speed express elevators in skyscrapers keep the acceleration relatively small, but maintain that acceleration long enough to reach high speeds. That way, the elevator can travel quickly to the upper floors without making the passengers feel too uncomfortable.

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CHECKPOINT 4.10 You are standing on a bathroom scale in an elevator that is moving downward. Nearing your stop, the elevator’s speed is decreasing. Is the scale reading greater or less than your weight?

4.11

AIR RESISTANCE

So far we have ignored the effect of air resistance on falling objects and projectiles. A skydiver relies on a parachute to provide a large force of air resistance (also called drag). Even with the parachute closed, drag is not negligible when the skydiver is falling rapidly. The drag force is similar to friction between two solid surfaces in that the direction of the force opposes the motion of the object through the air. However, in contrast to the force of friction, the magnitude of the drag force is strongly dependent on the speed of the object. In many cases, air drag is proportional to the square of the speed. Drag also depends on the size and shape of the object. Since the drag force increases as the speed increases, a falling object approaches an equilibrium situation in which the drag force is equal in magnitude to the weight but opposite in direction. The velocity at which this equilibrium occurs is called the object’s terminal velocity. (See text website for a more detailed treatment of drag.)

PHYSICS AT HOME Drop a basket-style paper coffee filter (or a cupcake paper) and a penny simultaneously from as close to the ceiling as you can safely do so. Air resistance on the penny is negligible unless it is dropped from a very high balcony. At the other extreme, the effect of air resistance on the coffee filter is very noticeable; it reaches its terminal speed almost immediately. Stack several (two to four) coffee filters together and drop them simultaneously with a single coffee filter. Why is the terminal speed higher for the stack? Crumple a coffee filter into a ball and drop it simultaneously with the penny. Air resistance on the coffee filter is now reduced, but still noticeable.

4.12

FUNDAMENTAL FORCES

One of the main goals of physics has been to understand the immense variety of forces in the universe in terms of the fewest number of fundamental laws. Physics has made great progress in this quest for unification; today all forces are understood in terms of just four fundamental interactions (Fig. 4.59). At the high temperatures present in the early universe, two of these interactions—the electromagnetic and weak forces—are now understood as the effects of a single electroweak interaction. The ultimate goal is to describe all forces in terms of a single interaction. Gravity You may be surprised to learn that gravity is by far the weakest of the fundamental forces. Any two objects exert gravitational forces on one another, but the force is tiny unless at least one of the masses is large. We tend to notice the relatively large gravitational forces exerted by planets and stars, but not the feeble gravitational

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Gravitation

Electromagnetism

Electricity

FUNDAMENTAL FORCES

Magnetism

Weak force Strong force

Gravity of the Sun and stars

Earth’s gravity

Figure 4.59 All forces result from just four fundamental forces: gravity, electromagnetism, and the weak and strong forces.

forces exerted by smaller objects, such as the gravitational force this book exerts on your body. Gravity has an unlimited range. The force gets weaker as the distance between two objects increases, but it never drops exactly to zero, no matter how far apart the objects get. Newton’s law of gravity is an early example of unification. Before Newton, people did not understand that the same kind of force that makes an apple fall from a tree also keeps the planets in their orbits around the Sun. A single law—Newton’s law of universal gravitation—describes both. Electromagnetism The electromagnetic force is unlimited in range, like gravity. It acts on particles with electric charge. The electric and magnetic forces were unified into a single theoretical framework in the nineteenth century. We study electromagnetic forces in detail in Part 3 of this book. Electromagnetism is the fundamental interaction that binds electrons to nuclei to form atoms and binds atoms together in molecules and solids. It is responsible for the properties of solids, liquids, and gases and forms the basis of the sciences of chemistry and biology. It is the fundamental interaction behind all macroscopic contact forces such as the frictional and normal forces between surfaces and forces exerted by springs, muscles, and the wind. The electromagnetic force is much stronger than gravity. For example, the electrical repulsion of two electrons at rest is about 1043 times as strong as the gravitational attraction between them. Macroscopic objects have a nearly perfect balance of positive and negative electric charge, resulting in a nearly perfect balance of attractive and repulsive electromagnetic forces between the objects. Therefore, despite the fundamental strength of the electromagnetic forces, the net electromagnetic force between two macroscopic objects is often negligibly small except when atoms on the two surfaces come very close to each other—what we think of as in contact. On a microscopic level, there is no fundamental difference between contact forces and other electromagnetic forces.

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CHAPTER 4 Force and Newton’s Laws of Motion

The Strong Force The strong force holds protons and neutrons together in the atomic nucleus. The same force binds quarks (a family of elementary particles) in combinations so they can form protons and neutrons and many more exotic subatomic particles. The strong force is the strongest of the four fundamental forces—hence its name—but its range is short: its effect is negligible at distances much larger than the size of an atomic nucleus (about 10−15 m). The Weak Force The range of the weak force is even shorter than that of the strong force (about 10−17 m). It is manifest in many radioactive decay processes.

Master the Concepts • A force is a push or a pull. Gravity and electromagnetic forces have unlimited range. All other forces exerted on macroscopic objects involve contact. Force is a vector quantity. • The SI unit of force is the newton: 1 N = 1 kg·m/s2. • The net force on a system is the vector sum of all the forces acting on it: ⃗ net = ∑ F ⃗ = F ⃗ 1 + F ⃗ 2 + ⋅ ⋅ ⋅ + F ⃗ n F

(4-2)

Since all the internal forces form interaction pairs, we need only sum the external forces. • Newton’s first law of motion: If zero net force acts on an object, then the object’s velocity does not change. Velocity is a vector whose magnitude is the speed at which the object moves and whose direction is the direction of motion. • Newton’s second law of motion relates the net force acting on an object to the object’s acceleration and its mass: ⃗ ∑F ⃗ a⃗ = ____ (4-4) m or ∑F = ma⃗ The acceleration is always in the same direction as the net force. Many problems involving Newton’s second law—whether equilibrium or nonequilibrium—can be solved by treating the x- and y-components of the forces and the acceleration separately:

∑Fx = max and ∑Fy = may ΣF a

(4-5) ΣF

due to some other object, but no forces acting on other objects. Up West

East

L Down D

T

W

• The magnitude of the gravitational force between two objects is Gm 1m 2 F = _______ (4-7) r2 where r is the distance between their centers. Each object is pulled toward the other’s center. • The weight of an object is the magnitude of the gravitational force acting on it. An object’s weight is proportional to its mass: W = mg [Eq. (4-10)], where g is the gravitational field strength. Near Earth’s surface, g ≈ 9.80 N/kg. • The normal force is a contact force perpendicular to the contact surfaces that pushes each object away from the other. N

a

• Newton’s third law of motion: In an interaction between two objects, each object exerts a force on the other. These two forces are equal in magnitude and opposite in direction. • A free-body diagram (FBD) includes vector arrows representing every force acting on the chosen object

W

• Friction is a contact force parallel to the contact surfaces. In a simplified model, the kinetic frictional force and the maximum static frictional force are proportional continued on next page

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Master the Concepts continued

to the normal force acting between the same contact surfaces. fs ≤ msN

(4-14)

f k = m kN

(4-15)

The static frictional force acts in the direction that tends to keep the surfaces from beginning to slide. The direction of the kinetic frictional force is in the direction that would tend to make the sliding stop. • An ideal cord pulls in the direction of the cord with forces of equal magnitude on the objects attached to its ends as long as no external force tangent to the cord is exerted on it anywhere between the ends. The tension of an ideal cord that runs through an ideal pulley is the same on both sides of the pulley.

Conceptual Questions 1. Explain the need for automobile seat belts in terms of Newton’s first law. 2. An American visitor to Finland is surprised to see heavy metal frames outside of all the apartment buildings. On Saturday morning the purpose of the frames becomes evident when several apartment dwellers appear, carrying rugs and carpet beaters to each frame. What role does the principle of inertia play in the rug beating process? Do you see a similarity to the role the principle of inertia plays when you throw a baseball? 3. You are lying on the beach after a dip in the ocean where the waves were buffeting you around. Is it true that there are now no forces acting on you? Explain. 4. A dog goes swimming at the beach and then shakes himself all over to get dry. What principle of physics aids in the drying process? Explain. 5. In an attempt to tighten the loosened steel head of a hammer, a carpenter holds the hammer vertically, raises it up, and then brings it down rapidly, hitting the bottom end of the wood handle on a twoby-four board. Explain how this tightens the head back onto the handle. 6. When a car begins to move forward, what force makes it do so? Remember that it has to be an external force; the internal forces

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• An object that is accelerating has an apparent weight that differs from its true weight. The apparent weight is equal to the normal force exerted by a supporting surface with the same acceleration. A helpful trick is to think of the apparent weight as the reading of a bathroom scale that supports the object. • The drag force exerted on an object moving through air opposes the motion of the object but, unlike kinetic friction, is strongly dependent on the object’s speed. When an object falls at its terminal velocity, the drag force is equal and opposite to the gravitational force, so the acceleration is zero. • At the fundamental level, there are four interactions: gravity, the strong and weak interactions, and the electromagnetic interaction. Contact forces are large-scale manifestations of many microscopic electromagnetic interactions.

all add to zero. How does the engine facilitate the propelling force? 7. Two cars are headed toward each other in opposite directions along a narrow country road. The cars collide head-on, crumpling up the hoods of both. Describe what happens to the car bodies in terms of the principle of inertia. Does the rear end of the car stop at the same time as the front end? 8. Can a body in free fall be in equilibrium? Explain. 9. (a) What assumptions do you make when you call the reading of a bathroom scale your “weight”? What does the scale really tell you? (b) Under what circumstances might the reading of the scale not be equal to your weight? 10. A freight train consists of an engine and several identical cars on level ground. Determine whether each of these statements is correct or incorrect and explain why. (a) If the train is moving at constant speed, the engine must be pulling with a force greater than the train’s weight. (b) If the train is moving at constant speed, the engine’s pull on the first car must exceed that car’s backward pull on the engine. (c) If the train is coasting, its inertia makes it slow down and eventually stop. 11. (a) Does a man weigh more at the North Pole or at the equator? (b) Does he weigh more at the top of Mt. Everest or at the base of the mountain? 12. What is the acceleration of an object thrown straight up into the air at the highest point of its motion? Does the answer depend on whether air resistance is negligible or not? Explain. 13. If a wagon starts at rest and pulls back on you with a force equal to the force you pull on it, as required by Newton’s third law, how is it possible for you to make the wagon start to move? Explain.

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14. You are standing on a bathroom scale in an elevator. In which of these situations must the scale read the same as when the elevator is at rest? Explain. (a) Moving up at constant speed. (b) Moving up with increasing speed. (c) In free fall (after the elevator cable has snapped). 15. A heavy ball hangs from a string attached to a sturdy wooden frame. A second string is attached to a hook on the bottom of the lead ball. You pull slowly and steadily on the lower string. Which string do you think will break first? Explain. 16. An SUV collides with a Mini Cooper convertible. Is the force exerted on the Mini by the SUV greater than, equal to, or less than the force exerted on the SUV by the Mini? Explain. 17. You are standing on one end of a light wooden raft that has floated 3 m away from the pier. If the raft is 6 m long by 2.5 m wide and you are standing on the raft end nearest to the pier, can you propel the raft back toward the pier where a friend is standing with a pole and hook trying to reach you? You have no oars. Make suggestions of what to do without getting yourself wet.

Multiple-Choice Questions

3m

6m

18. What does it mean when we refer to a cord as an “ideal cord” and a pulley as an “ideal pulley”? 19. If a feather and a lead brick are dropped simultaneously from the top of a ladder, the lead brick hits the ground first. What would happen if the experiment is repeated on the surface of the Moon? 20. A baseball is tossed straight up. Taking into consideration the force of air resistance, is the magnitude of the baseball’s acceleration zero, less than g, equal to g, or greater than g on the way up? At the top of the flight? On the way down? Explain. [Hint: The force of air resistance is directed opposite to the velocity. Assume in this case that its magnitude is less than the weight.] 21. Why might an elevator cable break during acceleration when lifting a lighter load than it normally supports at rest or at constant velocity? 22. If air resistance is ignored, what force(s) act on an object in free fall?

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23. The net force acting on an object is constant. Under what circumstances does the object move along a straight line? Under what circumstances does the object move along a curved path? 24. Pulleys and inclined planes are examples of simple machines. Explain what these machines do in Examples 4.10, 4.12, and 4.16 to make a task easier to perform. 25. For a problem about a crate sliding along an inclined plane, is it possible to choose the x-axis so that it is parallel to the incline? 26. A bird sits on a stretched clothesline, causing it to sag slightly. Is the tension in the line greatest where the bird sits, greater at either end of the line where it is attached to poles, or the same everywhere along the line? Treat the line as an ideal cord with negligible weight. 27. You decide to test your physics knowledge while going over a waterfall in a barrel. You take a baseball into the barrel with you and as you are falling vertically downward, you let go of the ball. What do you expect to see for the motion of the ball relative to the barrel? Will the ball fall faster than you and move toward the bottom of the barrel? Will it move slower than you and approach the top of the barrel, or will it hover apparently motionless within the falling barrel? Explain. [Warning: Do not try this.]

1. Interaction partners (a) are equal in magnitude and opposite in direction and act on the same object. (b) are equal in magnitude and opposite in direction and act on different objects. (c) appear in an FBD for a given object. (d) always involve gravitational force as one partner. (e) act in the same direction on the same object. 2. Within a given system, the internal forces (a) are always balanced by the external forces. (b) all add to zero. (c) are determined only by subtracting the external forces from the net force on the system. (d) determine the motion of the system. (e) can never add to zero. 3. A friction force is (a) a contact force that acts parallel to the contact surfaces. (b) a contact force that acts perpendicular to the contact surfaces. (c) a scalar quantity since it can act in any direction along a surface. (d) always proportional to the weight of an object. (e) always equal to the normal force between the objects.

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4. When a force is called a “normal” force, it is (a) the usual force expected given the arrangement of a system. (b) a force that is perpendicular to the surface of the Earth at any given location. (c) a force that is always vertical. (d) a contact force perpendicular to the contact surfaces between two solid objects. (e) the net force acting on a system. 5. Your car won’t start, so you are pushing it. You apply a horizontal force of 300 N to the car, but it doesn’t budge. What force is the interaction partner of the 300 N force you exert? (a) the frictional force exerted on the car by the road (b) the force exerted on you by the car (c) the frictional force exerted on you by the road (d) the normal force on you by the road (e) the normal force on the car by the road 6. Which of these is not a long-range force? (a) the force that makes raindrops fall to the ground (b) the force that makes a compass point north (c) the force that a person exerts on a chair while sitting (d) the force that keeps the Moon in its orbital path around the Earth 7. When an object is in translational equilibrium, which of these statements is not true? (a) The vector sum of the forces acting on the object is zero. (b) The object must be stationary. (c) The object has a constant velocity. (d) The speed of the object is constant. 8. To make an object start moving on a surface with friction requires (a) less force than to keep it moving on the surface. (b) the same force as to keep it moving on the surface. (c) more force than to keep it moving on the surface. (d) a force equal to the weight of the object. 9. A thin string that can support a weight of 35.0 N, but breaks under any larger weight, is attached to the ceiling of an elevator. How large a mass can be attached to the string if the initial acceleration as the elevator starts to ascend is 3.20 m/s2? (a) 3.57 kg (b) 2.69 kg (c) 4.26 kg (d) 2.96 kg (e) 5.30 kg 10. A woman stands on a bathroom scale in an elevator that is not moving. The scale reads 500 N. The elevator then moves downward at a constant velocity of 4.5 m/s. What

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does the scale read while the elevator descends with constant velocity? (a) 100 N (b) 250 N (c) 450 N (d) 500 N (e) 750 N 11. A 70.0-kg man stands on a bathroom scale in an elevator. What does the scale read if the elevator is slowing down at a rate of 3.00 m/s2 while descending? (a) 70 kg (b) 476 N (c) 686 N (d) 700 N (e) 896 N 12. A space probe leaves the solar system to explore interstellar space. Once it is far from any stars, when must it fire its rocket engines? (a) All the time, in order to keep moving. (b) Only when it wants to speed up. (c) When it wants to speed up or slow down. (d) Only when it wants to turn. (e) When it wants to speed up, slow down, or turn. 13. A small plane climbs with a constant velocity of 250 m/s at an angle of 28° with respect to the horizontal. Which statement is true concerning the magnitude of the net force on the plane? (a) It is equal to zero. (b) It is equal to the weight of the plane. (c) It is equal to the magnitude of the force of air resistance. (d) It is less than the weight of the plane but greater than zero. (e) It is equal to the component of the weight of the plane in the direction of motion. 14. Two blocks are connected by a light string passing over a pulley (see the figure and tutorial: pulley). The block with mass m1 slides on the frictionless horizontal surface, while the block with mass m2 hangs vertically. (m1 > m2.) The tension in the string is (a) zero. (b) less than m2g. (c) equal to m2g. (d) greater than m2g, but less than m1g. (e) equal to m1g. (f) greater than m1g.

m1

m2

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Problems Combination conceptual/quantitative problem Biological or medical application ✦ Challenging problem Blue # Detailed solution in the Student Solutions Manual Problems paired by concept 1 2 Text website interactive or tutorial

4.1 Force 1. A person is standing on a bathroom scale. Which of the following is not a force exerted on the scale: a contact force due to the floor, a contact force due to the person’s feet, the weight of the person, the weight of the scale? 2. A sack of flour has a weight of 19.8 N. What is its weight in pounds? 3. An astronaut weighs 175 lb. What is his weight in newtons? 4. Does the concept of a contact force apply to both a macroscopic scale and an atomic scale? Explain. 5. A force of 20 N is directed at an angle of 60° above the x-axis. A second force of 20 N is directed at an angle of 60° below the x-axis. What is the vector sum of these two forces? 6. Juan is helping his mother rearrange the living room furniture. Juan pushes on the armchair with a force of 30 N directed at an angle of 15° above a horizontal line while his mother pushes with a force of 40 N directed at an angle of 20° below the same horizontal. What is the vector sum of these two forces? ⃗ + B ⃗ +C ⃗ 7. In the drawing, what is the vector sum of forces A if each grid square is 2 N on a side?

9. Two of Robin Hood’s men are pulling a sledge loaded with some gold along a path that runs due north to their hideout. One man pulls his rope with a force of 62 N at an angle of 12° east of north and the other pulls with the same force at an angle of 12° west of north. Assume the ropes are parallel to the ground. What is the sum of these two forces on the sledge? 10. A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on either side of the canal. Each horse pulls with a force of 560 N at an angle of 15° with the centerline of the canal. Find the sum of the two forces exerted by the horses on the barge. 11. On her way to visit Grandmother, Red Riding Hood sat down to rest and placed her 1.2-kg basket of goodies beside her. A wolf came along, spotted the basket, and began to pull on the handle with a force of 6.4 N at an angle of 25° with respect to vertical. Red was not going to let go easily, so she pulled on the handle with a force of 12 N. If the net force on the basket is straight up, at what angle was Red Riding Hood pulling? 12. A parked automobile slips out of gear, rolls unattended down a slight incline, and then along a level road until it hits a stone wall. Draw an FBD to show the forces acting on the car while it is in contact with the wall. 13. Two objects, A and B, are acted on by the forces shown in the FBDs. Is the magnitude of the net force acting on object B greater than, less than, or equal to the magnitude of the net force acting on object A? Make a scale drawing on graph paper and explain the result. A 45°

45°

4N

4N B

45° 2N 2N

A

45° 2N 2N

N C

W

E

B

14. Find the magnitude and direction of the net force on the object in each of the FBDs for this problem.

S 10 N

⃗ + E ⃗ + F ⃗ 8. In the drawing, what is the vector sum of forces D if each grid square is 2 N on a side?

40 N (a)

18 N

F D W E

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10 N

N

10 N

10 N

10 N

E S

18 N

18 N

(b)

(c)

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PROBLEMS

15. A truck driving on a level highway is acted on by the following forces: a downward gravitational force of 52 kN (kilonewtons); an upward contact force due to the road of 52 kN; another contact force due to the road of 7 kN, directed east; and a drag force due to air resistance of 5 kN, directed west. What is the net force acting on the truck?

4.2 Inertia and Equilibrium: Newton’s First Law of Motion; 4.3 Net Force, Mass, and Acceleration: Newton’s Second Law of Motion 16. A sailboat, tied to a mooring with a line, weighs 820 N. The mooring line pulls horizontally toward the west on the sailboat with a force of 110 N. The sails are stowed away and the wind blows from the west. The boat is moored on a still lake—no water currents push on it. Draw an FBD for the sailboat and indicate the magnitude of each force. 17. A hummingbird is hovering motionless beside a flower. The blur of its wings shows that they are rapidly beating up and down. If the air pushes upward on the bird with a force of 0.30 N, what is the weight of the hummingbird? 18. You are pulling a suitcase through the airport at a constant speed. The handle of the suitcase makes an angle of 60° with respect to the horizontal direction. If you pull with a force of 5.0 N parallel to the handle, what is the contact force due to the floor acting on the suitcase? 19. A model sailboat is slowly sailing west across a pond at 0.33 m/s. A gust of wind blowing at 28° south of west gives the sailboat a constant acceleration of magnitude 0.30 m/s2 during a time interval of 2.0 s. (a) If the net force on the sailboat during the 2.0-s interval has magnitude 0.375 N, what is the sailboat’s mass? (b) What is the new velocity of the boat after the 2.0-s gust of wind? 20. A man is lazily floating on an air mattress in a swimming pool. If the weight of the man and air mattress together is 806 N, what is the upward force of the water acting on the mattress? 21. A bag of potatoes with weight 39.2 N is suspended from a string that exerts a force of 46.8 N. If the bag’s acceleration is upward at 1.90 m/s2, what is the mass of the potatoes? 22. A 2010-kg elevator moves with an upward acceleration of 1.50 m/s2. What is the force exerted by the cable on the elevator? 23. While an elevator of mass 2530 kg moves upward, the force exerted by the cable is 33.6 kN. (a) What is the acceleration of the elevator? (b) If at some point in the motion the velocity of the elevator is 1.20 m/s upward, what is the elevator’s velocity 4.00 s later?

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24. The vertical component of the acceleration of a sailplane is zero when the air pushes up against its wings with a force of 3.0 kN. (a) Assuming that the only forces on the sailplane are that due to gravity and that due to the air pushing against its wings, what is the gravitational force on the Earth due to the sailplane? (b) If the wing stalls and the upward force decreases to 2.0 kN, what is the acceleration of the sailplane? 25. A man lifts a 2.0-kg stone vertically with his hand at a constant upward velocity of 1.5 m/s. What is the magnitude of the total force of the man’s hand on the stone? 26. A man lifts a 2.0-kg stone vertically with his hand at a constant upward acceleration of 1.5 m/s2. What is the magnitude of the total force of the man’s hand on the stone? 27. What is the acceleration of an automobile of mass 1.40 × 103 kg when it is subjected to a forward force of 3.36 × 103 N? 28. A large wooden crate is pushed along a smooth, frictionless surface by a force of 100 N. The acceleration of the crate is measured to be 2.5 m/s2. What is the mass of the crate? 29. The forces on a small airplane (mass 1160 kg) in horizontal flight heading eastward are as follows: gravity = 16.000 kN downward, lift = 16.000 kN upward, thrust = 1.800 kN eastward, and drag = 1.400 kN westward. At t = 0, the plane’s speed is 60.0 m/s. If the forces remain constant, how far does the plane travel in the next 60.0 s? 30. While an elevator of mass 832 kg moves downward, the tension in the supporting cable is a constant 7730 N. Between t = 0 and t = 4.00 s, the elevator’s displacement is 5.00 m downward. What is the elevator’s speed at t = 4.00 s?

4.4 Interaction Pairs: Newton’s Third Law of Motion 31. A hanging potted plant is suspended by a cord from a hook in the ceiling. Draw an FBD for each of these: (a) the system consisting of plant, soil, and pot; (b) the cord; (c) the hook; (d) the system consisting of plant, soil, pot, cord, and hook. Label each force arrow using ⃗ ch would represent the force subscripts (for example, F exerted on the cord by the hook). 32. A bike is hanging from a hook in a garage. Consider the following forces: (a) the force of the Earth pulling down on the bike, (b) the force of the bike pulling up on the Earth, (c) the force of the hook pulling up on the bike, and (d) the force of the hook pulling down on the ceiling. Which two forces are equal and opposite because of Newton’s third law? Which two forces are equal and opposite because of Newton’s first law?

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33. A woman who weighs 600 N sits on a chair with her feet on the floor and her arms resting on the chair’s armrests. The chair weighs 100 N. Each armrest exerts an upward force of 25 N on her arms. The seat of the chair exerts an upward force of 500 N. (a) What force does the floor exert on her feet? (b) What force does the floor exert on the chair? (c) Consider the woman and the chair to be a single system. Draw an FBD for this system that includes all of the external forces acting on it. 34. A fisherman is holding a fishing rod with a large fish suspended from the line of the rod. Identify the forces acting on the rod and their interaction partners. 35. A fish is suspended by a line from a fishing rod. Choose two forces acting on the fish and describe the interaction partner of each.

41. A man weighs 0.80 kN on Earth. What is his mass in kilograms? 42. An astronaut stands at a position on the Moon such that Earth is directly over head and releases a Moon rock that was in her hand. (a) Which way will it fall? (b) What is the gravitational force exerted by the Moon on a 1.0-kg rock resting on the Moon’s surface? (c) What is the gravitational force exerted by the Earth on the same 1.0-kg rock resting on the surface of the Moon? (d) What is the net gravitational force on the rock? 43. Alex is on stage playing his bass guitar. Estimate the magnitude of the gravitational attraction between Alex and Pat, a fan who is standing 8 m from Alex. Alex has a mass of 55 kg and Pat has a mass of 40 kg. 44. The Space Shuttle carries a satellite in its cargo bay and places it into orbit around the Earth. Find the ratio of the Earth’s gravitational force on the satellite when it is on a launch pad at the Kennedy Space Center to the gravitational force exerted when the satellite is orbiting 6.00 × 103 km above the launch pad.

Problems 34 and 35 ✦36. A skydiver, who weighs 650 N, is falling at a constant speed with his parachute open. Consider the apparatus that connects the parachute to the skydiver to be part of the parachute. The parachute pulls upward with a force of 620 N. (a) What is the force of the air resistance acting on the skydiver? (b) Identify the forces and the interaction partners of each force exerted on the skydiver. (c) Identify the forces and interaction partners of each force exerted on the parachute. 37. Margie, who weighs 543 N, is standing on a bathroom scale that weighs 45 N. (a) With what force does the scale push up on Margie? (b) What is the interaction partner of that force? (c) With what force does the Earth push up on the scale? (d) Identify the interaction partner of that force. 38. Refer to Problem 36. Consider the skydiver and parachute to be a single system. What are the external forces acting on this system?

4.5 Gravitational Forces 39. (a) Calculate your weight in newtons. (b) What is the weight in newtons of 250 g of cheese? (c) Name a common object whose weight is about 1 N. 40. A young South African girl has a mass of 40.0 kg. (a) What is her weight in newtons? (b) If she came to the United States, what would her weight be in pounds as measured on an American scale? Assume g = 9.80 N/kg in both locations.

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45. How far above the surface of the Earth does an object have to be in order for it to have the same weight as it would have on the surface of the Moon? (Ignore any effects from the Earth’s gravity for the object on the Moon’s surface or from the Moon’s gravity for the object above the Earth.) 46. Find and compare the weight of a 65-kg man on Earth with the weight of the same man on (a) Mars, where g = 3.7 N/kg; (b) Venus, where g = 8.9 N/kg; and (c) Earth’s Moon, where g = 1.6 N/kg. 47. Find the altitudes above the Earth’s surface where Earth’s gravitational field strength would be (a) two thirds and (b) one third of its value at the surface. [Hint: First find the radius for each situation; then recall that the altitude is the distance from the surface to a point above the surface. Use proportional reasoning.] 48. During a balloon ascension, wearing an oxygen mask, you measure the weight of a calibrated 5.00-kg mass and find that the value of the gravitational field strength at your location is 9.792 N/kg. How high above sea level, where the gravitational field strength was measured to be 9.803 N/kg, are you located? 49. At what altitude above the Earth’s surface would your weight be half of what it is at the Earth’s surface? 50. (a) What is the magnitude of the gravitational force that the Earth exerts on the Moon? (b) What is the magnitude of the gravitational force that the Moon exerts on the Earth? See the inside front and back covers for necessary information.

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51. What is the approximate magnitude of the gravitational force between the Earth and the Voyager spacecraft when they are separated by 15 billion km? Each spacecraft has a mass of approximately 825 kg during the mission, although the mass at launch was 2100 kg because of expendable Titan-Centaur rockets. ✦52. In free fall, we assume the acceleration to be constant. Not only is air resistance ignored, but the gravitational field strength is assumed to be constant. From what height can an object fall to the Earth’s surface such that the gravitational field strength changes less than 1.000% during the fall?

4.6 Contact Forces 53. A book rests on the surface of the table. Consider the following four forces that arise in this situation: (a) the force of the Earth pulling on the book, (b) the force of the table pushing on the book, (c) the force of the book pushing on the table, and (d) the force of the book pulling on the Earth. The book is not moving. Which pair of forces must be equal in magnitude and opposite in direction even though they are not an interaction pair? 54. A crate full of artichokes rests on a ramp that is inclined 10.0° above the horizontal. Give the direction of the normal force and the friction force acting on the crate in each of these situations. (a) The crate is at rest. (b) The crate is being pushed and is sliding up the ramp. (c) The crate is being pushed and is sliding down the ramp. 55. Mechanical advantage is the ratio of the force required without the use of a simple machine to that needed when using the simple machine. Compare the force to lift an object with that needed to slide the same object up a frictionless incline and show that the mechanical advantage of the inclined plane is the length of the incline divided by the height of the incline (d/h in Fig. 4.25). 56. An 80.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 20.0° to the ground. (a) What is the normal force exerted on the crate by the ramp? (b) The interaction partner of this normal force has what magnitude and direction? It is exerted by what object on what object? Is it a contact or a long-range force? (c) What is the static frictional force exerted on the crate by the ramp? (d) What is the minimum possible value of the coefficient of static friction? (e) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. Find the magnitude and direction of the contact force.

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57. An 85-kg skier is sliding down a ski slope at a constant velocity. The slope makes an angle of 11° above the horizontal direction. (a) Ignoring any air resistance, what is the force of kinetic friction acting on the skier? (b) What is the coefficient of kinetic friction between the skis and the snow? Problems 58–60. A crate of potatoes of mass 18.0 kg is on a ramp with angle of incline 30° to the horizontal. The coefficients of friction are ms = 0.75 and mk = 0.40. Find the frictional force (magnitude and direction) on the crate if 58. 59. 60. 61.

62.

✦63.

64.

✦65.

the crate is at rest. the crate is sliding down the ramp. the crate is sliding up the ramp. You grab a book and give it a quick push across the top of a horizontal table. After a short push, the book slides across the table, and because of friction, comes to a stop. (a) Draw an FBD of the book while you are pushing it. (b) Draw an FBD of the book after you have stopped pushing it, while it is sliding across the table. (c) Draw an FBD of the book after it has stopped sliding. (d) In which of the preceding cases is the net force on the book not equal to zero? (e) If the book has a mass of 0.50 kg and the coefficient of friction between the book and the table is 0.40, what is the net force acting on the book in part (b)? (f) If there were no friction between the table and the book, what would the free-body diagram for part (b) look like? Would the book slow down in this case? Why or why not? (a) In Example 4.10, if the movers stop pushing on the safe, can static friction hold the safe in place without having it slide back down? (b) If not, what minimum force needs to be applied to hold the safe in place? A 3.0-kg block is at rest on a horizontal floor. If you push horizontally on the 3.0-kg block with a force of 12.0 N, it just starts to move. (a) What is the coefficient of static friction? (b) A 7.0-kg block is stacked on top of the 3.0-kg block. What is the magnitude F of the force, acting horizontally on the 3.0-kg block as before, that is required to make the two blocks start to move? A horse is trotting along pulling a sleigh through the snow. To move the sleigh, of mass m, straight ahead at a constant speed, the horse must pull with a force of magnitude T. (a) What is the net force acting on the sleigh? (b) What is the coefficient of kinetic friction between the sleigh and the snow? Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out

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some irregularities on the surface. The sanding block weighs 2.0 N and Brenda pushes on it with a force of 3.0 N at an angle of 30.0° with respect to the vertical, and angled toward the wall. Draw an FBD for the sanding block as it moves straight up the wall at a constant speed. What is the coefficient of kinetic friction between the wall and the block? 66. Four separate blocks are placed side by side in a left-toright row on a table. A horizontal force, acting toward the right, is applied to the block on the far left end of the row. Draw FBDs for (a) the second block on the left and for (b) the system of four blocks. 67. A box sits on a horizontal wooden ramp. The coefficient ✦ of static friction between the box and the ramp is 0.30. You grab one end of the ramp and lift it up, keeping the other end of the ramp on the ground. What is the angle between the ramp and the horizontal direction when the tutorial: box begins to slide down the ramp? ( crate on ramp) ✦68. In a playground, two slides have different angles of incline q 1 and q 2 (q 2 > q 1). A child slides down the first at constant speed; on the second, his acceleration down the slide is a. Assume the coefficient of kinetic friction is the same for both slides. (a) Find a in terms of q 1, q 2, and g. (b) Find the numerical value of a for q 1 = 45° and q 2 = 61°.

73. Two boxes with different masses are tied together on a frictionless ramp surface. What is the tension in each of the cords? 2.0 kg

1.0 kg

25°

74. A pulley is attached to the ceiling. Spring scale A is attached to the wall and a rope runs horizontally from it and over the pulley. The same rope is then attached to spring scale B. On the other side of scale B hangs a 120-N weight. What are the readings of the two scales A and B? The weights of the scales are negligible.

Pulley

A

B

120 N

4.7 Tension 69. A sailboat is tied to a mooring with a horizontal line. The wind is from the southwest. Draw an FBD and identify all the forces acting on the sailboat. 70. A towline is attached between a car and a glider. As the car speeds due east along the runway, the towline exerts a horizontal force of 850 N on the glider. What is the magnitude and direction of the force exerted by the glider on the towline? 71. In Example 4.14, find the tension in the coupling between cars 2 and 3. ( tutorial: towing a train) 72. A 200.0-N sign is suspended from a horizontal strut of negligible weight. The force exerted on the strut by the wall is horizontal. Draw an FBD to show the forces acting on the strut. Find the tension T in the diagonal cable supporting the strut.

75. Spring scale A is attached to the floor and a rope runs vertically upward, loops over the pulley, and runs down on the other side to a 120-N weight. Scale B is attached to the ceiling and the pulley is hung below it. What are the readings of the two spring scales, A and B? Neglect the weights of the pulley and scales.

B

Pulley

A T

120 N

30.0°

76. Two springs are connected in series so that spring scale A hangs from a hook on the ceiling and a second spring scale, B, hangs from the hook at the bottom of scale A.

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Apples weighing 120 N hang from the hook at the bottom of scale B. What are the readings on the upper scale A and the lower scale B? Ignore the weights of the scales.

the tension in each wire. ( tutorial: hanging picture) ✦80. A crow perches on a clothesline midway between two poles. Each end of the rope makes an angle of q below the horizontal where it connects to the pole. If the weight of the crow is W, what is the tension in the rope? Ignore the weight of the rope.

A

q

q B

77. A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is attached to the wall. The rope fastened to the wall makes a right angle with the wall. Ignore the masses of the rope and the pulley. Find (a) the tension in the rope from which the pulley hangs and (b) the angle q that the rope makes with the ceiling.

✦ 81. The drawing shows an elastic cord attached to two back teeth and stretched across a front tooth. The pur⃗ to the pose of this arrangement is to apply a force F front tooth. (The figure has been simplified by running the cord straight from the front tooth to the back teeth.) If the tension in the cord is 1.2 N, what are the magni⃗ applied to the front tude and direction of the force F tooth?

33°

33°

q 90°

zz

78. A 2.0-kg ball tied to a string fixed to the ceiling is pulled ⃗ Just before the ball is released to one side by a force F. and allowed to swing back and forth, (a) how large is ⃗ that is holding the ball in position and the force F (b) what is the tension in the string?

30.0°

F 2.0 kg

79. A 45-N lithograph is supported by two wires. One wire makes a 25° angle with the vertical and the other makes a 15° angle with the vertical. Find

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✦82. A cord, with a spring balance to measure forces attached midway along, is hanging from a hook attached to the ceiling. A mass of 10 kg is hanging from the lower end of the cord. The spring balance indicates a reading of 98 N for the force. Then two people hold the opposite ends of the same cord and pull against each other horizontally until the balance in the middle again reads 98 N. With what force must each person pull to attain this result? 83. Two blocks, masses m1 and m2, are connected by a ✦ massless cord. If the two blocks are pulled with a constant tension on a frictionless surface by applying a force of magnitude T2 to a second cord connected to m2, what is the ratio of the tensions in the two cords T1/T2 in terms of the masses? m1

T1

T2 m2

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4.8 Applying Newton’s Second Law 84. A 6.0-kg block, starting from rest, slides down a frictionless incline of length 2.0 m. When it arrives at the bottom of the incline, its speed is vf. At what distance from the top of the incline is the speed of the block 0.50 vf? 85. The coefficient of static friction between a block and a horizontal floor is 0.40, while the coefficient of kinetic friction is 0.15. The mass of the block is 5.0 kg. A horizontal force is applied to the block and slowly increased. (a) What is the value of the applied horizontal force at the instant that the block starts to slide? (b) What is the net force on the block after it starts to slide? 86. A 2.0-kg toy locomotive is pulling a 1.0-kg caboose. The frictional force of the track on the caboose is 0.50 N backward along the track. If the train’s acceleration forward is 3.0 m/s2, what is the magnitude of the force exerted by the locomotive on the caboose? 87. A block of mass m1 = 3.0 kg rests on a frictionless horizontal surface. A second block of mass m2 = 2.0 kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block. The blocks are released from rest. (a) Find the acceleration of the two blocks after they are released. (b) What is the velocity of the first block 1.2 s after the release of the blocks, assuming the first block does not run out of room on the table and the second block does not land on the floor? (c) How far has block 1 moved during the 1.2-s interval? (d) What is the displacement of the blocks from their initial positions 0.40 s after they are released?

the driver of the truck be concerned that the rope might break? 91. Two blocks are connected by a lightweight, flexible cord that passes over a frictionless pulley. If m1 = 3.6 kg and m2 = 9.2 kg, and block 2 is initially at rest 140 cm above the floor, how long does it take block 2 to reach the floor?

m2 m1

92. A 10.0-kg watermelon and a 7.00-kg pumpkin are attached to each other via a cord that wraps over a pulley, as shown. Friction is negligible everywhere in this tutorial: pulley) (a) Find the accelerasystem. ( tions of the pumpkin and the watermelon. Specify magnitude and direction. (b) If the system is released from rest, how far along the incline will the pumpkin travel in 0.30 s? (c) What is the speed of the watermelon after 0.20 s?

53.0°

30.0°

m1

m2

Problems 87 and 153 88. An engine pulls a train of 20 freight cars, each having a mass of 5.0 × 104 kg with a constant force. The cars move from rest to a speed of 4.0 m/s in 20.0 s on a straight track. Ignoring friction, what is the force with which the 10th car pulls the 11th one (at the middle of tutorial: school bus) the train)? ( 89. In Fig. 4.44, two blocks are connected by a lightweight, flexible cord that passes over a frictionless pulley. (a) If m1 = 3.0 kg and m2 = 5.0 kg, what are the accelerations of each block? (b) What is the tension in the cord? 90. A rope is attached from a truck to a 1400-kg car. The rope will break if the tension is greater than 2500 N. Ignoring friction, what is the maximum possible acceleration of the truck if the rope does not break? Should

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93. In the physics laboratory, a glider is released from rest on a frictionless air track inclined at an angle. If the glider has gained a speed of 25.0 cm/s in traveling 50.0 cm from the starting point, what was the angle of inclination of the track? Draw a graph of vx(t) when the positive x-axis points down the track. 94. A 10.0-kg block is released from rest on a frictionless ✦ track inclined at an angle of 55°. (a) What is the net force on the block after it is released? (b) What is the acceleration of the block? (c) If the block is released from rest, how long will it take for the block to attain a speed of 10.0 m/s? (d) Draw a motion diagram for the block. (e) Draw a graph of vx(t) for values of velocity between 0 and 10 m/s. Let the positive x-axis point down the track. ✦95. A box full of books rests on a wooden floor. The normal force the floor exerts on the box is 250 N. (a) You push horizontally on the box with a force of 120 N, but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor?

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(b) If you must push horizontally on the box with a force of at least 150 N to start it sliding, what is the coefficient of static friction? (c) Once the box is sliding, you only have to push with a force of 120 N to keep it sliding. What is the coefficient of kinetic friction? ✦ 96. A helicopter is lifting two crates simultaneously. One crate with a mass of 200 kg is attached to the helicopter by a cable. The second crate with a mass of 100 kg is hanging below the first crate and attached to the first crate by a cable. As the helicopter accelerates upward at a rate of 1.0 m/s2, what is the tension in each of the two cables?

4.10 Apparent Weight 97. Oliver has a mass of 76.2 kg. He is riding in an elevator that has a downward acceleration of 1.37 m/s2. With what magnitude force does the elevator floor push upward on Oliver? 98. While on an elevator, Jaden’s apparent weight is 550 N. When he is on the ground, the scale reading is 600 N. What is Jaden’s acceleration? 99. When on the ground, Ian’s weight is measured to be 640 N. When Ian is on an elevator, his apparent weight is 700 N. What is the net force on the system (Ian and the elevator) if their combined mass is 1050 kg? 100. Refer to Example 4.19. What is the apparent weight of the same passenger (weighing 598 N) in the following situations? In each case, the magnitude of the elevator’s acceleration is 0.50 m/s2. (a) After having stopped at the 15th floor, the passenger pushes the 8th floor button; the elevator is beginning to move downward. (b) The elevator is moving downward and is slowing down as it nears the 8th floor. 101.You are standing on a bathroom scale inside an elevator. Your weight is 140 lb, but the reading of the scale is 120 lb. (a) What is the magnitude and direction of the acceleration of the elevator? (b) Can you tell whether the elevator is speeding up or slowing down? 102. Yolanda, whose mass is 64.2 kg, is riding in an elevator that has an upward acceleration of 2.13 m/s2. What force does she exert on the floor of the elevator? 103. Felipe is going for a physical before joining the swim team. He is concerned about his weight, so he carries his scale into the elevator to check his weight while heading to the doctor’s office on the 21st floor of the building. If his scale reads 750 N while the elevator has an upward acceleration of 2.0 m/s2, what does the nurse measure his weight to be? ✦104. Luke stands on a scale in an elevator that has a constant acceleration upward. The scale reads 0.960 kN. When

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Luke picks up a box of mass 20.0 kg, the scale reads 1.200 kN. (The acceleration remains the same.) (a) Find the acceleration of the elevator. (b) Find Luke’s weight.

4.12 Fundamental Forces 105. Which of the fundamental forces has the shortest range, yet is responsible for producing the sunlight that reaches Earth? 106. Which of the fundamental forces governs the motion of planets in the solar system? Is this the strongest or the weakest of the fundamental forces? Explain. 107. Which of the following forces have an unlimited range: strong force, contact force, electromagnetic force, gravitational force? 108. Which of the following forces bind electrons to nuclei to form atoms: strong force, contact force, electromagnetic force, gravitational force? 109. Which of the fundamental forces binds quarks together to form protons, neutrons, and many exotic subatomic particles?

Comprehensive Problems 110. A car is driving on a straight, level road at constant speed. Draw an FBD for the car, showing the significant forces that act upon it. 111. A skier with a mass of 63 kg starts from rest and skis down an icy (frictionless) slope that has a length of 50 m at an angle of 32° with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to rest in a distance of 140 m along the horizontal path. (a) What is the speed of the skier at the bottom of the slope? (b) What is the coefficient of kinetic friction between the skier and the horizontal surface? 112. You want to push a 65-kg box up a 25° ramp. The coefficient of kinetic friction between the ramp and the box is 0.30. With what magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed? 113. An airplane is cruising along in a horizontal level flight at a constant velocity, heading due west. (a) If the weight of the plane is 2.6 × 104 N, what is the net force on the plane? (b) With what force does the air push upward on the plane? 114. A young boy with a broken leg is undergoing traction. (a) Find the magnitude of the total force of the traction apparatus applied to the leg, assuming the weight of the leg is 22 N and the weight hanging from the traction apparatus is also 22 N. (b) What is the horizontal

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component of the traction force acting on the leg? (c) What is the magnitude of the force exerted on the femur by the lower leg?

that each force makes with either the vertical or horizontal direction. (b) What is the tension in the rope? b = 10.0°

a = 5.0° Femur 30.0°

118. The readings of the two spring scales shown in the drawing are the same. (a) Explain why they are the same. [Hint: Draw free-body diagrams.] (b) What is the reading?

30.0°

Scale 22 N

115. When you hold up a 100-N weight in your hand, with your forearm horizontal and your palm up, the force exerted by your biceps is much larger than 100 N— perhaps as much as 1000 N. How can that be? What other forces are acting on your arm? Draw an FBD for the forearm, showing all of the forces. Assume that all the forces exerted on the forearm are purely vertical— either up or down.

550 N

550 N

Scale

Biceps 100 N 550 N

116. In the sport of curling, popular in Canada and Ireland, a player slides a 20.0-kg granite stone down a 38-mlong ice rink. Draw FBDs for the stone (a) while it sits at rest on the ice; (b) while it slides down the rink; (c) during a head-on collision with an opponent’s stone that was at rest on the ice.

117. A truck is towing a 1000-kg car at a constant speed up a hill that makes an angle of a = 5.0° with respect to the horizontal. A rope is attached from the truck to the car at an angle of b = 10.0° with respect to horizontal. Ignore any friction in this problem. (a) Draw an FBD showing all the forces on the car. Indicate the angle

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119. The tallest spot on Earth is Mt. Everest, which is 8850 m above sea level. If the radius of the Earth to sea level is 6370 km, how much does the gravitational field strength change between the sea level value at that location (9.826 N/kg) and the top of Mt. Everest? 120. By what percentage does the weight of an object change when it is moved from the equator at sea level, where the effective value of g is 9.784 N/kg, to the North Pole where g = 9.832 N/kg? 121. Two canal workers pull a barge along the narrow waterway at a constant speed. One worker pulls with a force of 105 N at an angle of 28° with respect to the forward motion of the barge and the other worker, on the opposite tow path, pulls at an angle of 38° relative to the barge motion. Both ropes are parallel to the ground. (a) With what magnitude force should the second worker pull to make the sum of the two forces be in the forward direction? (b) What is the magnitude of the force on the barge from the two tow ropes? 122. A large wrecking ball of mass m is resting against a wall. It hangs from the end of a cable that is attached at its upper end to a crane that is just touching the wall.

