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University Physics Volume 1
SENIOR CONTRIBUTING AUTHORS
SAMUEL J. LING, TRUMAN STATE UNIVERSITY JEFF SANNY, LOYOLA MARYMOUNT UNIVERSITY WILLIAM MOEBS, FORMERLY OF LOYOLA MARYMOUNT UNIVERSITY

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Contents
Preface 1
UNIT 1 MECHANICS
CHAPTER 1
Units and Measurement 5
Introduction 5 1.1 The Scope and Scale of Physics 6 1.2 Units and Standards 12 1.3 Unit Conversion 18 1.4 Dimensional Analysis 21 1.5 Estimates and Fermi Calculations 24 1.6 Significant Figures 27 1.7 Solving Problems in Physics 31 Chapter Review 34
CHAPTER 2
Vectors 43
Introduction 43 2.1 Scalars and Vectors 44 2.2 Coordinate Systems and Components of a Vector 55 2.3 Algebra of Vectors 65 2.4 Products of Vectors 74 Chapter Review 86
CHAPTER 3
Motion Along a Straight Line 99
Introduction 99 3.1 Position, Displacement, and Average Velocity 100 3.2 Instantaneous Velocity and Speed 104 3.3 Average and Instantaneous Acceleration 109 3.4 Motion with Constant Acceleration 117 3.5 Free Fall 129 3.6 Finding Velocity and Displacement from Acceleration 134 Chapter Review 138
CHAPTER 4
Motion in Two and Three Dimensions 149
Introduction 149 4.1 Displacement and Velocity Vectors 150 4.2 Acceleration Vector 157 4.3 Projectile Motion 162 4.4 Uniform Circular Motion 172 4.5 Relative Motion in One and Two Dimensions 179

Chapter Review 184

CHAPTER 5
Newton's Laws of Motion
Introduction 195 5.1 Forces 196 5.2 Newton's First Law 200 5.3 Newton's Second Law 204 5.4 Mass and Weight 215 5.5 Newton’s Third Law 217 5.6 Common Forces 223 5.7 Drawing Free-Body Diagrams Chapter Review 238

195 233

CHAPTER 6
Applications of Newton's Laws
Introduction 251 6.1 Solving Problems with Newton’s Laws 6.2 Friction 267 6.3 Centripetal Force 279 6.4 Drag Force and Terminal Speed 288 Chapter Review 297

251 252

CHAPTER 7
Work and Kinetic Energy 315
Introduction 315 7.1 Work 316 7.2 Kinetic Energy 325 7.3 Work-Energy Theorem 329 7.4 Power 333 Chapter Review 336

CHAPTER 8
Potential Energy and Conservation of Energy 347
Introduction 347 8.1 Potential Energy of a System 348 8.2