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The cable makes an angle of q with the wall. Ignoring friction between the ball and the wall, find the tension in the cable.

q

123. The figure shows the quadriceps and the patellar tendons attached to the patella (the kneecap). If the tension T in each tendon is 1.30 kN, what is (a) the magnitude and (b) the direction of the contact force ⃗ exerted on the patella by the femur? F T

Quadriceps tendon

37.0° Patella F

q

Femur

80.0°

Patellar tendon

Tibia

T

124. The coefficient of static friction between a block and a horizontal floor is 0.35, while the coefficient of kinetic friction is 0.22. The mass of the block is 4.6 kg and it is initially at rest. (a) What is the minimum horizontal applied force required to make the block start to slide? (b) Once the block is sliding, if you keep pushing on it with the same minimum starting force as in part (a), does the block move with constant velocity or does it accelerate? (c) If it moves with constant velocity, what is its velocity? If it accelerates, what is its acceleration? 125. Two blocks lie side by side on a frictionless table. The block on the left is of mass m; the one on the right is of mass 2m. The block on the right is pushed to the left with a force of magnitude F, pushing the other block in turn. What force does the block on the left exert on the block to its right? 126. A locomotive pulls a train of 10 identical cars, on a track that runs east-west, with a force of 2.0 × 106 N directed east. What is the force with which the last car to the west pulls on the rest of the train?

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127. The coefficient of static friction between a brick and a wooden board is 0.40 and the coefficient of kinetic friction between the brick and board is 0.30. You place the brick on the board and slowly lift one end of the board off the ground until the brick starts to slide down the board. (a) What angle does the board make with the ground when the brick starts to slide? (b) What is the acceleration of the brick as it slides down the board? 128. A woman of mass 51 kg is standing in an elevator. (a) If the elevator floor pushes up on her feet with a force of 408 N, what is the acceleration of the elevator? (b) If the elevator is moving at 1.5 m/s as it passes the fourth floor on its way down, what is its speed 4.0 s later? 129. In Fig. 4.15 an astronaut is playing shuffleboard on Earth. The puck has a mass of 2.0 kg. Between the board and puck the coefficient of static friction is 0.35 and of kinetic friction is 0.25. (a) If she pushes the puck with a force of 5.0 N in the forward direction, does the puck move? (b) As she is pushing, she trips and the force in the forward direction suddenly becomes 7.5 N. Does the puck move? (c) If so, what is the acceleration of the puck along the board if she maintains contact between puck and stick as she regains her footing while pushing steadily with a force of 6.0 N on the puck? (d) She carries her game to the Moon and again pushes a moving puck with a force of 6.0 N forward. Is the acceleration of the puck during contact more, the same, or less than on tutorial: rough table) Earth? Explain. ( ✦130. You want to hang a 15-N picture as in part (a) using some very fine twine that will break with more than 12 N of tension. Can you do this? What if you have it as illustrated in part (b) of the figure? 30° 50°

(a)

(b)

✦131. A roller coaster is towed up an incline at a steady speed of 0.50 m/s by a chain parallel to the surface of the incline. The slope is 3.0%, which means that the elevation increases by 3.0 m for every 100.0 m of horizontal distance. The mass of the roller coaster is 400.0 kg. Ignoring friction, what is the magnitude of the force exerted on the roller coaster by the chain? 132. A 320-kg satellite is in orbit around the Earth 16 000 km above the Earth’s surface. (a) What is the weight of the satellite when in orbit? (b) What was its weight when it was on the Earth’s surface, before being launched? (c) While it orbits the Earth, what force does the satellite exert on the Earth?

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✦133. The mass of the Moon is 0.0123 times that of the Earth. A spaceship is traveling along a line connecting the centers of the Earth and the Moon. At what distance from the Earth does the spaceship find the gravitational pull of the Earth equal in magnitude to that of the Moon? Express your answer as a percentage of the distance between the centers of the two bodies. ✦134. A model rocket is fired vertically from rest. It has a net acceleration of 17.5 m/s2. After 1.5 s, its fuel is exhausted and its only acceleration is that due to gravity. (a) Ignoring air resistance, how high does the rocket travel? (b) How long after liftoff does the rocket return to the ground? 135. The model rocket in Problem 134 has a mass of 87 g ✦ and you may assume the mass of the fuel is much less than 87 g. (a) What was the net force on the rocket during the first 1.5 s after liftoff? (b) What force was exerted on the rocket by the burning fuel? (c) What was the net force on the rocket after its fuel was spent? (d) The rocket’s vertical velocity was zero instantaneously when it was at the top of its trajectory. What were the net force and acceleration on the rocket at this instant? ✦ 136. A toy freight train consists of an engine and three identical cars. The train is moving to the right at constant speed along a straight, level track. Three spring scales are used to connect the cars as follows: spring scale A is located between the engine and the first car; scale B is between the first and second cars; scale C is between the second and third cars. (a) If air resistance and friction are negligible, what are the relative readings on the three spring scales A, B, and C? (b) Repeat part (a), taking air resistance and friction into consideration this time. [Hint: Draw an FBD for the car in the middle.] (c) If air resistance and friction together cause a force of magnitude 5.5 N on each car, directed toward the left, find the readings of scales A, B, and C. 137. Four identical spring scales, A, B, C, and D are used to ✦ hang a 220.0-N sack of potatoes. (a) Assume the scales have negligible weights and all four scales show the same reading. What is the reading of each scale? (b) Suppose that each scale has a weight of 5.0 N. If scales B and D show the same reading, what is the reading of each scale?

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A

C

B

D

138. A computer weighing 87 N rests on the horizontal surface of your desk. The coefficient of friction between the computer and the desk is 0.60. (a) Draw an FBD for the computer. (b) What is the magnitude of the frictional force acting on the computer? (c) How hard would you have to push on it to get it to start to slide across the desk? ✦139. A refrigerator magnet weighing 0.14 N is used to hold up a photograph weighing 0.030 N. The magnet attracts the refrigerator door with a magnetic force of 2.10 N. (a) Identify the interactions between the magnet and other objects. (b) Draw an FBD for the magnet, showing all the forces that act on it. (c) Which of these forces are long-range and which are contact forces? (d) Find the magnitudes of all the forces acting on the magnet. ✦140. A 50.0-kg crate is suspended between the floor and the ceiling using two spring scales, one attached to the ceiling and one to the floor. If the lower scale reads 120 N, what is the reading of the upper scale? Ignore the weight of the scales. 141. Spring scale A is attached to the ceiling. A 10.0-kg ✦ mass is suspended from the scale. A second spring scale, B, is hanging from a hook at the bottom of the 10.0-kg mass and a 4.0-kg mass hangs from the second spring scale. (a) What are the readings of the two scales if the masses of the scales are negligible? (b) What are the readings if each scale has a mass of 1.0 kg? ✦142. A crate of oranges weighing 180 N rests on a flatbed truck 2.0 m from the back of the truck. The coefficients of friction between the crate and the bed are ms = 0.30 and mk = 0.20. The truck drives on a straight, level highway at a constant 8.0 m/s. (a) What is the force of friction acting on the crate? (b) If the truck speeds up with an acceleration of 1.0 m/s2, what is the force of the friction on the crate? (c) What is the maximum acceleration the truck can have without the crate starting to slide? 143. A crate of books is to be put on a truck by rolling it up an incline of angle q using a dolly. The total mass of the crate and the dolly is m. Assume that rolling the dolly up the incline is the same as sliding it up a frictionless surface. (a) What is the magnitude of the horizontal force that must be applied just to hold the crate in place on the incline? (b) What horizontal force must be applied to roll the crate up at constant speed? (c) In order to start the dolly moving, it must be accelerated from rest. What horizontal force must be applied to give the crate an acceleration up the incline of magnitude a? ( tutorial: cart on ramp) ✦144. A toy cart of mass m1 moves on frictionless wheels as it is pulled by a string under tension T. A block of mass m2 rests on top of the cart. The coefficient of static friction between the cart and the block is m. Find the maximum tension T that will not cause the block to slide on the

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cart if the cart rolls on (a) a horizontal surface; (b) up a ✦ 148. A student’s head is bent over her physics book. The ramp of angle q above the horizontal. In both cases, the head weighs 50.0 N and is supported by the muscle ⃗ m exerted by the neck extensor muscles and by string is parallel to the surface on which the cart rolls. force F ⃗ c exerted at the atlantooccipital joint. the contact force F ✦145. A helicopter of mass M is lowering a truck of mass m ⃗ m is 60.0 N and is directed Given that the magnitude of F onto the deck of a ship. (a) At first, the helicopter and the 35 ° below the horizontal, find (a) the magnitude and truck move downward together (the length of the cable ⃗ (b) the direction of F . c doesn’t change). If their downward speed is decreasing at a rate of 0.10g, what is the tension in the cable? (b) As the truck gets close to the deck, the helicopter stops Fc moving downward. While it hovers, it lets out the cable so that the truck is still moving downward. If the truck’s f downward speed is decreasing at a rate of 0.10g, while 35° the helicopter is at rest, what is the tension in the cable? Fm ✦146. The coefficient of static friction between block A and a 50.0 N horizontal floor is 0.45 and the coefficient of static friction between block B and the floor is 0.30. The mass of each block is 2.0 kg and they are connected together by ⃗ pulling on block B is a cord. (a) If a horizontal force F slowly increased, in a direction parallel to the connecting cord, until it is barely enough to make the two blocks ⃗ at the instant start moving, what is the magnitude of F that they start to slide? (b) What is the tension in the cord ✦149. (a) If a spacecraft moves in a straight line between the connecting blocks A and B at that same instant? Earth and the Sun, at what point would the force of gravity on the spacecraft due to the Sun be as large as ✦147. Tamar wants to cut down a large, dead poplar tree with her chain saw, but she does not want it to fall onto the that due to the Earth? (b) If the spacecraft is close to, nearby gazebo. Yoojin comes to help with a long rope. but not at, this equilibrium point, does the net force on Yoojin, a physicist, suggests they tie the rope taut from the spacecraft tend to push it toward or away from the the poplar to the oak tree and then pull sideways on the equilibrium point? [Hint: Imagine the spacecraft a rope as shown in the figure. If the rope is 40.0 m long small distance d closer to the Earth and find out which and Yoojin pulls sideways at the midpoint of the rope gravitational force is stronger.] with a force of 360.0 N, causing a 2.00-m sideways ✦150. While trying to decide where to hang a framed picture, displacement of the rope at its midpoint, what force you press it against the wall to keep it from falling. The will the rope exert on the poplar tree? Compare this picture weighs 5.0 N and you press against the frame with pulling the rope directly away from the poplar with a force of 6.0 N at an angle of 40° from the vertiwith a force of 360.0 N and explain why the values are cal. (a) What is the direction of the normal force exerted different. [Hint: Until the poplar is cut through enough on the picture by your hand? (b) What is the direction to start falling, the rope is in equilibrium.] of the normal force exerted on the picture by the wall? (c) What is the coefficient of static friction between the wall and the picture? The frictional force exerted on the picture by the wall can have two possible direcDead tions. Explain why. poplar tree Gazebo Tamar

360.0 N

Yoojin pulling sideways

Oak tree

40.0 m Side view 40° Gazebo

Dead poplar

q

40.0 m 2.00 m

360.0 N Overhead view

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q

Oak tree

✦151. In a movie, a stuntman places himself on the front of a truck as the truck accelerates. The coefficient of friction between the stuntman and the truck is 0.65. The stuntman is not standing on anything but can “stick” to

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the front of the truck as long as the truck continues to accelerate. What minimum forward acceleration will keep the stuntman on the front of the truck? ✦152. An airplane of mass 2800 kg has just lifted off the runway. It is gaining altitude at a constant 2.3 m/s while the horizontal component of its velocity is increasing at a rate of 0.86 m/s2. Assume g = 9.81 m/s2. (a) Find the direction of the force exerted on the airplane by the air. (b) Find the horizontal and vertical components of the plane’s acceleration if the force due to the air has the same magnitude but has a direction 2.0° closer to the vertical than its direction in part (a). 153. In the figure with Problem 87, the block of mass m1 ✦ slides to the right with coefficient of kinetic friction mk on a horizontal surface. The block is connected to a hanging block of mass m2 by a light cord that passes over a light, frictionless pulley. (a) Find the acceleration of each of the blocks and the tension in the cord. (b) Check your answers in the special cases m1 > m2, and m1 = m2. (c) For what value of m2 (if any) do the two blocks slide at constant velocity? What is the tension in the cord in that case?

horizontal and vertical components gives the answer: the normal force is 750 N, up, and the frictional force is 110 N, to the left. The quantity msN is the maximum possible magnitude of the force of static friction for a surface. In this problem, the frictional force does not necessarily have the maximum possible magnitude. 4.9

(a) Normal

Static friction

Drag

Weight South

North

(b) Weight of the car = 11.0 kN; (c) 2.1 kN northward 4.10 (a) 110 N; (b) 230 N 4.11 3100 N 4.12

Answers to Practice Problems 4.1 (a) Fx = 49.1 N, Fy = 2.9 N; (b) F = 49.2 N; (c) 3.4° above the horizontal 4.2 0.5 kN downward 4.3 In the first case, the principle of inertia says that Negar tends to stay at rest with respect to the ground as the subway car begins to move forward, until forces acting on her (exerted by the strap and the floor) make her move forward. In the second case, Negar keeps moving forward with respect to the ground with constant speed as the subway car slows down, until forces acting on her make her slow down as well. 4.4 760 N, 81.7° above the –x-axis or 8.3° to the left of the +y-axis 4.5 The contact force exerted on the floor by the chest; 870 N, 59° below the rightward horizontal (+x-axis) 4.6 For m1 = m2 = 1000 kg and r = 4 m, F ≈ 4 μN, which is about the same magnitude as the weight of a mosquito. The claim that this tiny force caused the collision is ridiculous. 4.7 0.57 N or 0.13 lb 4.8 The chest is in equilibrium, so the net force on it is zero. Setting the net force equal to zero separately for the

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TB TC TC

TC = 902.0 N TB = 1804 N W = 1804 N

W

4.13 (a) 54 N; (b) 1.8 s 4.14 1.84 kN 4.15 Block 1: ∑F1y = T − m1g = 315 N − 255 N = 60 N; m1a1y = 60 N. Block 2: ∑F2y = m2g − T = 412 N − 315 N = 97 N; m2a2y = 97 N. 4.16 Impossible to pull the crate up with a single pulley. The entire weight of the crate would be supported by a single strand of cable and that weight exceeds the breaking strength of the cable. 4.17 2500 N 4.18 (a) down the incline; (b) up the incline; (c) 0.2 m/s2 down the incline 4.19 (a) 392 N; (b) 431 N

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ANSWERS TO CHECKPOINTS

Answers to Checkpoints 4.4 The two forces exerted by the two children on a toy cannot be interaction partners because they act on the same object (the toy), not on two different objects. Interaction partners act on different objects, one on each of the two objects that are interacting. The interaction partner of the force exerted by one child on the toy is the force that the toy exerts on that child. 4.5 The weight of the gear decreases as the value of g decreases. The mass of the gear does not change.

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145

4.6 One upward normal force on each leg due to the floor and one downward normal force on the desktop due to the laptop. 4.8 Yes. For motion along an incline, it simplifies the problem to choose one axis parallel to the incline and the other perpendicular to the incline. 4.10 Your velocity is downward and decreasing in magnitude, so your acceleration is upward. Then the upward normal force exerted on you by the scale must be greater than your weight. The scale reading is greater than your weight.

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CHAPTER

5

German athlete Susanne Keil throws the hammer during the German Athletics championships. Keil qualified for the 2004 Olympics in Athens with a 67.77-m throw.

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Circular Motion

In the track and field event called the hammer throw, the “hammer” is actually a metal ball (mass 4.00 kg for women or 7.26 kg for men) attached by a cable to a grip. The athlete whirls the hammer several times around while not leaving a circle of radius 2.1 m and then releases it. The winner is the athlete whose hammer lands the greatest distance away. How large a force does an athlete have to exert on the grip to whirl the massive hammer around in a circle? What kind of path does the hammer follow once it is released? (See pp. 155–156 for the answer.)

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5.1

• • • • •

Concepts & Skills to Review

gravitational forces (Section 4.5) Newton’s second law: force and acceleration (Sections 4.3 and 4.8) velocity and acceleration (Sections 2.2 and 2.3) apparent weight (Section 4.10) normal and frictional forces (Section 4.6)

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DESCRIPTION OF UNIFORM CIRCULAR MOTION

DESCRIPTION OF UNIFORM CIRCULAR MOTION

Ask someone to name the most important machine ever invented by humans and you are likely to get the wheel as a response. Rotating objects are so important to modern— and even not-so-modern—technology that we barely notice them. Examples include wheels on cars, bicycles, trains, and lawnmowers; propellers on airplanes and helicopters; CDs and DVDs; computer hard drives; the gears and hands of an analog clock; amusement park rides and centrifuges—the list is endless. Rotation of a Rigid Body To describe circular motion, we could use the familiar definitions of displacement, velocity, and acceleration. But much of the circular motion around us occurs in the rotation of a rigid object. A rigid body is one for which the distance between any two points of the body remains the same when the body is translated or rotated. When such an object rotates, every point on the object moves in a circular path. The radius of the path for any point is the distance between that point and the axis of rotation. When a compact disk spins inside a CD player, different points on the CD have different velocities and accelerations. The velocity and acceleration of a given point keep changing direction as the CD spins. It would be clumsy to describe the rotation of the CD by talking about the motion of arbitrary points on it. However, some quantities are the same for every point on the CD. It is much simpler, for instance, to say “the CD spins at 210 rpm” instead of saying “a point 6.0 cm from the rotation axis of the CD is moving at 1.3 m/s.” Angular Displacement and Angular Velocity To simplify the description of circular motion, we concentrate on angles instead of distances. If a CD spins through _14 of a turn, every point moves through the same angle (90°), but points at different radii move different linear distances. On the CD shown in Fig. 5.1, point 1 near the axis of rotation moves through a smaller distance than point 4 on the circumference. For this reason we define a set of variables that are analogous to displacement, velocity, and acceleration, but use angular measure instead of linear distance. Instead of displacement, we speak of angular displacement Δq, the angle through which the CD turns. A point on the CD moves along the circumference of a circle. As the point moves from the angular position q i to the angular position q f, a radial line drawn between the center of the circle and that point sweeps out an angle Δq = q f − q i, which is the angular displacement of the CD during that time interval (Fig. 5.2). Definition of angular displacement: Δq = q f − q i

(5-1)

The sign of the angular displacement indicates the sense of the rotation. The usual convention is that a positive angular displacement represents counterclockwise rotation and a negative angular displacement represents clockwise rotation. Counterclockwise and clockwise are only well defined for a particular viewing direction; counterclockwise rotation viewed from above is clockwise when viewed from below.

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In a rigid body, the distance between any two points is constant.

The abbreviation rpm means revolutions per minute.

CONNECTION: Equations (5-1) through (5-3) have a familiar form because w is the rate of change of q, just as velocity is the rate of change of position. + Counterclockwise − Clockwise

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CHAPTER 5 Circular Motion

Remember that the notation lim

Δt→0

indicates that Δq is the angular displacement during a very short time interval Δt (short enough that the ratio Δq /Δt doesn’t change significantly if we make the time interval even shorter).

The average angular velocity w av is the average rate of change of the angular displacement.

Definition of average angular velocity: Δq w av = ___ Δt

(5-2)

If we let the time interval Δt become shorter and shorter, we are averaging over smaller and smaller time intervals. In the limit Δt → 0, w av becomes the instantaneous angular velocity w.

4′ 3′ 2′ 1′

Definition of instantaneous angular velocity:

1 2 3 4

Δq w = lim ___ Δt→0 Δt

(5-3)

Figure 5.1 A CD rotates through _14 turn; points 1, 2, 3, and 4 travel through the same angle but different distances to reach their new positions, marked 1′, 2′, 3′, and 4′, respectively. rf

∆q qf

ri qi x

q f – q i = ∆q

Figure 5.2 Angular positions such as q i and q f are measured counterclockwise from a reference axis (usually the x-axis).

s = qr

r q r

Figure 5.3 Definition of the radian: angle q in radians is the arc length s divided by the radius r. The angle shown is 1 rad ≈ 57.3°.

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The angular velocity also indicates—through its algebraic sign—in what direction the CD is spinning. Since angular displacements can be measured in degrees or radians, angular velocities have units such as degrees/second, radians/second, degrees/day, and the like. Radian Measure You may be most familiar with measuring angles in degrees, but in many situations the most convenient measure is the radian. One such situation is when we relate the angular displacement or angular velocity of a rotating object with the distance traveled by, or the speed of, some point on the object. In Fig. 5.3, an angle q between two radii of a circle define an arc of length s. We say that q is the angle subtended by the arc. The arc length is proportional to both the radius of the circle and to the angle subtended. The angle q in radians is defined as q (in radians) = _rs (5-4) where r is the radius of the circle. Since an angle in radians is defined by the ratio of two lengths, it is dimensionless (a pure number). We use the term radians, abbreviated “rad,” to keep track of the angular measure used. Since “rad” is not a physical unit like meters or kilograms, it does not have to balance in Eq. (5-4). For the same reason, we can drop “rad” whenever there is no chance of being misunderstood. We can write w = 23 s−1 as long as context makes it clear that we mean 23 radians per second. In equations that relate linear variables to angular variables [such as Eq. (5-4)], think of r as the number of meters of arc length per radian of angle subtended. In other words, think of r as having units of meters per radian. Doing so, the radians cancel out in these equations. For example, if q = 2.0 rad and r = 1.2 m, then the arc length is m = 2.4 m s = q r = 2.0 rad × 1.2 ___ rad Since the arc length for an angle of 360° is the circumference of the circle, the radian measure of an angle of 360° is 2p r = 2p rad q = _sr = ____ r Therefore, the conversion factor between degrees and radians is 360° = 2p rad

(5-5)

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DESCRIPTION OF UNIFORM CIRCULAR MOTION

Example 5.1 Angular Speed of Earth Earth is rotating about its axis. What is its angular speed in rad/s? (The question asks for angular speed, so we do not have to worry about the direction of rotation.) Strategy The Earth’s angular velocity is constant, or nearly so. Therefore, we can calculate the average angular velocity for any convenient time interval and, in turn, the Earth’s instantaneous angular speed w . Solution It takes the Earth 1 day to complete one rotation, during which the angular displacement is 2p rad. More formally, during a time interval ∆t = 1 day, the angular displacement of the Earth is Δq = 2p rad. So the angular speed of the Earth is 2p rad/day, and then convert days to seconds. 1 day = 24 h = 24 h × 3600 s/h = 86 400 s 2p rad = 7.3 × 10−5 rad/s w = _______ 86 400 s

Discussion Notice that this problem is analogous to a problem in linear motion such as: “A car travels in a straight line at constant speed. In 3 h, it has traveled 192 mi. What is its velocity in m/s?” Just about everything in circular motion and rotation has this kind of analog—which means we can draw heavily on what we have already learned. Earth actually completes one rotation in 23.9345 h (see inside back cover) rather than in 24 h due to Earth’s motion around the Sun. This distinction would be important only if we needed a more precise value of w (more than two significant figures).

Practice Problem 5.1 Angular Speed of Venus Venus completes one rotation about its axis every 5816 h. What is the angular speed of the rotation of Venus in rad/s?

Relation Between Linear and Angular Speed For a point moving in a circular path of radius r, the linear distance traveled along the circular path during an angular displacement of Δq (in radians) is the arc length s where s = r|Δq | = r|q f − q i|

(angles in radians)

(5-6)

The point in question could be a point particle moving in a circular path, or it could be any point on a rotating rigid object. Since Eq. (5-6) comes directly from the definition of the radian, any equation derived from Eq. (5-6) is valid only when the angles are measured in radians. What is the linear speed at which the point moves? The average linear speed is the distance traveled divided by the time interval,

∆q = 2p rad Arctic Circle

Δq | s = r| _____ v av = __ (Δq in radians) Δt Δt

v

We recognize Δq /Δt as the average angular velocity wav. If we take the limit as Δt approaches zero, both average quantities (vav and w av) become instantaneous quantities. Therefore, the relationship between linear speed and angular speed is v = rw

(w in radians per unit time)

v

(5-7)

Equation (5-7) relates only the magnitudes of the linear and angular speeds. The direction of the velocity vector v⃗ is tangent to the circular path. For a rotating object, points farther from the axis move at higher linear speeds; they have a circle of bigger radius to travel and, therefore, cover more distance in the same time interval. For example, a person standing at the equator has a much higher linear speed due to Earth’s rotation than does a person standing at the Arctic Circle (see Fig. 5.4).

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Equator

Axis of rotation

Figure 5.4 A person standing at the Equator is moving much faster than another person standing at the Arctic Circle, but their angular speeds are the same.

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In uniform circular motion, speed is constant but velocity is not constant because the direction of the velocity is changing.

Period and Frequency When the speed of a point moving in a circle is constant, its motion is called uniform circular motion. Even though the speed of the point is constant, the velocity is not: the direction of the velocity is changing. This distinction is important when we find the acceleration of an object in uniform circular motion (Section 5.2). The time for the point to travel completely around the circle is called the period of the motion, T. The frequency of the motion, which is the number of revolutions per unit time, is defined as 1 f = __ (5-8) T since revolutions = _______________ 1 __________ second second/revolution

CHECKPOINT 5.1 1 If it takes ____ 7200 of a second for a computer hard drive to spin around once, what is its frequency?

The speed is the total distance traveled divided by the time taken, 2p r = 2p rf v = ____ T Then, for uniform circular motion v = 2p f w = __ r SI unit of frequency: 1 Hz = 1 rev/s

(w in radians per unit time)

(5-9)

where, in SI units, angular velocity w is measured in rad/s and frequency f is measured in hertz (Hz). The hertz is a derived unit equal to 1 rev/s. The dimensions of Eq. (5-9) are correct since both revolutions and radians are pure numbers. The physical dimensions on both sides are a number per second (s−1).

Example 5.2 Speed in a Centrifuge A centrifuge is spinning at 5400 rpm. (a) Find the period (in s) and frequency (in Hz) of the motion. (b) If the radius of the centrifuge is 14 cm, how fast (in m/s) is an object at the outer edge moving?

Strategy Remember that rpm means revolutions per minute. 5400 rpm is the frequency, but in a unit other than Hz. After a unit conversion, the other quantities can be found using the relations already discussed. Solution (a) First convert rpm to Hz: rev × _____ 1 min = 90 rev/s f = 5400 ____ min 60 s so the frequency is f = 90 Hz = 90 s−1. The period is T = 1/f = 0.011 s (b) To find the linear speed, we first find the angular speed in rad/s: rev × 2p ___ rad = 180p rad/s w = 90 ___ s rev continued on next page

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DESCRIPTION OF UNIFORM CIRCULAR MOTION

Example 5.2 continued

So w = 2p f = 180 p rad/s. The linear speed is v = w r = 180 p s−1 × 0.14 m = 79 m/s Discussion Notice that much of this problem was done with unit conversions. Instead of memorizing a formula such as w = 2p f, an understanding of where the formula came

from (in this case, that 2p radians correspond to one revolution) is more useful and less prone to error.

Practice Problem 5.2 Clothing in the Drier An automatic clothing drier spins at 51.6 rpm. If the radius of the drier drum is 30.5 cm, how fast is the outer edge of the drum moving?

Rolling Without Slipping: Rotation and Translation Combined When an object is rolling, it is both rotating and translating. The wheel rotates about an axle, but the axle is not at rest; it moves forward or backward. What is the relationship between the angular speed of the wheel and the linear speed of the axle? You might guess that v = w r is the answer. You would be right, as long as the object rolls without slipping or skidding. There is no fixed relationship between the linear and angular speeds of a wheel if it is allowed to skid or slip. When an impatient driver guns the engine the instant a traffic light turns green, the automobile wheels are likely to slip. The rubber sliding against the road surface makes the squealing sound and leaves tracks on the road. The driver could actually make the acceleration of the car greater by giving the engine less gas. When the wheels are skidding or slipping, kinetic friction propels the car forward instead of the potentially larger force of static friction. For a wheel that rolls without slipping, as the wheel turns through one complete rotation, the axle moves a distance equal to the circumference of the wheel (Fig. 5.5). Think of a paint roller leaving a line of paint as it rolls along a wall. After one complete

13.0 m/s

Hawk Z X

Hawk ZX

vaxle ∆q

65.0 cm

2p r (a)

Tire position after one revolution

s = r∆q (b)

Figure 5.5 (a) As a wheel of radius r that rolls without slipping turns through one complete revolution, the distance its axle moves is equal to the circumference of the wheel (2p r). (b) As a wheel rolls without slipping through an angle Δq, the distance the axle moves is equal to the arc length s.

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rotation, the same point on the roller wheel is touching the wall as was initially touching it. The length of the line of paint is 2p r. The elapsed time is T, so the axle’s speed is 2p r v axle = ____ T while the angular speed of the roller is 2p w = ___ T Thus, v axle = w r

(w in radians per unit time)

(5-10)

Example 5.3 Angular Speed of a Rolling Wheel Kevin is riding his motorcycle at a speed of 13.0 m/s. If the diameter of the rear tire is 65.0 cm, what is the angular speed of the rear wheel? Assume that it rolls without slipping. Strategy The given diameter of the tire enables us to find the circumference and, thus, the distance traveled in one revolution of the wheel. From the speed of the motorcycle we can find how many revolutions the tire must make per second. Solution During one revolution of the wheel, the motorcycle travels a distance equal to the tire’s circumference 2p r (Fig. 5.5). Then the time to make one revolution is T and the speed v is 2p r distance = ____ v = _______ T time Therefore, T = 2p r/v. For each revolution there is an angular displacement of Δq = 2p radians, so Δq 2p w = ____ = ___ Δt T Substituting T = 2p r/v and remembering that the radius is half the diameter, 2p = __ 13.0 m/s = 40.0 ___ v = __________ rad w = _____ s 2pr/v r (0.650 m)/2 Discussion Check: Time for one revolution is 2p rad = 0.157 s. _________ 40.0 rad/s

5.2

Time to travel a distance 2p r = 2.04 m is 2.04 m = 0.157 s. ________ 13.0 m/s Looks good. You could have obtained this answer immediately by looking back through the text for the equation w = v/r and plugging in numbers, but the solution here shows that you can re-create that equation. Here, and in many cases, there is no need to memorize a formula if you understand the concepts behind the formula. You are then less apt to make a mistake by forgetting a factor or constant in the equation, or by using an inappropriate formula. For another example, if an object moves along a straight line at a constant velocity, you know instantly that the displacement is the velocity times the time interval—not because you have memorized an equation (Δr⃗ = v⃗ Δt), but because you understand the concepts of displacement and velocity. This is the sort of internalization of scientific thinking that you will develop with more and more practice in problem solving.

Practice Problem 5.3 Rolling Drum A cylindrical steel drum is tipped over and rolled along the floor of a warehouse. If the drum has a radius of 0.40 m and makes one complete turn every 8.0 s, how long does it take to roll the drum 36 m?

RADIAL ACCELERATION

For a particle undergoing uniform circular motion, the magnitude of the velocity vector is constant, but its direction is continuously changing. At any instant of time, the direction of the instantaneous velocity is tangent to the path, as discussed in Section 3.2. Since the direction of the velocity continually changes, the particle has a nonzero acceleration.

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r1

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RADIAL ACCELERATION

r1 r2

r2

∆t → 0

|r1| = |r2|

v2

v1

v2

|v2| = |v1|

(a)

v1 v2

v1

∆v v1 + ∆v = v2

(b)

(c)

Figure 5.6 Uniform circular motion at constant speed. (a) The velocity vector is always tangent to the circular path and perpendicular to the radius at that point. (b) As the time interval between two velocity measurements decreases, the angle between the velocity vectors decreases. (c) The change in velocity (Δv⃗ ) is found by placing the tails of the two velocity vectors together. Then Δv⃗ is drawn from the tip of the initial velocity (v⃗1) to the tip of the final velocity (v⃗2) so that v⃗1 + Δv⃗ = v⃗2.

In Fig. 5.6a, two velocity vectors of equal magnitude are drawn tangent to a circular path of radius r, representing the velocity at two different times of an object moving around a circular path with constant speed. At any instant, the velocity vector is perpendicular to a radius drawn from the center of the circle to the position of the object. As the time between velocity measurements approaches zero, the radii become closer Δv ⃗ , we must first find the change together (Fig. 5.6b). To find the acceleration, a⃗ = lim ___ Δt→0 Δt in velocity Δv⃗ for a very short time interval. Figure 5.6c shows that as the time interval Δt approaches zero, the angle between the two velocities also approaches zero and Δv⃗ becomes perpendicular to the velocity. Since Δv⃗ is perpendicular to the velocity, it is directed along a radius of the circle. Inspection of Figs. 5.6b and 5.6c shows that Δv⃗ is radially inward (toward the center of the circle). Since the acceleration a⃗ has the same direction as Δv⃗ (in the limit Δt → 0), the acceleration is also directed radially inward (Fig. 5.7)—that is, along a radius of the circular path toward the center of the circle. The acceleration of an object undergoing uniform circular motion is often called the radial acceleration a⃗ r. The word radial here just reminds us of the direction of the acceleration. (A synonym for radial acceleration is centripetal acceleration. Centripetal means “toward the center.”)

CONNECTION: Radial acceleration is not a new kind of acceleration. The acceleration vector for an object moving in uniform circular motion is directed radially inward toward the center of the circle. In uniform circular motion, the direction of the acceleration is radially inward (toward the center of the circular path).

v1 v6

a1 a2

CHECKPOINT 5.2

v2

a6

Does a radial acceleration mean that the speed of the object is changing?

a3

v5 a5 a4

Magnitude of the Radial Acceleration To find the magnitude of the radial acceleration for uniform circular motion, we must find the change in velocity Δv⃗ for a time interval Δt in the limit Δt → 0. The velocity keeps the same magnitude but changes direction at a steady rate, equal to the angular velocity w . In a time interval Δt, the velocity v⃗ rotates through an angle equal to the angular displacement Δq = w Δt. During this time interval, the velocity vector sweeps out an arc of a circle of “radius” v (Fig. 5.8). In the limit Δt → 0, the

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v3 v4

Figure 5.7 In uniform circular motion, the acceleration is always directed toward the center of the circle, perpendicular to the velocity (see interactive: circular motion).

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v2

magnitude of Δv⃗ becomes equal to the arc length, since a very short arc approaches a straight line. Then

∆q

Δ v⃗ = arc length = radius of circle × angle subtended

v1

= v Δq = v w Δt

∆v

Figure 5.8 The velocity vector sweeps out an arc of a circle whose “length” is nearly equal to that of the chord Δv⃗.

Acceleration is the rate of change of velocity, so the magnitude of the radial acceleration is Δv ⃗ = vw (w in radians per unit time) a r = a⃗ = ____ (5-11) Δt where absolute value symbols are used with the vector quantities to indicate their magnitudes. Velocity and angular velocity are not independent; v = w r. It is usually most convenient to write the magnitude of the radial acceleration in terms of one or the other of these two quantities. So we write the radial acceleration in two other equivalent ways using v = w r: 2

v a r = __ r

or

a r = w 2r

(w in radians per unit time)

(5-12)

Note that Eqs. (5-11) and (5-12) assume that w is expressed in radians per unit time (normally rad/s, but rad/min or rad/h would be correct).

Example 5.4 A Spinning CD If a CD spins at 210 rpm, what is the radial acceleration of a point on the outer rim of the CD? The CD is 12 cm in diameter.

Then using Eq. (5-12), the radial acceleration is

Strategy From the number of revolutions per minute, we can find the frequency and the angular velocity. The angular velocity and the radius of the CD enable us to calculate the radial acceleration.

Discussion When finding the radial acceleration, use whichever form of Eq. (5-12) is more convenient. For rotating objects such as the spinning CD, it’s usually easiest to think in terms of the angular velocity. For an object moving around a circle, such as a satellite in orbit whose speed is known, it might be easier to use v2/r. Since the two equations are equivalent, either can be used in any situation.

Solution We convert 210 rpm into a frequency in revolutions per second (Hz). rev × ___ rev = 3.5 Hz 1 ____ min = 3.5 ___ f = 210 ____ s min 60 s For each revolution, the CD rotates through an angle of 2p radians. The angular velocity is rev = 7.0p rad/s radians × 3.5 ___ w = 2p f = 2p ______ rev s

a r = w 2r = (7.0p rad/s)2 × 0.060 m = 29 m/s2

Practice Problem 5.4 Radial Acceleration of a Point on an Old Record What is the radial acceleration of a point 25.4 cm from the center of a record that is rotating at 78 rpm on a turntable?

Applying Newton’s Second Law to Uniform Circular Motion Now that we know the magnitude and direction of the acceleration of any object in uniform circular motion, we can use Newton’s second law to relate the net force acting on the object to the speed and radius of its motion. The net force is found in the usual way: each of the individual forces acting on the object is identified and then the forces are added as vectors. Every force acting must be exerted by some other object. Resist the temptation to add in a new, separate force just because something moves in a circle. For an object to move in a circle at constant speed, real, physical forces such as gravity, tension, normal forces, and friction must act on it; these forces combine to produce a net force that has the correct magnitude and is always perpendicular to the velocity of the object.

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RADIAL ACCELERATION

Problem-Solving Strategy for an Object in Uniform Circular Motion 1. Begin as for any Newton’s second law problem: identify all the forces acting on the object and draw an FBD. 2. Choose perpendicular axes at the point of interest so that one is radial and the other is tangent to the circular path. 3. Find the radial component of each force. 4. Apply Newton’s second law as follows:

∑F r = ma r where ∑Fr is the radial component of the net force and the radial component of the acceleration is v2 = w 2r a r = __ r (For uniform circular motion, neither the net force nor the acceleration has a tangential component.)

Example 5.5 The Hammer Throw What force does the athlete exert on the grip? What path does the hammer follow after release?

An athlete whirls a 4.00-kg hammer six or seven times around and then releases it. Although the purpose of whirling it around several times is to increase the hammer’s speed, assume that just before the hammer is released, it moves at constant speed along a circular arc of radius 1.7 m. At the instant she releases the hammer, it is 1.0 m above the ground and its velocity is directed 40° above the horizontal. The hammer lands a horizontal distance of 74.0 m away. What force does the athlete apply to the grip just before she releases it? Ignore air resistance. Strategy After release, the only force acting on the hammer is gravity. The hammer moves in a parabolic trajectory like any other projectile. By analyzing the projectile motion of the hammer, we can find the speed of the hammer just

40°

Uniform x circular motion

Solution During its projectile motion, the initial velocity has magnitude vi (to be determined) and direction q = 40° above the horizontal. Choosing the +y-axis pointing up, the displacement of the hammer (in component form) is ∆x = 74.0 m and ∆y = −1.0 m (Fig. 5.9), the acceleration of the hammer is ax = 0 and ay = −g, and the initial velocity is vix = vi cos q and viy = vi sin q. Then, from Eqs. (4-8) and (4-9), Δx = (v i cos q ) Δt

and

Δy = (v i sin q ) Δt − _12 g(Δt)2

Projectile motion (parabolic trajectory) ∆ y = –1.0 m

Release point

y

after its release. Just before release, the forces acting on the hammer are the tension in the cable and gravity. We can relate the net force on the hammer to its radial acceleration, calculated from the speed and radius of its path. The problem becomes two subproblems, one dealing with circular motion and the other with projectile motion. The final velocity for the circular motion is the initial velocity for the projectile motion.

Figure 5.9 ∆ x = 74.0 m

Path of the hammer from just before its release until it hits the ground. (Distances are not to scale.) continued on next page

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Example 5.5 continued

Solving the left equation for Δt and substituting into the right equation gives

( v cos q )

1 g _______ Δx Δx Δy = v i sin q _______ − __ v i cos q

2

mg

i

After a bit of algebra, we can solve for vi. First we multiply 2 through by 2v i cos2 q : 2 2 x sin q _______ 2v i cos2q Δy = 2v i cos2q Δ cos q 2

2 v i cos2 q Δx − __________ g _______ 2 v i cos q

(

)

2

2

Now we solve for vi: ________________________

g(Δx)2 _______________________ 2Δx cosq sinq − 2Δy cos2q

√

________________________________________

=

acceleration of magnitude v2/r. Newton’s second law in the radial direction is 2

v i (2 Δy cos2q − 2 Δx cos q sin q ) = −g(Δx)2

√

FBD for the hammer just before its release. (Not to scale.)

mv ∑F r = T = ma r = ____ r

Subtracting the first term on the right side from both sides 2 and factoring out v i ,

vi =

Figure 5.10

T 2

9.80 m/s2 × (74.0 m)2 ______________________________________ 2(74.0 m) cos 40° sin 40° − 2(−1.0 m) cos2 40°

= 26.9 m/s The net force on the hammer can be found from Newton’s second law. The two forces acting on the hammer are due to the tension in the cable and to gravity (Fig. 5.10). We ignore the gravitational force, assuming that the hammer’s weight is small compared with the tension in the cable. Then the tension in the cable is the only significant force acting on the hammer. Assuming uniform circular motion, the cable pulls radially inward and causes a radial

Substituting numerical values, 4.00 kg × (26.9 m/s)2 T = __________________ = 1700 N 1.7 m The tension is much larger than the weight of the hammer (≈ 40 N), so the assumption that we could ignore the weight is justified. The athlete must apply a force of magnitude 1700 N—almost 400 lb—to the grip. Discussion This example demonstrates the cumulative nature of physics concepts. The basic concepts keep reappearing, to be used over and over and to be extended for use in new contexts. Part of the problem involves new concepts (radial acceleration); the rest of the problem involves old material (Newton’s second law, projectile motion, and tension in a cord).

Practice Problem 5.5 Rotating Carousel A horse located 8.0 m from the central axis of a rotating carousel moves at a speed of 6.0 m/s. The horse is at a fixed height (it does not move up and down). What is the net force acting on a child seated on this horse? The child’s weight is 130 N.

Example 5.6 Conical Pendulum Suppose you whirl a stone in a horizontal circle at a slow speed so that the weight of the stone is not negligible compared with the tension in the cord. Then the cord cannot be horizontal—the tension must have a vertical component to cancel the weight and leave a horizontal net force (Fig. 5.11). If the cord has length L, the stone has mass m, and the cord makes an angle f with the vertical direction, what is the constant angular speed of the stone? Strategy The net force must point toward the center of the circle, since the stone is in uniform circular motion.

With the stone in the position depicted in Fig. 5.11a, the direction of the net force is along the +x-axis. This time the tension in the cord does not pull toward the center, but the net force does. Solution Start by drawing an FBD (Fig. 5.11b). Now apply Newton’s second law in component form. The acceleration has components ax = w 2r and ay = 0. For the x-components,

∑F x = T sin f = ma x = mw 2r continued on next page

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Example 5.6 continued

Now we eliminate the tension: (mw 2L) cos f = mg

y

y

Solving for w , f

√

_______

g w = _______ L cos f

T f

L x

Discussion We should check the dimensions of the final expression. Since cos f is dimensionless,

√[

______

mg

L/T2 ] = ___ 1 ______ [L] [T]

x r = L sin f (a)

(b)

Figure 5.11 (a) A stone is whirled in a horizontal circle of radius r = L sin f. (b) An FBD for the stone.

Since the problem does not specify r, we must express r in terms of L and f. In Fig. 5.11a, the radius forms a right triangle with the cord and the y-axis. Then r = L sin f and

∑F x = T sin f = mw 2L sin f Therefore, T = mw 2L. For the y-components,

∑F y = T cos f − mg = ma y = 0 ⇒ T cos f = mg

5.3

which is correct for w (SI unit rad/s). Another check is to ask how w and f are related for a given length cord. As f increases toward 90°, the cord gets closer to horizontal and the radius increases. In our expression, as f increases, cos f decreases and, therefore, w increases, in accordance with experience: the stone would have to be whirled faster and faster to make the cord more nearly horizontal.

Conceptual Practice Problem 5.6 Conical Pendulum on the Moon Examine the result of Example 5.6 to see how w depends on g, all other things being equal. Where the gravitational field is weaker, do you have to whirl the stone faster or more slowly to keep the cord at the same angle f ? Is that in accord with your intuition?

UNBANKED AND BANKED CURVES

Unbanked Curves When you drive an automobile in a circular path along an unbanked roadway, friction acting on the tires due to the pavement acts to keep the automobile moving in a curved path. This frictional force acts sideways, toward the center of the car’s circular path (Fig. 5.12). The frictional force might also have a tangential component; for example, if the car is braking, a component of the frictional force makes the car slow down by acting backward (opposite to the car’s velocity). For now we assume that the car’s speed is constant and that the forward or backward component of the frictional force is negligibly small. As long as the tires roll without slipping, there is no relative motion between the bottom of the tires and the road, so it is the force of static friction that acts (see Section 4.6). If the car is in a skid, then it is the smaller force of kinetic friction that acts as the bottom portion of the tire slides along the pavement. As the speed of the car increases, or for slippery surfaces with low coefficients of friction, the static frictional force may not be enough to hold the car in its curved path. Banked Curves To help prevent cars from going into a skid or losing control, the roadway is often banked (tilted at a slight angle) around curves so that the outer portion of the road—the part farthest from the center of curvature—is higher than the ⃗ so inner portion. Banking changes the angle and magnitude of the normal force, N, that it has a horizontal component Nx directed toward the center of curvature (in the

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Application of radial acceleration and contact forces: banked roadways

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N N

fs fs

a

y

mg

W a

v

x (a)

(b)

(c)

Figure 5.12 (a) A car negotiating a curve at constant speed on an unbanked roadway. The car’s acceleration is toward the center of the circular path. (b) A head-on view of the same car. The center of the circular path is to the left as viewed here. ⃗ and f⃗s are shown acting on one tire, but they represent the total normal and frictional forces acting on The force vectors N all four tires. (c) FBD for the car. radial direction—see Fig. 5.13). Then we need no longer rely solely on friction to keep the car moving in a circular path as it negotiates the curve; this component of the normal force acts to help the car remain on the curved path. Figure 5.13 shows a banked road with the normal force, the gravitational force, and, in parts (b) and (c), the radial component of the normal force Nx. We choose the axes so that the x-axis is in the direction of the acceleration, which is to the left; the axes are not parallel and perpendicular to the incline.

Figure 5.13 (a) Head-on view of a car negotiating a curve at constant speed on a banked roadway. The car’s acceleration is toward the center of the circular path (to the left as viewed ⃗ represents the total norhere). N mal force acting on all four tires. The car moves at just the right speed so that the frictional force is zero. (b) Resolving the normal force into x- and y-components. (c) FBD for the car with the normal force represented by its components.

Nx Ny

N N

y

Ny

q

q

y

Nx q

x

x

W

Wy = −mg

a (a)

(b)

(c)

Example 5.7 A Possible Skid: Unbanked and Banked Curves A car is going around an unbanked curve at the recommended speed of 11 m/s (see Fig. 5.12). (a) If the radius of curvature of the path is 25 m and the coefficient of static friction between the rubber and the road is ms = 0.70, does

the car skid as it goes around the curve? (b) What happens if the driver ignores the highway speed limit sign and travels at 18 m/s? (c) What speed is safe for traveling around the curve if the road surface is wet from a recent rainstorm and the continued on next page

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Example 5.7 continued

coefficient of static friction between the wet road and the rubber tires is m s = 0.50? (d) For a car to safely negotiate the curve in icy conditions at a speed of 13 m/s, what banking angle would be required (see Fig. 5.13)? Strategy The force of static friction is the only horizontal force acting on the car when the curve is not banked. The maximum force of static friction, which depends on road conditions, determines the maximum possible radial acceleration of the car. Therefore, we can compare the radial acceleration necessary to go around the curve at the specified speeds with the maximum possible radial acceleration determined by the coefficient of static friction. For part (d), in icy conditions we cannot rely much on friction, but the normal force has a horizontal component when the road is banked. Solution (a) We find the radial acceleration required for a speed of 11 m/s: (11 m/s)2 v2 = ________ a r = __ = 4.8 m/s2 r 25 m In order to have that acceleration, the component of the net force acting toward the center of curvature must be 2

v ∑F r = ma r = m __ r The only force with a horizontal component is the static frictional force acting on the tires due to the road (see the FBD in Fig. 5.12c). Therefore, 2

v ∑F r = f s = m __ r We must check to make sure that the maximum frictional force is not exceeded: f s ≤ m sN Since N = mg, the car can go around the curve without skidding as long as 2

v ≤ m mg m __ s r Thus, the radial acceleration cannot exceed msg. That limits the car to speeds satisfying ____

v ≤ √ m sgr

(c) In part (a), we found that the car is limited to speeds satisfying ____

v ≤ √m sgr

With m s = 0.50, the maximum safe speed is ____________________

v max = √ m sgr = √ 0.50 × 9.80 m/s2 × 25 m = 11 m/s ____

which is the same maximum speed recommended by the road sign. The highway engineer knew what she was doing when she had the sign placed along the road. (d) Finally, we find the banking angle that would enable cars to travel around the curve at 13 m/s in icy conditions. Assuming that friction is negligible, the horizontal component of the normal force is the only horizontal force. With the x-axis pointing toward the center of curvature and the y-axis vertical (Fig. 5.13),

∑F x = N sin q = mv2/r

(1)

∑F y = N cos q − mg = 0

(2)

and

Dividing Eq. (1) by Eq. (2) gives mv2/r = __ N sin q = tan q = _____ v2 _______ mg rg N cos q 2 (13 m/s)2 −1 __ v q = tan rg = tan −1 ______________2 = 35° 25 m × 9.80 m/s

(3)

Discussion Notice that the mass of the car does not appear in Eq. (3); the same banking angle holds for a scooter, motorcycle, car, or tractor-trailer. Notice also that the banking angle depends on the square of the speed. Automobile racetracks and bicycle racetracks have highly banked road surfaces at hairpin curves to minimize skidding of the high-speed vehicles. However, a banking angle of 35° is far greater than those used in practice along public roadways. Careful drivers would not try to drive around this curve in icy conditions at 13 m/s. What do you think might happen in icy conditions to a car that is traveling very slowly along a road banked at such a steep angle? Highway curves are banked at slight angles to help drivers who are driving at reasonable speeds for the road conditions. They are not banked to save speed demons from their folly.

Substituting numerical values, ____________________

v ≤ √0.70 × 9.80 m/s2 × 25 m = 13 m/s Since 11 m/s is less than the maximum safe speed of 13 m/s, the car safely negotiates the curve. (b) At 18 m/s, the car moves at a speed higher than the maximum safe speed of 13 m/s. The frictional force cannot supply the radial acceleration needed for the car to go around the curve—the car goes into a skid.

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Practice Problem 5.7 A Bobsled Race A bobsled races down an icy hill and then comes on a horizontal curve, located 60.0 m from the bottom of the hill. The sled is traveling at 22.4 m/s (50 mph) as it approaches the curve that has a radius of curvature of 50.0 m. The curve is banked at an angle of 45° and the frictional force on the sled runners is negligible. Does the sled make it safely around the curve?

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Application of radial acceleration: banking angle of an airplane y Lx Ly

x L

If there is no friction between the road and the tires, then there is only one speed at which it is safe to drive around a given curve. With friction, there is a range of safe speeds. The static frictional force can have any magnitude from 0 to msN and it can be directed either up or down the bank of the road. When an airplane pilot makes a turn in the air, the pilot makes use of a banking angle. The airplane itself is tilted as if it were traveling over an inclined surface. Because of the shape of the wings, an aerodynamic force called lift acts upward when the plane is in level flight. To go around a turn, the wings are tilted; the lift force stays perpendicular to the wings and, therefore, now has a horizontal component (Fig. 5.14), just as the normal force has a horizontal component for a car on a banked curve. This component supplies the necessary radial acceleration, while the vertical component of the lift holds the plane up. Therefore, 2

mv L x = ma r = ____ r

and

L y = mg

where the x-axis is horizontal and the y-axis is vertical. The lift force is different in its physical origin from the normal force, but its components split up the same way, so a plane in a turn banks its wings at the same angle that a road would be banked for the same speed and radius of curvature. Of course, planes usually move much faster than cars and use large radii of curvature when they turn. ⃗ Figure 5.14 The lift force L is perpendicular to the wings of the plane. To turn, the pilot tilts the wings so a component of the lift force is directed toward the center of the circular path of the plane.

A plane can’t make a turn without tilting its wings. Why can a car turn on a flat road?

5.4

Application of radial acceleration: circular orbits m Fgrav

ME

CHECKPOINT 5.3

r

CIRCULAR ORBITS OF SATELLITES AND PLANETS

A satellite can orbit Earth in a circular path because of the long-range gravitational force on the satellite due to the Earth. The magnitude of the gravitational force on the satellite is Gm 1m 2 (2-6) F = _______ r2 where the universal gravitational constant is G = 6.67 × 10−11 N·m2/kg2. We can use Newton’s second law to find the speed of a satellite in circular orbit at constant speed. Let m be the mass of the satellite and ME be the mass of the Earth. The direction of the gravitational force on the satellite is always toward the center of the Earth, which is the center of the orbit (Fig. 5.15). Since gravity is the only force acting on the satellite, mM r

E ∑F r = G _____ 2

Figure 5.15 Satellite in orbit around Earth.

where r is the distance from the center of the Earth to the satellite. Then, from Newton’s second law, 2

mv ∑F r = ma r = ____ r Setting these equal, 2 mM E ____ = mv G _____ 2 r r

Solving for the speed yields

√

_____

GM E v = _____ (5-13) r Notice that the mass of the satellite does not appear in the equation for speed; it has been algebraically canceled. The greater inertia of a more massive satellite is overcome

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by a proportionally greater gravitational force acting on it. Thus, the speed of a satellite in a circular orbit does not depend on the mass of the satellite. Equation (5-13) also shows that satellites in lower orbits (smaller radii) have greater speeds. We have been discussing satellites orbiting Earth, but the same principles apply to the circular orbits of satellites around other planets and to the orbits of the planets around the Sun. For planetary orbits, the mass of the Sun would appear in Eq. (5-13) instead of the Earth’s mass, because the Sun’s gravitational pull keeps the planets in their orbits. The planetary orbits are actually ellipses (Fig. 5.16) instead of circles, although for most of the planets in the solar system the ellipses are nearly circular. Mercury is the exception; its orbit is markedly different from a circle. Earth (e = 0.017)

Sun The other focus for the orbit of Comet Tempel 1

Comet Tempel 1 (e = 0.519)

Figure 5.16 The shapes of two elliptical orbits around the Sun. (The sizes of the orbits are not to scale.) An ellipse looks like an elongated circle. The degree of elongation is measured by a quantity called the eccentricity e. A circle is a special case of an ellipse with e = 0. Most of the planetary orbits are nearly circular, with the exception of Mercury. The sum of the distances from any point on an ellipse to each of two fixed points (called the foci) is constant. The Sun is at one focus of each orbit. Since Earth’s orbit is nearly circular, the second focus is very near the Sun.

Example 5.8 Speed of a Satellite The Hubble Space Telescope is in a circular orbit 613 km above Earth’s surface. The average radius of the Earth is 6.37 × 103 km and the mass of Earth is 5.97 × 1024 kg. What is the speed of the telescope in its orbit?

where m is the mass of the telescope. Solving for the speed, we find

√

_____

√

GM E v = _____ r

_________________________________

Strategy We first need to find the orbital radius of the telescope. It is not 613 km; that is the distance from the surface of Earth to the telescope. We must add the radius of the Earth to 613 km to find the orbital radius, which is measured from the center of the Earth to the telescope. Then we use Newton’s second law, along with what we know about radial acceleration. Solution The radius of the telescope’s orbit is r = 6.13 × 102 km + 6.37 × 103 km = (0.613 + 6.37) × 103 km = 6.98 × 103 km The net force on the telescope is equal to the gravitational force, given by Newton’s law of gravity. Newton’s second law relates the net force to the acceleration. Both are directed radially inward. GmM

v=

6.67 × 10−11 N ⋅ m2/kg2 × 5.97 × 1024 kg ________________________________ 6.98 × 106 m v = 7550 m/s = 27 200 km/h

Discussion Any satellite orbiting Earth at an altitude of 613 km has this same speed, regardless of its mass.

Practice Problem 5.8 Speed of Earth in Its Orbit What is the speed of Earth in its approximately circular orbit about the Sun? The average Earth–Sun distance is 1.50 × 1011 m and the mass of the Sun is 1.987 × 1030 kg. Once you find the speed, use it along with the distance traveled by the Earth during one revolution about the Sun to calculate the time in seconds for one orbit.

2

mv E = ____ ∑F r = ______ 2 r r

Kepler’s Laws of Planetary Motion At the beginning of the seventeenth century, Johannes Kepler (1571–1630) proposed three laws to describe the motion of the planets. These laws predated Newton’s laws of motion and his law of gravity. They offered a far simpler description of planetary motion

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than anything that had been proposed previously. We turn history on its head and look at one of Kepler’s laws as a consequence of Newton’s laws. The fact that Newton could derive Kepler’s laws from his own work on gravity was seen as a confirmation of Newtonian mechanics. Kepler’s laws of planetary motion are • The planets travel in elliptical orbits (Fig. 5.16) with the Sun at one focus of the ellipse. • A line drawn from a planet to the Sun sweeps out equal areas in equal time intervals. • The square of the orbital period is proportional to the cube of the average distance from the planet to the Sun.

Application of radial acceleration: Kepler’s third law for a circular orbit

Kepler’s first law can be derived from the inverse square law of gravitational attraction. The derivation is a bit complicated, but for any two objects that have such an attraction, the orbit of one about the other is an ellipse, with the stationary object located at one focus. (Planetary orbits are also affected by gravitational interactions with other planets; Kepler’s laws ignore these small effects.) The circle is a special case of an ellipse where the two foci coincide. We discuss Kepler’s second law in Chapter 8. We can derive Kepler’s third law from Newton’s law of universal gravitation for the special case of a circular orbit. The gravitational force gives rise to the radial acceleration: GmM

2

mv Sun = ____ ∑F r = _______ 2 r r

Solving for v yields

√

______

GM Sun v = ______ r The distance traveled during one revolution is the circumference of the circle, which is equal to 2p r. The speed is the distance traveled during one orbit divided by the period: GM 2 r ____ v = √ ______ r = T ______ Sun

Now we solve for T:

p

√

______

r3 T = 2p ______ GM Sun Squaring both sides yields 2

4p r 3 = constant × r 3 T 2 = ______ GM Sun

Application of radial acceleration: geostationary orbits

(5-14)

Equation (5-14) is Kepler’s third law: the square of the period of a planet is directly proportional to the cube of the average orbital radius. Although Kepler’s laws were derived for the motion of planets, they apply to satellites orbiting the Earth as well. Many satellites, such as those used for communications, are placed in a geostationary (or geosynchronous) orbit—a circular orbit in Earth’s equatorial plane whose period is equal to Earth’s rotational period (Fig. 5.17). A satellite in geostationary orbit remains directly above a particular point on the equator; to observers on the ground, it seems to hover above that point without moving. Due to their fixed positions with respect to Earth’s surface, geostationary satellites are used as relay stations for communication signals. In Example 5.9, we find the speed of a geostationary satellite.

CHECKPOINT 5.4 Do all geostationary satellites, no matter their masses, have to be the same height above Earth? Explain.

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163

CIRCULAR ORBITS OF SATELLITES AND PLANETS

Figure 5.17 Geostationary satellite orbiting the Earth. The satellite has the same angular velocity as Earth, so it is always directly above point P.

Earth Communications satellite P

r

Example 5.9 Geostationary Satellite A 300.0-kg communications satellite is placed in a geostationary orbit 35,800 km above a relay station located in Kenya. What is the speed of the satellite in orbit? Strategy The period of the satellite is 1 d or approximately 24 h. To find the speed of the satellite in orbit we use Newton’s law of gravity and his second law of motion along with what we know about radial acceleration. Solution Let m be the mass of the satellite and let M E be the mass of the Earth. Gravity is the only force acting on the satellite in its orbit. From Newton’s law of universal gravitation, Newton’s second law, and the expression for radial acceleration, GmM

2

mv E = ____ ∑F r = ______ 2 r r

Solving for the speed yields

√

_____

GM E v = _____ r We must add the mean radius of the Earth, R E = 6.37 × 106 m, to the height of the satellite above the Earth’s surface to find the orbital radius. r = h + R E = 3.58 × 107 m + 0.637 × 107 m = 4.217 × 107 m Substituting numerical values into the speed equation,

√

________________________________

6.67 × 10−11 N⋅m2/kg2 × 5.97 × 1024 kg v = ________________________________ 4.217 × 107 m _______________

= √9.443 × 106 m2/s2

v = 3.07 × 103 m/s Discussion This result, an orbital speed of 3.07 km/s and a distance above Earth’s surface of 35,800 km, applies to all

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geostationary satellites. The mass of the satellite does not matter; it cancels out of the equations for orbital radius and for speed. If we were actually putting a satellite into orbit, we would use a more accurate value for the period. We should use a time of 23 h and 56 min, which is the length of a sidereal day—the time for Earth to complete one rotation about its axis relative to the fixed stars. The solar day, 24 h, is the period of time between the daily appearances of the Sun at its highest point in the sky. The fact that Earth moves around the Sun is what causes the difference between these two ways of measuring the length of a day. The error introduced by using the longer time is negligible in this problem. We can use Kepler’s third law to check the result. Examples 5.8 and 5.9 both concern circular orbits around the Earth. Is the square of the period proportional to the cube of the orbital radius? From Example 5.8, r1 = 6.98 × 103 km and 2p r 1 _________________ 2p × 6.98 × 103 km = 5810 s T 1 = ____ v = 7.55 km/s From the present example, r2 = 4.22 × 107 m and 3600 s = 86 400 s T 2 = 24 h × ______ 1h The ratio of the squares of the periods is

( )

(

)

T 2 2 _______ 2 ___ = 86 400 s = 221 T1 5810 s

The ratio of the cubes of the radii is 4.22 × 10 m = 221 ( __r ) = ( ___________ 6.98 × 10 m ) r2 1

3

7

3

6

Practice Problem 5.9 Orbital Radius of Venus The period of the orbit of Venus around the Sun is 0.615 Earth years. Using this information, find the radius of its orbit in terms of R, the radius of Earth’s orbit around the Sun.

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Example 5.10 Orbiting Satellites A satellite revolves about Earth with an orbital radius of r1 and speed v1. If an identical satellite were set into circular orbit with the same speed about a planet of mass three times that of Earth, what would its orbital radius be?

Now we apply Newton’s second law to the orbit of the second satellite about the planet of mass 3ME: 2 Gm × 3M E ____ mv 1 _________ = 2 r2 r2

G × 3M E r 2 = ________ 2 v1

Strategy We can apply Newton’s law of universal gravitation and set up a ratio to solve for the new orbital radius. Solution From Newton’s second law, the magnitude of the gravitational force on the satellite is equal to the satellite’s mass times the magnitude of its radial acceleration: 2 GmM E ____ mv 1 ______ = 2 r1 r1

where M E and m are the masses of Earth and of the satellite, respectively. Solving for r1 yields GM E r 1 = _____ 2 v1

The ratio of r2 to r1 is 2

r 2 __________ G × 3M E/v 1 __ =3 r1 = 2 GM E/v 1 Thus, r2 = 3r1. Discussion Notice that we did not rush to substitute numerical values for the constants G and ME into the equations. We took the ratio r2/r1 so that these constants cancel.

Practice Problem 5.10 Period of Lunar Lander A lunar lander is orbiting about the Moon. If the radius of its orbit is _13 the radius of Earth, what is the period of its orbit?

5.5

NONUNIFORM CIRCULAR MOTION

So far we have focused on uniform circular motion. Now we can extend the discussion to nonuniform circular motion, where the angular velocity changes with time. Figure 5.18a shows the velocity vectors v⃗ 1 and v⃗ 2 at two different times for an object moving in a circle with changing speed. In this case, the speed is increasing (v2 > v1). In Fig. 5.18b, we subtract v⃗ 1 from v⃗ 2 to find the change in velocity. In the limit Δt → 0, Δv⃗ does not become perpendicular to the velocity, as it did for uniform circular motion. Thus, the direction of the acceleration is not radial if the speed is changing. However, we can resolve the acceleration into tangential and radial components

Figure 5.18 Motion along a circular path with a changing speed: (a) the magnitude of velocity v⃗ 2 is greater than the magnitude of velocity v⃗ 1, (b) the direction of Δv⃗ is not radial when the speed is changing, and (c) components of a⃗ can be taken along a tangent to the curved path (at) and along a radius (ar).

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r1 ar r2 a v2

v1

at

a = ∆v ∆t

v1

∆v v2

(a)

v2 = v1 + ∆v

(b)

(c)

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NONUNIFORM CIRCULAR MOTION

165

(Fig. 5.18c). The radial component ar changes the direction of the velocity, and the tangential component at changes the magnitude of the velocity. Since these are perpendicular components of the acceleration, the magnitude of the acceleration is ______

√

2

2

a = ar + at

Using the same method as in Section 5.2 to find the radial acceleration, but working here with only the radial component of the acceleration, we find that 2

v = w 2r a r = __ r

(w in radians per unit time)

(5-12)

For circular motion, whether uniform or nonuniform, the radial component of the acceleration is given by Eq. (5-12). However, in uniform circular motion the radial component of the acceleration ar is constant in magnitude, but for nonuniform circular motion ar changes as the speed changes. Also still true for nonuniform circular motion is the relationship between speed and angular speed: v = r w

(5-7)

Many problems involving nonuniform circular motion are solved in the same way as for uniform circular motion. We find the radial component of the net force and then apply Newton’s second law along the radial direction:

∑F r = ma r

Problem-Solving Strategy for an Object in Nonuniform Circular Motion 1. Begin as for any Newton’s second law problem: Identify all the forces acting on the object and draw an FBD. 2. Choose perpendicular axes at the point of interest so that one axis is radial and the other is tangent to the circular path. 3. Find the radial component of each force. 4. Apply Newton’s second law along the radial direction:

∑F r = ma r where v2 = w 2r a r = __ r 5. If necessary, apply Newton’s second law to the tangential force components:

∑F t = ma t The tangential acceleration component at determines how the speed of the object changes.

CHECKPOINT 5.5 For an object in circular motion, what is it about the radial acceleration that distinguishes between uniform and nonuniform circular motion?

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Example 5.11 Vertical Loop-the-Loop A roller coaster includes a vertical circular loop of radius 20.0 m (Fig. 5.19a). What is the minimum speed at which the car must move at the top of the loop so that it doesn’t lose contact with the track? Strategy A roller coaster car moving around a vertical loop is in nonuniform circular motion; its speed decreases on the way up and increases on the way back down. Nevertheless, it is moving in a circle and has a radial acceleration component as given in Eq. (5-12) as long as it moves in a circle. The only forces acting on the car are gravity and the normal force of the track pushing the car. Even if frictional or drag forces are present, at the top of the loop they act in the tangential direction and, thus, do not contribute to the radial component of the net force. At the top of the loop, the track exerts a normal force on the car as long as the car

moves with a speed great enough to stay on the track. If the car moves too slowly, it loses contact with the track and the normal force is then zero. Solution The normal force exerted by the track on the car at the top pushes the car away from the track (downward); the normal force cannot pull up on the car. Then, at the top of the loop, the gravitational force and the normal force both point straight down toward the center of the loop. Figure 5.19b is an FBD for the car. From Newton’s second law, 2

mv top

∑F r = N + mg = ma r = _____ r or 2

mv top N = _____ r − mg continued on next page

vtop N

mg

(b) N

atop

abottom

mg (c) vbottom

(a)

Figure 5.19 (a) A roller coaster car on a vertical circular loop. At the bottom of the loop, the car’s acceleration a⃗ bottom points upward toward the center of the circle. At the top of the loop, the car’s acceleration a ⃗ top points downward. The magnitude of a⃗ top is smaller than that of a ⃗ bottom because the speed is smaller at the top than at the bottom. (b) FBD for the car at the top of the loop. The track is above the car, so the normal force on the car due to the track is downward. (c) FBD for the car at the bottom of the loop.

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NONUNIFORM CIRCULAR MOTION

Example 5.11 continued

where vtop stands for the speed at the top. In this expression, N stands for the magnitude of the normal force. Since N ≥ 0,

(

)

2

v top m ____ − g ≥ 0 r or __

Discussion If the car is going faster than 14 m/s at the top, its radial acceleration is larger. The track pushing on the car provides the additional net force component that results in a larger radial acceleration. The minimum speed occurs when gravity alone provides the radial acceleration at the top of the loop. In other words, ar = g at the top of the loop for minimum speed.

v top ≥ √gr

Imagine sending a roller coaster car around the loop many times with a slightly smaller speed at the top each time. As vtop ___ approaches √ gr , the normal force at the top gets smaller and ___ smaller. When v top = √ gr , the normal force just becomes zero at the top of the loop. Any slower and the car loses contact with the track before getting to the highest point and would fall off the track unless prevented from falling by a backup safety mechanism. Therefore, the minimum speed at the top is

Practice Problem 5.11 Normal Force at the Bottom of the Track If the speed of the roller coaster at the bottom of the loop is 25 m/s, what is the normal force exerted on the car by the track in terms of the car’s weight mg? (See Fig. 5.19c.)

________________

v top = √gr = √9.80 m/s2 × 20.0 m = 14.0 m/s __

PHYSICS AT HOME Go outside on a warm day and fill a bucket with water. Swing the bucket around in a vertical circle over your head. What, if anything, keeps the water in the bucket when the bucket is upside down over your head? Why doesn’t the water spill out? Do any upward forces act on the water at that point? [Hint: The FBD for the water when it is directly overhead is similar to the FBD for a roller coaster car at the top of a loop.]

Conceptual Example 5.12 Acceleration of a Pendulum Bob A pendulum is released from rest at point A (Fig. 5.20). Sketch qualitatively an FBD and the acceleration vector for the pendulum bob at points B and C.

D

A

C B

Figure 5.20 A pendulum swings to the right, starting from rest at point A.

Strategy Two forces appear on each FBD: gravity and the force due to the cord. The gravitational force is the same at both points (magnitude mg, direction down), but the force due to the cord varies in magnitude and in direction. Its direction is always along the cord. The net force on the bob is the sum of these two forces and its direction is the same as the direction of the acceleration. We can use what we know about the acceleration to guide us in drawing the forces. The pendulum bob moves along the arc of a circle, but not at constant speed. At any point, the radial component of the acceleration is ar = v2/r. Unless v = 0, the radial acceleration component is nonzero. As the pendulum bob swings toward the bottom (from A to B), its speed is increasing; as it rises on the other side, its speed is decreasing. When the speed is increasing, the tangential component of the acceleration at is in the same direction as the velocity. From B to D, the speed is decreasing and at is in the direction opposite to continued on next page

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Conceptual Example 5.12 continued

the velocity. At point B, the speed is neither increasing nor decreasing and at = 0. Solution and Discussion At point B, the tangential acceleration is zero, so the acceleration points in the radial direction: straight up (Fig. 5.21). The tension in the cord pulls straight up and gravity pulls down, so the tension must be larger than T the weight of the bob to give an a upward net force. The acceleration at point C has both tangential and radial components. The tangential B B acceleration is opposite to the mg velocity because the bob is (a) (b) slowing down. Figure 5.22 shows the tangential and radial Figure 5.21 acceleration components added (a) Acceleration of the bob to form the acceleration vector at point B. (b) FBD for the bob at B. a⃗ and the FBD for the bob.

5.6

ar

a

T

Figure 5.22

at C

C mg

(a) At point C, the bob has both tangential and radial acceleration components. (b) FBD for the bob at C.

(b)

(a)

When the two forces are added, they give a net force in the same direction as the acceleration vector.

Conceptual Practice Problem 5.12 Analysis of the Bob at Point D Sketch the FBD and the acceleration vector for the pendulum bob at point D, the highest point in its swing to the right.

TANGENTIAL AND ANGULAR ACCELERATION

An object in nonuniform circular motion has a changing speed and a changing angular velocity. To describe how the angular velocity changes, we define an angular acceleration. If the angular velocity is w 1 at time t1 and is w 2 at time t2, the change in angular velocity is Δw = w 2 − w 1 The time interval during which the angular velocity changes is Δt = t2 − t1. The average rate at which the angular velocity changes is called the average angular acceleration, a av. w2 − w1 Δw ___ a av = _______ t 2 − t 1 = Δt

(5-15)

As we let the time interval become shorter and shorter, a av approaches the instantaneous angular acceleration, a. Δw a = lim ___ Δt→0 Δt

(5-16)

If w is in units of rad/s, a is in units of rad/s2. The angular acceleration is closely related to the tangential component of the acceleration. The tangential component of velocity is v t = r w

(5-7)

Equation (5-7) gives us a way to relate tangential acceleration to the angular acceleration. The tangential acceleration is the rate of change of the tangential velocity, so Δv Δw at = ___t = r ___ Δt Δt Therefore,

| |

(in the limit Δt → 0) at = r a

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(5-17)

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5.6 TANGENTIAL AND ANGULAR ACCELERATION

Table 5.1

CONNECTION:

Relationships Between q, w, and a for Constant Angular Acceleration

Constant Acceleration Along x-Axis

Constant Angular Acceleration

Δvx = vfx − vix = ax Δt Δx = _12 (v fx + v ix) Δt

(2-9) (2-11)

Δw = w f − w i = a Δt Δq = _12 (w f + w i) Δt

(5-18) (5-19)

Δx = v ix Δt + _12 a x(Δt)2

(2-12)

Δq = w i Δt + _12 a (Δt)2

(5-20)

(2-13)

2 wf

2 fx

2 ix

v − v = 2a x Δx

−

2 wi

= 2a Δq

Because a is the rate of change of w , and w is the rate of change of q, the equations for constant a have the same form as those for constant ax.

(5-21)

Constant Angular Acceleration The mathematical relationships between q, w , and a are the same as the mathematical relationships between x, vx, and ax that we developed in Chapter 2. Each quantity is the instantaneous rate of change of the preceding quantity. For example, ax is the rate of change of vx and a is the rate of change of w. Because the mathematical relationships are the same, we can draw upon the skills and equations we developed to solve problems with constant acceleration ax. All we have to do is take the equations for constant acceleration and replace x with q, vx with w , and ax with a (see Table 5.1). Equation (5-18) is the definition of average angular acceleration, with a av replaced by a since the angular acceleration is constant. Constant a means that w changes linearly with time; therefore, the average angular velocity is halfway between the initial and final angular velocities for any time interval w av = _12 (w i + w f). Using this form for w av along with the definition of w av (w av = Δq /Δt) yields Eq. (5-19). Equations (5-20) and (5-21) can be derived from the preceding two relations in a manner analogous to the derivations of Eqs. (2-12) and (2-13) in Section 2.4.

CHECKPOINT 5.6 A centrifuge is “spinning up” with a constant angular acceleration. Can the radial acceleration of a sample in the centrifuge be constant? Explain.

Example 5.13 A Rotating Potter’s Wheel A potter’s wheel rotates from rest to 210 rpm in a time of 0.75 s. (a) What is the angular acceleration of the wheel during this time, assuming constant angular acceleration? (b) How many revolutions does the wheel make during this time interval? (c) Find the tangential and radial components of the acceleration of a point 12 cm from the rotation axis when the wheel is spinning at 180 rpm. Strategy We know the initial and final frequencies, so we can find the initial and final angular velocities. We also know the time it takes for the wheel to get to the final angular velocity. That is all we need to find the average angular

acceleration that, for constant angular acceleration, is equal to the instantaneous angular acceleration. To find the number of revolutions, we can find the angular displacement Δq in radians and then divide by 2p rad/rev. We can find the angular velocity at t = 0.75 s and use it to find the radial acceleration component. The tangential acceleration is calculated from a. continued on next page

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Example 5.13 continued

Since w i = 0, we can check _____

Solution (a) Initially the wheel is at rest, so the initial angular velocity is zero.

From the answers to (a) and (b),

w i = 0 rad/s

Converting 210 rpm to rad/s gives the final angular velocity: rev × ___ rad = 7.0p rad/s 1 ____ min × 2p ___ w f = 210 ____ min

60 s

w f = √ 2a Δq

rev

The angular acceleration is the rate of change of the angular velocity. Since a is constant, we can calculate it by finding the average angular acceleration for the time interval: w f − w i ____________ 7.0p rad/s = 29 rad/s2 a = _______ = 7.0p rad/s − 0 = _________ 0.75 s − 0 0.75 s tf − ti

(b) The angular displacement is Δq = _12 (w f + w i) Δt = _12 (7.0p rad/s + 0)(0.75 s) = 8.25 rad

_____

____________________

√2a Δq = √2 × 29 rad/s2 × 8.25 rad = 22 rad/s The original value for w f in rad/s was 7.0p rad/s. Since p ≈ 22/7, the check is successful.

Practice Problem 5.13 The London Eye The London Eye, a Ferris wheel on the banks of the Thames, has radius 67.5 m. At its cruising angular speed, it takes 30.0 min to make one complete revolution. Suppose that it takes 20.0 s to bring the wheel from rest to its cruising speed and that the angular acceleration is constant during startup. (a) What is the angular acceleration during startup? (b) What is the angular displacement of the wheel during startup?

Since 2p rad = one revolution, the number of revolutions is 8.25 rad = 1.3 rev _________ 2p rad/rev (c) At 180 rpm, the angular velocity is rev × ___ rad = 6.0p rad/s 1 ____ min × 2p ___ w = 180 ____ s rev min

60

The radial acceleration component is a r = w 2r = (6.0p rad/s)2 × 0.12 m = 43 m/s2 and the tangential acceleration component is a t = a r = 29 rad/s2 × 0.12 m = 3.5 m/s2 Discussion A quick check involves another of the equations for constant acceleration: The London Eye

w f − w i = 2a Δq 2

2

5.7 Application of apparent weight and circular motion: apparent weightlessness of orbiting astronauts

APPARENT WEIGHT AND ARTIFICIAL GRAVITY

You are no doubt familiar with pictures of astronauts “floating” while in orbit around the Earth. It seems as if the astronauts are weightless. To be truly weightless, the force of gravity acting on the astronauts due to Earth would have to be zero, or at least close to zero. Is it? We can calculate the weight of an astronaut in orbit. The orbital altitude for the space shuttle is typically about 600 km above the Earth. Then the orbital radius is 600 km + 6400 km = 7000 km. Comparing the astronaut’s weight in orbit to his or her weight on Earth’s surface, GM Em ________ 2 W orbit _________ (R E + h)2 ________ RE (6400 km)2 ______ __________ = = 0.84 = = W surface GM Em (R + h)2 (7000 km)2 ______ 2

E

RE

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171

APPARENT WEIGHT AND ARTIFICIAL GRAVITY

The weight in orbit is 0.84 times the weight on the surface. The astronaut weighs less but certainly isn’t weightless! Then why does the astronaut seem to be weightless? Recall Section 4.10 on the apparent weightlessness of someone unfortunate enough to be in an elevator when the cable snaps. In that situation, the elevator and the passenger both have the same acceleration (a⃗ = g⃗). Similarly, the astronaut has the same acceleration as the space shuttle, which is equal to the local gravitational field g⃗. Apparent weightlessness occurs when a⃗ = g⃗, where g⃗ is the local gravitational field. Application: Artificial Gravity In order for astronauts to spend long periods of time living in a space station without the deleterious effects of apparent weightlessness, artificial gravity would have to be created on the station. Many science fiction novels and movies feature ring-shaped space stations that rotate in order to create artificial gravity for the occupants. In a rotating space station, the acceleration of an astronaut is inward (toward the rotation axis), but the apparent gravitational field is outward. Therefore, the ceiling of rooms on the station are closest to the rotation axis and the floor is farthest away (Fig. 5.23). The centrifuge is a device that creates artificial gravity on a smaller scale. Centrifuges are common not only in scientific and medical laboratories but also in everyday life. The first successful centrifuge was used to separate cream from milk in the 1880s. Water drips out of sopping wet clothes due to the pull of gravity when the clothes are hung on a clothesline, but the water is removed much faster by the artificial gravity created in the spin cycle of a washing machine. The human body can be adversely affected not only by too little artificial gravity, but also by too much. Stunt pilots have to be careful about the accelerations to which they subject their bodies. An acceleration of about 3g can cause temporary blindness due to an inadequate supply of oxygen to the retina; the heart has difficulty pumping blood up to the head due to the blood’s increased apparent weight. Larger accelerations can cause unconsciousness. Pressurized flight suits enable pilots to sustain accelerations up to about 5g.

Figure 5.23 A rotating space station from the movie 2001: A Space Odyssey. Note jogger in the upper half running on the floor.

Example 5.14 Stunt Pilot Dave wants to practice vertical circles for a flying show exhibition. (a) What must the minimum radius of the circle be to ensure that his acceleration at the bottom does not exceed 3.0g? The speed of the plane is 78 m/s at the bottom of the circle. (b) What is Dave’s apparent weight at the bottom of the circular path? Express your answer in terms of his true weight. a

Strategy For the minimum radius, we use the maximum possible radial acceleration since a r = v2/r. For the maximum radial acceleration, the tangential acceleration must______ be zero (Fig. 5.24)—the magnitude of the acceleration is a = √a 2r + a 2t . Therefore, the radial acceleration component has magnitude 3.0g at the bottom. To find Dave’s apparent weight, we do not need to use the numerical value of the radius found in part (a); we already know that his acceleration is upward and has magnitude 3.0g. Solution (a) The magnitude of the radial acceleration is a r = v2/r Solving for the radius, 2

Figure 5.24 Velocity and acceleration vectors for the plane at the bottom of the circle.

v

2

v v = ____ r = __ a r 3.0g (78 m/s)2 = ____________2 = 210 m 3.0 × 9.8 m/s continued on next page

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Example 5.14 continued

(b) Dave’s apparent weight is the magnitude of the normal force of the plane pushing up on him. Let the y-axis point upward. The normal force is up and the gravitational force is down (Fig. 5.25). Then N

Discussion It might have been tempting to jump to the conclusion that an acceleration of 3.0g means that his apparent weight is 3.0mg. But is his apparent weight zero when his acceleration is zero? No.

y

∑F y = N − mg = ma y x

Practice Problem 5.14 Weight

where ay = +3.0g. Therefore, W ′ = N = m(g + a y) = 4.0mg

mg

Figure 5.25 FBD for Dave.

His apparent weight is 4.0 times his true weight.

Astronaut’s Apparent

What is the apparent weight of a 730-N astronaut when her spaceship has an acceleration of magnitude 2.0g in the following two situations: (a) just above the surface of Earth, acceleration straight up; (b) far from any stars or planets?

Application of Apparent Weight to Objects at Rest with Respect to Earth’s Surface Due to Earth’s rotation, the effective value of g measured in a coordinate system attached to Earth’s surface is slightly less than the true value of the gravitational field strength (see Section 4.5). The net force of an object placed on a scale is not zero because the object has a radial acceleration a r = w 2r directed toward Earth’s axis of rotation (Fig. 5.26). This relatively small effect is greatest where r is greatest—at the equator, where the effective value of g is about 0.3% smaller than the true value of g.

Arctic Circle a

Equator

Figure 5.26 An object at rest with respect to Earth’s surface has a radial acceleration due to Earth’s rotation. The angular frequency w is the same everywhere, so the radial acceleration ar = w 2r is proportional to the distance from the axis of rotation.

a

Axis of rotation

Master the Concepts • The angular displacement Δq is the angle through which an object has turned. Positive and negative angular displacements indicate rotation in different directions. Conventionally, positive represents counterclockwise motion.

rf

• Average angular velocity: ∆q qf

ri

q2 − q1 Δ q ___ w av = ______ t 2 − t 1 = Δt

qi x

q f – q i = ∆q

(5-2)

• Average angular acceleration: w 2 − w 1 ____ Δw a av = _______ t 2 − t 1 = Δt

(5-15)

continued on next page

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CONCEPTUAL QUESTIONS

f = 1/T

(5-8)

w = v/r = 2p f

(5-9)

Master the Concepts continued

• The instantaneous angular velocity and acceleration are the limits of the average quantities as Δt → 0. • A useful measure of angle is the radian: 2p radians = 360° Using radian measure for q , the arc length s of a circle of radius r subtended by an angle q is s=qr

(q in radian measure)

where the SI unit of angular velocity is rad/s and that of frequency is rev/s = Hz. • A rolling object is both rotating and translating. If the object rolls without skidding or slipping, then v axle = r w

(5-4) Hawk ZX

(w in radians per unit time)

Hawk Z X

• Using radian measure for w , the speed of an object in circular motion (including a point on a rotating object) is v = r w

(5-10)

vaxle ∆q

(5-7)

• Using radian measure for a , the tangential acceleration component is related to the angular acceleration by a t = r a

(a in radians per time2)

s = r∆q

(5-17)

• An object moving in a circle has a radial acceleration component given by v2 = w 2r (w in radians per unit time) (5-12) a r = __ r • The tangential and radial acceleraar tion components are two perpendicular components of the acceleration a at vector. The radial acceleration component changes the direction of the velocity and the tangential acceleraa = ∆v ∆t tion component changes the speed. • Uniform circular motion means that v and w are constant. In uniform circular motion, the time to complete one revolution is constant and is called the period T. The frequency f is the number of revolutions completed per second.

• Kepler’s third law says that the square of the period of a planetary orbit is proportional to the cube of the orbital radius: T 2 = constant × r 3

• For constant angular acceleration, we can use equations analogous to those we developed for constant acceleration ax:

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Δw = w f − w i = a Δt

(5-18)

Δq = _12 (w f + w i) Δt

(5-19)

Δq = w i Δt + _12 a (Δt)2

(5-20)

2 wf

Conceptual Questions 1. Is depressing the “accelerator” (gas pedal) of a car the only way that the driver can make the car accelerate (in the physics sense of the word)? If not, what else can the driver do to give the car an acceleration? 2. Two children ride on a merry-go-round. One is 2 m from the axis of rotation and the other is 4 m from it. Which child has the larger (a) linear speed, (b) acceleration, (c) angular speed, and (d) angular displacement? 3. Explain why the orbital radius and the speed of a satellite in circular orbit are not independent. 4. In uniform circular motion, is the velocity constant? Is the acceleration constant? Explain. 5. In uniform circular motion, the net force is perpendicular to the velocity and changes the direction of the velocity but not the speed. If a projectile is launched horizontally,

(5-14)

6.

7.

8.

9.

−

2 wi

= 2a Δq

(5-21)

the net force (ignoring air resistance) is perpendicular to the initial velocity, and yet the projectile gains speed as it falls. What is the difference between the two situations? The speed of a satellite in circular orbit around a planet does not depend on the mass of the satellite. Does it depend on the mass of the planet? Explain. A flywheel (a massive disk) rotates with constant angular acceleration. For a point on the rim of the flywheel, is the tangential acceleration component constant? Is the radial acceleration component constant? Explain why the force of gravity due to the Earth does not pull the Moon in closer and closer on an inward spiral until it hits Earth’s surface. When a roller coaster takes a sharp turn to the right, it feels as if you are pushed toward the left. Does a force push you to the left? If so, what is it? If not, why does there seem to be such a force?

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CHAPTER 5 Circular Motion

10. Is there anywhere on Earth where a bathroom scale reads your true weight? If so, where? Where does your apparent weight due to Earth’s rotation differ most from your true weight? 11. A physics teacher draws a cutaway view of a car rounding a banked curve as a rectangle atop a right triangle. A student draws a coordinate system based on the drawing. Is there another choice of axes that would make the problem easier to solve? 12. A bridal party is at a rehearsal dinner. The best Brandy snifter man challenges the bridegroom to pick up an olive using only a brandy snifOlive ter. How does the groom accomplish this task?

Multiple-Choice Questions 1. A spider sits on a turntable that is rotating at a constant 33 rpm. The acceleration a⃗ of the spider is (a) greater the closer the spider is to the central axis. (b) greater the farther the spider is from the central axis. (c) nonzero and independent of the location of the spider on the turntable. (d) zero. y

B

A C

x Earth

D

Multiple-Choice Questions 2–5 and Problem 36 Questions 2–5: A satellite in orbit travels around the Earth in uniform circular motion. In the figure, the satellite moves counterclockwise (ABCDA). Answer choices: (a) +x (b) +y (c) −x (d) −y (e) 45° above +x (toward +y) (f) 45° below +x (toward −y) (g) 45° above −x (toward +y) (h) 45° below −x (toward −y) 2. What is the direction of the satellite’s instantaneous velocity at point D?

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3. What is the direction of the satellite’s average velocity for one quarter of an orbit, starting at C and ending at D? 4. What is the direction of the satellite’s average acceleration for one half of an orbit, starting at C and ending at A? 5. What is the direction of the satellite’s instantaneous acceleration at point C? 6. Two satellites are in orbit around Mars with the same orbital radius. Satellite 2 has twice the mass of satellite 1. The radial acceleration of satellite 2 has (a) twice the magnitude of the radial acceleration of satellite 1. (b) the same magnitude as the radial acceleration of satellite 1. (c) half the magnitude of the radial acceleration of satellite 1. (d) four times the magnitude of the radial acceleration of satellite 1. Questions 7–8: A boy swings in a tire swing. Answer choices: (a) At the highest point of the motion (b) At the lowest point of the motion (c) At a point neither highest nor lowest (d) It is constant. 7. When is the tension in the rope the greatest? 8. When is the tangential acceleration the greatest? Questions 9–10 concern these three statements: (1) Its acceleration is constant. (2) Its radial acceleration component is constant in magnitude. (3) Its tangential acceleration component is constant in magnitude. 9. An object is in uniform circular motion. Identify the correct statement(s). (a) 1 only (b) 2 only (c) 3 only (d) 1, 2, and 3 (e) 2 and 3 (f) 1 and 2 (g) 1 and 3 (h) None of them 10. An object is in nonuniform circular motion with constant angular acceleration. Identify the correct statement(s). (Use the same answer choices as Question 9.) 11. An astronaut is out in space far from any large bodies. He uses his jets to start spinning, then releases a baseball he has been holding in his hand. Ignoring the gravitational force between the astronaut and the baseball, how would you describe the path of the baseball after it leaves the astronaut’s hand? (a) It continues to circle the astronaut in a circle with the same radius it had before leaving the astronaut’s hand. (b) It moves off in a straight line. (c) It moves off in an ever-widening arc.

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PROBLEMS

12. An object moving in a circle at a constant speed has an acceleration that is (a) in the direction of motion (b) toward the center of the circle (c) away from the center of the circle (d) zero

chord. (The angle is measured by determining the angle between two tangents 100 ft apart; since each tangent is perpendicular to a radius, the angles are the same.) In modern railroad construction, track curvature is kept below 1.5°. What is the radius of curvature of a “1.5° curve”? [Hint: Since the angle is small, the length of the chord is approximately equal to the arc length along the curve.]

Problems

✦ Blue # 1

2

Combination conceptual/quantitative problem Biological or medical application Challenging problem Detailed solution in the Student Solutions Manual Problems paired by concept Text website interactive or tutorial

5.1 Description of Uniform Circular Motion 1. A carnival swing is fixed on the end of an 8.0-m-long beam. If the swing and beam sweep through an angle of 120°, what is the distance through which the riders move? 2. A soccer ball of diameter 31 cm rolls without slipping at a linear speed of 2.8 m/s. Through how many revolutions has the soccer ball turned as it moves a linear distance of 18 m? 3. Find the average angular speed of the second hand of a clock. 4. Convert these to radian measure: (a) 30.0°, (b) 135°, (c) _14 revolution, (d) 33.3 revolutions. 5. A bicycle is moving at 9.0 m/s. What is the angular speed of its tires if their radius is 35 cm? ( tutorial: car tire) 6. An elevator cable winds on a drum of radius 90.0 cm that is connected to a motor. (a) If the elevator is moving down at 0.50 m/s, what is the angular speed of the drum? (b) If the elevator moves down 6.0 m, how many revolutions has the drum made? 7. Grace is playing with her dolls and decides to give them a ride on a merry-go-round. She places one of them on an old record player turntable and sets the angular speed at 33.3 rpm. (a) What is their angular speed in rad/s? (b) If the doll is 13 cm from the center of the spinning turntable platform, how fast (in m/s) is the doll moving? 8. A wheel is rotating at a rate of 2.0 revolutions every 3.0 s. Through what angle, in radians, does the wheel rotate in 1.0 s? ✦ 9. In the construction of railroads, it is important that curves be gentle, so as not to damage passengers or freight. Curvature is not measured by the radius of curvature, but in the following way. First a 100.0-ft-long chord is measured. Then the curvature is reported as the angle subtended by two radii at the endpoints of the

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100 ft q

5.2 Radial Acceleration 10. Verify that all three expressions for radial acceleration (vw , v2/r, and w 2r) have the correct dimensions for an acceleration. 11. An apparatus is designed to study insects at an acceleration of magnitude 980 m/s2 (= 100g). The apparatus consists of a 2.0-m rod with insect containers at either end. The rod rotates about an axis perpendicular to the rod and at its center. (a) How fast does an insect move when it experiences a radial acceleration of 980 m/s2? (b) What is the angular speed of the insect? ( tutorial: centrifuge)

2.0 m

12. The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out. (a) What force keeps the people from falling out the bottom of the cylinder? (b) If the coefficient of friction is 0.40 and the cylinder has a radius of 2.5 m, what is the minimum angular speed of the cylinder so that the people don’t fall out? (Normally the operator runs it considerably faster as a safety measure.)

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CHAPTER 5 Circular Motion

13. Objects that are at rest relative to Earth’s surface are in circular motion due to Earth’s rotation. What is the radial acceleration of an African baobab tree located at the equator? ✦14. Earth’s orbit around the Sun is nearly circular. The period is 1 yr = 365.25 d. (a) In an elapsed time of 1 d, what is Earth’s angular displacement? (b) What is the change in Earth’s velocity, Δv⃗? (c) What is Earth’s average acceleration during 1 d? (d) Compare your answer for (c) to the magnitude of Earth’s instantaneous radial acceleration. Explain. 15. A 0.700-kg ball is on the end of a rope that is 1.30 m 70.0° in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole’s symmetry axis. The rope makes an angle of 70.0° with respect to the vertical. What is the tangen- Axis of rotation tial speed of the ball? 16. A child’s toy has a 0.100-kg ball attached to two strings, A and B. The strings are also attached to a stick and the ball swings around the stick along a A circular path in a horizontal plane. Both strings are 15.0 cm long and make an angle of 30.0° with B respect to the horizontal. (a) Draw an FBD for the ball showing the tension forces and the gravitational force. (b) Find the magnitude of the tension in each string when the ball’s angular speed is 6.00p rad/s. 17. A child swings a rock of mass m in a horizontal circle ✦ using a rope of length L. The rock moves at constant speed v. (a) Ignoring gravity, find the tension in the rope. (b) Now include gravity (the weight of the rock is no longer negligible, although the weight of the rope still is negligible). What is the tension in the rope? Express the tension in terms of m, g, v, L, and the angle q that the rope makes with the horizontal. ( tutorial: skip rope) ✦18. A conical pendulum consists of a bob (mass m) attached to a string (length L) swinging in a horizontal circle (Fig. 5.11). As the string moves, it sweeps out the area of a cone. The angle that the string makes with the vertical is f. (a) What is the tension in the string? (b) What is the period of the pendulum?

5.3 Unbanked and Banked Curves 19. A curve in a stretch of highway has radius R. The road is unbanked. The coefficient of static friction between the tires and road is ms. (a) What is the fastest speed that a car can safely travel around the curve? (b) Explain

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20.

21.

22.

23.

what happens when a car enters the curve at a speed greater than the maximum safe speed. Illustrate with an interactive: banked curve) FBD. ( A highway curve has a radius of 825 m. At what angle should the road be banked so that a car traveling at 26.8 m/s (60 mph) has no tendency to skid sideways on the road? [Hint: No tendency to skid means the frictional force is zero.] A curve in a highway has radius of curvature 120 m and is banked at 3.0°. On a day when the road is icy, what is the safest speed to go around the curve? A roller coaster car of mass 320 kg (including passengers) travels around a horizontal curve of radius 35 m. Its speed is 16 m/s. What is the magnitude and direction of the total force exerted on the car by the track? A velodrome is built for use in the Olympics. The radius of curvature of the surface is 20.0 m. At what angle should the surface be banked for cyclists moving at 18 m/s? (Choose an angle so that no frictional force is needed to keep the cyclists in their circular path. Large banking angles are used in velodromes.)

24. A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is not banked. The mass of the car is 1400 kg. (a) What is the frictional force on the car? (b) Does the frictional force necessarily have magnitude msN? Explain. ✦25. A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0°. The mass of the car is 1400 kg. (a) What is the frictional force on the car? (b) At what speed could you drive around this curve so that the force of friction is zero? 26. A curve in a stretch of highway has radius R. The road is ✦ banked at angle q to the horizontal. The coefficient of static friction between the tires and road is ms. What is the fastest speed that a car can travel through the curve? 27. An airplane is flying at constant speed v in a horizontal circle of radius r. The lift force on the wings due to the air is perpendicular to the wings. At what angle to the vertical must the wings be banked to fly in this circle? ( tutorial: plane in turn)

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PROBLEMS

✦28. A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.0 m/s without any friction. When a car is going 20.0 m/s on this curve, what minimum coefficient of static friction is needed if the car is to navigate the curve without slipping?

5.4 Circular Orbits of Satellites and Planets 29. What is the average linear speed of the Earth about the Sun? 30. The orbital speed of Earth about the Sun is 3.0 × 104 m/s and its distance from the Sun is 1.5 × 1011 m. The mass of Earth is approximately 6.0 × 1024 kg and that of the Sun is 2.0 × 1030 kg. What is the magnitude of the force exerted by the Sun on Earth? [Hint: Two different methods are possible. Try both.] 31. Two satellites are in circular orbits around Jupiter. One, with orbital radius r, makes one revolution every 16 h. The other satellite has orbital radius 4.0r. How long does the second satellite take to make one revolution around Jupiter? 32. The Hubble Space Telescope orbits Earth 613 km above Earth’s surface. What is the period of the telescope’s orbit? 33. Io, one of Jupiter’s satellites, has an orbital period of 1.77 d. Europa, another of Jupiter’s satellites, has an orbital period of about 3.54 d. Both moons have nearly circular orbits. Use Kepler’s third law to find the distance of each satellite from Jupiter’s center. Jupiter’s mass is 1.9 × 1027 kg. 34. A spy satellite is in circular orbit around Earth. It makes one revolution in 6.00 h. (a) How high above Earth’s surface is the satellite? (b) What is the satellite’s acceleration? 35. Mars has a mass of about 6.42 × 1023 kg. The length of a day on Mars is 24 h and 37 min, a little longer than the length of a day on Earth. Your task is to put a satellite into a circular orbit around Mars so that it stays above one spot on the surface, orbiting Mars once each Mars day. At what distance from the center of the planet should you place the satellite? ✦36. A satellite travels around Earth in uniform circular motion at an altitude of 35 800 km above Earth’s surface. The satellite is in geosynchronous orbit (that is, the time for it to complete one orbit is exactly 1 d). In the figure with Multiple-Choice Questions 2–5, the satellite moves counterclockwise (ABCDA). State directions in terms of the x- and y-axes. (a) What is the satellite’s instantaneous velocity at point C? (b) What is the satellite’s average velocity for one quarter of an orbit, starting at A and ending at B? (c) What is the satellite’s average acceleration for one quarter of an orbit, starting at A and ending at B? (d) What is the satellite’s instantaneous acceleration at point D?

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177

✦37. A spacecraft is in orbit around Jupiter. The radius of the orbit is 3.0 times the radius of Jupiter (which is RJ = 71 500 km). The gravitational field at the surface of Jupiter is 23 N/kg. What is the period of the spacecraft’s orbit? [Hint: You don’t need to look up any more data about Jupiter to solve the problem.]

5.5 Nonuniform Circular Motion 38. A roller coaster has a vertical loop with radius 29.5 m. With what minimum speed should the roller coaster car be moving at the top of the loop so that the passengers do not lose contact with the seats? 39. A pendulum is 0.80 m long and the bob has a mass of 1.0 kg. At the bottom of its swing, the bob’s speed is 1.6 m/s. (a) What is the tension in the string at the bottom of the swing? (b) Explain why the tension is greater than the weight of the bob. 40. A 35.0-kg child swings on a rope with a length of 6.50 m that is hanging from a tree. At the bottom of the swing, the child is moving at a speed of 4.20 m/s. What is the tension in the rope? 41. A car approaches the top of a hill that is shaped like a vertical circle with a radius of 55.0 m. What is the fastest speed that the car can go over the hill without losing contact with the ground?

5.6 Tangential and Angular Acceleration 42. A child pushes a merry-go-round from rest to a final angular speed of 0.50 rev/s with constant angular acceleration. In doing so, the child pushes the merry-goround 2.0 revolutions. What is the angular acceleration of the merry-go-round? 43. A cyclist starts from rest and pedals so that the wheels make 8.0 revolutions in the first 5.0 s. What is the angular acceleration of the wheels (assumed constant)? 44. During normal operation, a computer’s hard disk spins at 7200 rpm. If it takes the hard disk 4.0 s to reach this angular velocity starting from rest, what is the average angular acceleration of the hard disk in rad/s2? 45. Derive Eq. 5-20 from Eqs. 5-18 and 5-19. [Hint: See the derivation of Eq. (2-12) in Section 2.4.] 46. Derive Eq. 5-21 from Eqs. 5-18 and 5-19. ✦47. A pendulum is 0.800 m long and the bob has a mass of 1.00 kg. When the string makes an angle of q = 15.0° with the vertical, the bob is moving at 1.40 m/s. Find the tangential and radial acceleration components and the tension in the string. [Hint: Draw an FBD q for the bob. Choose the x-axis to be tangential to the motion of the bob and the y-axis to be radial. Apply Newton’s second law.] Problems 47 and 48

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✦48. Find the tangential acceleration of a freely swinging pendulum when it makes an angle q with the vertical. 49. A turntable reaches an angular speed of 33.3 rpm in 2.0 s, starting from rest. (a) Assuming the angular acceleration is constant, what is its magnitude? (b) How many revolutions does the turntable make during this time interval? 50. A wheel’s angular acceleration is constant. Initially its angular velocity is zero. During the first 1.0-s time interval, it rotates through an angle of 90.0°. (a) Through what angle does it rotate during the next 1.0-s time interval? (b) Through what angle during the third 1.0-s time interval? 51. A car that is initially at rest moves along a circular path with a constant tangential acceleration component of 2.00 m/s2. The circular path has a radius of 50.0 m. The initial position of the car is at the far west location on the circle and the initial velocity is to the north. (a) After the car has traveled _14 of the circumference, what is the speed of the car? (b) At this point, what is the radial acceleration component of the car? (c) At this same point, what is the total acceleration of the car? 52. A disk rotates with constant angular acceleration. The initial angular speed of the disk is 2p rad/s. After the disk rotates through 10p radians, the angular speed is 7p rad/s. (a) What is the magnitude of the angular acceleration? (b) How much time did it take for the disk to rotate through 10p radians? (c) What is the tangential acceleration of a point located at a distance of 5.0 cm from the center of the disk? 53. In a Beams ultracentrifuge, the rotor is suspended magnetically in a vacuum. Since there is no mechanical connection to the rotor, the only friction is the air resistance due to the few air molecules in the vacuum. If the rotor is spinning with an angular speed of 5.0 × 105 rad/s and the driving force is turned off, its spinning slows down at an angular rate of 0.40 rad/s2. (a) How long does the rotor spin before coming to rest? (b) During this time, through how many revolutions does the rotor spin? 54. The rotor of the Beams ultracentrifuge (see Problem 53) is 20.0 cm long. For a point at the end of the rotor, find the (a) initial speed, (b) tangential acceleration component, and (c) maximum radial acceleration component.

5.7 Apparent Weight and Artificial Gravity 55. If a washing machine’s drum has a radius of 25 cm and spins at 4.0 rev/s, what is the strength of the artificial gravity to which the clothes are subjected? Express your answer as a multiple of g. 56. A space station is shaped like a ring and rotates to simulate gravity. If the radius of the space station is 120 m, at what frequency must it rotate so that it simulates Earth’s gravity? [Hint: The apparent weight of the astronauts tutomust be the same as their weight on Earth.] ( rial: space station)

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57. A biologist is studying growth in space. He wants to simulate Earth’s gravitational field, so he r ar positions the plants on a g rotating platform in the geff spaceship. The distance Axis of rotation of each plant from the central axis of rotation is r = 0.20 m. What angular speed is required? ✦58. A biologist is studying plant growth and wants to simulate a gravitational field twice as strong as Earth’s. She places the plants on a horizontal rotating table in her laboratory on Earth at a distance of 12.5 cm from the axis of rotation. What angular speed will give the plants an effective gravitational field g⃗ eff, whose magnitude is 2.0g? [Hint: Remember to account for Earth’s gravitational field as well as the artificial gravity when finding the apparent weight.] ✦59. Objects that are at rest relative to the Earth’s surface are in circular motion due to Earth’s rotation. (a) What is the radial acceleration of an object at the equator? (b) Is the object’s apparent weight greater or less than its weight? Explain. (c) By what percentage does the apparent weight differ from the weight at the equator? (d) Is there any place on Earth where a bathroom scale reading is equal to your true weight? Explain. 60. A person of mass M stands on a bathroom scale inside a Ferris wheel compartment. The Ferris wheel has radius R and angular velocity w. What is the apparent weight of the person (a) at the top and (b) at the bottom? 61. A person rides a Ferris wheel that turns with constant angular velocity. Her weight is 520.0 N. At the top of the ride her apparent weight is 1.5 N different from her true weight. (a) Is her apparent weight at the top 521.5 N or 518.5 N? Why? (b) What is her apparent weight at the bottom of the ride? (c) If the angular speed of the Ferris wheel is 0.025 rad/s, what is its radius? ✦62. Objects that are at rest relative to Earth’s surface are in circular motion due to Earth’s rotation. What is the radial acceleration of a painting hanging in the Prado Museum in Madrid, Spain, at a latitude of 40.2° North? (Note that the object’s radial acceleration is not directed toward the center of the Earth.) Madrid

40.2° N Madrid, Spain

0° Equator

Three-dimensional view

40.2° Equator

Rotation axis Cross-sectional view

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COMPREHENSIVE PROBLEMS

63. A rotating flywheel slows down at a constant rate due to friction in its bearings. After 1 min, its angular velocity has diminished to 0.80 of its initial value w. At the end of the third minute, what is the angular velocity in terms of the initial value?

Comprehensive Problems 64. The Earth rotates on its own axis once per day (24.0 h). What is the tangential speed of the summit of Mt. Kilimanjaro (elevation 5895 m above sea level), which is located approximately on the equator, due to the rotation of the Earth? The equatorial radius of Earth is 6378 km. 65. A trimmer for cutting weeds and grass near trees and borders has a nylon cord of 0.23-m length that whirls about an axle at 660 rad/s. What is the linear speed of the tip of the nylon cord? 66. A high-speed dental drill is rotating at 3.14 × 104 rad/s. Through how many degrees does the drill rotate in 1.00 s? 67. A jogger runs counterclockwise around a path of radius 90.0 m at constant speed. He makes 1.00 revolution in 188.4 s. At t = 0, he is heading due east. (a) What is the jogger’s instantaneous velocity at t = 376.8 s? (b) What is his instantaneous velocity at t = 94.2 s? 68. Two gears A and B are in contact. The radius of gear A is twice that of gear B. (a) When A’s angular velocity is 6.00 Hz counterclockwise, what is B’s angular velocity? (b) If A’s radius to the tip of the teeth is 10.0 cm, what is the linear speed of a point on the tip of a gear tooth? What is the linear speed of a point on the tip of B’s gear tooth?

A

B

Problems 68 and 69 69. If gear A in Problem 68 has an initial frequency of 0.955 Hz and an angular acceleration of 3.0 rad/s2, how many rotations does each gear go through in 2.0 s? ✦70. The time to sunset can be estimated by holding out your arm, holding your fingers horizontally in front of your eyes, and counting the number of fingers that fit between the horizon and the setting Sun. (a) What is the angular speed, in radians per second, of the Sun’s apparent circular motion around the Earth? (b) Estimate the angle subtended by one finger held at arm’s length. (c) How long in minutes does it take the Sun to “move” through this same angle? 71. In the professional videotape recording system known as quadriplex, four tape heads are mounted on the

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circumference of a drum of radius 2.5 cm that spins at 1500 rad/s. (a) At what speed are the tape heads moving? (b) Why are moving tape heads used instead of stationary ones, as in audiotape recorders? [Hint: How fast would the tape have to move if the heads were stationary?] 72. The Milky Way galaxy rotates about its center with a period of about 200 million yr. The Sun is 2 × 1020 m from the center of the galaxy. How fast is the Sun moving with respect to the center of the galaxy? 73. A small body of mass 0.50 kg is attached by a 0.50-mlong cord to a pin set into the surface of a frictionless table top. The body moves in a circle on the horizontal surface with a speed of 2.0p m/s. (a) What is the magnitude of the radial acceleration of the body? (b) What is the tension in the cord? 74. Two blocks, one with mass m1 = 0.050 kg and one with mass m2 = 0.030 kg, are connected to one another by a string. The inner block is connected to a central pole by another string as shown in the figure with r1 = 0.40 m and r2 = 0.75 m. When the blocks are spun around on a horizontal frictionless surface at an angular speed of 1.5 rev/s, what is the tension in each of the two strings?

m2 m1

r1 r2

✦75. What’s the fastest way to make a U-turn at constant speed? Suppose that you need to make a 180° turn on a circular path. The minimum radius (due to the car’s steering system) is 5.0 m, while the maximum (due to the width of the road) is 20.0 m. Your acceleration must never exceed 3.0 m/s2 or else you will skid. Should you use the smallest possible radius, so the distance is small, or the largest, so you can go faster without skidding, or something in between? What is the minimum possible time for this U-turn? 76. The Milky Way galaxy rotates about its center with a period of about 200 million yr. The Sun is 2 × 1020 m from the center of the galaxy. (a) What is the Sun’s radial acceleration? (b) What is the net gravitational force on the Sun due to the other stars in the Milky Way? 77. Bacteria swim using a corkscrew-like helical flagellum that rotates. For a bacterium with a flagellum that has a pitch of 1.0 μm that rotates at 110 rev/s, how fast could it swim if there were no “slippage” in the medium in which it is swimming? The pitch of a helix is the distance between “threads.” 78. You place a penny on a turntable at a distance of 10.0 cm from the center. The coefficient of static friction between the penny and the turntable is 0.350. The turntable’s

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angular acceleration is 2.00 rad/s2. How long after you turn on the turntable will the penny begin to slide off of the turntable? A coin is placed on a turntable that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the turntable is 0.1, how far from the center of the turntable can the coin be placed without having it slip off? Grace, playing with her dolls, pretends the turntable of an old phonograph is a merry-go-round. The dolls are 12.7 cm from the central axis. She changes the setting from 33.3 rpm to 45.0 rpm. (a) For this new setting, what is the linear speed of a point on the turntable at the location of the dolls? (b) If the coefficient of static friction between the dolls and the turntable is 0.13, do the dolls stay on the turntable? Your car’s wheels are 65 cm in diameter and the wheels are spinning at an angular velocity of 101 rad/s. How fast is your car moving in kilometers per hour (assume no slippage)? In an amusement park rocket ride, cars are suspended from 4.25-m cables attached to rotating arms at a distance of 6.00 m from the axis of rotation. The cables swing out at an angle of 45.0° when the ride is operating. What is the angular speed of rotation?

4.25 m

4.25 m

45.0° 6.00 m

6.00 m

45.0°

83. Centrifuges are commonly used in biological laboratories for the isolation and maintenance of cell preparations. For cell separation, the centrifugation conditions are typically 1.0 × 103 rpm using an 8.0-cm-radius rotor. (a) What is the radial acceleration of material in the centrifuge under these conditions? Express your answer as a multiple of g. (b) At 1.0 × 103 rpm (and with a 8.0-cm rotor), what is the net force on a red blood cell whose mass is 9.0 × 10−14 kg? (c) What is the net force on a virus particle of mass 5.0 × 10−21 kg under the same conditions? (d) To pellet out virus particles and even to separate large molecules such as proteins, superhigh-speed centrifuges called ultracentrifuges are used in which the rotor spins in a vacuum to reduce heating due to friction. What is the radial acceleration inside an ultracentrifuge at 75 000 rpm with an 8.0-cm rotor? Express your answer as a multiple of g. ✦84. You take a homemade “accelerometer” to an amusement park. This accelerometer consists of a metal nut attached to a string and connected to a protractor, as

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shown in the figure. While riding a roller coaster that is moving at a uniform speed around a circular path, you hold up the accelerometer and notice that the string is making an angle of 55° with respect to the vertical with the nut pointing away from the center of the circle, as shown. (a) What is the radial acceleration of the roller coaster? (b) What is your radial acceleration expressed as a multiple of g? (c) If the roller coaster track is turn- Center of 55° ing in a radius of roller coaster’s 80.0 m, how fast are circular path you moving? 85. Massimo, a machinist, is cutting threads for a bolt on a lathe. He wants the bolt to have 18 threads per inch. If the cutting tool moves parallel to the axis of the wouldbe bolt at a linear velocity of 0.080 in./s, what must the rotational speed of the lathe chuck be to ensure the correct thread density? [Hint: One thread is formed for each complete revolution of the chuck.] 86. In Chapter 19 we will see that a charged particle can undergo uniform circular motion when acted on by a magnetic force and no other forces. (a) For that to be true, what must be the angle between the magnetic force and the particle’s velocity? (b) The magnitude of the magnetic force on a charged particle is proportional to the particle’s speed, F = kv. Show that two identical charged particles moving in circles at different speeds in the same magnetic field must have the same period. (c) Show that the radius of the particle’s circular path is proportional to the speed. ✦87. Find the orbital radius of a geosynchronous satellite. Do not assume the speed found in Example 5.9. Start by writing an equation that relates the period, radius, and speed of the orbiting satellite. Then apply Newton’s second law to the satellite. You will have two equations with two unknowns (the speed and radius). Eliminate the speed algebraically and solve for the radius.

Answers to Practice Problems 5.1 3.001 × 10−7 rad/s 5.2 1.65 m/s 5.3 1.9 min 5.4 17 m/s2 5.5 60 N toward the center of the circular path 5.6 More slowly 5.7 No 5.8 29.7 km/s; 3.17 × 107 s 5.9 0.723R 5.10 2.44 h 5.11 4.2mg

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ANSWERS TO CHECKPOINTS

5.12 Acceleration is purely tangential:

T

D a

D mg

5.13 (a) 1.75 × 10−4 rad/s2; (b) 0.0349 rad (2.00°) 5.14 (a) 2200 N; (b) 1500 N

Answers to Checkpoints 5.1 7200 Hz 5.2 No, for uniform circular motion the direction of the velocity vector is continuously changing but the magnitude of the velocity (the speed) is unchanged.

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5.3 The car has friction between the road and the tires to exert a horizontal force that causes the radial acceleration. 5.4 To be geosynchronous the satellites must have an orbital period of 1 d. The only quantities that affect the period are the mass of Earth and the radial distance from Earth’s center. These quantities are the same for all satellites no matter the mass. 5.5 For nonuniform circular motion, the direction and the magnitude of the velocity are both changing. There are tangential and radial components to the acceleration. The magnitude of the radial component changes as the speed changes. For uniform circular motion, the magnitude of the velocity is constant but the direction changes. The radial acceleration is constant in magnitude (and the tangential acceleration is zero). 5.6 The radial acceleration cannot be constant because the radius r is constant but the angular velocity w is changing ar = w 2r.

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REVIEW & SYNTHESIS: CHAPTERS 1–5

Review & Synthesis: Chapters 1−5 Review Exercises 1. From your knowledge of Newton’s second law and dimensional analysis, find the units (in SI base units) of the spring constant k in the equation F = kx, where F is a force and x is a distance. 2. Harrison traveled 2.00 km west, then 5.00 km in a direction 53.0° south of west, then 1.00 km in a direction 60.0° north of west. (a) In what direction, and for how far, should Harrison travel to return to his starting point? (b) If Harrison returns directly to his starting point with a speed of 5.00 m/s, how long will the return trip take? 3. (a) How many center-stripe road reflectors, separated by 17.6 yd, are required along a 2.20-mile section of curving mountain roadway? (b) Solve the same problem for a road length of 3.54 km with the markers placed every 16.0 m. Would you prefer to be the highway engineer in a country with a metric system or U.S. customary units? 4. A baby was spitting up after nursing and the pediatrician prescribed Zantac syrup to reduce the baby’s stomach acid. The prescription called for 0.75 mL to be taken twice a day for a month. The pharmacist printed a label for the bottle of syrup that said “3/4 tsp. twice a day.” By what factor was the baby overmedicated before the error was discovered at the baby’s next office visit two weeks later? [Hint: 1 tsp = 4.9 mL.] 5. Mike swims 50.0 m with a speed of 1.84 m/s, then turns around and swims 34.0 m in the opposite direction with a speed of 1.62 m/s. (a) What is his average speed? (b) What is his average velocity? 6. You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, it accelerates because the engines are at full throttle and because there is a catapult that propels the jet forward. You begin to wonder how much force is supplied by the catapult. You look on the Web and find that the flight deck of an aircraft carrier is about 90 m long, that an F-14 has a mass of 33 000 kg, that each of the two engines supplies 27 000 lb of force, and that the takeoff speed of such a plane is about 160 mi/h. Estimate the average force on the jet due to the catapult. 7. On April 15, 1999, a Korean cargo plane crashed due to a confusion over units. The plane was to fly from Shanghai, China, to Seoul, Korea. After take-off the plane climbed to 900 m. Then the first officer was instructed by the Shanghai tower to climb to 1500 m and maintain that altitude. The captain, after reaching 1450 m, twice asked the first officer at what altitude they should fly. He was twice told incorrectly they were to be at 1500 ft. The captain pushed the control column quickly forward and started a steep descent. The plane could not recover from the dive and crashed. How much above the correct altitude did the captain think they were when he started

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his rapid descent and lost control? (It turns out that aircraft altitudes are given in feet throughout the world except in China, Mongolia, and the former Soviet states where meters are used.) Paula swims across a river that is 10.2 m wide. She can swim at 0.833 m/s in still water, but the river flows with a speed of 1.43 m/s. If Paula swims in such a way that she crosses the river in as short a time as possible, how far downstream is she when she gets to the opposite shore? Peter is collecting paving stones from a quarry. He harnesses two dogs, Sandy and Rufus, in tandem to the ⃗ at a 15° angle to the loaded cart. Sandy pulls with force F north of east; Rufus pulls with 1.5 times the force of Sandy and at an angle of 30.0° south of east. Use a ruler and protractor to draw the force vectors to scale (choose a simple scale, such as 2.0 cm ↔ F ). Find the sum of the two force vectors graphically. Measure its length and find the magnitude of the sum from the scale used and the direction with the protractor. Will the cart stay on the road that runs directly west to east? A tire swing hangs at a 12° angle to the vertical when a stiff breeze is blowing. In terms of the tire’s weight W, (a) what is the magnitude of the horizontal force exerted on the tire by the wind? (b) What is the tension in the rope supporting the tire? Ignore the weight of the rope. An astronaut of mass 60.0 kg and a small asteroid of mass 40.0 kg are initially at rest with respect to the space station. The astronaut pushes the asteroid with a constant force of magnitude 250 N for 0.35 s. Gravitational forces are negligible. (a) How far apart are the astronaut and the asteroid 5.00 s after the astronaut stops pushing? (b) What is their relative speed at this time? In the fairy tale, Rapunzel, the beautiful maiden let her long golden hair hang down from the tower in which she was held prisoner so that her prince could use her hair as a climbing rope to climb the tower and rescue her. (a) Estimate how much force is required to pull a strand of hair out of your head. (b) There are about 105 hairs growing out of Rapunzel’s head. If the prince has a mass of 60 kg, estimate the average force pulling on each strand of hair. Will Rapunzel be bald by the time the prince reaches the top of the 30-m tower? Marie slides a paper plate with a slice of pizza across a horizontal table to her friend Jaden. The coefficient of friction between the table and plate is 0.32. If the pizza must travel 44 cm to get from Marie to Jaden, what initial speed should Marie give the plate of pizza so that it just stops when it gets to Jaden? Two wooden crates with masses as shown are tied together by a horizontal cord. Another cord is tied to the first crate and it is pulled with a force of 195 N at an angle of 20.0°, as shown. Each crate has a coefficient of

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REVIEW & SYNTHESIS: CHAPTERS 1–5

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before they hit the flat ground at the bottom of the cliff. kinetic friction of 0.550. 14.0 kg 9.00 kg 20.0° (b) Illustrate your answer by calculating the final speeds Find the tension in the for three rocks thrown in the specified directions with rope between the crates initial speeds of 10.0 m/s from a cliff that is 15.00 m and the magnitude of high. [Hint: Remember that the speed is the magnitude the acceleration of the system. of the velocity vector.] A boy has stacked two blocks on the floor so that a 5.00-kg block is on top of a 2.00-kg block. (a) If the 22. You are watching the Super Bowl where your favorite coefficient of static friction between the two blocks is team is leading by a score of 21 to 20. The other team is 0.400 and the coefficient of static friction between lining up to try to kick the winning field goal. You the bottom block and the floor is 0.220, with what watched their kicker warm up and you saw that he could minimum force should the boy push horizontally on kick the football with a velocity of 21 m/s. He lines up for the upper block to make both blocks start to slide a 45-yd kick. You watch as he kicks the ball at an angle of together along the floor? (b) If he pushes too hard, the 35° above the horizontal. Assuming he kicks the ball top block starts to slide off the lower block. What is the straight and with the same speed as during the warmup, maximum force with which he can push without that will the ball clear the 10-ft-high goal post, or will your happening if the coefficient of kinetic friction between favorite team win the Super Bowl? the bottom block and the floor is 0.200? 23. A coin is placed on a turntable 13.0 cm from the center. A binary star consists of two stars of masses M1 and The coefficient of static friction between the coin and the 4.0M1 a distance d apart. Is there any point where the turntable is 0.110. Once the turntable is turned on, its gravitational field due to the two stars is zero? If so, angular acceleration is 1.20 rad/s2. How long will it take where is that point? until the coin begins to slide? Two boys are trying to break a cord. Gerardo says they ✦24. Carlos and Shannon are sledding down a snow-covered should each pull in opposite directions on the two ends; slope that is angled at 12° below the horizontal. When Stefan says they should tie the cord to a pole and both sliding on snow, Carlos’s sled has a coefficient of pull together on the opposite end. Which plan is more friction m k = 0.10; Shannon has a “supersled” with likely to work? m k = 0.010. Carlos takes off down the slope starting from rest. When Carlos is 5.0 m from the starting point, Fish don’t move as fast as you might think. A small trout Shannon starts down the slope from rest. (a) How far has a top swimming speed of only about 2 m/s, which is have they traveled when Shannon catches up to Carlos? about the speed of a brisk walk (for a human, not a fish!). (b) How fast is Shannon moving with respect to Carlos It may seem to move faster because it is capable of large as she passes by? accelerations—it can dart about, changing its speed or direction very quickly. (a) If a trout starts from rest and 25. A proposed “space elevator” consists of a cable going accelerates to 2 m/s in 0.05 s, what is the trout’s average all the way from the ground to a space station in geoacceleration? (b) During this acceleration, what is the synchronous orbit (always above the same point on average net force on the trout? Express your answer as a Earth’s surface). Elevator “cars” would climb the cable multiple of the trout’s weight. (c) Explain how the trout to transport cargo to outer space. Consider a cable congets the water to push it forward. nected between the equator and a space station at height A spotter plane sees a school of tuna swimming at a H above the surface. Ignore the mass of the cable*. steady 5.00 km/h northwest. The pilot informs a fishing (a) Find the height H. (b) Suppose there is an elevator trawler, which is just then 100.0 km due south of the car of mass 100 kg sitting halfway up at height H/2. fish. The trawler sails along a straight-line course and What tension T would be required in the cable to hold intercepts the tuna after 4.0 h. How fast did the trawler the car in place? Which part of the cable would be under move? [Hint: First find the velocity of the trawler relatension (above the car or below it)? tive to the tuna.] 26. Anthony is going to drive a flat-bed truck up a hill that Julia is delivering newspapers. Suppose she is driving at makes an angle of 10° with respect to the horizontal 15 m/s along a straight road and wants to drop a paper out direction. A 36.0-kg package sits in the back of the truck. the window from a height of 1.00 m so it slides along the The coefficient of static friction between the package shoulder and comes to rest in the customer’s driveway. At what horizontal distance before the driveway should she drop the paper? The coefficient of kinetic friction between *More realistically, the mass of the cable is one of the primary the newspaper and the ground is 0.40. Ignore air resisengineering challenges of a space elevator. The cable is so long that tance and assume no bouncing or rolling. it would have a very large mass and would have to withstand an Three rocks are thrown from a cliff with the same initial enormous tension to support its own weight. The cable would need speeds but in different directions: one straight down, to be supported by a counterweight positioned beyond the geosynone straight up, and one horizontally. Ignore air resischronous orbit. Some believe carbon nanotubes hold the key to tance. (a) Compare the speeds of the three rocks just producing a cable with the required properties.

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REVIEW & SYNTHESIS: CHAPTERS 1–5

and the truck bed is 0.380. What is the maximum acceleration the truck can have without the package falling off the back? A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.0 m/s without any friction. On a cold day when the street is icy, the coefficient of static friction between the tires and the road is 0.120. What is the slowest speed the car can go around this curve without sliding down the bank? You want to lift a heavy box with a mass of 98.0 kg using the twopulley system as shown. With what minimum force do you have to pull down on the rope in order to lift the box at a constant velocity? One pulley is attached to the ceiling and one 98.0 kg to the box. At time t = 0, block A of mass 0.225 kg and block B of mass 0.600 kg rest on a horizontal frictionless surface a distance 3.40 m apart, with block A located to the left of block B. A horizontal force of 2.00 N directed to the right is applied to block A for a time interval Δt = 0.100 s. During the same time interval, a 5.00-N horizontal force directed to the left is applied to block B. How far from B’s initial position do the two blocks meet? How much time A B 3.40 m has elapsed from t = 0 until the blocks meet? A hamster of mass 0.100 kg gets onto his 20.0-cmdiameter exercise wheel and runs along inside the wheel for 0.800 s until its frequency of rotation is 1.00 Hz. (a) What is the tangential acceleration of the wheel, assuming it is constant? (b) What is the normal force on the hamster just before he stops? The hamster is at the bottom of the wheel during the entire 0.800 s. A pellet is fired from a toy cannon with a velocity of 12 m/s directed 60° above the horizontal. After 0.10 s, a second identical pellet is fired with the same initial velocity. After an additional 0.15 s have passed, what is the velocity of the first pellet with respect to the second? Ignore air resistance. A crate is sliding down a frictionless ramp that is inclined at 35.0°. (a) If the crate is released from rest, how far does it travel down the incline in 2.50 s if it does not get to the bottom of the ramp before the time has elapsed? (b) How fast is the crate moving after 2.50 s of travel? The invention of the cannon in the fourteenth century made the catapult unnecessary and ended the safety of castle walls. Stone walls were no match for balls shot from cannons. Suppose a cannonball of mass 5.00 kg is launched from a height of 1.10 m, at an angle of elevation of 30.0° with an initial velocity of 50.0 m/s, toward a castle wall of height 30 m and located 215 m away from the cannon. (a) The range of a projectile is defined as the horizontal distance traveled when the projectile returns to its

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original height. Derive an equation for the range in terms of vi, g, and angle of elevation q. (b) What will be the range reached by the projectile, if it is not intercepted by the wall? (c) If the cannonball travels far enough to hit the wall, find the height at which it strikes. ✦34. Two blocks are connected by a lightweight, flexible cord that passes over a single frictionless pulley. If m1 >> m2, find (a) the acceleration of each block and (b) the tension in the cord. 35. A runner runs three-quarters of the way around a circular track of radius 60.0 m, when she collides with another runner and trips. (a) How far had the runner traveled on the track before the collision? (b) What was the magnitude of the displacement of the runner from her starting position when the accident occurred? 36. A solar sailplane is going from Earth to Mars. Its sail is oriented to give a solar radiation force of 8.00 × 102 N. The gravitational force due to the Sun is 173 N and the gravitational force due to the Earth is 1.00 × 102 N. All forces are in the plane formed by Earth, Sun, and sailplane. The mass of the sailplane is 14 500 kg. (a) What is the net force (magnitude and direction) acting on the sailplane? (b) What is the acceleration of the sailplane? 8.00 × 102 N

(vectors not to scale) Solar sailplane 30.0° 173 N Sun

90° 1.00 × 102 N Earth

37. A star near the visible edge of a galaxy travels in a uniform circular orbit. It is 40 000 ly (light-years) from the galactic center and has a speed of 275 km/s. (a) Estimate the total mass of the galaxy based on the motion of the star? [Hint: For this estimate, assume the total mass to be concentrated at the galactic center and relate it to the gravitational force on the star.] (b) The total visible mass (i.e., matter we can detect via electromagnetic radiation) of the galaxy is 1011 solar masses. What fraction of the total mass of the galaxy is visible*, according to this estimate? ✦38. One of the tricky things about learning to sail is distinguishing the true wind from the apparent wind. When you are on a sailboat and you feel the wind on your face, you are experiencing the apparent wind—the motion of *In many galaxies the stars appear to have roughly the same orbital speed over a large range of distances from the center. A popular hypothesis to explain such galaxy rotation velocities is the existence of dark matter—matter that we cannot detect via electromagnetic radiation. Dark matter is thought to account for the majority of the mass of some galaxies and nearly a fourth of the total mass of the universe.

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REVIEW & SYNTHESIS: CHAPTERS 1–5

the air relative to you. The true wind is Velocity of (1) the speed and direcboat relative to water tion of the air relative to the water while the apparent wind is the speed and direction of the (2) air relative to the sailboat. The figure shows three different directions for the true wind along with one possible sail ori(3) entation as indicated by the position of the boom attached to the mast. (a) In each case, draw a vector diagram to establish the magnitude and direction of the apparent wind. (b) In which of the three cases is the apparent wind speed greater than the true wind speed? (Assume that the speed of the boat relative to the water is less than the true wind speed.) (c) In which of the three cases is the direction of the apparent wind direction forward of the true wind? [“Forward” means coming from a direction more nearly straight ahead. For example, (1) is forward of (2), which is forward of (3).]

MCAT Review The section that follows includes MCAT exam material and is reprinted with permission of the Association of American Medical Colleges (AAMC).

Read the paragraph and then answer the following four questions: The study of the flight of projectiles has many practical applications. The main forces acting on a projectile are air resistance and gravity. The path of a projectile is often approximated by ignoring the effects of air resistance. Gravity is then the only force acting on the projectile. When air resistance is ⃗R, is introduced. FR is included in the analysis, another force, F proportional to the square of the velocity, v. The direction of the air resistance is exactly opposite the direction of motion. The equation for air resistance is FR = bv2, where b is a proportionality constant that depends on such factors as the density of the air and the shape of the projectile. Air resistance was studied by launching a 0.5-kg projectile from a level surface. The projectile was launched with a speed of 30 m/s at a 40° angle to the surface. (Note: Assume air resistance is present unless otherwise specified.) 1. What is the magnitude of the vertical acceleration of the projectile immediately after it is launched? (Note: vy = vertical velocity component.) A. −g + (bvvy) B. −g − (bvvy) C. −g + (bvvy)/(0.5 kg) D. −g − (bvvy)/(0.5 kg)

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2. Approximately what horizontal distance does the projectile travel before returning to the elevation from which it was launched? (Note: Assume that the effects of air resistance are negligible.) A. 45 m B. 60 m C. 90 m D. 120 m 3. What is the magnitude of the horizontal component of air resistance on the projectile at any point during flight? (Note: vx = horizontal velocity component.) A. (bvvx) cos 40° B. (bvvx)/2 C. (bvvx) sin 40° D. bvvx 4. How does the amount of time it takes a projectile to reach its maximum height compare to the time it takes to fall from its maximum height back to the ground? (Note: b is greater than zero.) A. The times are the same. B. The time to reach its maximum height is greater. C. The time to fall back to the ground is greater. D. Either can be greater depending on the magnitude of b. Read the paragraph and then answer the following questions: A raft is constructed from wood and used in a river that varies in depth, width, and current at several points along its length. The river at point A has a current of 2 m/s, a width of 200 m, and an average depth of 3 m. 5. Near point A, the raft is rowed at a constant velocity of 2 m/s relative to the river current and perpendicular to it. How far does the raft travel before it reaches the other side? A. 224 m B. 250 m C. 283 m D. 400 m 6. A rower at point A rows the raft at 3 m/s relative to the river current and wants to end up directly across the river from the point of origin. At what angle to the shore should the rower direct the raft? A. cos−1 _53 B. cos−1 _25 C. cos−1 _32 D. cos−1 _23 7. A rock is dropped from a cliff that is 100 m above ground level. How long does it take the rock to reach the ground? (Note: Use g = 10 m/s2.) A. 4.5 s B. 10 s C. 14 s D. 20 s

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CHAPTER

6

Conservation of Energy

As a kangaroo hops along, the maximum height of each hop might be around 2.8 m. This height is only slightly higher than that achieved by an Olympic high jumper, but the kangaroo is able to achieve this height hop after hop as it travels with a horizontal velocity of 15 m/s or more. What features of kangaroo anatomy make this feat possible? It cannot simply be a matter of having more powerful leg muscles. If it were, the kangaroo would have to consume large amounts of energy-rich food to supply the muscles with enough chemical energy for each jump, but in reality a kangaroo’s diet consists largely of grasses that are poor in energy content. (See p. 210 for the answer.)

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6.1 THE LAW OF CONSERVATION OF ENERGY

• • • • •

gravitational forces (Section 4.5) Newton’s second law: force and acceleration (Sections 4.3–4.8) components of vectors (Section 3.2) circular orbits (Section 5.4) area under a graph (Sections 2.2 and 2.3)

6.1

Concepts & Skills to Review

THE LAW OF CONSERVATION OF ENERGY

Until now, we have relied on Newton’s laws of motion to be the fundamental physical laws used to analyze the forces that act on objects and to predict the motion of objects. Now we introduce another physical principle: the conservation of energy. A conservation law is a physical principle that identifies some quantity that does not change with time. Conservation of energy means that every physical process leaves the total energy in the universe unchanged. Energy can be converted from one form to another, or transferred from one place to another. If we are careful to account for all the energy transformations, we find that the total energy remains the same.

Conservation law: a physical law that identifies a quantity that does not change with time.

The Law of Conservation of Energy The total energy in the universe is unchanged by any physical process: total energy before = total energy after. “Turn down the thermostat—we’re trying to conserve energy!” In ordinary language, conserving energy means trying not to waste useful energy resources. In the scientific meaning of conservation, energy is always conserved no matter what happens. When we “produce” or “generate” electric energy, for instance, we aren’t creating any new energy; we’re just converting energy from one form into another that’s more useful to us. Conservation of energy is one of the few universal principles of physics. No exceptions to the law of conservation of energy have been found. Conservation of energy is a powerful tool in the search to understand nature. It applies equally well to radioactive decay, the gravitational collapse of a star, a chemical reaction, a biological process such as respiration, and to the generation of electricity by a wind turbine (Fig. 6.1). Think about the energy conversions that make life possible. Green plants use photosynthesis to convert the energy they receive from the Sun into stored chemical energy. When animals eat the plants, that stored energy enables motion, growth, and maintenance of body temperature. Energy conservation governs every one of these processes. Choosing Between Alternative Solution Methods Some problems can be solved using either energy conservation or Newton’s second law. Usually the energy method is easier. We often don’t know the details of all the forces acting on an object, making a direct application of Newton’s second law difficult. Using conservation of energy enables us to solve some of these problems more easily. When deciding which of these two approaches to use to solve a problem, try using energy conservation first. If the energy method does not lead to the solution, then try Newton’s second law.

Figure 6.1 At a California “wind farm,” these wind turbines convert the energy of motion of the air into electric energy.

Historical Development of the Principle of Energy Conservation While many scientists contributed to the development of the law of conservation of energy, the law’s first clear statement was made in 1842 by the German surgeon Julius Robert von Mayer (1814–1878). As a ship’s physician on a voyage to what is now Indonesia, Mayer had noticed that the sailors’ venous blood was a much deeper red in the tropics than it was in Europe. He concluded that less oxygen was being used because they didn’t need to “burn” as much fuel to keep the body warm in the warmer climate.

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CHAPTER 6 Conservation of Energy

Table 6.1

Some Common Forms of Energy

Form of Energy

Brief Description

Translational kinetic Elastic

Energy of translational motion (Chapter 6) Energy stored in a “springy” object or material when it is deformed (Chapter 6)* Energy of gravitational interactions (Chapter 6) Energy of rotational motion (Chapter 8)* Energy of the oscillatory motions of atoms and molecules in a substance caused by a mechanical wave passing through it (Chapters 11 and 12)* Energies of motion and interaction of atoms and molecules in solids, liquids, and gases, related to our sensation of temperature (Chapters 14 and 15)* Energy of interaction of electric charges and currents; energy of electromagnetic fields, including electromagnetic waves such as light (Chapters 14, 17–22) The total energy of a particle of mass m when it is at rest, given by Einstein’s famous equation E = mc2 (Chapters 26, 29, and 30) Energies of motion and interaction of electrons in atoms and molecules (Chapter 28)* Energies of motion and interaction of protons and neutrons in atomic nuclei (Chapters 29 and 30)

Gravitational Rotational kinetic Vibrational, acoustic, seismic Internal

Electromagnetic

Rest

Figure 6.2 The stored chemi-

Chemical

cal energy in food enables a weightlifter to lift the barbell over her head.

Nuclear

*Not a fundamental form of energy; made up of microscopic kinetic or electromagnetic energy.

In 1843, the English physicist James Prescott Joule (1818–1889), whose “day job” was running the family brewery, performed precise experiments to show that gravitational potential energy could be converted into a previously unrecognized form of energy (internal energy). It had previously been thought that forces like friction “use up” energy. Thanks to Mayer, Joule, and others, we now know that friction converts mechanical forms of energy into internal energy and that total energy is always conserved.

Forms of Energy Kinetic energy: energy of motion. Potential energy: stored energy due to interaction.

Translation: motion of an object in which any point of the object moves with the same velocity as any other point. (That is, the object does not rotate or change shape.)

Energy comes in many different forms (Fig. 6.2). Table 6.1 summarizes the main forms of energy discussed in this text and indicates the principal chapters that discuss each one. At the most fundamental level, there are only three kinds of energy: energy due to motion (kinetic energy), stored energy due to interaction (potential energy), and rest energy. To apply the energy conservation principle, we need to learn how to calculate the amount of each form of energy. There isn’t one formula that applies to all. Fortunately, we don’t have to learn about all of them at once. This chapter focuses on three forms of macroscopic mechanical energy (kinetic energy, gravitational potential energy, and elastic potential energy). For now, we use energy conservation as a tool to understand the translational motion of objects, but we do not consider rotational motion or changes in the internal energy of an object. We assume that these moving objects are perfectly rigid, so every point on the object moves through the same displacement.

6.2

WORK DONE BY A CONSTANT FORCE

To apply the principle of energy conservation, we need to learn how energy can be converted from one form to another. We begin with an example. Suppose the trunk in Fig. 6.3a weighs 220 N and must be lifted a height h = 4.0 m. To lift it at constant speed, Rosie must exert a force of 220 N on the rope, assuming an ideal pulley and rope. (We

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6.2 WORK DONE BY A CONSTANT FORCE

(a) Single pulley Initial

Figure 6.3 (a) Rosie moves a

(b) Two pulleys Final

Initial

Final

1

1 2 2

d

d

mg

d

1– 2 mg

trunk into her dorm room through the window. (b) The two-pulley system makes it easier for Rosie to lift the trunk: the force she must exert is halved. Is she getting something for nothing, or does she still have to do the same amount of work to lift the trunk?

2d

mg 1– 2 mg

ignore for now the brief initial time when she pulls with more than 220 N to accelerate the trunk from rest to its constant speed and the brief time she pulls with less than 220 N to let it come to rest.) As discussed in Example 4.12, she would only have to exert half the force (110 N) if she were to use the two-pulley system of Fig. 6.3b. She doesn’t get something for nothing, though. To lift the trunk 4.0 m, the sections of rope on both sides of pulley 2 must be shortened by 4.0 m, so Rosie must pull an 8.0-m length of rope. The two-pulley system enables her to pull with half the force, but now she must pull the rope through twice the distance. Notice that the product of the magnitude of the force and the distance is the same in both cases: 220 N × 4.0 m = 110 N × 8.0 m = 880 N⋅m = W This product is called the work (W) done by Rosie on the rope. Work is a scalar quantity; it does not have a direction. The same symbol W is often used for the weight of an object. To avoid confusion, we write mg for weight and let W stand for work. Don’t be misled by the many different meanings the word work has in ordinary conversation. We talk about doing homework, or going to work, or having too much work to do. Not everything we call “work” in conversation is work as defined in physics. The SI unit of work and energy is the newton-meter (N·m), which is given the name joule (symbol: J). Using either method, Rosie must do 880 J of work on the rope to lift the trunk. When we say that Rosie does 880 J of work, we mean that Rosie supplies 880 J of energy—the amount of energy required to lift the trunk 4.0 m. Work is an energy transfer that occurs when a force acts on an object that moves. Rosie does no work on the rope while she holds it in one place because the displacement is zero. She can just as well fasten it and walk away (Fig. 6.4). If there is no displacement, no work is done and no energy is transferred. Why then does she get tired if she holds the rope in place for a long time? Although Rosie does no work on the rope when holding it in place, work is done inside her body by muscle fibers, which have to contract and expand continually to maintain tension in the muscle. This internal work converts chemical energy into internal energy—the muscle warms up—but no energy is transferred to the trunk.

Figure 6.4 While the trunk is held in place by tying the rope, no work is done and no energy transfers occur.

SI unit of work and energy is the joule: 1 J = 1 N·m. Work: an energy transfer that occurs when a force acts on an object that moves.

Work Done by a Force not Parallel to the Displacement The force that Rosie exerts on the rope is in the same direction as the displacement of that end of the rope. More generally, how much work is done by a constant force that is at some angle to the displacement? It turns out that only the component of the force in the direction of the displacement

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CHAPTER 6 Conservation of Energy

F q F cos q

∆r

x

Figure 6.5 The work done by the force of the towrope on the water-skier during a displacement Δr⃗ is (F cos q ) Δr, where ⃗ (F cos q ) is the component of F in the direction of Δr⃗.

The scalar product (or dot product) of two vectors is defined by the ⃗ ⋅ B ⃗ = AB cos q, where q equation A ⃗ and B ⃗ when is the angle between A they are drawn starting at the same point. The special name and notation are used because this pattern occurs often in physics and mathematics. Work can be expressed ⃗ ⋅Δr⃗. using the scalar product: W = F See Appendix A.8 for more information on the scalar product.

does work. So, in general, the work done by a constant force is defined as the product of the magnitude of the displacement and the component of the force in the direction of the displacement. If q represents the angle between the force and displacement vectors when they are drawn starting at the same point, then the force component in the direction of the displacement is F cos q (Fig. 6.5). Therefore, work done by a constant force on an object can be written W = F Δr cos q, where F is the magnitude of the force and Δr is the magnitude of the displacement of the object. ⃗ acting on an object whose displacement is Δr⃗: Work done by a constant force F W = F Δ r cos q

(6-1)

⃗ and Δr⃗) (q is the angle between F If we choose the x-axis parallel to the displacement, then the component of the force in the direction of the displacement is Fx = F cos q, so W = Fx Δx. Alternatively, we can identify Δr cos q in Eq. (6-1) as the component of the displacement in the direction of the force (Fig. 6.6). Therefore, if we choose the x-axis parallel to the force, then the component of the displacement in the direction of the force is Δx and W = Fx Δx, as before: ⃗ acting on an object whose displacement is Δr⃗: Work done by a constant force F W = Fx Δx

(6-2)

⃗ and/or Δr⃗ parallel to the x-axis) (F ⃗ and Δr⃗ is less Work Can Be Positive, Negative, or Zero When the angle between F than 90°, cos q in Eq. (6-1) is positive, so the work done by the force is positive (W > 0). ⃗ and Δr⃗ is greater than 90°, cos q is negative and the work done If the angle between F by the force is negative (W < 0). Pay careful attention to the algebraic sign when calculating work. For example, the rope pulls Rosie’s trunk in the direction of its displacement, so q = 0 and cos q = 1; the rope does positive work on the trunk. At the same time, gravity pulls downward in the direction opposite to the displacement, so q = 180° and cos q = −1; gravity does negative work on the trunk. If the force is perpendicular to the displacement, q = 90° and cos 90° = 0, so the work done is zero. For example, the normal force exerted by a stationary surface on a sliding object does no work because it is perpendicular to the displacement of the object (Fig. 6.7a). Even if the surface is curved, at any instant the normal force is perpendicular to the velocity of the object. During a short time interval, then, the normal force is perpendicular to the displacement Δr⃗ = v⃗ Δt (Fig. 6.7b), so the normal force still does zero work. On the other hand, if the surface exerting the normal force is moving, then the normal force can do work. In Fig. 6.7c, the normal force exerted by the forklift on the pallet does positive work as it lifts the pallet.

q

Figure 6.6 The work done by the force of gravity on the hang glider during a displacement Δr⃗ is F(Δr cos q), which is F times the component of Δr⃗ in the ⃗ direction of F.

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F Αr

∆r cos q

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6.2 WORK DONE BY A CONSTANT FORCE

N

191

∆r

N

N

∆r = v∆t

∆r

(a)

(b)

(c)

Figure 6.7 (a) The normal force does no work because it is perpendicular to the displacement. (b) Even while sliding on a curved surface, the direction of the normal force is always perpendicular to the displacement during a short Δt, so it does no work. (c) The normal force that the forklift exerts on the pallet does work; it is not perpendicular to the displacement.

v v

Fg

Fg

A

P

T Fg

v

v (a)

(b)

(c)

Figure 6.8 (a) The tension in the string of a pendulum is always perpendicular to the velocity of the pendulum bob, so the string does no work on the bob. (b) No matter where the satellite is in its circular orbit, it experiences a gravitational force directed toward the center of the Earth. This force is always perpendicular to the satellite’s velocity; thus, gravity does no work on the satellite. (c) In an elliptical orbit, the gravitational force is not always perpendicular to the velocity. As the satellite moves counterclockwise in its orbit from point P to point A, gravity does negative work; from A to P, gravity does positive work. No work is done by the tension in the string on a swinging pendulum bob because the tension is always perpendicular to the velocity of the bob (Fig. 6.8a). Similarly, no work is done by the Earth’s gravitational force on a satellite in circular orbit (Fig. 6.8b). In a circular orbit, the gravitational force is always directed along a radius from the satellite to the center of the Earth. At every point in the orbit, the gravitational force is perpendicular to the velocity of the satellite (which is tangent to the circular orbit). By contrast, gravity does work on a satellite in a noncircular orbit (Fig. 6.8c). Only at points A and P are the gravitational force and the satellite’s velocity perpendicular. Wherever the angle between the gravitational force and the velocity is less than 90°, gravity is doing positive work, increasing the satellite’s kinetic energy by making it move faster. Wherever the angle between the gravitational force and the velocity is greater than 90°, gravity is doing negative work, decreasing the satellite’s kinetic energy by slowing it down.

Application of work: elliptical orbits

CHECKPOINT 6.2 A force is applied to a moving object, but no work is done. How is that possible?

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CHAPTER 6 Conservation of Energy

Example 6.1 Antique Chest Delivery A valuable antique chest, made in 1907 by Gustav Stickley, is to be moved into a truck. The weight of the chest is 1400 N. To get the chest from the ground onto the truck bed, which is 1.0 m higher, the movers must decide what to do. Should they lift it straight up, or should they push it up their 4.0-mlong ramp? Assume they push the chest on a wheeled dolly, which in a simplified model is equivalent to sliding it up a frictionless ramp. (a) Find the work done by the movers on the chest if they lift it straight up 1.0 m at constant speed. (b) Find the work done by the movers on the chest if they slide the chest up the 4.0-m-long frictionless ramp at constant speed by pushing parallel to the ramp. (c) Find the work done by gravity on the chest in each case. (d) Find the work done by the normal force of the ramp on the chest. Assume that all forces are constant. Strategy To calculate work, we use either Eq. (6-1) or Eq. (6-2), whichever is easier. For (a) and (b), we must calculate the force exerted by the movers. Drawing the FBD helps us calculate the forces. The ramp is a simple machine— just as for Rosie’s pulleys, the ramp cannot reduce the amount of work that must be done, so we expect the work done by the movers to be the same in both cases (ignoring friction). We Fm expect the work done by gravity to be negative in both cases, since the chest is moving up while gravity pulls down. The normal force due to the mg ramp is perpendicular to the displacement, so it does zero work on the chest. Since more than one Figure 6.9 force does work on the chest, FBD for the chest as we use subscripts to clarify the movers lift it straight up at conwhich work is being calculated.

Given: Weight of chest mg = 1400 N; length of ramp d = 4.0 m; height of ramp h = 1.0 m To find: Work done by movers on the chest Wm and work done by gravity on the chest Wg in the two cases; work done by the normal force on the chest WN. Solution (a) The displacement is 1.0 m straight up. The ⃗ m equal in magnitude movers must exert an upward force F to the weight of the chest to move it at constant speed (Fig. 6.9). The work done to lift it 1.0 m is Wm = Fm Δr cos q = 1400 N × 1.0 m × cos 0 = +1400 J ⃗ m and Δr⃗ are in the same direction where q = 0 because F (upward). (b) Figure 6.10 shows a sketch of the situation. We take the x-axis along the inclined ramp and the y-axis perpendicular to the ramp and resolve the gravitational force into its x- and y-components (Fig. 6.11a). Figure 6.11b is the FBD for the chest. Sliding along at constant speed, the chest’s acceleration is zero, so the x-components of the forces add to zero. N +y

y x

′ Fm

mg sin f

+x

f f

mg cos f

mg (a)

(b)

Figure 6.11 (a) Resolving mg⃗ into x- and y-components; (b) FBD for the chest.

stant speed.

continued on next page

′ Fm

N

4.0 m

mg

1.0 m

f

Figure 6.10 An antique chest is pushed up a ramp into a truck.

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Example 6.1 continued

The x-component of the gravitational force acts in the ⃗ m acts in –x-direction and the force exerted by the movers F′ the +x-direction. [The prime symbol indicates that the force exerted by the movers is different from what it was in part (a).]

along the y-axis is Δy = h = 1.0 m. The work done by gravity is the same for the two cases. Using Eq. (6-2), Wg = Fgy Δy = −mg Δy = −1400 N × 1.0 m = −1400 J

∑Fx = F m′ − mg sin f = 0

(d) The normal force of the ramp on the chest does no work because it acts in a direction perpendicular to the displacement of the chest.

From the right triangle formed by the ramp, the ground, and the truck bed in Fig. 6.12:

WN = N Δr cos 90° = 0

height of truck bed h sin f = _________________ = __ distance along ramp d We can now solve for F m ′: mgh Fm ′ = mg sin f = ____ d The force and displacement are in the same direction, so q = 0: mgh ′ d cos 0 = ____ × d × 1 = mgh = +1400 J Wm = F m d The work done by the movers is the same as in (a). (c) In both cases, the force of gravity has magnitude mg and acts downward. Choosing the y-axis so it now points upward, Fgy = −mg. In both cases, the component of the displacement Figure 6.12

4.0 m 1.0 m

f

Finding the angle of the incline.

Discussion Since d, the length of the ramp, cancels when multiplying the force times the distance, the work done by the movers is the same for any length ramp (as long as the height is the same). Using the ramp, the movers apply one quarter the force over a displacement that is four times larger. With a real ramp, friction acts to oppose the motion of the chest, so the movers would have to do more than 1400 J of work to slide the chest up the ramp. There’s no getting around it; if the movers want to get that chest into the truck, they’re going to have to do at least 1400 J of work.

Practice Problem 6.1 Bicycling Uphill A bicyclist climbs a 2.0-km-long hill that makes an angle of 7.0° with the horizontal. The total weight of the bike and the rider is 750 N. How much work is done on the bike and rider by gravity?

Total Work When several forces act on an object, the total work is the sum of the work done by each force individually: W = W + W + … + WN (6-3) total

1

2

Total work is sometimes called net work because the work done by each force can be positive, negative, or zero, so the total work is often smaller than the work done by any one of the forces. Because we assume a rigid object with no rotational or internal motion, another way to calculate the total work is to find the work done by the net force as if there were a single force acting: Wtotal = Fnet Δr cos q

(6-4)

To show that these two methods give the same result, let’s choose the x-axis in the direction of the displacement. Then the work done by each individual force is the x-component of the force times Δx. From Eq. (6-3), W = F Δx + F Δx + … + FNx Δx total

1x

2x

Factoring out the Δx from each term, Wtotal = (F1x + F2x + … + FNx) Δx = ( ∑Fx ) Δx ∑Fx is the x-component of the net force. In Eq. (6-4), Fnet cos q is the component of the net force in the direction of the displacement, which is the x-component of the net force. The two methods give the same total work.

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CHAPTER 6 Conservation of Energy

Example 6.2 Fun on a Sled Diane pulls a sled along a snowy path on level ground with her little brother Jasper riding on the sled (Fig. 6.13). The total mass of Jasper and the sled is 26 kg. The cord makes a 20.0° angle with the ground. As a simplified model, assume that the force of friction on the sled is determined by m k = 0.16, even though the surfaces are not dry (some snow melts as the runners slide along it). Find (a) the work done by Diane and (b) the work done by the ground on the sled while the sled moves 120 m along the path at a constant 3 km/h. (c) What is the total work done on the sled?

To find the tension, we need to eliminate the unknown normal force N. Equation (2) also involves the normal force N. We multiply Eq. (2) by m k,

Strategy (a,b) To find the work done by a force on an object, we need to know the magnitudes and directions of the force and of the displacement of the object. The sled’s acceleration is zero, so the vector sum of all the external forces (gravity, friction, rope tension, and the normal force) is zero. We draw the FBD and use Newton’s second law to find the tension in the rope and the force of kinetic friction on the sled. Then we apply Eq. (6-1) or Eq. (6-2) to find the work done by each. (c) We have two methods to find the total work. We’ll use Eq. (6-3) to calculate the total work and Eq. (6-4) as a check.

0.16 × 26 kg × 9.80 m/s2 = _______________________ = 41 N 0.16 × sin 20.0° + cos 20.0°

Solution (a) The FBD is shown in Fig. 6.14. The x- and y-axes are parallel and perpendicular to the ground, respectively. After resolving the tension into its components (Fig. 6.15), Newton’s second law with zero acceleration yields

∑Fx = +T cos q − fk = 0

(1)

∑Fy = +T sin q − mg + N = 0

(2)

where T is the tension and q = 20.0°. The force of kinetic friction is fk = mkN

(4)

Adding Eqs. (3) and (4) eliminates N. Then we solve for T. T cos q + mkT sin q − mkmg = 0 mkmg T = _____________ mksin q + cos q

Now that we know the tension, we find the work done by Diane. The component of the tension T acting parallel to the displacement is Tx = T cos q and the displacement is Δx = 120 m. The work done by Diane is WT = (T cos q )Δx = 41 N × cos 20.0° × 120 m = +4600 J (b) The force on the sled due to the ground has two components: N and fk. The normal force does no work since it is perpendicular to the displacement of the sled. Friction acts in a direction opposite to the displacement, so the angle between the force and displacement is 180°. The work done by friction is Wf = fk Δx cos 180° = −fk Δx From Eq. (1), fk = T cos q Therefore, the work done by the ground—the work done by the frictional force—is

Substituting this into Eq. (1) T cos q − mkN = 0

mk T sin q − mkmg + mkN = 0

Wf = −fk Δx = −(T cos q ) Δx

(3)

Except for the negative sign, Wf is the same as W T: W f = −4600 J. T

N

T

v = 3 km/h T 20.0°

20.0° fk

q

T sin q

T cos q q = 20.0°

Displacement = 120 m

mg

Figure 6.13

Figure 6.14

Jasper being pulled on a sled.

FBD.

Figure 6.15 Resolving the tension into x- and y-components. continued on next page

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6.3

KINETIC ENERGY

195

Example 6.2 continued

(c) The tension and friction are the only forces that do work on the sled. The normal force and gravity are both perpendicular to the displacement, so they do zero work. Wtotal = WT + Wf = 4600 J + (−4600 J) = 0 Discussion To check (c), note that the sled travels with constant velocity, so the net force acting on it is zero. W total = F netΔr cos q = 0. The speed (3 km/h) was not used in the solution. Assuming that the frictional force on the sled is independent of speed, Diane would exert the same force to pull the sled at

any constant speed. Then the work she does is the same for a 120-m displacement. At a higher speed, though, she would have to do that amount of work in a shorter time interval.

Practice Problem 6.2 A Different Angle Find the tension if Diane pulls at an angle q = 30.0° instead of 20.0°, assuming the same coefficient of friction. What is the work done by Diane on the sled in this case for a 120-m displacement? Explain how the tension can be greater but the work done by Diane smaller.

Work Done by Dissipative Forces The work done by kinetic friction was calculated in Example 6.2 according to a simplified model of friction. In this model, when friction truly does −4600 J of work on the sled, it transfers 4600 J of energy from the sled to the ground’s internal energy—the ground warms up a bit. In reality, 4600 J of energy is converted into internal energy shared between the ground and the sled—both the ground and the sled warm up a little. So the 4600 J is not all transferred to the ground; some stays in the sled but is converted to a different form of energy. Rather than saying friction does −4600 J of work, a more accurate statement is that friction dissipates 4600 J of energy. Dissipation is the conversion of energy from an organized form to a disorganized form such as the kinetic energy associated with the random motions of the atoms and molecules within an object, which is part of the object’s internal energy. As a practical matter, we usually are not concerned with where the internal energy appears. When we can calculate the work done by friction using Eq. (6-1), we get the correct amount of energy dissipated; we just don’t know how much of it is transferred to the stationary surface and how much remains in the sliding object. This is how we apply the term work to kinetic friction or to other dissipative forces such as air resistance. (In Chapters 14 and 15, when we study internal energy in detail, we will look at situations in which we do care where the internal energy appears.)

6.3

KINETIC ENERGY

⃗ net acts on a rigid object of mass m during a displacement Suppose a constant net force F Δr⃗. Choosing the x-axis in the direction of the net force, the total work done on the object is Wtotal = Fnet Δx where Δx is the x-component of the displacement. Newton’s second law tells us that ⃗ net = ma⃗, so F Wtotal = max Δx

(6-5)

Since the acceleration is constant, we can use any of the equations for constant acceleration from Chapter 2. From Eq. (2-13), v 2fx − v 2ix = 2ax Δ x or ax Δx = _12 (v 2fx − v 2ix) Substituting into Eq. (6-5) yields Wtotal = _12 m(v 2fx − v 2ix)

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Since the net force is in the x-direction, ay and az are both zero. Only the x-component of the velocity changes; vy and vz are constant. As a result, v 2f − v 2i = (v 2fx + v 2fy + v 2fz) − (v 2ix + v 2iy + v 2iz) = v 2fx − v 2ix Therefore, the total work done is Wtotal = _12 m(v 2f − v 2i ) = _12 mv 2f − _12 mv 2i The total work done is equal to the change in the quantity _12 mv2, which is called the object’s translational kinetic energy (symbol K). (Often we just say kinetic energy if it is understood that we mean translational kinetic energy.) Translational kinetic energy is the energy associated with motion of the object as a whole; it does not include the energy of rotational or internal motion. Translational kinetic energy:

Relation between total work and kinetic energy

K = _12 mv2

(6-6)

Wtotal = ΔK

(6-7)

Work-kinetic energy theorem:

Kinetic energy is a scalar quantity and is always positive if the object is moving or zero if it is at rest. Kinetic energy is never negative, although a change in kinetic energy can be negative. The kinetic energy of an object moving with speed v is equal to the work that must be done on the object to accelerate it to that speed starting from rest. When the total work done is positive, the object’s speed increases, increasing the kinetic energy. When the total work done is negative, the object’s speed decreases, decreasing the kinetic energy.

Conceptual Example 6.3 Collision Damage Why is the damage caused by an automobile collision so much worse when the vehicles involved are moving at high speeds? Strategy When a collision occurs, the kinetic energy of the automobiles gets converted into other forms of energy. We can use the kinetic energy as a rough measure of how much damage can be done in a collision. Solution and Discussion Suppose we compare the kinetic energy of a car at two different speeds: 60.0 mi/h and 72.0 mi/h (which is 20.0% greater than 60.0 mi/h). If kinetic energy were proportional to speed, then a 20.0% increase in speed would mean a 20.0% increase in kinetic energy. However, since kinetic energy is proportional to the square of the speed, a 20.0% speed increase causes an increase in kinetic energy greater than 20.0%. Working by proportions, we can find the percent increase in kinetic energy:

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_1 mv 2 K2 _____ 72.0 mi/h 2 = 1.44 ___ = _21 22 = ________ K1 60.0 mi/h mv 2

1

(

)

Therefore, a 20.0% increase in speed causes a 44% increase in kinetic energy. What seems like a relatively modest difference in speed makes a lot of difference when a collision occurs.

Practice Problem 6.3 with a Stone Wall

Two Different Cars Collide

Suppose a sports utility vehicle and a small electric car both collide with a stone wall and come to a dead stop. If the SUV mass is 2.5 times that of the small car and the speed of the SUV is 60.0 mph while that of the other car is 40.0 mph, what is the ratio of the kinetic energy changes for the two cars (SUV to small car)?

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Example 6.4 Bungee Jumping A bungee jumper makes a jump in the Gorge du Verdon in southern France. The jumping platform is 182 m above the bottom of the gorge. The jumper weighs 780 N. If the jumper falls to within 68 m of the bottom of the gorge, how much work is done by the bungee cord on the jumper during his descent? Ignore air resistance. Strategy Ignoring air resistance, only two forces act on the jumper during the descent: gravity and the tension in the cord. Since the jumper has zero kinetic energy at both the highest and lowest points of the jump, the change in kinetic energy for the descent is zero. Therefore, the total work done by the two forces on the jumper must equal zero. Solution Let Wg and Wc represent the work done on the jumper by gravity and by the cord. Then Wtotal = Wg + Wc = ΔK = 0 The work done by gravity is

Then the work done by gravity is Wg = −(780 N) × (−114 m) = +89 kJ The work done by the cord is Wc = Wtotal − Wg = −89 kJ. Discussion The work done by gravity is positive, since the force and the displacement are in the same direction (downward). If not for the negative work done by the cord, the jumper would have a kinetic energy of 89 kJ after falling 114 m. The length of the bungee cord is not given, but it does not affect the answer. At first the jumper is in free fall as the cord plays out to its full length; only then does the cord begin to stretch and exert a force on the jumper, ultimately bringing him to rest again. Regardless of the length of the cord, the total work done by gravity and by the cord must be zero since the change in the jumper’s kinetic energy is zero.

Practice Problem 6.4 The Bungee Jumper’s Speed

Wg = Fy Δy = −mg Δy where the weight of the jumper is mg = 780 N. With y = 0 at the bottom of the gorge, the vertical component of the displacement is

Suppose that during the jumper’s descent, at a height of 111 m above the bottom of the gorge, the cord has done −21.7 kJ of work on the jumper. What is the jumper’s speed at that point?

Δy = yf − yi = 68 m − 182 m = −114 m

CHECKPOINT 6.3 Kinetic energy and work are related. Can kinetic energy ever be negative? Can work ever be negative?

6.4

GRAVITATIONAL POTENTIAL ENERGY (1)

Gravitational Potential Energy When Gravitational Force Is Constant Toss a stone up with initial speed vi. Ignoring air resistance, how high does the stone go? We can solve this problem with Newton’s second law, but let’s use work and energy instead. The stone’s initial kinetic energy is Ki = _12 mv 2i . For an upward displacement Δy, gravity does negative work W grav = −mg Δy. No other forces act, so this is the total work done on the stone. The stone is momentarily at rest at the top, so K f = 0. Then Wgrav = Kf − Ki 1 mv 2 ⇒ −mg Δy = − __ i 2

v2 Δy = ___i 2g

From the standpoint of energy conservation, where did the stone’s initial kinetic energy go? If total energy cannot change, it must be “stored” somewhere. Furthermore,

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The symbol for potential energy is U.

CHAPTER 6 Conservation of Energy

the stone gets its kinetic energy back as it falls from its highest point to its initial position, so the energy is stored in a way that is easily recovered as kinetic energy. Stored energy due to the interaction of an object with something else (here, Earth’s gravitational field) that can easily be recovered as kinetic energy is called potential energy (symbol U). The change in gravitational potential energy when an object moves up or down is the negative of the work done by gravity: Change in gravitational potential energy: ΔUgrav = −Wgrav

(6-8)

If the gravitational field is uniform, the work done by gravity is Wgrav = Fy Δy = −mg Δy where the y-axis points up. Therefore, Change in gravitational potential energy: ΔUgrav = mg Δy

(6-9)

(uniform g⃗, y-axis up) Equation (6-9) holds even if the object does not move in a straight-line path. Significance of the Negative Sign in Eq. (6-8) When the stone moves up, Δy is positive. The gravitational force and the displacement of the stone are in opposite directions, so the work done by gravity is negative, gravity is taking away kinetic energy and adding it to its stored potential energy, so the potential energy increases (Fig. 6.16a). If the stone moves down, Δy is negative. The work done by gravity is positive; gravity is giving back kinetic energy by depleting its storage of potential energy, so the potential energy decreases (Fig. 6.16b).

CHECKPOINT 6.4 A stone is tossed straight up in the air and is moving upward. (a) Does the gravitational potential energy increase, decrease, or stay the same? (b) What about the kinetic energy? (c) What force, if any, does work on the stone once it leaves the hand of the one who threw it?

More potential energy

Final position

More potential energy

+y

mg

Figure 6.16 (a) When the stone moves up, the gravitational potential energy increases. (b) When the stone moves down, the gravitational potential energy decreases.

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Less potential energy (a)

∆y

∆U < 0

Initial position

mg

+y

mg ∆y

∆U > 0

Initial position

Less potential energy +x (b)

Final position

mg

+x

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199

Other Forms of Potential Energy In addition to gravitational potential energy, other kinds of potential energy include elastic potential energy (Section 6.7) and electric potential energy (Chapter 17). Forces that have potential energies associated with them are called conservative forces, for reasons we explain shortly. Not every force has an associated potential energy. For instance, there is no such thing as “frictional potential energy.” When kinetic friction does work, it converts energy into a disorganized form that is not easily recoverable as kinetic energy. Mechanical Energy The total work done on an object can always be written as the sum of the work done by conservative forces (Wcons) plus the work done by nonconservative forces (Wnc). Since the total work is equal to the change in the object’s kinetic energy [Eq. (6-7)], Wtotal = Wcons + Wnc = ΔK

⇒

Wnc = ΔK − Wcons

(6-10)

Following the same reasoning we used for gravity [see Eq. (6-8)], the change in the total potential energy is equal to the negative of the work done by the conservative forces: ΔU = −Wcons

(6-11)

Combining Eqs. (6-10) and (6-11) yields Wnc = ΔK + ΔU = ΔEmech

(6-12)

or (Ki + Ui) + Wnc = (Kf + Uf) The sum of the kinetic and potential energies (K + U) is called the mechanical energy (Emech). Wnc is equal to the change in mechanical energy. When finding the change in mechanical energy, do not include the work done by conservative forces. Conservative forces such as gravity do not change the mechanical energy; they just change one form of mechanical energy into another. Work done by conservative forces is already accounted for by the change in potential energy. The term conservative force comes from a time before the general law of conservation of energy was understood and when no forms of energy other than mechanical energy were recognized. Back then, it was thought that certain forces conserved energy and others did not. Now we believe that total energy is always conserved. Nonconservative forces do not conserve mechanical energy, but they do conserve total energy.

Mechanical energy: the sum of the kinetic and potential energies

Conservation of Mechanical Energy When nonconservative forces do no work, mechanical energy is conserved: Ei = Ef

Example 6.5 Rock Climbing in Yosemite A team of climbers is rappelling down steep terrain in the Yosemite valley (Fig. 6.17). Mei-Ling (mass 60.0 kg) slides down a line starting from rest 12.0 m above a horizontal shelf. If she lands on the shelf below with a speed of 2.0 m/s,

calculate the energy dissipated by the kinetic frictional forces acting between her and the line. The local value of g is 9.78 N/kg. Ignore air resistance. continued on next page

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Example 6.5 continued

Given: mass of climber, m = 60.0 kg; ∆y = −12.0 m; vi = 0 m/s and vf = 2.0 m/s, just before stopping; local field strength g = 9.78 N/kg. To find: change in mechanical energy ΔE. Solution Wnc = ΔEmech = ΔK + ΔU, so we need to calculate the changes in kinetic and potential energy. Mei-Ling’s kinetic energy is initially zero since she starts at rest. The change in her kinetic energy is ΔK = _12 mv 2f − _12 mv 2i = _12 mv 2f − 0 = _12 (60.0 kg) × (2.0 m/s)2 = +120 J 12.0 m

The change in her potential energy is ΔU = mg Δy = 60.0 kg × 9.78 m/s2 × (0 − 12.0 m) = −7040 J The work done by friction is

v

ΔEmech = ΔK + ΔU = 120 J + (−7040 J) = −6920 J The amount of energy dissipated by friction (converted from mechanical energy into internal energy) is 6920 J. Fortunately, Mei-Ling is wearing gloves, so her hands don’t get burned. Discussion If the line had broken when Mei-Ling was at the top, her final kinetic energy would have been +7040 J— disastrously large since it corresponds to a final speed of Figure 6.17

√ √ ___

Mei-Ling rappelling downward from a position 12.0 m above a shelf.

Strategy The forces acting on Mei-Ling are gravity and kinetic friction (Fig. 6.18). The only force whose work is not included in the change in potential energy is the work done by kinetic friction. Therefore, the change in the mechanical energy, ΔK + ΔU, is equal to the work done by fk friction. Since we know Mei-Ling’s initial and final speeds as well as her mass, mg we can calculate the change in her kinetic energy. From the change in Figure 6.18 height, we can calculate the change in FBD for Mei-Ling. potential energy.

_______

7040 J = 15.3 m/s K = _______ v = ___ _1 m 30.0 kg 2 Instead, kinetic friction reduces her final kinetic energy to a manageable +120 J (which corresponds to a final speed of 2.0 m/s). Mei-Ling can absorb this much kinetic energy safely by landing on the shelf while bending her knees.

Practice Problem 6.5 by Air Resistance

Energy Dissipated

A ball thrown straight up at an initial speed of 14.0 m/s reaches a maximum height of 7.6 m. What fraction of the ball’s initial kinetic energy is dissipated by air resistance as the ball moves upward?

Choosing Where the Potential Energy Is Zero Notice that when we apply Eq. (6-12), only the change in potential energy enters the calculation. Therefore, we can always assign the value of the potential energy for any one position. Most often, we choose some convenient position and assign it to have zero potential energy. Once that choice is made, the potential energy of every other configuration is determined by Eq. (6-11). For gravitational potential energy in a uniform gravitational field, we usually choose the potential energy to be zero at some convenient place: on the floor, on a table,

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or at the top of a ladder. After assigning y = 0 to that place, the potential energy at any other place is U = mgy. Gravitational potential energy: Ugrav = mgy

(6-13)

(uniform g⃗, y-axis up, assign U = 0 to y = 0) Potential energy is then positive above y = 0 and negative below it. There is no special significance to the sign of the potential energy. What matters is the sign of the potential energy change.

Example 6.6 A Quick Descent A ski trail makes a vertical descent of 78 m. A novice skier, unable to control his speed, skis down this trail and is lucky enough not to hit any trees. What is his speed at the bottom of the trail, ignoring friction and air resistance? Strategy When nonconservative forces do no work, Wnc = ΔEmech = 0 and mechanical energy does not change. A

skilled skier can control his speed by, in effect, controlling how much work the frictional force does on the skis. Here we assume no friction or air resistance. Then the only forces acting on the skier are the normal force and gravity (Fig. 6.19). The normal force does no work, since it is always perpendicular to the skier’s velocity, so Wnc = 0. continued on next page

N v N v mg N 78 m

v

mg

N

v

mg

mg

Energy

Potential energy

Kinetic energy

Figure 6.19 The final speed of the skier depends only on the initial and final altitudes if no friction acts.

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Example 6.6 continued

Solution Because Wnc = 0, the mechanical energy does not change: Ki + Ui = Kf + Uf If we choose the y-axis up and y = 0 at the bottom of the hill, yi = 78 m and yf = 0. Then Ui = mgyi

and

Uf = 0

If the skier starts with zero kinetic energy, then Ki = 0 and Kf = _12 mv 2f . Setting the mechanical energies equal, 0 + mgyi = _12 mv 2f + 0 Solving for the final speed vf, ____

__________________

vf = √ 2gyi = √2 × 9.80 m/s2 × 78 m = 39 m/s Discussion Notice that the solution did not depend on the detailed shape of the path. If the slope were constant (Fig. 6.20), we could use Newton’s second law to find the skier’s acceleration and then the change in velocity:

∑Fx = mg sin q = max ⇒ ax = g sin q From Eq. (2-13), v 2fx − v 2ix _______ v 2fx h = _____ Δx = _______ = 2ax 2g sin q sin q

⇒

____

vfx = √ 2gh

This method shows that N the final speed does not depend on the angle of the slope, but the energy mg sin q method shows that the y final speed is the same for any shape path, not mg cos q just for constant slopes. q x On the other hand, the time that it takes the skier Figure 6.20 to reach the bottom does FBD for the skier on a constant depend on the length and slope. contour of the trail. A final speed of 39 m/s (87 mi/h) is dangerously fast. In reality, friction and air resistance would do negative work on the skier, so the final speed would be smaller.

Practice Problem 6.6 Speeding Roller Coaster A roller coaster is hauled to the top of the first hill of the ride by a motorized chain drive. After that, the train of cars is released and no more energy is supplied by an external motor. The cars are moving at 4.0 m/s at the top of the first hill, 35.0 m above the ground. How fast are they moving at the top of the second hill, 22.0 m above the ground? Ignore friction and air resistance.

where h = 78 m.

Recognizing a Conservative Force In Example 6.6, the final speed doesn’t depend on the shape of the trail: it could have been a steep descent, or a long gradual one, or have a complicated profile with varying slope. It could even be a vertical descent—the final speed is the same for free fall off a 78-m-high building. Any time the work done by a force is independent of path—that is, the work depends only on the initial and final positions—the force is conservative. We depend on the path-independence of the work done to define the potential energy in Eq. (6-11). Energy stored as potential energy by a conservative force during a displacement from point A to point B can be recovered as kinetic energy. We can simply reverse displacement to get all of the energy back: ΔUB →A = −ΔUA →B. The work done by friction, air resistance, and other contact forces does depend on path, so these forces cannot have potential energies associated with them. We cannot use friction to store energy in a form that is completely recoverable as kinetic energy.

6.5

GRAVITATIONAL POTENTIAL ENERGY (2)

The expressions for gravitational potential energy developed in Section 6.4 apply when the gravitational force is constant (or nearly constant). If the gravitational force is not constant, such as when a satellite is placed into orbit around the Earth, Eqs. (6-9) and (6-13) cannot be used. Instead, we need to use an expression for gravitational potential

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energy that corresponds to Newton’s law of universal gravitation. Recall that the magnitude of the gravitational force that one body exerts on another is Gm1m2 F = ______ r2

(2-6)

where r is the distance between the centers of the bodies. The corresponding expression for gravitational potential energy in terms of the distance between two bodies is Gravitational potential energy: Gm1m2 U = − ______ r

(6-14)

(assign U = 0 when r = ∞) A graph showing the gravitational potential energy as a function of r is shown in Fig. 6.21. Note that we have assigned the potential energy to be zero at infinite separation (U = 0 when r = ∞). Why this choice? Simply put, any other choice would mean adding a constant term to the expression for U. This constant term would always subtract out of our equations, which involve only changes in potential energy. This choice (U = 0 when r = ∞) means that the gravitational potential energy is negative for any finite value of r, because potential energy decreases as the bodies get closer together and increases as they get farther apart. Does Eq. (6-14) Contradict Eq. (6-9)? Calculus is used to derive Eq. (6-14), but we can verify that it is consistent with Eq. (6-9) without using calculus. For a very small displacement from ri to rf = ri + Δy (Fig. 6.22), the potential energy change given by Eq. (6-14) must reduce to the constant-force case:

(

) (

GMEm GMEm ΔU = Uf − Ui = − ______ − − ______ ri ri + Δy

)

Rearranging and factoring out the common factors GMEm and then rewriting with a common denominator, ri + Δy − ri 1 − ______ 1 ΔU = GMEm __ = GMEm _________ ri(ri + Δy) ri ri + Δy

(

)

(6-15)

U(r)

r

Figure 6.21 Gravitational potential energy as a function of r, the distance between the centers of the two bodies. The potential energy increases as the distance increases.

∆y

r

For values of Δy that are small compared with ri, ri + Δy ≈ ri. Making that approximation in the denominator of Eq. (6-15),

( )

GME ΔU = m _____ Δy r 2i

Figure 6.22 An object at a (Δy 0 and Krot = 0; (b) Ktr = 0 and Krot > 0; (c) Krot = _25 Ktr .

Acceleration of Rolling Objects What is the acceleration of a ball rolling down an incline? Figure 8.35 shows the forces acting on the ball. Static friction is the force that makes the ball rotate; if there were no friction, instead of rolling, the ball would just slide down the incline. This is true because friction is the only force acting that yields a

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nonzero torque about the rotation axis through the ball’s center of mass. Gravity gives zero torque because it acts at the axis, so the lever arm is zero. The normal force points directly at the axis, so its lever arm is also zero. The frictional force f⃗ provides a torque

N

fs

t = rf

q

where r is the ball’s radius. An analysis of the forces and torques combined with Newton’s second law in both forms enables us to calculate the acceleration of the ball in Example 8.13.

mg

Figure 8.35 Forces acting on a ball rolling downhill.

Example 8.13 Acceleration of a Rolling Ball Calculate the acceleration of a solid ball rolling down a slope inclined at an angle q to the horizontal (Fig. 8.36a).

Solution Since the net torque is

Strategy The net torque is related to the angular acceleration by ∑t = Ia, Newton’s second law for rotation. Similarly, the net force acting on the ball gives the acceleration of the ⃗ = ma⃗ . The axis of rotation is through center of mass: ∑F CM the ball’s cm. As already discussed, neither gravity nor the normal force produce a torque about this axis; the net torque is ∑t = rf, where f is the magnitude of the frictional force. One problem is that the force of friction is unknown. We must resist the temptation to assume that f = msN; there is no reason to assume that static friction has its maximum possible magnitude. We do know that the two accelerations, translational and rotational, are related. We know that vCM and w are proportional since r is constant. To stay proportional they must change in lock step; their rates of change, aCM and a, are proportional to each other by the same factor of r. Thus, aCM = a r. This connection should enable us to eliminate f and solve for the acceleration. Since the speed of a ball after rolling a certain distance was found to be independent of the mass and radius of the ball in Example 8.12, we expect the same to be true of the acceleration.

the angular acceleration is

∑t = rf ∑t rf a = ___ = __ I I

(1)

where I is the ball’s rotational inertia about its cm. Figure 8.36b shows the forces along the incline acting on the ball. The acceleration of the cm is found from Newton’s second law. The component of the net force acting along the incline (in the direction of the acceleration) is

∑Fx = mg sin q − f = maCM

(2)

Because the ball is rolling without slipping, the acceleration of the cm and the angular acceleration are related by aCM = a r Now we try to eliminate the unknown frictional force f from the previous equations. Solving Eq. (1) for f gives Ia f = ___ r Substituting this into Eq. (2), we get Ia = ma mg sin q − ___ CM r

N f r h

(a)

x

v

mg sin q

d q

Now to eliminate a, we can substitute a = aCM /r:

mg cos q

IaCM = maCM mg sin q − ____ r2

q

q mg ΣFx = mg sin q – f (b)

Figure 8.36 (a) A ball rolling downhill. (b) FBD for the ball, with the gravitational force resolved into components perpendicular and parallel to the incline.

Solving for aCM, g sin q aCM = _________ 1 + I/(mr 2) For a solid sphere, I = _25 mr 2, so g sin q __ = 5 g sin q aCM = ______ 7 1 + _25 continued on next page

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Example 8.13 continued

Discussion The acceleration of an object sliding down an incline without friction is a = g sin q. The acceleration of the rolling ball is smaller than g sin q due to the frictional force directed up the incline. We can check the answer using the result of Example 8.12. The ball’s acceleration is constant. If the ball starts from rest as in Fig. 8.36a, after it has rolled a distance d, its speed v is

√(

__________

)

g sin q v = √ 2ad = 2 ______ d 1+b ____

8.8

where b = _25 . The vertical drop during this time is h = d sin q, so

√

_____

2gh v = _____ 1+b

Practice Problem 8.13 Cylinder

Acceleration of a Hollow

Calculate the acceleration of a thin hollow cylindrical shell rolling down a slope inclined at an angle q to the horizontal.

ANGULAR MOMENTUM

Newton’s second law for translational motion can be written in two ways: ⃗ Δp ⃗ = lim ___ ⃗ = ma⃗ (constant mass) (general form) or ∑F ∑F Δt→0

Δt

In Eq. (8-9) we wrote Newton’s second law for rotation as ∑t = Ia, which applies only when I is constant—that is, for a rigid body rotating about a fixed axis. A more general form of Newton’s second law for rotation uses the concept of angular momentum (symbol L). The net external torque acting on a system is equal to the rate of change of the angular momentum of the system. ΔL ∑t = lim ___ (8-13) Δt→0 Δt

CONNECTION: Note the analogy with Δp⃗ ⃗ = lim ___ ∑F Δt→0 Δt

The angular momentum of a rigid body rotating about a fixed axis is the rotational inertia times the angular velocity, which is analogous to the definition of linear momentum (mass times velocity): CONNECTION:

Angular momentum: L = Iw

(8-14)

(rigid body, fixed axis)

Note the analogy with p ⃗ = mv⃗. See the Master the Concepts section for a complete table of these analogies.

Either Eq. (8-13) or Eq. (8-14) can be used to show that the SI units of angular momentum are kg·m2/s. For a rigid body rotating around a fixed axis, angular momentum doesn’t tell us anything new. The rotational inertia is constant for such a body since the distance ri between every point on the object and the axis stays the same. Then any change in angular momentum must be due to a change in angular velocity w : IΔw = I lim ___ Δw = Ia ΔL = lim ____ ∑t = lim ___ Δt→0 Δt Δt→0 Δt Δt→0 Δt Conservation of Angular Momentum However, Eq. (8-13) is not restricted to rigid objects or to fixed rotation axes. In particular, if the net external torque acting on a

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system is zero, then the angular momentum of the system cannot change. This is the law of conservation of angular momentum: Conservation of angular momentum can be applied to any system if the net external torque on the system is zero (or negligibly small).

CONNECTION: Another conservation law

Application of angular momentum: figure skater

Conservation of angular momentum: If ∑t = 0, Li = Lf

(8-15)

Here Li and Lf represent the angular momentum of the system at two different times. Conservation of angular momentum is one of the most basic and fundamental laws of physics, along with the two other conservation laws we have studied so far (energy and linear momentum). For an isolated system, the total energy, total linear momentum, and total angular momentum of the system are each conserved. None of these quantities can change unless some external agent causes the change. With conservation of energy, we add up the amounts of the different forms of energy (such as kinetic energy and gravitational potential energy) to find the total energy. The conservation law refers to the total energy. By contrast, linear momentum and angular momentum cannot be added to find the “total momentum.” They are entirely different quantities, not two forms of the same quantity. They even have different dimensions, so it would be impossible to add them. Conservation of linear momentum and conservation of angular momentum are separate laws of physics. Changing Rotational Inertia In this section, we restrict our consideration to cases where the axis of rotation is fixed but where the rotational inertia is not necessarily constant. One familiar example of a changing rotational inertia occurs when a figure skater spins (Fig. 8.37). To start the spin, the skater glides along with her arms outstretched and then begins to rotate her body about a vertical axis by pushing against the ice with a skate. The push of the ice against the skate provides the external torque that gives the skater her initial angular momentum. Initially the skater’s arms and the leg not in contact with the ice are extended away from her body. The mass of the arms and leg when extended contribute more to her rotational inertia than they do when held close to the body. As the skater spins, she pulls her arms and leg close and straightens her body to decrease her rotational inertia. As she does, her angular velocity increases dramatically in such a way that her angular momentum stays the same.

Figure 8.37 Figure skater Lucinda Ruh at the (a) beginning and (b) end of a spin. Her angular velocity is much higher in (b) than in (a).

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(a)

(b)

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CHECKPOINT 8.8 If the skater then extends her arms and leg back to their initial configuration, does her angular velocity decrease back to its initial value, ignoring friction?

Many natural phenomena can be understood in terms of angular momentum. In a hurricane, circulating air is sucked inward by a low-pressure region at the center of the storm (the eye). As the air moves closer and closer to the axis of rotation, it circulates faster and faster. An even more dramatic example is the formation of a pulsar. Under certain conditions, a star can implode under its own gravity, forming a neutron star (a collection of tightly packed neutrons). If the Sun were to collapse into a neutron star, its radius would be only about 13 km. If a star is rotating before its collapse, then as its rotational inertia decreases dramatically, its angular velocity must increase to keep its angular momentum constant. Such rapidly rotating neutron stars are called pulsars because they emit regular pulses of x-rays, at the same frequency as their rotation, that can be detected when they reach Earth. Some pulsars rotate in only a few thousandths of a second per revolution.

Applications of angular momentum: hurricanes and pulsars

Example 8.14 Mouse on a Wheel A 0.10-kg mouse is perched at point B on the rim of a 2.00-kg wagon wheel that rotates freely in a horizontal plane at 1.00 rev/s (Fig. 8.38). The mouse crawls to point A at the center. Assume the mass of the wheel is concentrated at the rim. What is the frequency of rotation in rev/s when the mouse arrives at point A? Strategy Assuming that frictional torques are negligibly small, there is no external torque acting on the mouse/wheel system. Then the angular momentum of the mouse/wheel system must be conserved; it takes an external torque to change angular momentum. The mouse and wheel exert torques on one another, but these internal torques only transfer some angular momentum between the wheel and the mouse without changing the total angular momentum. We can think of the system as initially being a rigid body with rotational inertia Ii. When the mouse reaches the center, we think of the system as a rigid body with a different rotational

inertia If. The mouse changes the rotational inertia of the mouse/wheel system by moving from the outer rim, where its mass makes the maximum possible contribution to the rotational inertia, to the rotation axis, where its mass makes no contribution to the rotational inertia. Solution Initially, all of the mass of the system is at a distance R from the rotation axis, where R is the radius of the wheel. Therefore, Ii = (M + m)R2 where M is the mass of the wheel and m is the mass of the mouse. After the mouse moves to the center of the wheel, its mass contributes nothing to the rotational inertia of the system: If = MR2 From conservation of angular momentum, Ii wi = If wf Substituting the rotational inertias and w = 2p f,

B

(M + m)R2 × 2p fi = MR2 × 2p ff

R

Factors of 2p R2 cancel from each side, leaving (M + m)fi = Mff

A

Solving for ff, Figure 8.38 Mouse on a rotating wheel.

2.10 kg M + m f = _______ (1.00 rev/s) = 1.05 rev/s ff = ______ M i 2.00 kg continued on next page

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CHAPTER 8 Torque and Angular Momentum

Example 8.14 continued

Discussion Conservation laws are powerful tools. We do not need to know the details of what happens as the mouse crawls along the spoke from the outer edge of the wheel; we need only look at the initial and final conditions. A common mistake in this sort of problem is to assume that the initial rotational kinetic energy is equal to the final rotational kinetic energy. This is not true because the mouse crawling in toward the center expends energy to do so. In

other words, the mouse converts some internal energy into rotational kinetic energy.

Practice Problem 8.14 Kinetic Energy

Change in Rotational

What is the percentage change in the rotational kinetic energy of the mouse/wheel system?

Angular Momentum in Planetary Orbits Application of angular momentum: planetary orbits

Conservation of angular momentum applies to planets orbiting the Sun in elliptical orbits. Kepler’s second law says that the orbital speed varies in such a way that the planet sweeps out area at a constant rate (Fig. 8.39a). In Problem 104, you can show that Kepler’s second law is a direct result of conservation of angular momentum, where the angular momentum of the planet is calculated using an axis of rotation perpendicular to the plane of the orbit and passing through the Sun. When the planet is closer to the Sun, it moves faster; when it is farther away, it moves more slowly. Conservation of angular momentum can be used to relate the orbital speeds and radii at two different points in the orbit. The same applies to satellites and moons orbiting planets. v⊥ v Sun

Figure 8.39 The planet’s speed varies such that it sweeps out equal areas in equal time intervals. The eccentricity of the planetary orbit is exaggerated for clarity.

q Planet

Sun r

va Aphelion

Perihelion

vp (b)

(a)

Example 8.15 Earth’s Orbital Speed At perihelion (closest approach to the Sun), Earth is 1.47 × 108 km from the Sun and its orbital speed is 30.3 km/s. What is Earth’s orbital speed at aphelion (greatest distance from the Sun), when it is 1.52 × 108 km from the Sun? Note that at these two points Earth’s velocity is perpendicular to a radial line from the Sun (see Fig. 8.39a).

Earth’s rotational inertia, we treat it as a point particle since its radius is much less than its distance from the axis of rotation.

Strategy We take the axis of rotation through the Sun. Then the gravitational force on Earth points directly toward the axis; with zero lever arm, the torque is zero. With no other external forces acting on the Earth, the net external torque is zero. Earth’s angular momentum about the rotation axis through the Sun must therefore be conserved. To find

where m is Earth’s mass and r is its distance from the Sun. The angular velocity is v⊥ w = __ r where v⊥ is the component of the velocity perpendicular to a radial line from the Sun. At the two points under consideration,

Solution The rotational inertia of the Earth is I = mr 2

continued on next page

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8.9 THE VECTOR NATURE OF ANGULAR MOMENTUM

Example 8.15 continued

v⊥ = v. As the distance from the Sun r varies, its speed v must vary to conserve angular momentum: Iiw i = Ifw f By substitution, vi vf 2 2 __ mr i × __ ri = mr f × rf or rivi = rfvf

(1)

Practice Problem 8.15 Puck on a String

Solving for vf, 1.47 × 108 km × 30.3 km/s = 29.3 km/s vf = ri/rf vi = ____________ 1.52 × 108 km Discussion Earth moves slower at a point farther from the Sun. This is what we expect from energy conservation. The potential energy is greater at aphelion than at perihelion.

8.9

Since the mechanical energy of the orbit is constant, the kinetic energy must be smaller at aphelion. Equation (1) implies that the orbital speed and orbital radius are inversely proportional, but strictly speaking this equation only applies to the perihelion and aphelion. At a general point in the orbit, the perpendicular component v⊥ is inversely proportional to r (see Fig. 8.39b). The orbits of Earth and most of the other planets are nearly circular so that q ≈ 0° and v⊥ ≈ v.

A puck on a frictionless, horizontal air table is attached to a string that passes down through a hole in the table. Initially the puck moves at 12 cm/s in a circle of radius 24 cm. If the string is pulled through the hole, reducing the radius of the puck’s circular motion to 18 cm, what is the new speed of the puck?

THE VECTOR NATURE OF ANGULAR MOMENTUM

Until now we have treated torque and angular momentum as scalar quantities. Such a treatment is adequate in the cases we have considered so far. However, the law of conservation of angular momentum applies to all systems, including rotating objects whose axis of rotation changes direction. Torque and angular momentum are actually vector quantities. Angular momentum is conserved in both magnitude and direction in the absence of external torques. An important special case is that of a symmetrical object rotating about an axis of symmetry, such as the spinning disk in Fig. 8.40. The magnitude of the angular momentum of such an object is L = Iw. The direction of the angular momentum vector points along the axis of rotation. To choose between the two directions along the axis, a righthand rule is used. Align your right hand so that, as you curl your fingers in toward your palm, your fingertips follow the object’s rotation; then your thumb points in the direction of ⃗L.

L

L

Figure 8.40 Right-hand rule for finding the direction of the angular momentum of a spinning disk.

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CHAPTER 8 Torque and Angular Momentum

Figure 8.41 Spinning like a

N

top, the Earth maintains the direction of its angular momentum due to rotation as it revolves around the Sun (not to scale).

Autumnal equinox in northern hemisphere S

L L

L Sun Winter solstice in northern hemisphere

L

Summer solstice in northern hemisphere

Vernal equinox in northern hemisphere

Application of angular momentum: the gyroscope

For rotation about a fixed axis, the net torque is also along the axis of rotation, in the direction of the change in angular momentum it causes. The sign convention we have used up to now for angular momentum and torque gives the sign of the z-component of the vector quantity, where the z-axis points toward the viewer (out of the page). A disk with a large rotational inertia can be used as a gyroscope. When the gyroscope spins at a large angular velocity, it has a large angular momentum. It is then difficult to change the orientation of the gyroscope’s rotation axis, because to do so requires changing its angular momentum. To change the direction of a large angular momentum requires a correspondingly large torque. Thus, a gyroscope can be used to maintain stability. Gyroscopes are used in guidance systems in airplanes, submarines, and space vehicles to maintain a constant direction in space. The same principle explains the great stability of rifle bullets and spinning tops. A rifle bullet is made to spin as it passes through the rifle’s barrel. The spinning bullet then keeps its correct orientation—nose first—as it travels through the air. Otherwise, a small torque due to air resistance could make the bullet turn around randomly, greatly increasing air resistance and undermining accuracy. A properly thrown football is made to spin for the same reasons. A spinning top can stay balanced for a long time, while the same top soon falls over if it is not spinning. The Earth’s rotation gives it a large angular momentum. As the Earth orbits the Sun, the axis of rotation stays in a fixed direction in space. The axis points nearly at Polaris (the North Star), so even as the Earth rotates around its axis, Polaris maintains its position in the northern sky. The fixed direction of the rotation axis gives us the regular progression of the seasons (Fig. 8.41).

A Classic Demonstration A demonstration often done in physics classes is for a student to hold a spinning bicycle wheel while standing on a platform that is free to rotate. The wheel’s rotation axis is initially horizontal (Fig. 8.42a). Then the student repositions the wheel so that its axis of rotation is vertical (Fig. 8.42b). As he repositions the wheel, the platform begins to rotate opposite to the wheel’s rotation. If we assume no friction acts to resist rotation of the platform, then the platform continues to rotate as long as the wheel is held with its axis vertical. If the student returns the wheel to its original orientation, the rotation of the platform stops. The platform is free to rotate about a vertical axis. As a result, once the student steps onto the platform, the vertical component Ly of the angular momentum of the system (student + platform + wheel) is conserved. The horizontal components of ⃗L are not conserved. The platform is not free to rotate about any horizontal axis since the floor can exert external torques to keep it from doing so. In vector language, we would say that only the vertical component of the external torque is zero, so only the vertical component of angular momentum is conserved. Initially Ly = 0 since the student and the platform have zero angular momentum and the wheel’s angular momentum is horizontal. When the wheel is repositioned so that it

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MASTER THE CONCEPTS

Figure 8.42 A demonstration Lwheel

of angular momentum conservation.

y

y

x

x

Lwheel

Platform at rest Lplatform + student (b)

(a)

spins with an upward angular momentum (Ly > 0), the rest of the system (the student and the platform) must acquire an equal magnitude of downward angular momentum (Ly < 0) so that the vertical component of the total angular momentum is still zero. Thus, the platform and student rotate in the opposite sense from the rotation of the wheel. Since the platform and student have more rotational inertia than the wheel, they do not spin as fast as the wheel, but their vertical angular momentum is just as large. The student and the wheel apply torques to each other to transfer angular momentum from one part of the system to the other. These torques are equal and opposite and they have both vertical and horizontal components. As the student lifts the wheel, he feels a strange twisting force that tends to rotate him about a horizontal axis. The platform prevents the horizontal rotation by exerting unequal normal forces on the student’s feet. The horizontal component of the torque is so counterintuitive that, if the student is not expecting it, he can easily be thrown from the platform!

Master the Concepts • The rotational kinetic energy of a rigid object with rotational inertia I and angular velocity w is Krot = _12 Iw 2

(8-1)

In this expression, w must be measured in radians per unit time. • Rotational inertia is a measure of how difficult it is to change an object’s angular velocity. It is defined as: N

2

I = ∑ mir i i=1

(8-2)

where ri is the perpendicular distance between a particle of mass mi and the rotation axis. The rotational inertia depends on the location of the rotation axis. • Torque measures the effectiveness of a force for twisting or turning an object. It can be calculated in two equivalent ways: either as the product of the perpendicular component of the force with the shortest distance between the rotation axis and the point of application of the force t = ±rF⊥

(8-3) continued on next page

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CHAPTER 8 Torque and Angular Momentum

• Newton’s second law for rotation is

Master the Concepts continued

∑t = Ia

(8-9) where radian measure must be used for a. A more general form is

or as the product of the magnitude of the force with its lever arm (the perpendicular distance between the line of action of the force and the axis of rotation) t = ±r⊥F

ΔL ∑t = lim ___ Δt→0 Δt

(8-4)

where L is the angular momentum of the system. • The total kinetic energy of a body that is rolling without slipping is the sum of the rotational kinetic energy about an axis through the cm and the translational kinetic energy:

F Axis

F⊥

q q

r r⊥

2

90°

r⊥

K = _12 Mv CM + _12 ICMw 2

t = rF sin q r⊥ = r sin q F⊥ = F sin q 90° F⊥

Axis

(8-13)

(8-11)

• The angular momentum of a rigid body rotating about a fixed axis is the rotational inertia times the angular velocity:

F

L = Iw (8-14) • The law of conservation of angular momentum: if the net external torque acting on a system is zero, then the angular momentum of the system cannot change.

q

r

If ∑t = 0, L i = L f • A force whose perpendicular component tends to cause rotation in the CCW direction gives rise to a positive torque; a force whose perpendicular component tends to cause rotation in the CW direction gives rise to a negative torque. • The work done by a constant torque is the product of the torque and the angular displacement: W = t Δq (Δq in radians)

(8-6)

• The conditions for translational and rotational equilibrium are ⃗ = 0 and ∑t = 0 ∑F

(8-8)

The rotation axis can be chosen arbitrarily when calculating torques in equilibrium problems. Generally, the best place to choose the axis is at the point of application of an unknown force so that the unknown force does not appear in the torque equation.

Conceptual Questions 1. In Fig. 8.2b, where should the doorknob be located to make the door easier to open? 2. Explain why it is easier to drive a wood screw using a screwdriver with a large diameter handle rather than one with a thin handle. 3. Why is it easier to push open a swinging door from near the edge away from the hinges rather than in the middle of the door?

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(8-15)

• This table summarizes the analogous quantities and equations in translational and rotational motion. Translation

Rotation

m

I t

⃗ F a⃗ ⃗ = ma⃗ ∑F Δx W = FxΔx v⃗

a ∑t = Ia Δq W = t Δq w

K = _12 mv2

K = _12 Iw 2

p ⃗ = mv⃗

L = Iw

Δp ⃗ ⃗ = lim ___ ∑F Δt→0 Δt ⃗ = 0, p If ∑F ⃗ is conserved

4. A book measures 3 cm by 16 cm by 24 cm. About which of the axes shown in the figure is its rotational inertia smallest? 5. A body in equilibrium has only two forces acting on it. We found in Section 4.2 that the

ΔL ∑t = lim ___ Δt→0 Δt If ∑t = 0, L is conserved

Axis 1

3 cm

Axis 2

24 cm Axis 3 16 cm

Conceptual Question 4

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CONCEPTUAL QUESTIONS

6.

7.

8.

9.

10.

forces must be equal in magnitude and opposite in direction in order to give a translational net force of zero. What else must be true of the two forces for the body to be in equilibrium? [Hint: Consider the lines of action of the forces.] Why do many helicopters have a small propeller attached to the tail that rotates in a vertical plane? Why is this attached at the tail rather than somewhere else? [Hint: Most of the helicopter’s mass is forward, in the cab.] In the “Pinewood Derby,” Cub Scouts construct cars and then race them down an incline. Some say that, everything else being equal (friction, drag coefficient, same wheels, etc.), a heavier car will win; others maintain that the weight of the car does not matter. Who is right? Explain. [Hint: Think about the fraction of the car’s kinetic energy that is rotational.] A large barrel lies on its side. In order to roll it across the floor, you F apply a horizontal force, Axis as shown in the figure. If the applied force points toward the axis of rotation, which runs down the center of the barrel through the center of mass, it produces zero torque about that axis. How then can this applied force make the barrel start to roll? Animals that can run fast always have thin legs. Their leg muscles are concentrated close to the hip joint; only tendons extend into the lower leg. Using the concept of rotational inertia, explain how this helps them run fast. Part (a) of the figure shows a simplified model of how the triceps muscle connects to the forearm. As the angle q is changed, the tendon wraps around a nearly circular arc. Explain how this is much more effective than if the tendon is connected as in part (b) of the figure. [Hint: Look at the lever arm as q changes.] Triceps muscle q

Tendon connects here

(a)

horizontal? What about for other angles between the upper arm and the forearm? Consider also the rotational inertia of the forearm about the elbow and of the entire arm about the shoulder.

Flexor (biceps)

Flexor

(a)

(b)

Question 11 Axis 12. In Section 8.6, it was asserted that the sum of all the internal F12 torques (that is, the torques due to interm1 nal forces) acting on a rigid object is zero. m2 The figure shows two particles in a rigid F21 object. The particles ⃗ 12 and exert forces F ⃗ 21 on each other. These forces are directed along a line F that joins the two particles. Explain why the torques due to these two forces must be equal and opposite even though the forces are applied at different points (and, therefore, possibly different distances from the axis). 13. A playground merry-go-round (Fig. 8.5) spins with negligible friction. A child moves from the center out to the rim of the merry-go-round platform. Let the system be the merry-go-round plus the child. Which of these quantities change: angular velocity of the system, rotational kinetic energy of the system, angular momentum of the system? Explain your answer. 14. The figure shows a balancing toy with weights extending on either side. The toy is extremely stable. It can be pushed quite far off center one way or the other but it does not fall over. Explain why it is so stable.

q

(b)

Question 10 11. Part (a) of the figure shows a simplified model of how the biceps muscle enables the forearm to support a load. What are the advantages of this arrangement as opposed to the alternative shown in part (b), where the flexor muscle is in the forearm instead of in the upper arm? Are the two equally effective when the forearm is

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15. Explain why the posture taken by defensive football linemen makes them more difficult to push out of the way. Consider both the height of the center of gravity and the size of the support base (the area on the ground bounded

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16.

17.

18.

19.

20.

CHAPTER 8 Torque and Angular Momentum

by the hands and feet touching the ground). In order to knock a person over, what has to happen to the center of gravity? Which do you think needs a more complex neurological system for maintaining balance: four legged animals or humans? The center of gravity of the upper body of a bird is located below the hips; in a human, the center of CG gravity of the upper body is located well above the hips. Since the upper body is supported by the hips, are birds or humans more stable? Consider what happens if the upper body is displaced a little so that CG its center of gravity is not directly above or below the hips. In what direction does the torque due to gravity tend to make the upper body rotate about an axis through the hips? An astronaut wants to remove a bolt from a satellite in orbit. He positions himself so that he is at rest with respect to the satellite, then pulls out a wrench and attempts to remove the bolt. What is wrong with his method? How can he remove the bolt? Your door is hinged to close automatically after being opened. Where is the best place to put a wedge-shaped door stopper on a slippery floor in order to hold the door open? Should it be placed close to the hinge or far from it? You are riding your bicycle and approaching a rather steep hill. Which gear should you use to go uphill, a low gear or a high gear? With a low gear the wheel rotates less than with a high gear for one rotation of the pedals. One way to find the center of gravity of an irregular flat object is to suspend it from CG various points so that it is free to rotate. When the object hangs in

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equilibrium, a vertical line is drawn downward from the support point. After drawing lines from several different support points, the center of gravity is the point where the lines all intersect. Explain how this works. 21. One of the effects of significant global warming would be the melting of part or all of the polar ice caps. This, in turn, would change the length of the day (the period of the Earth’s rotation). Explain why. Would the day get longer or shorter?

Multiple-Choice Questions 1. A heavy box is resting a on the floor. You would b like to push the box to c P tip it over on its side, using the minimum force possible. Which of the force vectors in the diagram shows the correct location and direction of the force? The forces have equal horizontal components. Assume enough friction so that the box does not slide; instead it rotates about point P. 2. When both are expressed in terms of SI base units, torque has the same units as (a) angular acceleration (b) angular momentum (c) force (d) energy (e) rotational inertia (f) angular velocity Questions 3–4: A uniform solid cylinder rolls without slipping down an incline. At the bottom of the incline, the speed, v, of the cylinder is measured and the translational and rotational kinetic energies (Ktr, Krot) are calculated. A hole is drilled through the cylinder along its axis and the experiment is repeated; at the bottom of the incline the cylinder now has speed v′ and translational and rotational kinetic energies K ′tr and K ′rot. 3. How does the speed of the cylinder compare with its original value? (a) v′ < v (b) v′ = v (c) v′ > v (d) Answer depends on the radius of the hole drilled. 4. How does the ratio of rotational to translational kinetic energy of the cylinder compare to its original value? K K K K′rot ____ K′rot ____ K′rot ____ < rot (b) ____ = rot > rot (a) ____ (c) ____ Ktr K′tr Ktr K′tr K′tr Ktr (d) Answer depends on the radius of the hole drilled. 5. The SI units of angular momentum are kg⋅m rad rad (c) _____ (a) ___ (b) ___ s s2 s2 kg⋅m2 kg⋅m2 kg⋅m (d) _____ (e) _____ (f) _____ 2 s s s

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PROBLEMS

6. Which of the forces in the figure produces the largest magnitude torque about the rotation axis indicated? (a) 1 (b) 2 (c) 3 (d) 4 1 2 3 Axis

4

Multiple-Choice Questions 6–8 7. Which of the forces in the figure produces a CW torque about the rotation axis indicated? (a) 3 only (b) 4 only (c) 1 and 2 (d) 1, 2, and 3 (e) 1, 2, and 4 8. Which pair of forces in the figure might produce equal magnitude torques with opposite signs? (a) 2 and 3 (b) 2 and 4 (c) 1 and 2 (d) 1 and 3 (e) 1 and 4 (f) 3 and 4 9. A high diver in midair pulls her legs inward toward her chest in order to rotate faster. Doing so changes which of these quantities: her angular momentum L, her rotational inertia I, and her rotational kinetic energy Krot? (a) L only (b) I only (c) Krot only (d) L and I only (e) I and Krot only (f) all three 10. A uniform bar of mass m is supported by a pivot at its top, about which the bar can swing like a q pendulum. If a force F is applied perpendicularly to the lower end F of the bar as in the diagram, how big must F be in order to hold the bar in equilibrium at an angle q from the vertical? (a) 2mg (b) 2mg sin q (c) (mg/2) sin q (d) 2mg cos q (e) (mg/2) cos q (f) mg sin q

Problems

✦ Blue # 1

2

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Combination conceptual/quantitative problem Biological or medical application Challenging problem Detailed solution in the Student Solutions Manual Problems paired by concept Text website interactive or tutorial

8.1 Rotational Kinetic Energy and Rotational Inertia 1. Verify that _12 Iw 2 has dimensions of energy. 2. What is the rotational inertia of a solid iron disk of mass 49 kg, with a thickness of 5.00 cm and radius of 20.0 cm, about an axis through its center and perpendicular to it? 3. A bowling ball made for a child has half the radius of an adult bowling ball. They are made of the same material (and therefore have the same mass per unit volume). By what factor is the (a) mass and (b) rotational inertia of the child’s ball reduced compared with the adult ball? 4. Find the rotational inertia of the y A system of point particles shown in the figure assuming the system rotates about the (a) x-axis, B x (b) y-axis, (c) z-axis. The z-axis C is perpendicular to the xy-plane and points out of the page. Point particle A has a mass of 200 g and is located at (x, y, z) = (−3.0 cm, 5.0 cm, 0), point particle B has a mass of 300 g and is at (6.0 cm, 0, 0), and point particle C has a mass of 500 g and is at (−5.0 cm, −4.0 cm, 0). (d) What are the x- and y-coordinates of the center of mass of the system? 5. Four point masses of 3.0 kg each are arranged in a square on massless rods. The length of a side of the square is 0.50 m. What is the rotational inertia for rotation about an axis (a) passing through masses B and C? (b) passing through masses A and C? (c) passing through the center of the square and perpendicular to the plane of the square? A

B

A

B

0.50 m D

0.50 m (a)

C

A

B 0.50 m

0.50 m D

0.50 m (b)

C

D

0.50 m

C

(c)

6. How much work is done by the motor in a CD player to make a CD spin, starting from rest? The CD has a diameter of 12.0 cm and a mass of 15.8 g. The laser scans at a constant tangential velocity of 1.20 m/s. Assume that the music is first detected at a radius of 20.0 mm from the center of the disk. Ignore the small circular hole at the CD’s center. 7. Find the ratio of the rotational inertia of the Earth for rotation about its own axis to its rotational inertia for rotation about the Sun. 8. A bicycle has wheels of radius 0.32 m. Each wheel has a rotational inertia of 0.080 kg·m2 about its axle. The total mass of the bicycle including the wheels and the rider is 79 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?

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9. In many problems in previous chapters, cars and other objects that roll on wheels were considered to act as if they were sliding without friction. (a) Can the same assumption be made for a wheel rolling by itself ? Explain your answer. (b) If a moving car of total mass 1300 kg has four wheels, each with rotational inertia of 0.705 kg·m2 and radius of 35 cm, what fraction of the total kinetic energy is rotational? 10. A centrifuge has a rotational inertia of 6.5 × 10−3 kg·m2. How much energy must be supplied to bring it from rest to 420 rad/s (4000 rpm)?

8.2 Torque 11. A mechanic turns a wrench using a force of 25 N at a distance of 16 cm from the rotation axis. The force is perpendicular to the wrench handle. What magnitude torque does she apply to the wrench? 12. The pull cord of a lawnmower engine is wound around a drum of radius 6.00 cm. While the cord is pulled with a force of 75 N to start the engine, what magnitude torque does the cord apply to the drum? 13. A child of mass 40.0 kg is sitting on a horizontal seesaw at a distance of 2.0 m from the supporting axis. What is the magnitude of the torque about the axis due to the weight of the child? 14. A 124-g mass is placed on one pan of a balance, at a point 25 cm from the support of the balance. What is the magnitude of the torque about the support exerted by the mass? 15. A uniform door weighs 50.0 N and is 1.0 m wide and 2.6 m high. What is the magnitude of the torque due to the door’s own weight about a horizontal axis perpendicular to the door and passing through a corner? 16. A tower outside the Houses of Parliament in London has a famous clock commonly referred to as Big Ben, the name of its 13-ton chiming bell. The hour hand of each clock face is 2.7 m long and has a mass of 60.0 kg. Assume the hour hand to be a uniform rod attached at one end. (a) What is the torque on the clock mechanism due to the weight of one of the four hour hands when the clock strikes noon? The axis of rotation is perpendicular to a clock face and through the center of the clock. (b) What is the torque due to the weight of one hour hand about the same axis when the clock tolls 9:00 a.m.? ✦17. Any pair of equal and opposite forces acting on the same object is called a couple. Consider the couple in part (a) of the figure. The rotation axis is perpendicular to the page and passes through point P. (a) Show that the net torque due to this couple is equal to Fd, where d is the distance between the lines of action of the two forces. Because the distance d is independent of the location of the rotation axis, this shows that the torque is the same for any rotation axis. (b) Repeat for the couple in part (b) of the figure. Show that the torque is still

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Fd if d is the perpendicular distance between the lines of action of the forces. F

F

P

P

d x1

d x1

F F x2

x2

(a)

(b)

18. A 46.4-N force is (a) (b) applied to the 43.0° 1.26 m outer edge of a (c) door of width Axis 1.26 m in such a way that it acts (a) perpendicular to the door, (b) at an angle of 43.0° with respect to the door surface, (c) so that the line of action of the force passes through the axis of the door hinges. Find the torque for these three cases. 19. A trap door, of length and width 1.65 m, is held open at an angle of 65.0° with respect to the floor. A rope is attached to the raised edge of the door and fastened to the wall behind 65.0° the door in such a position that the rope pulls perpendicularly to the trap door. If the mass of the trap door is 16.8 kg, what is the torque exerted on tutorial: deck hatch) the trap door by the rope? ( 20. A weightless rod, 10.0 m long, supports three weights as shown. Where is its center of gravity? 5.0 kg

15.0 kg

10.0 kg

0.0

5.0 m

10.0 m

2.00 m 21. A door weighing 300.0 N measures 2.00 m × 3.00 m and is of uniform density; that is, the 0.25 m mass is uniformly 3.00 m distributed throughout the volume. A doorknob is 5.0 N attached to the door 300.0 N as shown. Where is the center of gravity if the doorknob weighs 5.0 N and is located 0.25 m from the edge? ✦22. A plate of uniform thickness is shaped as shown. Where is the center of gravity? Assume the origin (0, 0) is

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located at the lower left corner of the plate; the upper left corner is at (0, s) and upper right corner is at (s, s). s 0.50s s

0.50s

0.50s

8.3 Calculating Work Done from the Torque 23. A stone used to grind wheat into flour is turned through 12 revolutions by a constant force of 20.0 N applied to the rim of a 10.0-cm-radius shaft connected to the wheel. How much work is done on the stone during the 12 revolutions? 24. The radius of a wheel is 0.500 m. A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude 5.00 N, unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. (a) How much rope unwinds while the wheel makes 1.00 revolution? (b) How much work is done by the rope on the wheel during this time? (c) What is the torque on the wheel due to the rope? (d) What is the angular displacement Δq, in radians, of the wheel during 1.00 revolution? (e) Show that the numerical value of the work done is equal to the product t Δq. ✦ 25. A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at the effective radius of the flywheel). (a) How much work is done to bring this wheel from rest to a speed of 120 rpm in a time interval of 30.0 s? (b) What is the applied torque on the flywheel (assumed constant)? ✦26. A Ferris wheel rotates because a motor exerts a torque on the wheel. The radius of the London Eye, a huge observation wheel on the banks of the Thames, is 67.5 m and its mass is 1.90 × 106 kg. The cruising angular speed of the wheel is 3.50 × 10−3 rad/s. (a) How much work does the motor need to do to bring the stationary wheel up to cruising speed? [Hint: Treat the wheel as a hoop.] (b) What is the torque (assumed constant) the motor needs to provide to the wheel if it takes 20.0 s to reach the cruising angular speed?

8.4 Rotational Equilibrium 27. A rod is being used as a FA lever as shown. The fulcrum is 1.2 m from the 2.4 m load and 2.4 m from the applied force. If the load 1.2 m has a mass of 20.0 kg, what force must be applied to lift the load? 28. A weight of 1200 N rests on a lever at a point 0.50 m from a support. On the same side of the support, at a

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distance of 3.0 m from it, an 3.0 m upward force with magnitude F is F applied. Ignore the weight of the board itself. If the system is in 1200 N equilibrium, what is F? 0.50 m 29. A sculpture is 4.00 m tall and has its center of gravity located 1.80 m above the center of its base. The base is a square with a side of 1.10 m. To what CG angle q can the q sculpture be tipped before it falls over? 1.80 m tutorial: filing ( cabinet) 1.10 m 30. A house painter is standing on a uniform, horizontal platform that is held in equilibrium by two cables attached to supports on the roof. The painter has a mass of 75 kg and the mass of the platform is 20.0 kg. The distance from the left end of the platform to where the FL painter is standing FR is d = 2.0 m and the total length of the platform is 5.0 m. (a) How large is the force exerted by the left-hand cable on the platform? (b) How large is the d force exerted by the right-hand cable? 31. Four identical uni0.1667 m 0.0833 m form metersticks are 0.3333 m stacked on a table as shown. Where is the 0.8600 m x-coordinate of the cm of the metersticks if the origin is chosen at the left end of the lowest stick? Why does the system balance? ✦32. A uniform diving board, of length 5.0 m and mass 55 kg, is supported at two points; one support is located 3.4 m from the end of the board and the second is at 4.6 m from the end (see Fig. 8.19). What are the forces acting on the board due to the two supports when a diver of mass 65 kg stands at the end of the board over the water? Assume that these forces are vertical. ( tutorial: plank) [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.] ✦33. A house painter stands 3.0 m above the ground on a 5.0-m-long ladder that leans against the wall at a point

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4.7 m above the ground. The painter weighs 680 N and the ladder weighs 120 N. Assuming no friction between the house and the upper end of the ladder, find the force of friction that the driveway exerts on the bottom of the interladder. ( active: ladder; tutorial: ladder)

Wall Painter

4.7 m Ladder 3.0 m

Driveway

✦34. A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber’s waist 15 cm to the right of his center of gravity. If the climber weighs 770 N, find 25° (a) the tension in the rope and (b) the magnitude and direction of the contact CG force exerted by the 91 cm wall on the climber’s 106 cm feet. 35. A sign is supported by a uniform horizontal boom of length 3.00 m T and weight 80.0 N. A 35° cable, inclined at an Hinge angle of 35° with the 80.0 N boom, is attached at 1.50 m a distance of 2.38 m 120.0 N 2.38 m from the hinge at the 3.00 m wall. The weight of the sign is 120.0 N. What is the tension in the cable and what are the horizontal and vertical forces Fx and Fy exerted on the boom by the hinge? Comment on the magnitude of Fy. T 36. A boom of mass m supports a steel girder of weight W hanging from its end. One end of the mg q boom is hinged at the W floor; a cable attaches to the other end of the boom and pulls horizontally on it. The boom makes an angle q with the horizontal. Find the tension in the cable as a function of m, W, q, and g. Comment on the tension at q = 0 and q = 90°.

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37. You are asked to hang a uniform beam and sign using a cable that q has a breaking strength of 417 N. The store 0.80 m owner desires that it hang out over the sidewalk as shown. The 1.00 m sign has a weight of 200.0 N and the beam’s weight is 50.0 N. The beam’s length is 1.50 m and the sign’s dimensions are 1.00 m horizontally × 0.80 m vertically. What is the minimum angle q that you can have between the beam and cable? 38. Refer to Problem 37. You chose an angle q of 33.8°. An 8.7-kg cat has climbed onto the beam and is walking from the wall toward the point where the cable meets the beam. How far can the cat walk before the cable breaks? 39. A man is doing push-ups. He has a mass of 68 kg and his center of gravity is located at a horizontal distance of 0.70 m from his palms and 1.00 m from his feet. Find the forces exerted by the floor on his palms and feet. CG

0.70 m

1.00 m

8.5 Equilibrium in the Human Body 40. Your friend balances a package with mass m = 10 kg on top of his head while standing. The mass of his upper body is M = 55 kg (about 65% of his total mass). Because the spine is vertical rather than horizontal, the ⃗ s in Fig 8.32) force exerted by the sacrum on the spine (F is directed approximately straight up and the force ⃗ b) is negligibly small. exerted by the back muscles (F ⃗ s. Find the magnitude of F 41. Find the tension in the Achilles tendon and the force that the tibia exerts on the ankle joint when a person who weighs 750 N supports himself on the ball of one foot. The normal force N = 750 N pushes up on the ball of the foot on one side of the ankle joint, while the Achilles tendon pulls up on the foot on the other side of the joint. Gastrocnemiussoleus muscles FAchilles

Tibia FTibia

Achilles tendon N

Calcaneus (heel bone)

12.8 cm 4.60 cm

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42. In the movie Terminator, Arnold Schwarzenegger lifts ✦45. someone up by the neck and, with both arms fully extended and horizontal, holds the person off the ground. If the person being held weighs 700 N, is 60 cm from the shoulder joint, and Arnold has an anatomy analogous to that in Fig. 8.30, what force must each of the deltoid muscles exert to perform this task? 43. Find the force exerted by the biceps muscle in holding a 1-L milk carton (weight 9.9 N) with the forearm parallel to the floor. Assume that the hand is 35.0 cm from the elbow and that the upper arm is 30.0 cm long. The elbow is bent at a right angle and one tendon of the biceps is attached to the forearm at a position 5.00 cm from the elbow, while the other tendon is attached at 30.0 cm from the elbow. The weight of the forearm and empty hand is 18.0 N and the center of gravity of the forearm is at a distance of 16.5 cm from the elbow. ✦ 46.

Fb 30.0 cm

CG

9.9 N 5.00 cm 16.5 cm

18.0 N 35.0 cm

44. A person is doing leg lifts with 3.0-kg ankle weights. She is sitting in a chair with her legs bent at a right angle initially. The quadriceps muscles are attached to the patella via a tendon; the patella is connected to the tibia by the patellar tendon, which attaches to bone 10.0 cm below the knee joint. Assume that the tendon pulls at an angle of 20.0° with respect to the lower leg, regardless of the position of the lower leg. The lower leg has a mass of 5.0 kg and its center of gravity is 22 cm below the knee. The ankle weight is 41 cm from the knee. If the person lifts one leg, find the force exerted by the patellar tendon to hold the leg at an angle of (a) 30.0° and (b) 90.0° with respect to the vertical. Quadriceps muscle 20.0° Patella

Patellar tendon 10.0 cm

Femur Tibia 22 cm 41 cm

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One day when your friend from Problem 40 is picking up a package, you notice that he bends at the waist to pick it up rather than keeping his back straight and bending his knees. You suspect that the lower back pain he complains about is caused by the large force ⃗ s in Fig. 8.32) when he lifts on his lower vertebrae (F objects in this way. Suppose that when the spine is horizontal, the back muscles exert a force ⃗ Fb as in Fig. 8.32 (44 cm from the sacrum and at an angle of 12° to the horizontal). Assume that the cm of his upper body (including the arms) is at its geometric center, 38 cm from the sacrum. Find the hori⃗ s when your friend is holding zontal component of F a 10-kg package at a distance of 76 cm from his ⃗ s found sacrum. Compare this with the magnitude of F in Problem 40. A man is trying to lift 60.0 kg off the floor by bending at the waist (see Fig. 8.32). Assume that the man’s upper body weighs 455 N and the upper body’s center of gravity is 38 cm from the sacrum (tailbone). (a) If, when bent over, the hands are a horizontal distance of 76 cm from the sacrum, what torque must be exerted by the erector spinae muscles to lift 60.0 kg off the floor? (The axis of rotation passes through the sacrum, as shown in Fig. 8.32.) (b) When bent over, the erector spinae muscles are a horizontal distance of 44 cm from the sacrum and act at a 12° angle above the horizontal. What force ⃗ b in Fig. 8.32) do the erector spinae muscles need to (F exert to lift the weight? (c) What is the component of this force that compresses the spinal column?

8.6 Rotational Form of Newton’s Second Law 47. Verify that the units of the rotational form of Newton’s second law [Eq. (8-9)] are consistent. In other words, show that the product of a rotational inertia expressed in kg·m2 and an angular acceleration expressed in rad/s2 is a torque expressed in N·m. 48. A spinning flywheel has rotational inertia I = 400.0 kg·m2. Its angular velocity decreases from 20.0 rad/s to zero in 300.0 s due to friction. What is the frictional torque acting? 49. A turntable must spin at 33.3 rpm (3.49 rad/s) to play an old-fashioned vinyl record. How much torque must the motor deliver if the turntable is to reach its final angular speed in 2.0 revolutions, starting from rest? The turntable is a uniform disk of diameter 30.5 cm and mass 0.22 kg. 50. A lawn sprinkler has three spouts that spray water, each 15.0 cm 15.0 cm long. As the water is sprayed, the sprinkler turns around in a circle. The sprinkler has a total rotational inertia of

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51.

52.

53.

54.

55.

✦56.

CHAPTER 8 Torque and Angular Momentum

9.20 × 10−2 kg·m2. If the sprinkler starts from rest and takes 3.20 s to reach its final speed of 2.2 rev/s, what force does each spout exert on the sprinkler? A chain pulls tangentially on a 40.6-kg uniform cylindrical gear with a tension of 72.5 N. The chain is attached along the outside radius of the gear at 0.650 m from the axis of rotation. Starting from rest, the gear takes 1.70 s to reach its rotational speed of 1.35 rev/s. What is the total frictional torque opposing the rotation of the gear? Four masses are arranged A B A 4.0 kg as shown. They are con0.75 m B 3.0 kg nected by rigid, massless Axis C 5.0 kg rods of lengths 0.75 m D 2.0 kg and 0.50 m. What torque D C must be applied to cause 0.50 m an angular acceleration of 0.75 rad/s2 about the axis shown? A bicycle wheel, of radius 0.30 m and mass 2 kg (concentrated on the rim), is rotating at 4.00 rev/s. After 50 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces? A playground merry-go-round (see Fig. 8.5), made in the shape of a solid disk, has a diameter of 2.50 m and a mass of 350.0 kg. Two children, each of mass 30.0 kg, sit on opposite sides at the edge of the platform. Approximate the children as point masses. (a) What torque is required to bring the merry-go-round from rest to 25 rpm in 20.0 s? (b) If two other bigger children are going to push on the merry-go-round rim to produce this acceleration, with what force magnitude must each child push? ( tutorial: roundabout) Two children standing on opposite sides of a merry-goround (see Fig. 8.5) are trying to rotate it. They each push in opposite directions with forces of magnitude 10.0 N. (a) If the merry-go-round has a mass of 180 kg and a radius of 2.0 m, what is the angular acceleration of the merry-go-round? (Assume the merry-go-round is a uniform disk.) (b) How fast is the merry-go-round rotating after 4.0 s? Refer to Atwood’s machine (Example 8.2). (a) Assuming that the cord does not slip as it passes around the pulley, what is the relationship between the angular acceleration of the pulley (a) and the magnitude of the linear acceleration of the blocks (a)? (b) What is the net torque on the pulley about its axis of rotation in terms of the tensions T1 and T2 in the left and right sides of the cord? (c) Explain why the tensions cannot be equal if m1 ≠ m2. (d) Apply Newton’s second law to each of the blocks and Newton’s second law for rotation to the pulley. Use these three equations to solve for a, T1, and T2. (e) Since the blocks move with constant acceleration, use the result of Example 8.2 along with the constant 2 2 acceleration equation v fy − v iy = 2ay Δy to check your answer for a.

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Axis ✦ 57. Derive the rotational form of Newton’s second law as follows. Consider a Fi rigid object that conmi ri sists of a large number N of particles. Let Fi, mi, and ri represent the tangential component of the net force acting on the ith particle, the mass of that particle, and the particle’s distance from the axis of rotation, respectively. (a) Use Newton’s second law to find ai, the particle’s tangential acceleration. (b) Find the torque acting on this particle. (c) Replace ai with an equivalent expression in terms of the angular acceleration a. (d) Sum the torques due to all the particles and show that N

∑ t i = Ia

i=1

8.7 The Motion of Rolling Objects 58. A solid sphere is rolling without slipping or sliding down a board that is tilted at an angle of 35° with respect to the horizontal. What is its acceleration? 59. A solid sphere is released from rest and allowed to roll down a board that has one end resting on the floor and is tilted at 30° with respect to the horizontal. If the sphere is released from a height of 60 cm above the floor, what is the sphere’s speed when it reaches the lowest end of the board? 60. A hollow cylinder, a uniform solid sphere, and a uniform solid cylinder all have the same mass m. The three objects are rolling on a horizontal surface with identical translational speeds v. Find their total kinetic energies in terms of m and v and order them from smallest to largest. 61. A solid sphere of mass 0.600 kg rolls without slipping along a horizontal surface with a translational speed of 5.00 m/s. It comes to an incline that makes an angle of 30° with the horizontal surface. Ignoring energy losses due to friction, to what vertical height above the horizontal surface does the sphere rise on the incline? 62. A bucket of water with a mass of 2.0 kg is attached to a rope that is wound around a cylinder. The cylinder has a mass of 3.0 kg and is mounted horizontally on frictionless bearings. The bucket is released from rest. (a) Find its speed after it has fallen through a distance of 0.80 m. What are (b) the tension in the rope and (c) the acceleration of the bucket? 63. A 1.10-kg bucket is tied to a rope that is wrapped around a pole mounted horizontally on frictionless bearings. The cylindrical pole has a diameter of 0.340 m and a mass of 2.60 kg. When the Problems 62 and 63

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✦64.

✦ 65.

✦66.

67.

✦68.

bucket is released from rest, how long will it take to fall to the bottom of the well, a distance of 17.0 m? A uniform solid cylinder rolls without slipping down an incline. A hole is drilled through the cylinder along its axis. The radius of the hole is 0.50 times the (outer) radius of the cylinder. (a) Does the cylinder take more or less time to roll down the incline now that the hole has been drilled? Explain. (b) By what percentage does drilling the hole change the time for the cylinder to roll down the incline? ( tutorial: rolling) A solid sphere of radius R and mass M slides without h r friction down a loop-the-loop track. The sphere starts Problems 65 and 66 from rest at a height of h above the horizontal. Assume that the radius of the sphere is small compared to the radius r of the loop. (a) Find the minimum value of h in terms of r so that the sphere remains on the track all the way around the loop. (b) Find the minimum value of h if, instead, the sphere rolls without slipping on the track. A hollow cylinder, of radius R and mass M, rolls without slipping down a loop-the-loop track of radius r. The cylinder starts from rest at a height h above the horizontal section of track. What is the minimum value of h so that the cylinder remains on the track all the way around the loop? If the hollow cylinder of Problem 66 is replaced with a solid sphere, will the minimum value of h increase, decrease, or remain the same? Once you think you know the answer and can explain why, redo the calculation to find h. The string in a yo-yo is wound around an axle of radius 0.500 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.200 kg and outer radius 2.00 cm. Starting from rest, it rotates and falls a distance of 1.00 m (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle. (a) What is the speed of the yo-yo when it reaches the distance of 1.00 m? (b) How long does it take to fall? [Hint: The translational and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.]

72.

73.

74.

75.

76.

77.

78.

✦79.

wheel. If the radius of the wheel is 2.6 m and it is rotating at 350 rpm, what is the magnitude of its angular momentum? The angular momentum of a spinning wheel is 240 kg·m2/s. After application of a constant braking torque for 2.5 s, it slows and has a new angular momentum of 115 kg·m2/s. What is the torque applied? How long would a braking torque of 4.00 N·m have to act to just stop a spinning wheel that has an initial angular momentum of 6.40 kg·m2/s? A figure skater is spinning at a rate of 1.0 rev/s with her arms outstretched. She then draws her arms in to her chest, reducing her rotational inertia to 67% of its original value. What is her new rate of rotation? A skater is initially spinning at a rate of 10.0 rad/s with a rotational inertia of 2.50 kg·m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to 1.60 kg·m2? A uniform disk with a mass of 800 g and radius 17.0 cm is rotating on frictionless bearings with an angular speed of 18.0 Hz when Jill drops a 120-g clod of clay on a point 8.00 cm from the center of the disk, where it sticks. What is the new angular speed of the disk? A spoked wheel with a radius of 40.0 cm and a mass of 2.00 kg is mounted horizontally on frictionless bearings. JiaJun puts his 0.500-kg guinea pig on the outer edge of the wheel. The guinea pig begins to run along the edge of the wheel with a speed of 20.0 cm/s with respect to the ground. What is the angular velocity of the wheel? Assume the spokes of the wheel have negligible mass. A diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.00 complete rotations in 1.33 s. If his rotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board? The rotational inertia for a diver in a pike position is about 15.5 kg·m2; it is only 8.0 kg·m2 in a tuck position. (a) If the diver gives himself an initial angular momentum of 106 kg·m2/s as he jumps off the board, how many turns can he make when jumping off a 10.0-m platform

8.8 Angular Momentum 69. A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a frequency of 0.550 rev/s. What is its angular momentum? Assume the turntable is a uniform disk. 70. Assume the Earth is a uniform solid sphere with radius of 6.37 × 106 m and mass of 5.97 × 1024 kg. Find the magnitude of the angular momentum of the Earth due to rotation about its axis. 71. The mass of a flywheel is 5.6 × 104 kg. This particular flywheel has its mass concentrated at the rim of the

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(a)

(b)

Problem 79. (a) Mark Ruiz in the tuck position. (b) Gregory Louganis in the pike position.

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in a tuck position? (b) How many in a pike position? [Hint: Gravity exerts no torque on the person as he falls; assume he is rotating throughout the 10.0-m dive.] 80. Consider the merry-go-round of Practice Problem 8.1. The child is initially standing on the ground when the merry-go-round is rotating at 0.75 rev/s. The child then steps on the merry-go-round. How fast is the merry-goround rotating now? By how much did the rotational kinetic energy of the merry-go-round and child change?

8.9 The Vector Nature of Angular Momentum Problems 81 and 82. A solid cylindrical disk is to be used as a stabilizer in a ship. By using a massive disk rotating in the hold of the ship, the captain knows that a large torque is required to tilt its angular momentum vector. The mass of the disk to be used is 1.00 × 105 kg and it has a radius of 2.00 m. ✦81. If the cylinder rotates at 300.0 rpm, what is the magnitude of the average torque required to tilt its axis by 60.0° in a time of 3.00 s? [Hint: Draw a vector diagram of the initial and final angular momenta.] 82. How should the disk be oriented to prevent rocking from side to side and from bow to stern? Does this orientation make it difficult to steer the ship? Explain.

✦87. A gymnast is performing a giant swing on the high bar. In a simplified model of the giant swing, assume that the gymnast keeps his arms and body straight as he swings all the way around the upper bar. Problem 87. Notice that the Assume also that angular speed is much greater the gymnast does no at the bottom of the swing. work during the swing. With what angular speed should he be moving at the bottom of the giant swing in order to make it all the way around? The distance from the bar to his feet is 2.0 m and his center of gravity is 1.0 m from his feet. ✦88. The 12.2-m crane weighs 18 kN and is lifting a 67-kN load. The hoisting cable (tension T1) passes over a pulley at the top of the crane and attaches to an electric winch in the cab. The pendant cable (tension T2), which supports the crane, is fixed to the top of the crane. Find the ten⃗ p at the pivot. sions in the two cables and the force F

T2

Comprehensive Problems 83. The Moon’s distance from Earth varies between 3.56 × 105 km at perigee and 4.07 × 105 km at apogee. What is the ratio of its orbital speed around Earth at perigee to that at apogee? 84. A ceiling fan has four blades, each with a mass of 0.35 kg and a length of 60 cm. Model each blade as a rod connected to the fan axle at one end. When the fan is turned on, it takes 4.35 s for the fan to reach its final angular speed of 1.8 rev/s. What torque was applied to the fan by the motor? Ignore torque due to the air. 85. The distance from the center of the breastbone to a man’s hand, with the arm outstretched and horizontal to the floor, is 1.0 m. The man is holding a 10.0-kg dumbbell, oriented vertically, in his hand, with the arm horizontal. What is the torque due to this weight about a horizontal axis through the breastbone perpendicular to his chest? 86. A uniform rod of length L is free to pivot around an axis through its upper end. If it is released from rest when horizontal, at what speed is the lower end moving at its lowest point? [Hint: The gravitational potential energy change is determined by the change in height of the center of gravity.]

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T1

40.0° T1 10.0° 5.0° .2

12

m

18 kN 67 kN

89. A collection of objects is set to rolling, without slipping, down a slope inclined at 30°. The objects are a solid sphere, a hollow sphere, a solid cylinder, and a hollow cylinder. A frictionless cube is also allowed to slide down the same incline. Which one gets to the bottom first? List the others in the order they arrive at the finish line. 90. A uniform cylinder with a radius of 15 cm has been attached to two cords and the cords are wound around it and hung from the ceiling. The cylinder is released from rest and the cords unwind as the cylinder descends. (a) What is the acceleration of the r cylinder? (b) If the mass of the cylinder is 2.6 kg, what is the tension in each cord?

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91. A modern sculpture has a large 95. A flat object in the xy-plane is free to rotate about horizontal spring, with a spring the z-axis. The gravitational field is uniform in the constant of 275 N/m, that is −y-direction. Think of the object as a large number of attached to a 53.0-kg piece of particles with masses mi located at coordinates (xi, yi), uniform metal at its end and as in the figure. (a) Show that the torques on the partiholds the metal at an angle of cles about the z-axis can be written ti = −ximig. (b) Show 50.0° 50.0° above the horizontal that if the center of gravity is located at (xCG, yCG), the direction. The other end of the total torque due to gravity on the object must be metal is wedged into a corner Σti = −xCGMg, where M is the total mass of the object. as shown. By how much has the spring stretched? (c) Show that xCG = xCM. (This same line of reasoning can be applied to objects that are not flat and to other ✦92. A painter (mass axes of rotation to show that yCG = yCM and zCG = zCM.) 61 kg) is walking along a trestle, consisting of y3 a uniform plank m3g y1 (mass 20.0 kg, m g y6 1 length 6.00 m) y2 m6g balanced on two y5 m2g sawhorses. Each m5g 0.28 m sawhorse is y4 1.40 m 1.40 m 6.00 m m4g placed 1.40 m from an end of x1 x2 x3 x4 x5 x6 the plank. A paint bucket (mass 4.0 kg, diameter 28 cm) Axis of rotation is placed as close as possible to the right-hand edge of the perpendicular to page plank while still having the whole bucket in contact with the plank. (a) How close to the right-hand edge of the 96. The operation of the Princeton Tokomak Fusion Test plank can the painter walk before tipping the plank and Reactor requires large bursts of energy. The power spilling the paint? (b) How close to the left-hand edge needed exceeds the amount that can be supplied by the can the same painter walk before causing the plank to utility company. Prior to pulsing the reactor, energy is tip? [Hint: As the painter walks toward the right-hand stored in a giant flywheel of mass 7.27 × 105 kg and edge of the plank and the plank starts to tip clockwise, rotational inertia 4.55 × 106 kg·m2. The flywheel rotates what is the force acting upward on the plank from the at a maximum angular speed of 386 rpm. When the left-hand sawhorse support?] stored energy is needed to operate the reactor, the fly✦93. An experimental flywheel, used to store energy and wheel is connected to an electrical generator, which conreplace an automobile engine, is a solid disk of mass verts some of the rotational kinetic energy into electric 200.0 kg and radius 0.40 m. (a) What is its rotational energy. (a) If the flywheel is a uniform disk, what is its inertia? (b) When driving at 22.4 m/s (50 mph), the fully radius? (b) If the flywheel is a hollow cylinder with its energized flywheel is rotating at an angular speed of mass concentrated at the rim, what is its radius? (c) If the 3160 rad/s. What is the initial rotational kinetic energy flywheel slows to 252 rpm in 5.00 s, what is the average of the flywheel? (c) If the total mass of the car is power supplied by the flywheel during that time? 1000.0 kg, find the ratio of the initial rotational kinetic ✦97. A box of mass 42 kg sits 42 kg energy of the flywheel to the translational kinetic energy on top of a ladder. Ignorof the car. (d) If the force of air resistance on the car is ing the weight of the 670.0 N, how far can the car travel at a speed of ladder, find the tension 22.4 m/s (50 mph) with the initial stored energy? Ignore in the rope. Assume that Rope losses of mechanical energy due to means other than air the rope exerts horizonh resistance. tal forces on the ladder 0.50h ✦94. (a) Assume the Earth is a uniform solid sphere. Find the at each end. [Hint: Use a 75° 75° kinetic energy of the Earth due to its rotation about its symmetry argument; axis. (b) Suppose we could somehow extract 1.0% of then analyze the forces 1.26 m the Earth’s rotational kinetic energy to use for other and torques on one side purposes. By how much would that change the length of of the ladder.] the day? (c) For how many years would 1.0% of the ✦98. A person is trying to lift a ladder of mass 15 kg and Earth’s rotational kinetic energy supply the world’s length 8.0 m. The person is exerting a vertical force on 21 energy usage (assume a constant 1.0 × 10 J per year)? the ladder at a point of contact 2.0 m from the center of

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99.

100.

101.

✦102.

103.

✦104.

CHAPTER 8 Torque and Angular Momentum

gravity. The opposite end of the ladder rests on the floor. (a) When the ladder makes an angle of 60.0° with the floor, what is this vertical force? (b) A person tries to help by lifting the ladder at the point of contact with the floor. Does this help the person trying to lift the ladder? Explain. A crustacean (Hemisquilla ensigera) rotates its anterior limb to strike a mollusk, intending to break it open. The limb reaches an angular velocity of 175 rad/s in 1.50 ms. We can approximate the limb as a thin rod rotating about an axis perpendicular to one end (the joint where the limb attaches to the crustacean). (a) If the mass of the limb is 28.0 g and the length is 3.80 cm, what is the rotational inertia of the limb about that axis? (b) If the extensor muscle is 3.00 mm from the joint and acts perpendicular to the limb, what is the muscular force required to achieve the blow? A block of mass m2 m1 I hangs from a rope. Pulley The rope wraps around a pulley of rotational inertia I and m2 then attaches to a second block of mass m1, which sits on a frictionless table. What is the acceleration of the blocks when they are released? A 2.0-kg uniform flat disk is thrown into the air with a linear speed of 10.0 m/s. As it travels, the disk spins at 3.0 rev/s. If the radius of the disk is 10.0 cm, what is the magnitude of its angular momentum? A hoop of 2.00-m circumference is rolling down an inclined plane of length 10.0 m in a time of 10.0 s. It started out from rest. (a) What is its angular velocity when it arrives at the bottom? (b) If the mass of the hoop, concentrated at the rim, is 1.50 kg, what is the angular momentum of the hoop when it reaches the bottom of the incline? (c) What force(s) supplied the net torque to change the hoop’s angular momentum? Explain. [Hint: Use a rotation axis through the hoop’s center.] (d) What is the magnitude of this force? A large clock has a second hand with a mass of 0.10 kg concentrated at the tip of the pointer. (a) If the length of the second hand is 30.0 cm, what is its angular momentum? (b) The same clock has an hour hand with a mass of 0.20 kg concentrated at the tip of the pointer. If the hour hand has a length of 20.0 cm, what is its angular momentum? A planet moves around the Sun in an elliptical orbit (see Fig. 8.39). (a) Show that the external torque acting on the planet about an axis through the Sun is zero. (b) Since the torque is zero, the planet’s angular momentum is constant. Write an expression for the planet’s angular momentum in terms of its mass m, its distance r from the Sun, and its angular velocity w. (c) Given r and w, how much area is swept out during a short time

gia04535_ch08_260-315.indd 308

Δt? [Hint: Think of the area as a fraction of the area of a circle, like a slice of pie; if Δt is short enough, the radius of the orbit during that time is nearly constant.] (d) Show that the area swept out per unit time is constant. You have just proved Kepler’s second law! ✦105. A 68-kg woman stands straight with both feet flat on the floor. Her center of gravity is a horizontal distance of 3.0 cm in front of a line that connects her two ankle joints. The Achilles tendon attaches the calf muscle to the foot a distance of 4.4 cm behind the ankle joint. If the Achilles tendon is inclined at an angle of 81° with respect to the horizontal, find the force that each calf muscle needs to exert while she is standing. [Hint: Consider the equilibrium of the part of the body above the ankle joint.] 106. A merry-go-round (radius R, rotational inertia Ii) spins with negligible friction. Its initial angular velocity is w i. A child (mass m) on the merry-go-round moves from the center out to the rim. (a) Calculate the angular velocity after the child moves out to the rim. (b) Calculate the rotational kinetic energy and angular momentum of the system (merry-go-round + child) before and after. 107. Since humans are generally not symmetrically shaped, the height of our center of gravity is generally not half of our height. One way to determine the location of the center of gravity is shown in the diagram. A 2.2-m-long uniform plank is supported by two bathroom scales, one at either end. Initially the scales each read 100.0 N. A 1.60-m-tall student then lies on top of the plank, with the soles of his feet directly above scale B. Now scale A reads 394.0 N and scale B reads 541.0 N. (a) What is the student’s weight? (b) How far is his center of gravity from the soles of his feet? (c) When standing, how far above the floor is his center of gravity, expressed as a fraction of his height? CG

mpg A

msg

B

x1

x2

Problem 107 ✦108. A spool of thread of mass m rests on a plane inclined at R angle q. The end of r the thread is tied as shown. The outer radius of the spool q is R and the inner radius (where the thread is wound) is r. The rotational

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ANSWERS TO PRACTICE PROBLEMS

inertia of the spool is I. Give all answers in terms of m, q, R, r, I, and g. (a) If there is no friction between the spool and the incline, describe the motion of the spool and calculate its acceleration. (b) If the coefficient of friction is large enough to keep the spool from slipping, calculate the magnitude and direction of the frictional force. (c) What is the minimum possible coefficient of friction to keep the spool from slipping in part (b)? 109. A bicycle travels up an incline at constant FC r2 velocity. The magnir1 tude of the frictional force due to the road on the rear wheel is f = 3.8 N. The upper f section of chain pulls on the sprocket wheel, which is attached to the rear ⃗ C . The lower section of chain is wheel, with a force F slack. If the radius of the rear wheel is 6.0 times the radius of the sprocket wheel, what is the magnitude of ⃗ C with which the chain pulls? the force F ✦110. A circus roustabout is attaching the circus tent to the top of the main support post of length L when the post suddenly breaks at the base. The worker’s weight is negligible relative to that of the uniform post. What is the speed with which L the roustabout reaches the ground if (a) he jumps at the instant he hears the post crack or (b) if he clings to the post and rides to the ground with it? (c) Which is the safest course of action for the roustabout? ✦111. A student stands on a platform that is free to rotate and holds two dumbbells, each at a distance of 65 cm from his central axis. Another student gives him a push and starts the system of student, dumbbells, and platform rotating at 0.50 rev/s. The student on the platform then pulls the dumbbells in close to his chest so that they are each 22 cm from his central axis. Each dumbbell has a mass of 1.00 kg and the rotational inertia of the student, platform, and dumbbells is initially 2.40 kg·m2. Model each arm as a uniform rod of mass 3.00 kg with one end at the central axis; the length of the arm is initially 65 cm and then is reduced to 22 cm. What is his new rate 65 cm 65 cm of rotation? 22 cm

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309

112. A person places his hand palm downward on a scale and pushes down on the scale 38 cm until it reads 96 N. 2.5 cm The triceps muscle 96 N is responsible for this arm extension force. Find the force exerted by the triceps muscle. The bottom of the triceps muscle is 2.5 cm to the left of the elbow joint and the palm is pushing at approximately 38 cm to the right of the elbow joint. 113. The posture of small Body Leg Leg animals may preFwind vent them from q being blown over by mg the wind. For example, with wind blowing from the side, a small insect stands with bent legs; the more bent the legs, the lower the body and the smaller the angle q. The wind exerts a force on the insect, which causes a torque about the point where the downwind feet touch. The torque due to the weight of the insect must be equal and opposite to keep the insect from being blown over. For example, the drag force on a blowfly due to a sideways wind is Fwind = cAv2, where v is the velocity of the wind, A is the cross-sectional area on which the wind is blowing, and c ≈ 1.3 N·s2·m−4. (a) If the blowfly has a cross-sectional side area of 0.10 cm2, a mass of 0.070 g, and crouches such that q = 30.0°, what is the maximum wind speed in which the blowfly can stand? (Assume that the drag force acts at the center of gravity.) (b) How about if it stands such that q = 80.0°? (c) Compare to the maximum wind velocity that a dog can withstand, if the dog stands such that q = 80.0°, has a cross-sectional area of 0.030 m2, and weighs 10.0 kg. (Assume the same value of c.) ✦114. (a) Redo Example 8.7 to find an algebraic solution for d in terms of M, m, ms, L, and q. (b) Use this expression to show that placing the ladder at a larger angle q (that is, more nearly vertical) enables the person to climb farther up the ladder without having it slip, all other things being equal. (c) Using the numerical values from Example 8.7, find the minimum angle q that enables the person to climb all the way to the top of the ladder.

Answers to Practice Problems 2 8.1 390 kg·m _____________ 2m2 gh 8.2 v = ____________ m1 + m2 + I/R2

√

8.3 53 N; 8.4 N·m 8.4 −65 N·m

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8.5 8.3 J 8.6 left support, downward; right support, upward 8.7 0.27 8.8 57 N, downward 8.9 It must lie in the same vertical plane as the two ropes holding up the rings. Otherwise, the gravitational force would have a nonzero lever arm with respect to a horizontal axis that passes through the contact points between his hands and the rings; thus, gravity would cause a net torque about that axis. 8.10 460 N 8.11 (a) 2380 rad; (b) 3.17 kJ; (c) 1.34 N·m 8.12 solid ball, _27 ; hollow ball, _25 8.13 _12 g sin q 8.14 5% increase 8.15 16 cm/s

Answers to Checkpoints

8.2 The longer handle lets you push at a greater distance from the rotation axis. Thus, you can exert a larger torque. 8.4 Yes in both cases. Torque depends not only on the magnitude and direction of the force but also on the point where the force is applied. Two forces that do not add to zero can produce torques that add to zero due to different lever arms. Then the net torque is zero and the net torque nonzero; the object is in rotational equilibrium but not in translational equilibrium. Similarly, two forces that add to zero can have different lever arms and produce torques that do not add to zero. In this case the net force is zero and the net torque is nonzero; the object is in translational equilibrium but not in rotational equilibrium. 8.7 (a) falling without spinning; (b) spinning about a fixed axis; (c) rolling without slipping along a surface 8.8 Yes. If friction is negligible, the external torque is zero so her angular momentum does not change. Extending her arms and leg makes her rotational inertia increase back to its initial value, so her angular velocity decreases to its initial value.

8.1 Rotational inertia involves distances from masses to the rotation axis; distances along the rotation axis are irrelevant. Another way to see it: cut the cylinder or disk into a large number of thin disks with the same radius. Each thin disk has rotational inertia Ii = _12 mi R2. Now add up the rotational inertias of the thin disks: I = ∑Ii = ∑_12 mi R2 = _1 R2 ∑m = _1 MR2. 2 2 i

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Review & Synthesis: Chapters 6–8 Review Exercises 1. A spring scale in a French market is calibrated to show the mass of vegetables in grams and kilograms. (a) If the marks on the scale are 1.0 mm apart for every 25 g, what maximum extension of the spring is required to measure up to 5.0 kg? (b) What is the spring constant of the spring? [Hint: Remember that the scale really measures force.] 2. Plot a graph of this data for a spring resting horizontally on a table. Use your graph to find (a) the spring constant and (b) the relaxed length of the spring. Force (N)

0.200

0.450

0.800

1.500

Spring length (cm)

13.3

15.0

17.3

22.0

11.

12.

3. A pendulum consists of a bob of mass m attached to the end of a cord of length L. The pendulum is released from a point at a height of L/2 above the lowest point of the swing. What is the tension in the cord as the bob passes the lowest point? 4. How much energy is expended by an 80.0-kg person in climbing a vertical distance of 15 m? Assume that mus13. cles have an efficiency of 22%; that is, the work done by the muscles to climb is 22% of the energy expended. 5. Ugonna stands at the top of an incline and pushes a 100-kg crate to get it started sliding down the incline. The crate slows to a halt after traveling 1.50 m along the incline. (a) If the initial speed of the crate was 2.00 m/s and the 14. angle of inclination is 30.0°, how much energy was dissipated by friction? (b) What is the coefficient of sliding friction? 6. A packing carton slides down an inclined plane of angle 30.0° and of incline length 2.0 m. If the initial speed of the carton is 4.0 m/s directed down the incline, what is the speed at the bottom? Ignore friction. 7. A child’s playground swing is supported by chains that are 4.0 m long. If the swing is 0.50 m above the ground and moving at 6.0 m/s when the chains are vertical, what is ✦15. the maximum height of the swing? 8. A block slides down a plane that is inclined at an angle of 53° with respect to the horizontal. If the coefficient of kinetic friction is 0.70, what is the acceleration of the block? 9. Gerald wants to know how fast he can throw a ball, so he hangs a 2.30-kg target on a rope from a tree. He picks up a 0.50-kg ball of putty and throws it horizontally against the 16. target. The putty sticks to the target and the putty and target swing up a vertical distance of 1.50 m from its original position. How fast did Gerald throw the ball of putty? 10. A hollow cylinder rolls without slipping or sliding along a horizontal surface toward an incline. If the cylinder’s

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speed is 3.00 m/s at the base of the incline and the angle of inclination is 37.0°, how far along the incline does the cylinder travel before coming to a stop? A grinding wheel, with a mass of 20.0 kg and a radius of 22.4 cm, is a uniform cylindrical disk. (a) Find the rotational inertia of the wheel about its central axis. (b) When the grinding wheel’s motor is turned off, friction causes the wheel to slow from 1200 rpm to rest in 60.0 s. What torque must the motor provide to accelerate the wheel from rest to 1200 rpm in 4.00 s? Assume that the frictional torque is the same regardless of whether the motor is on or off. An 11-kg bicycle is moving with a linear speed of 7.5 m/s. Each wheel can be modeled as a thin hoop with a mass of 1.3 kg and a diameter of 70 cm. The bicycle is stopped in 4.5 s by the action of brake pads that squeeze the wheels and slow them down. The coefficient of friction between the brake pads and a wheel is 0.90. There are four brake pads altogether; assume they apply equal magnitude normal forces on the wheels. What is the normal force applied to a wheel by one of the brake pads? A 0.185-kg spherical steel ball is used in a pinball machine. The ramp is 2.05 m long and tilted at an angle of 5.00°. Just after a flipper hits the ball at the bottom of the ramp, the ball has an initial speed of 2.20 m/s. What is the speed of the ball when it reaches the top of the pinball machine? A rotating star collapses under the influence of gravitational forces to form a pulsar. The radius of the star after collapse is 1.0 × 10−4 times the radius before collapse. There is no change in mass. In both cases, the mass of the star is uniformly distributed in a spherical shape. Find the ratios of the (a) angular momentum, (b) angular velocity, and (c) rotational kinetic energy of the star after collapse to the values before collapse. (d) If the period of the star’s rotation before collapse is 1.0 × 107 s, what is its period after collapse? A 0.122-kg dart is fired from a gun with a speed of 132 m/s horizontally into a 5.00-kg wooden block. The block is attached to a spring with a spring constant of 8.56 N/m. The coefficient of kinetic friction between the block and the horizontal surface it is resting on is 0.630. After the dart embeds itself into the block, the block slides along the surface and compresses the spring. What is the maximum compression of the spring? A 5.60-kg uniform door is 0.760 m wide by 2.030 m high, and is hung by two hinges, one at 0.280 m from the top and one at 0.280 m from the bottom of the door. If the vertical components of the forces on each of the two hinges are identical, find the vertical and horizontal force components acting on each hinge due to the door.

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17. Consider the apparatus shown in the figure (not to scale). The pulley, which can be treated as a uniform disk, has a mass of 60.0 g and a radius of 3.00 cm. The spool also has a radius of 3.00 cm. The rotational inertia of the spool, axle, and paddles about their axis of rotation is 0.00140 kg·m2. The block has a mass of 0.870 kg and is released from rest. After the block has fallen a distance of 2.50 m, it has a speed of 3.00 m/s. How much energy has been delivered to the fluid in the beaker? Pulley

22.

Spool Axle Paddles

23.

24. 18. It is the bottom of the ninth inning at a baseball game. The score is tied and there is a runner on second base when the batter gets a hit. The 85-kg base runner rounds third base and is heading for home with a speed of 8.0 m/s. Just before he reaches home plate, he crashes into the opposing team’s catcher, and the two players slide together along the base path toward home plate. The catcher has a mass of 95 kg and the coefficient of friction between the players and the dirt on the base path is 0.70. How far do the catcher and base runner slide? 19. Pendulum bob A has half the mass of pendulum bob B. Each bob is tied to a string that is 5.1 m long. When bob A is held with its string horizontal and then released, it swings down and, once bob A’s string is vertical, it collides elastically with bob B. How high do the bobs rise after the collision? 20. During a game of marbles, the “shooter,” ✦ 40° a marble with three times the mass of the q other marbles, has a speed of 3.2 m/s just before it hits one of the marbles. The other marble bounces off the shooter in an elastic collision at an angle of 40°, as shown, and the shooter moves off at an angle q. Determine (a) the speed of the shooter after the collision, (b) the speed of the marble after the collision, and (c) the angle q. 21. At the beginning of a scene in an action movie, the 78.0-kg star, Indianapolis Jones, will stand on a ledge 3.70 m above the ground and the 55.0-kg heroine, Georgia Smith, will stand on the ground. Jones will swing down on a rope, grab Smith around the waist, and continue swinging until they come to rest on another ledge on the other side of the set. At what height above the ground should the second ledge be placed? Assume that Jones and Smith remain nearly upright during the

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25.

26.

27.

swing so that their cms are always the same distance above their feet. A uniform disk is rotated about its symmetry axis. The disk goes from rest to an angular speed of 11 rad/s in a time of 0.20 s with a constant angular acceleration. The rotational inertia and radius of the disk are 1.5 kg·m2 and 11.5 cm, respectively. (a) What is the angular acceleration during the 0.20-s interval? (b) What is the net torque on the disk during this time? (c) After the applied torque stops, a frictional torque remains. This torque has an associated angular acceleration of 9.8 rad/s2. Through what total angle q (starting from time t = 0) does the disk rotate before coming to rest? (d) What is the speed of a point halfway between the rim of the disk and its rotation axis 0.20 s after the applied torque is removed? A block is released from rest and slides down an incline. The coefficient of sliding friction is 0.38 and the angle of inclination is 60.0°. Use energy considerations to find how fast the block is sliding after it has traveled a distance of 30.0 cm along the incline. A uniform solid cylinder rolls without slipping or sliding down an incline. The angle of inclination is 60.0°. Use energy considerations to find the cylinder’s speed after it has traveled a distance of 30.0 cm along the incline. A block of mass 2.00 kg slides eastward along a frictionless surface with a speed of 2.70 m/s. A chunk of clay with a mass of 1.50 kg slides southward on the same surface with a speed of 3.20 m/s. The two objects collide and move off together. What is their velocity after the collision? An ice-skater, with a mass of 60.0 kg, 60.0 kg glides in a circle of radius 1.4 m with a tangential speed of 6.0 m/s 6.0 m/s. A second skater, with a mass of 30.0 kg, glides on the same circular 30.0 kg 2.0 m/s path with a tangential speed of 2.0 m/s. At an instant of time, both skaters grab the ends of a lightweight, rigid set of rods, set at 90° to each other, that can freely rotate about a pole, fixed in place on the ice. (a) If each rod is 1.4 m long, what is the tangential speed of the skaters after they grab the rods? (b) What is the direction of the angular momentum before and after the skaters “collide” with the rods? In a motor, a flywheel (solid disk of radius R and mass M) is rotating with angular velocity w i. When the clutch is released, a second disk (radius r and mass m) initially not rotating is brought into frictional contact with the flywheel. The two disks spin around the same axle with frictionless bearings. After a short time, friction between the two disks brings them to a common angular velocity. (a) Ignoring external influences, what is the final angular velocity?

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28.

29.

30.

✦31.

(b) Does the total angular momentum of the two change? If so, account for the change. If not, explain why it does not. (c) Repeat (b) for the rotational kinetic energy. A child’s toy is made of a 12.0-cm-radius rotating wheel that picks up 1.00-g pieces of candy in a pocket at its lowest point, brings the candy to the top, then releases it. The frequency of rotation is 1.60 Hz. (a) How far from its starting point does the candy land? (b) What is the radial acceleration of the candy when it is on the wheel? A Vulcan spaceship has a mass of 65 000 kg and a Romulan spaceship is twice as massive. Both have engines that generate the same total force of 9.5 × 106 N. (a) If each spaceship fires its engine for the same amount of time, starting from rest, which will have the greater kinetic energy? Which will have the greater momentum? (b) If each spaceship fires its engine for the same distance, which will have the greater kinetic energy? Which will have the greater momentum? (c) Calculate the energy and momentum of each spaceship in parts (a) and (b), ignoring any change in mass due to whatever is expelled by the engines. In part (a), assume that the engines are fired for 100 s. In part (b), assume that the engines are fired for 100 m. Two blocks of masses m1 m2 and m2, resting on fricm1 tionless inclined planes, are connected by a massf q less rope passing over an ideal pulley. Angle f = 45.0° and angle q = 36.9°; mass m1 is 6.00 kg and mass m2 is 4.00 kg. (a) Using energy conservation, find how fast the blocks are moving after they travel 2.00 m along the inclines. (b) Now solve the same problem using Newton’s second law. [Hint: First find the acceleration of each of the blocks. Then find how fast either block is moving after it travels 2.00 m along the incline with constant acceleration.] A particle, constrained to move along the x-axis, has a total mechanical energy of −100 J. The potential energy of the particle is shown in the graph. At time t = 0, the particle is located at x = 5.5 cm and is moving to the left. (a) What is the particle’s potential energy at t = 0? What is its kinetic energy at this time? (b) What are the particle’s total, potential, and kinetic energies when it is at x = 1 cm and moving to the right? (c) What is the particle’s kinetic energy when it is at x = 3 cm and moving to the left? (d) Describe the motion of this particle starting at t = 0. U (J) –100

1

3

5.5

11

sends it flying horizontally toward a window. The lawnmower blade can be modeled as a thin rod with a mass of 2.0 kg and a length of 50 cm rotating about its center. The stone impacts the blade near one end and is ejected with a velocity perpendicular to the rotation axis and the blade at the moment of collision. As a result of the impact, the blade slows from 60 rev/s to 55 rev/s. The window is 1.00 m in height, and its center is located 10.0 m away and at the same height as the lawnmower. (a) With what speed is the stone shot out by the mower? [Hint: The external force due to the lawnmower’s drive shaft on the system (blade + stone) cannot be ignored during the collision, but the external torque about the shaft can be ignored. The angular momentum of the stone just after impact can be calculated from its tangential velocity and its distance from the rotation axis.] (b) Ignoring air resistance, will the stone hit the window?

10.0 m

v

1.00 m

33. A person on a bicycle (combined total mass 80.0 kg) starts from rest and coasts down a hill to the bottom 20.0 m below. Each wheel can be treated as a hoop with mass 1.5 kg and radius 40 cm. Ignore friction and air resistance. (a) Find the speed of the bike at the bottom. (b) Would the speed at the bottom be the same for a less massive rider? Explain. 34. Tarzan wants to swing on a vine across a river. He is standing on a ledge 3.00 m above the water’s edge, and the river is 5.00 m wide. The vine is attached to a tree branch that is 8.00 m directly above the opposite edge of the river. Initially the vine makes a 60.0° angle with the vertical as he is holding it. He swings across starting from rest, but unfortunately the vine breaks when the vine is 20.0° from the vertical. (a) Assuming Tarzan weighs 900.0 N, what was the tension in the vine just before it broke? (b) Does he land safely on the other side of the river?

60.0⬚

13.5 x (cm)

–300

8.00 m

–550 3.00 m

✦32. You are mowing the lawn on a hill near your house when the lawnmower blade strikes a stone of mass 100 g and

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35. A boy of mass 60 kg is sledding down a 70-m slope starting from rest. The slope is angled at 15° below the horizontal. After going 20 m along the slope he passes his friend, who jumps on the sled. The friend has a mass of 50 kg and the coefficient of kinetic friction between the sled and the snow is 0.12. Ignoring the mass of the sled, find their speed at the bottom. 36. You want to throw a banana to a monkey hanging from a branch as shown in the figure. The banana has a mass of 200 g and the monkey has a mass of 3.00 kg. The monkey is startled and drops from the branch the moment you throw the banana. Ignore air resistance. (a) In what direction should you aim the banana so the monkey catches it in the air? (b) Explain why your answer to part (a) is the same for different values of the banana’s launch speed. (c) If the monkey catches the banana at the point indicated in the figure, what was the banana’s initial speed? (d) What is the horizontal distance d to the spot where the monkey lands?

1.67 m 3.33 m 3.00 m 2.00 m d

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REVIEW & SYNTHESIS: CHAPTERS 6–8

8. What is the magnitude of the force pushing each friction pad onto the wheel? A. 10 N B. 25 N C. 40 N D. 50 N 9. Which of the following is closest to the radial acceleration of the part of the wheel that passes between the friction pads? A. 10 m/s2 B. 20 m/s2 C. 40 m/s2 D. 50 m/s2 10. If the wheel has a kinetic energy of 30 J when the cyclist stops pedaling, how many rotations will it make before coming to rest? A. Less than 1 B. Between 1 and 2 C. Between 2 and 3 D. Between 3 and 4 11. What is the difference between the average mechanical power output of the cyclist in the passage and the power dissipated by the wheel at the friction pads? A. 5 W B. 10 W C. 20 W D. 27 W 12. Which of the following actions would most likely increase the fraction of the cyclist’s mechanical power output that is dissipated by the wheel at the friction pads? A. Reducing the force on the friction pads and pedaling at the same rate B. Maintaining the same force on the friction pads and pedaling at a slower rate C. Maintaining the same force on the friction pads and pedaling at a faster rate D. Increasing the force on the friction pads and pedaling at the same rate

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13. Which of the following is the best estimate of the number of liters of oxygen the cyclist in the passage would consume in the 20 min of activity? A. 25 L B. 30 L C. 45 L D. 50 L 14. During a second workout, the cyclist reduces the force on the friction pads by 50%, then pedals for two times the previous distance in _12 the previous time. How does the amount of energy dissipated by the pads in the second workout compare with energy dissipated in the first workout? A. One-eighth as much B. One-half as much C. Equal D. Two times as much 15. What is the ratio of the distance moved by a pedal to the distance moved by a point on the wheel located at a radius of 0.3 m in the same amount of time? A. 0.25 B. 0.5 C. 1 D. 2 16. A cyclist’s average metabolic rate during a workout is 500 W. If the cyclist wishes to expend at least 300 kcal (1 kcal = 4186 J) of energy, how long must the cyclist exercise at this rate? A. 0.6 min B. 3.6 min C. 36.0 min D. 41.9 min 17. If the friction pads are moved to a location 0.4 m from the center of the wheel, how does the amount of work done on the wheel, per revolution, change? A. It decreases by 25% B. It stays the same C. It increases by 33% D. It increases by 78%

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CHAPTER

9

Fluids

A hippopotamus in Kruger National Park, South Africa, wants to feed on the vegetation growing on the bottom of a pond. When the hippo wades into the pond, it floats. How does a hippopotamus get its floating body to sink to the bottom of a pond? (See p. 329 for the answer.)

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9.2

• • • •

conservation of energy (Chapter 6) force as rate of change of momentum (Section 7.3) conservation of momentum in collisions (Sections 7.7 and 7.8) equilibrium (Section 4.2)

9.1

Concepts & Skills to Review

STATES OF MATTER

Ordinary matter is usually classified into three familiar states or phases: solids, liquids, and gases. Solids tend to hold their shapes. Many solids are quite rigid; they are not easily deformed by external forces because forces due to neighboring atoms hold each atom in a particular position. Although the atoms vibrate around fixed equilibrium positions, they do not have enough energy to break the bonds with their neighbors. To bend an iron bar, for example, the arrangement of the atoms must be altered, which is not easy to do. A blacksmith heats iron in a forge to loosen the bonds between atoms so that he can bend the metal into the required shape. In contrast to solids, liquids and gases do not hold their shapes. A liquid flows and takes the shape of its container and a gas expands to fill its container. Fluids—both liquids and gases—are easily deformed by external forces. This chapter deals mainly with properties that are common to both liquids and gases. The atoms or molecules in a fluid do not have fixed positions, so a fluid does not have a definite shape. An applied force can easily make a fluid flow; for instance, the squeezing of the heart muscle exerts an applied force that pumps blood through the blood vessels of the body. However, this squeezing does not change the volume of the blood by much. In many situations we can think of liquids as incompressible—that is, as having a fixed volume that is impossible to change. The shape of the liquid can be changed by pouring it from a container of one shape into a container of a different shape, but the volume of the liquid remains the same. In most liquids, the atoms or molecules are almost as closely packed as those in the solid phase of the same material. The intermolecular forces in a liquid are almost as strong as those in solids, but the molecules are not locked in fixed positions as they are in solids. That is why the volume of the liquid can remain nearly constant while the shape is easily changed. Water is one of the exceptions: in cold water, the molecules in the liquid phase are actually more closely packed than those in the solid phase (ice). Gases, on the other hand, cannot be characterized by a definite volume nor by a definite shape. A gas expands to fill its container and can easily be compressed. The molecules in a gas are very far apart compared to the molecules in liquids and solids. The molecules are almost free of interactions with each other except when they collide.

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317

PRESSURE

Fluids (liquids and gases) are materials that flow.

Gases are much more compressible than liquids.

PRESSURE

Microscopic Origin of Pressure A static fluid does not flow; it is everywhere at rest. In the study of fluid statics (hydrostatics), we also assume that any solid object in contact with the fluid—whether a vessel containing the fluid or an object submerged in the fluid—is at rest. The atoms or molecules in a static fluid are not themselves static; they are continually moving. The motion of people bouncing up and down and bumping into each other in a mosh pit gives you a rough idea of the motion of the closely packed atoms or molecules in a liquid; in gases, the atoms or molecules are much farther apart than in liquids, so they travel greater distances between collisions.

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CHAPTER 9 Fluids

y

y pix pi pi

piy x pf

x pf

pfy = piy

pfx = –pix (a)

(b)

Figure 9.1 (a) A single fluid molecule bouncing off a container wall. (b) In this elastic collision, the y-component of the momentum is unchanged, while the x-component reverses direction.

The force due to a static fluid on a surface is always perpendicular to the surface.

Fluid pressure is caused by collisions of the fast-moving atoms or molecules of a fluid. When a single molecule hits a container wall and rebounds, its momentum changes due to the force exerted on it by the wall. Figure 9.1a shows a molecule of a fluid within a container making an elastic collision with one of the container walls. In this case, the y-component of momentum is unchanged, while the x-component reverses direction (Fig. 9.1b). The momentum change is in the +x-direction, which occurs because the wall exerts a force to the right on the molecule. By Newton’s third law, the molecule exerts a force to the left on the wall during the collision. If we consider all the molecules colliding with this wall, on average they exert no force on the wall in the ± y-direction, but all exert a force in the −x-direction. The frequent collisions of fluid molecules with the walls of the container cause a net force pushing outward on the walls.

PHYSICS AT HOME Drop a very tiny speck of dust or lint into a container of water and push the speck below the surface. The motion of the speck—called Brownian motion—is easily observed as it is pushed and bumped about randomly by collisions with water molecules. The water molecules themselves move about randomly, but at much higher speeds than the speck of dust due to their much smaller mass.

Definition of Pressure A static fluid exerts a force on any surface with which it comes in contact; the direction of the force is perpendicular to the surface (Fig. 9.2). A static fluid cannot exert a force parallel to the surface. If it did, the surface would exert a force on the fluid parallel to the surface, by Newton’s third law. This force would make the fluid flow along the surface, contradicting the premise that the fluid is static. The average pressure of a fluid at points on a planar surface is Definition of average pressure: F Pav = __ A

Figure 9.2 Forces due to a static fluid acting on the walls of the container and on a submerged object.

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(9-1)

where F is the magnitude of the force acting perpendicularly to the surface and A is the area of the surface. By imagining a tiny surface at various points within the fluid and measuring the force that acts on it, we can define the pressure at any point within the fluid. In the limit of a small area A, P = F/A is the pressure P of the fluid. Pressure is a scalar quantity; it does not have a direction. The force acting on an object submerged in a fluid—or on some portion of the fluid itself—is a vector quantity; its direction is perpendicular to the contact surface. Pressure is defined as a scalar because, at a given location in the fluid, the magnitude of the force per unit area is the same for any orientation of the surface. The molecules in a static fluid are moving in random directions; there can be no preferred direction since that would constitute fluid flow. There is no reason that a surface would have a greater number of collisions, or collisions with more energetic molecules, for one particular surface orientation compared with any other orientation. The SI unit for pressure is the newton per square meter (N/m2), which is named the pascal (symbol Pa) after the French scientist Blaise Pascal (1623–1662). Another commonly used unit of pressure is the atmosphere (atm). One atmosphere is the average air pressure at sea level. The conversion factor between atmospheres and pascals is 1 atm = 101.3 kPa Other units of pressure in common use are introduced in Section 9.5.

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9.2

319

PRESSURE

Example 9.1 Pressure due to Stiletto-Heeled Shoes A young woman weighing 534 N (120 lb) walks to her bedroom while wearing tennis shoes. She then gets dressed for her evening date, putting on her new stiletto-heeled dress shoes. The area of the heel section of her tennis shoe is 60.0 cm2 and the area of the heel of her dress shoe is 1.00 cm2. For each pair of shoes, find the average pressure caused by the heel making contact with the floor when her entire weight is supported by one heel.

The average pressure is the woman’s weight divided by the area of the heel. For the tennis shoe: 534 N F = ____________ = 8.90 × 104 N/m2 = 89.0 kPa P = __ A 6.00 × 10−3 m2

Strategy The average pressure is the force applied to the floor divided by the contact area. The force that the heel exerts on the floor is 534 N. To keep the units straight, we convert the areas from square centimeters to square meters.

Discussion In atmospheres, these pressures are 0.879 atm and 52.7 atm, respectively. The pressure due to the dress shoe is 60 times the pressure due to the tennis shoe since the 1 same force is spread over __ the area. 60

Solution To convert the area of the tennis shoe heel and the dress shoe heel from cm2 to m2, we multiply by the 1 m 2. For the tennis shoe heel: conversion factor ______ 102 cm

(

)

(

)

(

)

1 m 2 = 6.00 × 10−3 m2 A = 60.0 cm2 × ______ 102 cm For the dress shoe heel: 1 m 2 = 1.00 × 10−4 m2 A = 1.00 cm2 × ______ 102 cm

For the stilettos: 534 N = 5.34 × 106 N/m2 = 5.34 MPa P = ____________ −4 2 1.00 × 10 m

Practice Problem 9.1 Dress Shoe Heel

Pressure from an Ordinary

Fortunately for floor manufacturers, and for women’s feet, stiletto heels are out of fashion more often than they are in fashion. Suppose that a woman’s dress shoes have heels that are each 4.0 cm2 in area. Find the pressure on the floor, when the entire weight is on a single heel, for such a shoe worn by the same woman as in Example 9.1. Find the factor by which this pressure exceeds the pressure from the tennis shoe heel.

Atmospheric Pressure On the surface of the Earth, we live at the bottom of an ocean of fluid called air. The forces exerted by air on our bodies and on surfaces of other objects may be surprisingly large: 1 atm is approximately 10 N/cm2 of surface area, or nearly 15 lb/in2. We are not crushed by this pressure because most of the fluids in our bodies are at approximately the same pressure as the air around us. As an analogy, consider a sealed bag of potato chips. Why is the bag not crushed by the air pushing in on all sides? Because the air inside the bag is at the same pressure and pushes out on the sides of the bag. The pressure of the fluids inside our cells matches the pressure of the surrounding fluids pushing in on the cell membranes, so the cells do not rupture. By contrast, the blood pressure in the arteries is as much as 20 kPa higher than atmospheric pressure. The strong, elastic arterial walls are stretched by the pressure of the blood inside; the walls squeeze the arterial blood to keep its higher pressure from being transmitted to other fluids in the body. Changing weather conditions cause variations of approximately 5% in the actual value of air pressure at sea level; 101.3 kPa (1 atm) is only the average value. Air pressure also decreases with increasing elevation. (In Section 9.4, we study the effect of gravity on fluid pressure in detail.) The average air pressure in Leadville, Colorado, the highest incorporated city in the United States (elevation 3100 m), is 70 kPa. Some Tibetans live at altitudes of over 5000 m, where the average air pressure is only half its value at sea level. In problems, please assume that the atmospheric pressure is 1 atm unless the problem states otherwise.

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9.3

Figure 9.3 Forces acting on a cube of fluid.

PASCAL’S PRINCIPLE

If the weight of a static fluid is negligible (as, for example, in a hydraulic system under high pressure), then the pressure must be the same everywhere in the fluid. Why? In Fig. 9.3, imagine the submerged cube to be composed of the same fluid as its surroundings. Ignoring the fluid’s weight, the only forces acting on the cubical piece of fluid are those due to the surrounding fluid pushing inward. The forces pushing on each pair of opposite sides of the cube must be equal in magnitude, since the fluid inside the cube is in equilibrium. Therefore, the pressure must be the same on both sides. Since we can extend this argument to any size and shape piece of fluid, the fluid pressure must be the same everywhere in a weightless, static fluid. More generally, when the weight of the fluid is not negligible, the pressure is not the same everywhere. In this case, analysis of the forces acting on a piece of fluid (see Conceptual Question 15) leads to a more general result called Pascal’s principle.

Pascal’s Principle d2

A2

F1 d1

A change in pressure at any point in a confined fluid is transmitted everywhere throughout the fluid.

F2 A1

Hydraulic fluid

Figure 9.4 Simplified diagram of a hydraulic lift. Notice that piston 1 has to move a great distance (d1) to lift the truck a much smaller distance (d2). In a real hydraulic lift, piston 1 is usually replaced by a pump that draws fluid from a reservoir and pushes it into the hydraulic system.

Applications of Pascal’s Principle: Hydraulic Lifts, Brakes, and Controls When a truck needs to have its muffler replaced, it is lifted into the air by a mechanism called a hydraulic lift (Fig. 9.4). A force is exerted on a liquid by a piston with a relatively small area; the resulting increase in pressure is transmitted everywhere throughout the liquid. Then the truck is lifted by the fluid pressure on a piston of much larger area. The upward force on the truck is much larger than the force applied to the small piston. Pascal’s principle has many other applications, such as the hydraulic brakes in cars and trucks and the hydraulic controls in airplanes. To analyze the forces in the hydraulic lift, let force F1 be applied to the small piston of area A1, causing a pressure increase: F ΔP = ___1 A1 A truck is supported by a piston of much larger area A2 on the other side of the lift. The increase in pressure due to the small piston is transmitted everywhere in the liquid. Ignoring the weight of the fluid (or assuming the two pistons to be at the same height), the force F2 exerted by the fluid on the large piston is related to F1 by F1 ___ F ___ = 2 A1 A2

CONNECTION: As for levers, systems of pulleys, and other simple machines, the hydraulic lift reduces the applied force needed to perform a task, but the work done is the same.

Since A2 is larger than A1, the force exerted on the large piston (F2) is larger than the force applied to the small piston (F1). We are not getting something for nothing; just as for the two-pulley systems discussed in Section 6.2, the advantage of the smaller force applied to the small piston is balanced by a greater distance it must be moved. The small piston has to move a long distance d1 while the large piston moves a short distance d2. Assuming the liquid to be incompressible, the volume of fluid displaced by each piston is the same, so A1d1 = A2d2 The displacements of the pistons are inversely proportional to their areas, while the forces are directly proportional to the areas; then the product of force and displacement is the same: F F1 ___ × A1d1 = ___2 × A2d2 A1 A2

⇒

F1d1 = F2d2

The work (force times displacement) done by moving the small piston equals the work done by the large piston in raising the truck.

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9.4 THE EFFECT OF GRAVITY ON FLUID PRESSURE

Example 9.2 The Hydraulic Lift In a hydraulic lift, if the radius of the smaller piston is 2.0 cm and the radius of the larger piston is 20.0 cm, what weight can the larger piston support when a force of 250 N is applied to the smaller piston? Strategy According to Pascal’s principle, the pressure increases the same amount at every point in the fluid. A natural way to work is in terms of proportions since the forces are proportional to the areas of the pistons. Solution Since the pressure on the two pistons increases by the same amount, F1 ___ F ___ = 2 A1 A2 Equivalently, the forces are proportional to the areas: F2 ___ A ___ = 2 F1 A1

F2 = 100F1 = 25 000 N = 25 kN Discussion One common error in this sort of problem is to think of the area and the force as a tradeoff—in other words, that the piston with the large area has the small force and vice versa. Since the pressures are the same, the force exerted by the fluid on either piston is proportional to the piston’s area. We make the piston that lifts the truck large because we know the force on it will be large, in direct proportion to its area.

Practice Problem 9.2 Principle

Application of Pascal’s

Consider the hydraulic lift of Example 9.2. (a) What is the increase in pressure caused by the 250-N force on the small piston? (b) If the larger piston moves 5.0 cm, how far does the smaller piston move?

The ratio of the radii is r2/r1 = 10, so the ratio of the areas is A2/A1 = (r2/r1)2 = 100. Then the weight that can be supported is

9.4

THE EFFECT OF GRAVITY ON FLUID PRESSURE

On a drive through the mountains or on a trip in a small plane, the feeling of our ears popping is evidence that pressure is not the same everywhere in a static fluid. Gravity makes fluid pressure increase as you move down and decrease as you move up. To understand more about this variation, we must first define the density of a fluid. Density The density of a substance is its mass per unit volume. The Greek letter r (rho) is used to represent density. The density of a uniform substance of mass m and volume V is m r = __ V

Density of a uniform substance: its mass divided by its volume.

(9-2)

The SI units of density are kilograms per cubic meter: kg/m3. For a nonuniform substance, Eq. (9-2) defines the average density. Table 9.1 lists the densities of some common substances. Note that temperatures and pressures are specified in the table. For solids and liquids, density is only weakly dependent on temperature and pressure. On the other hand, gases are highly compressible, so even a relatively small change in temperature or pressure can change the density of a gas significantly. Pressure Variation with Depth due to Gravity Now, using the concept of density, we can find how pressure increases with depth due to gravity. Suppose we have a glass beaker containing a static liquid of uniform density r. Within this liquid, imagine a cylinder of liquid with cross-sectional area A and height d (Fig. 9.5a). The mass of the liquid in this cylinder is m = rV

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Table 9.1

Gases Hydrogen Helium Steam (100°C) Nitrogen Air (20°C) Air (0°C) Oxygen Carbon dioxide A

Densities of Common Substances (at 0°C and 1 atm unless otherwise indicated) Density (kg/m3) 0.090 0.18 0.60 1.25 1.20 1.29 1.43 1.98

Liquids

Density (kg/m3)

Gasoline Ethanol Oil Water (0°C) Water (3.98°C) Water (20°C) Seawater Blood (37°C) Mercury

680 790 800–900 999.87 1000.00 1001.80 1025 1060 13 600

d

(a)

Solids Polystyrene Cork Wood (pine) Wood (oak) Ice Wood (ebony) Bone Concrete Quartz, granite Aluminum Iron, steel Copper Lead Gold Platinum

Density (kg/m3) 100 240 350–550 600–900 917 1000–1300 1500–2000 2000 2700 2702 7860 8920 11 300 19 300 21 500

P1A

where the volume of the cylinder is y mg

V = Ad The weight of the cylinder of liquid is therefore

P2A

(b)

Figure 9.5 Applying Newton’s second law to a cylinder of liquid tells us how pressure increases with increasing depth. (a) A cylinder of liquid of height d and area A. (b) Vertical forces on the cylinder of liquid.

mg = ( rAd )g The vertical forces acting on this column of liquid are shown in Fig. 9.5b. The pressure at the top of the cylinder is P1 and the pressure at the bottom is P2. Since the liquid in the column is in equilibrium, the net vertical force acting on it must be zero by Newton’s second law:

∑Fy = P2A − P1A − rAdg = 0 Dividing by the common factor A and rearranging yields: Pressure variation with depth in a static fluid with uniform density: P2 = P1 + rgd

(9-3)

where point 2 is a depth d below point 1 Since we can imagine a cylinder anywhere we choose within the liquid, Eq. (9-3) relates the pressure at any two points in a static liquid where point 2 is a depth d below point 1. For gases, Eq. (9-3) can be applied as long as the depth d is small enough that changes in the density due to gravity are negligible. Since liquids are nearly incompressible, Eq. (9-3) holds to great depths in liquids. For a liquid that is open to the atmosphere, suppose we take point 1 at the surface and point 2 a depth d below. Then P1 = Patm, so the pressure at a depth d below the surface is Pressure at a depth d below the surface of a liquid open to the atmosphere: P = Patm + rgd

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(9-4)

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9.4 THE EFFECT OF GRAVITY ON FLUID PRESSURE

CHECKPOINT 9.4 Pressure in a static fluid depends on vertical position. Can it also depend on horizontal position? Explain.

Example 9.3 A Diver A diver swims to a depth of 3.2 m in a freshwater lake. What is the increase in the force pushing in on her eardrum, compared to what it was at the lake surface? The area of the eardrum is 0.60 cm2.

where A = 0.60 cm2 = 6.0 × 10−5 m2. Then

Strategy We can find the increase in pressure at a depth of 3.2 m and then find the corresponding increase in force on the eardrum. If the force on the eardrum at the surface is P1A and the force at a depth of 3.2 m is P2A, then the increase in the force is (P2 − P1)A.

Discussion A force also pushes outward on the eardrum due to the pressure inside the ear canal. If the diver descends rapidly so that the pressure inside the ear canal does not change, then a 1.9-N net force due to fluid pressure pushes inward on the eardrum. When the diver’s ear “pops,” the pressure inside the ear canal increases to equal the fluid pressure outside the eardrum, so that the net force due to fluid pressure on the eardrum is zero.

Solution The increase in pressure depends on the depth d and the density of water. From Table 9.1, the density of water is 1000 kg/m3 to two significant figures for any reasonable temperature. P2 − P1 = rgd ΔP = 1000 kg/m3 × 9.8 m/s2 × 3.2 m = 31.4 kPa The increase in force on the eardrum is

ΔF = (3.14 × 104 Pa) × (6.0 × 10−5 m2) = 1.9 N

Practice Problem 9.3 Depth

Limits on Submarine

A submarine is constructed so that it can safely withstand a pressure of 1.6 × 107 Pa. How deep may this submarine descend in the ocean if the average density of seawater is 1025 kg/m3?

ΔF = ΔP × A

Conceptual Example 9.4 The Hydrostatic Paradox Three vessels have different shapes, but the same base area and the same weight when empty (Fig. 9.6). The vessels are filled with water to the same level and then the air is pumped out. The top surface of the water is then at a low pressure that, for simplicity, we assume to be zero. (a) Are the water pressures at the bottom of each vessel the same? If not, which is largest and which is smallest? (b) If the three vessels containing water are weighed on a scale, do they give the same reading? If not, which weighs the most and which weighs the least? (c) If the water exerts the same downward force on the bottom of each vessel, is that force equal to the weight of water in the vessel? Is there a paradox here?

A

B

C d

Figure 9.6 Three differently shaped vessels filled with water to same level.

[Hint: Think about the forces due to fluid pressure on the sides of the containers; do they have vertical components?] continued on next page

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Conceptual Example 9.4 continued

The volume of water in the cylinder is V = p r 2d, so Solution and Discussion (a) The water at the bottom of each vessel is the same depth d below the surface. Water at the surface of each vessel is at a pressure Psurface = 0. Therefore, the pressures at the bottom must be equal: P = Psurface + rgd = rgd (b) The weight of each filled vessel is equal to the weight of the vessel itself plus the weight of the water inside. The vessels themselves have equal weights, but vessel A holds more water than C, whereas vessel B holds less water than C. Vessel A weighs the most and vessel B weighs the least. (c) Each container supports the water inside by exerting an upward force equal in magnitude to the weight of the water. By Newton’s third law, the water exerts a downward force on the container of the same magnitude. Figure 9.7 shows the forces acting on each container due to the water. In vessel C, the horizontal forces on any two diametrically opposite points on the walls of the container are equal and opposite; thus, the net force on the container walls is zero. The force on the bottom is F = PA = ( rgd )(p r2)

A

B

C d

Figure 9.7

F = rgV = ( rV)g = mg The force on the bottom of vessel C is equal to the weight of the water, as expected. However, the force on the bottom of vessel A is less than the weight of the water in the container, while the force on the bottom of vessel B is greater than the weight of the water. Then how can the water be in equilibrium? In vessel A, the forces on the container walls have downward components as well as horizontal components. The horizontal components of the forces on any two diametrically opposite points are equal and opposite, so the horizontal components add to zero. The sum of the downward components of the forces on the walls and the downward force on the bottom of the container is equal to the weight of the water. Similarly, the forces on the walls of vessel B have upward components. In each case, the total force on the bottom and sides of the container due to the water is equal to the weight of the water.

Conceptual Practice Problem 9.4 Is Pressure Determined by Column Height? Figure 9.8 shows a vessel with two points marked at the bottom of the water in the vessel. A narrow column of water is drawn above each point. (a) Is the pressure at point 2, P2, the same as 1 2 the pressure at point 1, P1, even though the column of water above point 2 is not Figure 9.8 as tall? (b) Does P = Patm + rgd imply Two different points that P2 < P1? Explain. on the bottom of an open vessel.

Forces exerted on the containers by the water.

9.5 Units of pressure: 1 atm = 101.3 kPa = 1.013 bar = 14.7 lb/in2 = 760.0 mm Hg = 760.0 torr = 29.9 in Hg

MEASURING PRESSURE

Many other units are used for pressure besides atmospheres and pascals. In the United States, the pressure in an automobile tire is measured in pounds per square inch (lb/in2). Weather bureaus record atmospheric pressure in bars or millibars. In the United States, television weather reports and home barometers measure pressure in inches of mercury. One atmosphere is equal to approximately 1 bar (1000 millibars), 76 cm of mercury, or 29.9 in. of mercury. Blood pressure, the difference between the pressure in the blood and atmospheric pressure, is measured in millimeters of mercury (mm Hg), also called the torr. Inches or millimeters of mercury may seem like strange units for pressure: how can a force per unit area be equal to a distance (so many mm Hg)? There is an assumption inherent in using these pressure units that we can understand by studying the mercury manometer.

Manometer A mercury manometer consists of a vertical U-shaped tube, containing some mercury, with one side typically open to the atmosphere and the other connected to a vessel

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containing a gas whose pressure we want to measure. Figure 9.9 shows the manometer before it is connected to such a vessel. When both sides of the manometer are open to the atmosphere, the mercury levels are the same. Now we connect an inflated balloon to the left side of the U-tube (Fig. 9.10). If the gas in the balloon is at a higher pressure than the atmosphere, the gas pushes the mercury down on the left side and forces it up on the right side. The density of a gas is small compared to the density of mercury, so every point within the gas is assumed to be at the same pressure no matter what the depth. At point B, the mercury pushes on the gas with the same magnitude force with which the gas pushes on the mercury, so point B is at the same pressure as the gas. Since point B′ is at the same height within the mercury as point B, the pressure at B′ is the same as at B. Point C is at atmospheric pressure. The pressure at B is PB = PB = PC + rgd

Open to the atmosphere

A′

A Starting level

B′

B Hg

Figure 9.9 A mercury

where r is the density of mercury. The difference in the pressures on the two sides of the manometer is ΔP = PB − PC = rgd

(9-5)

Thus, the difference in mercury levels d is a measure of the pressure difference— commonly reported in millimeters of mercury (mm Hg). The pressure measured when one side of the manometer is open is the difference between atmospheric pressure and the gas pressure rather than the absolute pressure of the gas. This difference is called the gauge pressure, since it is what most gauges (not just manometers) measure:

manometer open on both sides. Points A and A′ are both at atmospheric pressure. Any two points (such as B and B′) at the same height within the fluid are at the same pressure: PB = PB′. Open to the atmosphere

Gas

Gauge pressure: Pgauge = Pabs − Patm

C

(9-6) d

3

Since the density of mercury is 13 600 kg/m , 1.00 mm Hg can be converted to pascals by substituting d = 1.00 mm in Eq. (9-5):

B

1.00 mm Hg = rgd = (13 600 kg/m3)(9.80 m/s2)(0.00100 m) = 133 Pa The liquid in a manometer may be something other than mercury, such as water or oil. Equation (9-5) still applies, as long as we use the correct density r of the liquid in the manometer.

B′ Hg

Figure 9.10 The manometer connected on one side to a container of gas at a pressure greater than atmospheric pressure.

Example 9.5 The Mercury Manometer A manometer is attached to a container of gas to determine its pressure. Before the container is attached, both sides of the manometer are open to the atmosphere. After the container is attached, the mercury on the side attached to the gas container rises 12 cm above its previous level. (a) What is the gauge pressure of the gas in Pa? (b) What is the absolute pressure of the gas in Pa? Strategy The mercury column is higher on the side connected to the container of gas, so we know that the pressure of the enclosed gas is lower than atmospheric pressure. We need to

find the difference in levels of the mercury columns on the two sides. Careful: It is not 12 cm! If one side went up by 12 cm, then the other side has gone down by 12 cm, since the same volume of mercury is contained in the manometer. Solution (a) The difference in the mercury levels is 24 cm (Fig. 9.11). Since the mercury on the gas side went up, the absolute pressure of the gas is lower than atmospheric pressure. Therefore, the gauge pressure of the gas is less than zero. The gauge pressure in Pa is Pgauge = rgd continued on next page

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Example 9.5 continued

(b) The absolute pressure of the gas is where the “depth” is d = −24 cm (the mercury is 24 cm higher on the gas side). Then

P = Pgauge + Patm = −32 kPa + 101 kPa = 69 kPa

Pgauge = 13 600 kg/m3 × 9.8 m/s2 × (−0.24 m) = −32 kPa

Discussion As a check, the manometer tells us directly that the gauge pressure of the gas is −240 mm Hg. Converting to pascals gives Open to the atmosphere

−240 mm Hg × 133 Pa/mm Hg = −32 kPa

Practice Problem 9.5 Manometer

Gas 12 cm

Figure 9.11 24 cm

12 cm

Hg

When a container of gas is attached to one side of the manometer, one side goes down 12 cm and the other side goes up 12 cm.

Column Heights in

A mercury manometer is connected to a container of gas. (a) The height of the mercury column on the side connected to the gas is 22.0 cm (measured from the bottom of the manometer). What is the height of the mercury column on the open side if the gauge pressure is measured to be 13.3 kPa? (b) If the gauge pressure of the gas doubles, what are the new heights of the two columns?

Barometer A manometer can act as a barometer—a device to measure atmospheric pressure. Instead of attaching a container with a gas to one end of the manometer, attach a container and a vacuum pump. Pump the air out of the container to get as close to a vacuum—zero pressure—as possible. Then the atmosphere pushes down on one side and pushes the fluid up on the other side toward the evacuated container. Figure 9.12 shows a barometer in which the vacuum is not created by a vacuum pump. The barometer was invented by Evangelista Torricelli (1608–1647), an assistant to Galileo, in the 1600s; in his honor, one millimeter of mercury is called one torr.

Vacuum (P = 0)

d Atmospheric pressure A B

Hg

Figure 9.12 A simple barometer. A tube, of length greater than 76 cm and closed at one end, is filled with mercury. The tube is then inverted into an open container of mercury. Some mercury flows down from the tube into the bowl. The space left at the top of the tube is nearly a vacuum because nothing is left but a negligible amount of mercury vapor. Points A and B are at the same level in the mercury and, therefore, are both at atmospheric pressure since the bowl is open to the air. The distance d from A to the top of the mercury column in the closed tube is a measure of the atmospheric pressure (often called barometric pressure because it is measured with a barometer).

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PHYSICS AT HOME When you next have a drink with a straw, insert the straw into the drink and place your finger over the upper opening of the straw so that no more air can enter the straw. Raise the lower end of the straw up out of your drink. Does the liquid in the straw flow back down into your glass? What do you suppose is holding the liquid in place? Make an FBD on your paper napkin. Some air is trapped between your finger and the top of the liquid in the straw; that air exerts a downward force on the liquid of magnitude P1A (Fig. 9.13). A downward gravitational force mg also acts on the liquid. The air at the bottom of the straw exerts an upward force on the liquid of magnitude PatmA; this upward force is what holds the liquid in place. Because the liquid does not pour out of the straw, but instead is in equilibrium,

P1A

mg

PatmA = P1A + mg Thus, the pressure P1 of the air trapped above the liquid must be less than atmospheric pressure. How did P1 become less than atmospheric pressure? As you pulled the straw up and out, the liquid in the straw falls a bit, expanding the volume available to the air trapped above the liquid. When a gas expands under conditions like this, its pressure decreases. When you remove your finger from the top of the straw, air can get in at the top of the straw. Then the pressures above and below the liquid are equal, so the gravitational force pulls the liquid down and out of the straw.

Patm A

Figure 9.13 Force acting on the liquid inside a straw.

Sphygmomanometer Blood pressure is measured with a sphygmomanometer (Fig. 9.14). The oldest kind of sphygmomanometer consists of a mercury manometer on one side attached to a closed bag—the cuff. The cuff is wrapped around the upper arm at the level of the heart and is then pumped up with air. The manometer measures the gauge pressure of the air in the cuff. At first, the pressure in the cuff is higher than the systolic pressure—the maximum pressure in the brachial artery that occurs when the heart contracts. The cuff pressure squeezes the artery closed and no blood flows into the forearm. A valve on the cuff is then opened to allow air to escape slowly. When the cuff pressure decreases to just below the systolic pressure, a little squirt of blood flows past the constriction in the artery with each heartbeat. The sound of turbulent blood flow past the constriction can be heard through the stethoscope. As air continues to escape from the cuff, the sound of blood flowing through the constriction in the artery continues to be heard. When the pressure in the cuff reaches the diastolic pressure in the artery—the minimum pressure that occurs when the heart muscle is relaxed—there is no longer a constriction in the artery, so the pulsing sounds cease. The gauge pressures for a healthy heart are nominally around 120 mm Hg (systolic) and 80 mm Hg (diastolic).

9.6

Figure 9.14 A sphygmomanometer being used to measure blood pressure.

THE BUOYANT FORCE

When an object is immersed in a fluid, the pressure on the lower surface of the object is higher than the pressure on the upper surface. The difference in pressures leads to an upward net force acting on the object due to the fluid pressure. If you try to push a beach ball underwater, you feel the effects of the buoyant force pushing the ball back up. It takes a rather large force to hold such an object completely underwater; the instant you let go, the object pops back up to the surface.

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Application of the manometer: measuring blood pressure

CONNECTION: The buoyant force is not a new kind of force exerted by a fluid; it is the sum of forces due to fluid pressure.

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F1

d

Consider a rectangular solid immersed in a fluid of uniform density r (Fig. 9.15a). For each vertical face (left, right, front, and back), there is a face of equal area opposite it. The forces on these two faces due to the fluid are equal in magnitude since the areas and the average pressures are the same. The directions are opposite, so the forces acting on the vertical faces cancel in pairs. Let the top and bottom surfaces each have area A. The force on the lower face of the block is F2 = P2A; the force on the upper face is F1 = P1A. The total force on the block due to the fluid, called the buoyant force FB, is upward since F2 > F1 (Fig. 9.15b). ⃗ B = F ⃗ 1 + F ⃗ 2 F FB = (P2 − P1)A

F2

Since P2 − P1 = rgd, the magnitude of the buoyant force can be written (a)

Buoyant force: FB F2

FB = r gdA = r gV

(9-7)

F1 (b)

Figure 9.15 (a) Forces due to fluid pressure on the top and bottom of an immersed rectangular solid. (b) The buoyant force is ⃗ and F ⃗ . Since the sum of F 2 1 ⃗ ⃗ F2 > F1 , the net force due to fluid pressure is upward.

| | | |

mg FB

Figure 9.16 Forces acting on a floating ice cube. The ice cube ⃗ + mg⃗ = 0. is in equilibrium, so F B

where V = Ad is the volume of the block. Note that r V is the mass of the volume V of the fluid that the block displaces. Thus, the buoyant force on the submerged block is equal to the weight of an equal volume of fluid, a result called Archimedes’ principle.

Archimedes’ Principle A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object.

Archimedes’ principle applies to a submerged object of any shape even though we derived it for a rectangular block. Why? Imagine replacing an irregular submerged object with enough fluid to fill the object’s place. This “piece” of fluid is in equilibrium, so the buoyant force must be equal to its weight. The buoyant force is the net force exerted on the “piece” of fluid by the surrounding fluid, which is identical to the buoyant force on the irregular object since the two have the same shape and surface area. The same argument can be used to show that if an object is only partly submerged, the buoyant force is still equal to the weight of fluid displaced. Equation (9-7) applies as long as V is the part of the object’s volume below the fluid surface rather than the entire volume of the object. Net Force due to Gravity and Buoyancy The net force due to gravity and buoyancy acting on an object totally or partially immersed in a fluid (Fig. 9.16) is ⃗ = mg⃗ + F ⃗ F B The force of gravity on an object of volume Vo and average density ro is W = mg = ro gVo and the buoyant force is FB = rf gVf where Vf and rf are the volume of fluid displaced and the fluid density, respectively. Choosing up to be the +y-direction, the net force due to gravity and buoyancy is Fy = rf gVf − ro gVo

(9-8)

Here Fy can be positive or negative, depending on which density is larger. Imagine releasing a pebble and an air bubble underwater. The pebble’s average density is greater

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9.6 THE BUOYANT FORCE

than the density of water, so the net force on it is downward; the pebble sinks. An air bubble’s average density is less than the density of water, so the net force is upward, causing the bubble to rise toward the surface of the water. If the object is completely submerged, the volumes of the object and the displaced fluid are the same and Fy = ( rf − ro )gV If r o < r f, the object floats with only part of its volume submerged. In equilibrium, the object displaces a volume of fluid whose weight is equal to the object’s weight. At that point the gravitational force equals the buoyant force and the object floats. Setting Fy = 0 in Eq. (9-8) yields rf gVf = ro gVo

which can be rearranged as: V ro ___f = __ Vo rf On the left side of this equation is the fraction of the object’s volume that is submerged; it is equal to the ratio of the density of the object to the density of the fluid. Specific Gravity This ratio of densities is called the specific gravity of the material when rf is the density of water at 4°C. Specific gravity is without units because it is a ratio of two densities. Water at 4°C is chosen as the reference material because at that temperature, the density of water is a maximum (at atmospheric pressure). The gram was originally defined as the mass of one cubic centimeter of water at 4°C. Thus, water at 4°C has a density of 1 g/cm3 (1000 kg/m3). The specific gravity of seawater is 1.025, which means that seawater has a density of 1.025 g/cm3 (1025 kg/m3).

Specific gravity: r r = __________ S.G. = _____ rwater 1000 kg/m3

(9-9)

Blood tests often include determination of the specific gravity of the blood— normally around 1.040 to 1.065. A reading that is too low may indicate anemia, since the presence of red blood cells increases the average density of the blood. Before taking blood from a donor, a drop of the blood is placed in a solution of known density. If the drop does not sink, it is not safe for the donor to give blood because the concentration of red blood cells is too low. Urinalysis also includes a specific gravity measurement (normally 1.015 to 1.030); too high a value indicates an abnormally high concentration of dissolved salts, which can signal a medical problem. Freighters, aircraft carriers, and cruise ships float, although they are made from steel and other materials that are more dense than seawater. When a ship floats, the buoyant force acting on the ship is equal to the ship’s weight. A ship is constructed so that it displaces a volume of seawater larger than the volume of the steel and other construction materials. The average density of the ship is its weight divided by its total volume. A large part of a ship’s interior is filled with air. All of the “empty” space contributes to the total volume; the resulting average density is less than that of seawater, allowing the ship to float. Now we can understand how a hippopotamus can sink to the bottom of a pond: it can expel some of the air in its body by exhaling. Exhalation increases the average density of the hippopotamus so that it is just slightly above the density of the water; thus, it sinks. (An armadillo does just the opposite: it swallows air, inflating its stomach and intestines, to increase the buoyant force for a swim across a large lake.) When the hippo needs to breathe, it swims back up to the surface.

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Applications of specific gravity measurements in medicine

Application of Archimedes’ principle: how a ship can float How can the hippo sink?

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Example 9.6 The Golden (?) Falcon A small statue in the shape of a falcon has a weight of 24.1 N. The owner of the statue claims it is made of solid gold. When the statue is completely submerged in a container brimful of water, the weight of the water that spills over the top and into a bucket is 1.25 N. Find the density and specific gravity of the metal. Is the density consistent with the claim that the falcon is solid gold? Strategy When the statue is completely submerged, it displaces a volume V of water equal to its own volume. The weight of the displaced water is equal to the buoyant force. Let msg = 24.1 N represent the weight of the statue (in terms of its mass ms) and let mwg = 1.25 N represent the weight of the water. Solution The specific gravity of the statue is ms /V ___ m r S.G. = ___s = _____ = s rw mw /V mw Rather than calculate the masses in kilograms, we recognize that a ratio of masses is equal to the ratio of the weights: ms g ______ 24.1 N S.G. = ____ mw g = 1.25 N = 19.3

From Table 9.1, the statue has the correct density; it may possibly be gold. Discussion According to legend, this method to determine the specific gravity of a solid was discovered by Archimedes in the third century b.c.e. King Hieron II asked Archimedes to find a way to check whether his crown was made of pure gold—without melting down the crown, of course! Archimedes came up with his method while he was taking a bath; he noticed the water level rising as he got in and connected the rising water level with the volume of water displaced by his body. In his excitement, he jumped out of the bath and ran naked through the streets of Siracusa (a city in Sicily) shouting “Eureka!”

Practice Problem 9.6 Substance

Identifying an Unknown

An unknown solid substance has a weight of 142.0 N. The object is suspended from a scale and hung so that it is completely submerged in water (but not touching bottom). The scale reads 129.4 N. Find the specific gravity of the object and determine whether the substance could be anything listed in Table 9.1.

The density of the statue is rs = S.G. × rw = 19.3 × 1000 kg/m3 = 1.93 × 104 kg/m3

Example 9.7 Hidden Depths of an Iceberg What percentage of a floating iceberg’s volume is above water? The specific gravity of ice is 0.917 and the specific gravity of the surrounding seawater is 1.025. Strategy The ratio of the density of ice to the density of seawater tells us the ratio of the volume of ice that is submerged in the seawater to the total volume of the iceberg. The rest of the ice is above the water. Solution We could calculate the densities of seawater and of ice in SI units from their specific gravities, but that is unnecessary; the ratio of the specific gravities is equal to the ratio of the densities: S.G.ice rice /rwater rice _________ = ___________ = ______ r r r S.G.seawater seawater / water seawater

The fraction of the iceberg’s volume that is submerged is equal to the ratio of the densities of ice and seawater. Thus, the ratio of the volume submerged to the total volume of ice is Vsubmerged S.G.ice rice ________ = _________ = ______ r Vice S.G. seawater seawater 0.917 = 0.895 = _____ 1.025 89.5% of the ice is below the surface of the water, leaving only 10.5% above the surface. Discussion An alternative solution does not depend on remembering that the ratio of the volumes is equal to the continued on next page

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Example 9.7 continued

ratio of the densities. The buoyant force is equal to the weight of a volume V submerged of water: buoyant force = rseawater Vsubmerged g The weight of the iceberg is mg = r iceV ice g. From Newton’s second law, the buoyant force must be equal in magnitude to the weight of the iceberg when it is floating in equilibrium: rseawater Vsubmerged g = r Vice g

gradually fill up the ponds and lakes from the bottom. It would not form on top of lakes and remain there. The consequences for fish and other bottom dwellers of solidly frozen lakes would be catastrophic. The water below the surface layer of ice formed in winter remains just above freezing so that the fish are able to survive.

Practice Problem 9.7 Versus Seawater

Floating in Freshwater

ice

or

Vsubmerged rice ________ = ______ Vice rseawater

The fact that ice floats is of great importance for the balance of nature. If ice were more dense than water, it would

If the average density of a human being is 980 kg/m3, what fraction of a human body floats above water in a freshwater pond and what fraction floats above seawater in the ocean? The specific gravity of seawater is 1.025.

Conceptual Example 9.8 A Hovering Fish How is it that a fish is able to hover almost motionless in one spot—until some attractive food is spotted and, with a flip of the tail, off it swims after the food? Fish have a thin-walled bladder, called a swim bladder, located under the spinal column. The swim bladder contains a mixture of oxygen and nitrogen obtained from the blood of the fish. How does the swim bladder help the fish keep the buoyant and gravitational forces balanced so that it can hover? Solution and Discussion If the fish’s average density is greater than that of the surrounding water, it will sink; if its average density is smaller than that of the water, it will rise. By varying the volume of the swim bladder, the fish is able to vary its overall volume and, thus, its average density. By adjusting its average density to match the density

of the surrounding water, the fish can remain suspended in position. The fish can also adjust the volume of the bladder when it wants to rise or sink.

Conceptual Practice Problem 9.8 The Diving Beetle A diving beetle traps a bubble of air under its wings. While under the water, the beetle uses the air in the bubble to breathe, gradually exchanging the oxygen for carbon dioxide. (a) What does the beetle do to the air bubble so that it can dive under the water? (b) Once under water, what does the beetle do so that it can rise to the surface? [Hint: Treat the beetle and the air bubble as a single system. How can the beetle change the buoyant force acting on the system?]

Buoyant Forces on Objects Immersed in a Gas Gases such as air are fluids and exert buoyant forces just as liquids do. The buoyant force due to air is often negligible if an object’s average density is much larger than the density of air. To see a significant buoyant force in air, we must use an object with a small average density. A hot air balloon has an opening at the bottom and a burner for heating the air within (Fig. 9.17). Many molecules of the heated air escape through the opening, decreasing the balloon’s average density. When the balloon is less dense on average than the surrounding air, it rises; at higher altitudes, the surrounding air becomes less and less dense. At some particular altitude, the buoyant force is equal in magnitude to the weight of the balloon. Then, by Newton’s second law, the net force on the balloon is zero. The balloon is in stable equilibrium at this altitude: if the balloon rises a bit, it experiences a net force downward, while if the balloon sinks down a bit, it is pushed back upward.

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Application of buoyant forces: hot air balloons

Figure 9.17 The buoyant force due to the outside air keeps these balloons aloft.

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9.7

FLUID FLOW

Types of Fluid Flow The study of moving fluids is a wonderfully complex subject. To illustrate some important ideas in less complex situations, we limit our study at first to fluids flowing under special conditions. One difference between moving fluids and static fluids is that a moving fluid can exert a force parallel to any surface over or past which it flows; a static fluid cannot. Since the moving fluid exerts a force against a surface, the surface must also exert a force on the fluid. This viscous force opposes the flow of the fluid; it is the counterpart to the kinetic frictional force between solids. An external force must act on a viscous fluid (and thereby do work) to keep it flowing. Viscosity is considered in Section 9.9. Until then, we consider only nonviscous fluids—fluid flow where the viscous forces are negligibly small. We also ignore surface tension, which is considered in Section 9.11. Fluid flow can be characterized as steady or unsteady. When the flow is steady, the velocity of the fluid at any point is constant in time. The velocity is not necessarily the same everywhere, but at any particular point, the velocity of the fluid passing that point remains constant in time. The density and pressure at any point in a steadily flowing fluid are also constant in time. Steady flow is laminar. The fluid flows in neat layers so that each small portion of fluid that passes a particular point follows the same path as every other portion of fluid that passes the same point. The path that the fluid follows, starting from any point, is called a streamline (Fig. 9.18). The streamlines may curve and bend, but they cannot cross each other; if they did, the fluid would have to “decide” which way to go when it gets to such a point. The direction of the fluid velocity at any point must be tangent to the streamline passing through that point. Streamlines are a convenient way to depict fluid flow in a sketch.

Figure 9.18 A wind tunnel shows the streamlines in the flow of air past a car.

The Ideal Fluid The special case that we consider first is the flow of an ideal fluid. An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity. Under some conditions, real fluids can be modeled as (nearly) ideal, but not under all conditions. The flow of an ideal fluid is governed by two principles: the continuity equation and Bernoulli’s equation. The continuity equation is an expression of conservation of mass for an incompressible fluid: since no fluid is created or destroyed, the total mass of the fluid must be constant. Bernoulli’s equation, discussed in Section 9.8, is a form of the energy conservation law applied to fluid flow. Together, these two equations enable us to predict the flow of an ideal fluid.

The Continuity Equation We start by deriving the continuity equation, which relates the speed of flow to the cross-sectional area of the fluid. Suppose an incompressible fluid flows into a pipe of nonuniform cross-sectional area under conditions of steady flow. In Fig. 9.19, the fluid on the left moves at speed v1. During a time Δt, the fluid travels a distance x1 = v1 Δt If A1 is the cross-sectional area of this section of pipe, then the mass of water moving past point 1 in time Δt is Δm1 = rV1 = r A1 x1 = r A1 v1 Δt x2 x1 A1

v1 1

Figure 9.19 An incompressible fluid flowing horizontally through a nonuniform pipe.

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A2

v2

2 ∆ m1 ∆ m2

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Figure 9.20 Streamlines in a v1

pipe of varying cross-sectional area. Streamlines are closer together where the fluid velocity is larger and farther apart where the velocity is smaller.

v2

During this same time interval, the mass of fluid moving past point 2 is Δm2 = rV2 = r A2 x2 = r A2 v2 Δt But, if the flow is steady, the mass passing through one section of pipe in time interval Δt must pass through any other section of the pipe in the same time interval. Therefore, Δm1 = Δm2

or

r A1 v1 Δt = r A2 v2 Δt

(9-10)

The quantity r Av is the mass flow rate of the fluid: Mass flow rate: Δm = r Av ___ Δt

(SI unit: kg/s)

(9-11)

Since the time intervals Δt are the same, Eq. (9-11) says that the mass flow rate past any two points is the same. Since the density of an incompressible fluid is constant, the volume flow rate past any two points must also be the same: Volume flow rate: ΔV = Av ___ Δt

(SI unit: m3/s)

(9-12)

The continuity equation for an incompressible fluid equates the volume flow rates past two different points: Continuity equation for incompressible fluid: A1 v1 = A2 v2

(9-13)

The same volume of fluid that enters the pipe in a given time interval exits the pipe in the same time interval. Where the radius of the tube is large, the speed of the fluid is small; where the radius is small, the fluid speed is large. A familiar example is what happens when you use your thumb to partially block the end of a garden hose to make a jet of water. The water moves past your thumb, where the cross-sectional area is smaller, at a greater speed than it moves in the hose. Similarly, water traveling along a river speeds up, forming rapids, when the riverbed narrows or is partially blocked by rocks and boulders. Streamlines are closer together where the fluid flows faster and farther apart where it flows more slowly (Fig. 9.20). Thus, streamlines help us visualize fluid flow. The fluid velocity at any point is tangent to a streamline through that point.

PHYSICS AT HOME The continuity equation applies to an ideal fluid even if it is not flowing through a pipe. Turn on a faucet so that the water flows out in a moderate stream (Fig. 9.21). The falling water is in free fall, accelerated by gravity until it hits the sink below. As the water falls, its speed increases. The stream of water gradually narrows as it falls so that the product of speed and cross-sectional area is constant, as predicted by the continuity equation.

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Figure 9.21 Demonstrating the continuity equation at a bathroom sink. Notice that the stream of water is narrower where the flow speed is faster.

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CHAPTER 9 Fluids

Example 9.9 Speed of Blood Flow The heart pumps blood into the aorta, which has an inner radius of 1.0 cm. The aorta feeds 32 major arteries. If blood in the aorta travels at a speed of 28 cm/s, at approximately what average speed does it travel in the arteries? Assume that blood can be treated as an ideal fluid and that the arteries each have an inner radius of 0.21 cm. Strategy Since we have assumed blood to be an ideal fluid, we can apply the continuity equation. The main tube (the aorta) is connected to multiple tubes (the major arteries), so this problem seems to be more complicated than a single pipe with a constriction in it. What matters here is the total cross-sectional area into which the blood flows. Solution We start by finding the cross-sectional area of the aorta A1 = p r 2aorta and then the total cross-sectional area of the arteries 2

A2 = 32p r artery

A1 v1 = A2 v2 A p × (0.010 m)2 v2 = v1 ___1 = 0.28 m/s × _______________2 = 0.20 m/s A2 32p × (0.0021 m) Discussion The blood flow slows in the arteries because the total cross-sectional area is greater than that of the aorta alone. From the arteries, the blood travels to the many capillaries of the body. Each capillary has a tiny cross-sectional area, but there are so many of them that the blood flow slows greatly once it reaches the capillaries—allowing time for the exchange of oxygen, carbon dioxide, and nutrients between the blood and the body tissues.

Practice Problem 9.9 Hosing Down a Wastebasket A garden hose fills a 32-L wastebasket in 120 s. The opening at the end of the hose has a radius of 1.00 cm. (a) How fast is the water traveling as it leaves the hose? (b) How fast does the water travel if half the exit area is obstructed by placing a finger over the opening?

Now we apply the continuity equation and solve for the unknown speed.

9.8

The Bernoulli effect: Fluid flows faster where the pressure is lower.

BERNOULLI’S EQUATION

The continuity equation relates the flow velocities of an ideal fluid at two different points, based on the change in cross-sectional area of the pipe. According to the continuity equation, the fluid must speed up as it enters a constriction (Fig. 9.22) and then slow down to its original speed when it leaves the constriction. Using energy ideas, we will show that the pressure of the fluid in the constriction (P2) cannot be the same as the pressure before or after the constriction (P1). For horizontal flow the speed is higher where the pressure is lower. This principle is often called the Bernoulli effect. The Bernoulli effect can seem counterintuitive at first; isn’t rapidly moving fluid at high pressure? For instance, if you were hit with the fast-moving water out of a firehose, you would be knocked over easily. The force that knocks you over is indeed due to fluid pressure; you would justifiably conclude that the pressure was high. However, the pressure is not high until you slow down the water by getting in its way. The rapidly moving water in the jet is, in fact, approximately at atmospheric pressure (zero gauge pressure), but when you stop the water, its pressure increases dramatically. Let’s find the quantitative relationship between pressure changes and flow speed changes for an ideal fluid. In Fig. 9.23, the shaded volume of fluid flows to the right.

Figure 9.22 A small volume of fluid speeds up as it moves into a constriction (position A) and then slows down as it moves out of the constriction (position B).

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P1 v1

P1 A a

P2

v2 v2 > v1

B

v1

a

P2 < P 1

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∆x2

Figure 9.23 Applying conservation of energy to the flow of an ideal fluid. The shaded volume of fluid in (a) is flowing to the right; (b) shows the same volume of fluid a short time later.

v2

∆x1

P2

v1

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y2

P1 y1 (a) ∆x2 ∆x1 A2 A1 (b)

If the left end moves a distance Δx1, then the right end moves a distance Δx2. Since the fluid is incompressible, A1 Δx1 = A2 Δx2 = V Work is done by the neighboring fluid during this flow. Fluid behind (to the left) pushes forward, doing positive work, while fluid ahead pushes backward, doing negative work. The total work done on the shaded volume by neighboring fluid is W = P1 A1 Δx1 − P2 A2 Δx2 = (P1 − P2 )V Since no dissipative forces act on an ideal fluid, the work done is equal to the total change in kinetic and gravitational potential energy. The net effect of the displacement is to move a volume V of fluid from height y1 to height y2 and to change its speed from v1 to v2. The energy change is therefore ΔE = ΔK + ΔU = _12 m(v 22 − v 21) + mg(y2 − y1 )

CONNECTION: Bernoulli’s equation is a restatement of the principle of energy conservation applied to the flow of an ideal fluid.

where the +y-direction is up. Substituting m = rV and equating the work done on the fluid to the change in its energy yields (P1 − P2 ) V = _12 rV(v 22 − v 21) + rVg(y2 − y1 ) Dividing both sides by V and rearranging yields Bernoulli’s equation, named after Swiss mathematician Daniel Bernoulli (1700–1782), but first derived by fellow Swiss mathematician Leonhard Euler (pronounced like oiler, 1707–1783). Bernoulli’s equation (for ideal fluid flow): P1 + rgy1 + _12 r v 21 = P2 + rgy2 + _12 r v 22 (or P + rgy + _12 r v2 = constant)

(9-14)

Bernoulli’s equation relates the pressure, flow speed, and height at two points in an ideal fluid. Although we derived Bernoulli’s equation in a relatively simple situation, it applies to the flow of any ideal fluid as long as points 1 and 2 are on the same streamline. Each term in Bernoulli’s equation has units of pressure, which in the SI system is Pa or N/m2. Since a joule is a newton-meter, the pascal is also equal to a joule per cubic

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CHAPTER 9 Fluids

meter (J/m3). Each term represents work or energy per unit volume. The pressure is the work done by the fluid on the fluid ahead of it per unit volume of flow. The kinetic energy per unit volume is _12 r v2 and the gravitational potential energy per unit volume is rgy. (Text website tutorial: energies)

CHECKPOINT 9.8 Discuss Bernoulli’s equation in two special cases: (a) horizontal flow (y1 = y2) and (b) a static fluid (v1 = v2 = 0).

Example 9.10 Torricelli’s Theorem A barrel full of rainwater has a spigot near the bottom, at a depth of 0.80 m beneath the water surface. (a) When the spigot is directed horizontally (Fig. 9.24a) and is opened, how fast does the water come out? (b) If the opening points upward (Fig. 9.24b), how high does the resulting “fountain” go? ( tutorial: waterfall) Strategy The water at the surface is at atmospheric pressure. The water emerging from the spigot is also at atmospheric pressure since it is in contact with the air. If the pressure of the emerging water were different than that of the air, the stream would expand or contract until the pressures were equal. We apply Bernoulli’s equation to two points: point 1 at the water surface and point 2 in the emerging stream of water. Solution (a) Since P1 = P2, Bernoulli’s equation is 2

Point 1 is 0.80 m above point 2, so

d=? v

g(y1 − y2 ) = _12 v 22 __________

v2 = √ 2g(y1 − y2 ) = 4.0 m/s

Discussion The result of part (b) is called Torricelli’s theorem. In reality, the fountain does not reach as high as the original water level; some energy is dissipated due to viscosity and air resistance.

Practice Problem 9.10 Fluid in Free Fall (b)

Figure 9.24 Full barrel of rainwater with open spigot (a) horizontal and (b) upward.

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After dividing through by r, we solve for v2:

2

2 (a)

2

rgy1 = rgy2 + _12 rv 2

The “fountain” goes right back up to the top of the water in the barrel!

0.80 m v

Since the cross-sectional area of the spigot A2 is much smaller than the area of the top of the barrel A1, the speed of the water at the surface v1 is negligibly small compared with v2. Setting v1 = 0, Bernoulli’s equation reduces to

rgy1 = rgy2

y1 − y2 = 0.80 m 1

v1 A1 = v2 A2

(b) Now take point 2 to be at the top of the fountain. Then v2 = 0 and Bernoulli’s equation reduces to

2

rgy1 + _12 rv 1 = rgy2 + _12 rv 2

1

The speed of the emerging water is v2. What is v1, the speed of the water at the surface? The water at the surface is moving slowly, since the barrel is draining. The continuity equation requires that

Verify that the speed found in part (a) is the same as if the water just fell 0.80 m straight down. That shouldn’t be too surprising since Bernoulli’s equation is an expression of energy conservation.

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Example 9.11 The Venturi Meter A Venturi meter (Fig. 9.25) measures fluid speed in a pipe. A constriction (of cross-sectional area A2) is put in a pipe of normal cross-sectional area A1. Two vertical tubes, open to the atmosphere, rise from two points, one of which is in the constriction. The vertical tubes function like manometers, enabling the pressure to be determined. From this information the flow speed in the pipe can be determined. Suppose that the pipe in question carries water, A1 = 2.0A2, and the fluid heights in the vertical tubes are h1 = 1.20 m and h2 = 0.80 m. (a) Find the ratio of the flow speeds v2/v1. (b) Find the gauge pressures P1 and P2. (c) Find the flow speed v1 in the pipe. Strategy Neither of the two flow speeds is given. We need more than Bernoulli’s equation to solve this problem. Since we know the ratio of the areas, the continuity equation gives us the ratio of the speeds. The height of the water in the vertical tubes enables us to find the pressures at points 1 and 2. The fluid pressure at the bottom of each vertical tube is the same as the pressure of the moving fluid just beneath each tube—otherwise, water would flow into or out of the vertical tubes until the pressure equalized. The water in the vertical tubes is static, so the gauge pressure at the bottom is P = rgd. Once we have the ratio of the speeds and the pressures, we apply Bernoulli’s equation. Solution (a) From the continuity equation, the product of flow speed and area must be the same at points 1 and 2. Therefore, A v __2 = ___1 = 2.0 v1 A 2 The water flows twice as fast in the constriction as in the rest of the pipe.

(b) The gauge pressures are: P1 = rgh1 = 1000 kg/m3 × 9.80 N/kg × 1.20 m = 11.8 kPa P2 = rgh2 = 1000 kg/m3 × 9.80 N/kg × 0.80 m = 7.8 kPa (c) Now we apply Bernoulli’s equation. We can use gauge pressures as long as we do so on both sides—in effect we are just subtracting atmospheric pressure from both sides of the equation: 2

2

P1 + rgy1 + _12 r v 1 = P2 + rgy2 + _12 r v 2 Since the tube is horizontal, y1 ≈ y2 and we can ignore the small change in gravitational potential energy density rgy. Then 2

2

P1 + _12 r v 1 = P2 + _12 r v 2 We are trying to find v1, so we can eliminate v2 by substituting v2 = 2.0v1: 2

P1 + _12 r v 1 = P2 + _12 r (2.0v1 )2 Simplifying, 2

P1 − P2 = 1.5r v 1

√

_________________

11 800 Pa − 7800 Pa = 1.6 m/s v1 = _________________ 1.5 × 1000 kg/m3 Discussion The assumption that y1 ≈ y2 is fine as long as the pipe radius is small compared with the difference between the static water heights (40 cm). Otherwise, we would have to account for the different y values in Bernoulli’s equation. One subtle point: recall that we assumed that the fluid pressure at the bottom of the vertical tubes was the same as the pressure of the moving fluid just beneath. Does that contradict Bernoulli’s equation? Since there is an abrupt change in fluid speed, shouldn’t there be a significant difference in the pressures? No, because these points are not on the same streamline.

h1 h2

1

2 v1

A1

Figure 9.25

v2 A2 Streamlines

Practice Problem 9.11 Garden Hose Water flows horizontally through a garden hose of radius 1.0 cm at a speed of 1.4 m/s. The water shoots horizontally out of a nozzle of radius 0.25 cm. What is the gauge pressure of the water inside the hose?

Venturi meter.

Application of Bernoulli’s Principle: Arterial Flutter and Aneurisms Suppose an artery is narrowed due to buildup of plaque on its inner walls. The flow of blood through the constriction is similar to that shown in Fig. 9.22. Bernoulli’s equation tells us that the pressure P2 in the constriction is lower than the pressure elsewhere.

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Figure 9.26 Streamlines showing the airflow past an airplane wing in a wind tunnel.

CHAPTER 9 Fluids

The arterial walls are elastic rather than rigid, so the lower pressure allows the arterial walls to contract a bit in the constriction. Now the flow velocity is even higher and the pressure even lower. Eventually the artery wall collapses, shutting off the flow of blood. Then the pressure builds up, reopens the artery, and allows blood to flow. The cycle of arterial flutter then begins again. The opposite may happen where the arterial wall is weak. Blood pressure pushes the artery walls outward, forming a bulge called an aneurism. The lower flow speed in the bulge is accompanied by a higher blood pressure, which enlarges the aneurism even more (see Problem 88). Ultimately the artery may burst from the increased pressure. Application of Bernoulli’s Principle: Airplane Wings How does an airplane wing generate lift? Figure 9.26 is a sketch of some streamlines for air flowing past an airplane wing in a wind tunnel. The streamlines bend, showing that the wing deflects air downward. By Newton’s third law (or conservation of momentum), if the wing pushes downward on the air, the air also pushes upward on the wing. This upward force on the wing is lift. However, the situation is not as simple as air “bouncing” off the bottom of the wing—note that air passing above the wing is also deflected downward. We can use Bernoulli’s equation to get more insight into the generation of lift. (Bernoulli’s equation applies in an approximate way to moving air. Even though air is not incompressible, for subsonic flight the density changes are small enough to be ignored.) If the air exerts a net upward force on the wing, the air pressure must be lower above the wing than beneath the wing. In Fig. 9.26, the streamlines above the wing are closer together than beneath the wing, showing that the flow speed above the wing is faster than it is beneath. This observation confirms that the pressure is lower above the wing, because where the pressure is lower, the flow speed is faster.

9.9 CONNECTION: Kinetic friction makes a sliding object slow down unless an applied force balances the force of friction. Similarly, viscous forces oppose the flow of a fluid. Steady flow of a viscous fluid requires an applied force to balance the viscous forces. The applied force is due to the pressure difference.

VISCOSITY

Bernoulli’s equation ignores viscosity (fluid friction). According to Bernoulli’s equation, an ideal fluid can continue to flow in a horizontal pipe at constant velocity on its own, just as a hockey puck would slide across frictionless ice at constant velocity without anything pushing it along. However, all real fluids have some viscosity; to maintain flow in a viscous fluid, we have to apply an external force since viscous forces oppose the flow of the fluid (Fig. 9.27). A pressure difference between the ends of the pipe must be maintained to keep a real liquid moving through a horizontal pipe. The pressure difference is important—in everything from blood flowing through arteries to oil pumped through a pipeline. To visualize viscous flow in a tube of circular cross section, imagine the fluid to flow in cylindrical layers, or shells. If there were no viscosity, all the layers would move at the same speed (Fig. 9.28a). In viscous flow, the fluid speed depends on the distance from the tube walls (Fig. 9.28b). The fastest flow is at the center of the tube. Layers closer to the wall of the tube move more slowly. The outermost layer of fluid, which is in contact with the tube, does not move. Each layer of fluid exerts viscous forces on the Direction of flow

Figure 9.27 (a) To maintain viscous flow, a net force due to fluid pressure (P1 − P2) A must be applied in the direction of flow to balance the viscous force Fv due to the pipe, which opposes flow. (b) The pressure in the fluid decreases from P1 at the left end to P2 at the right end.

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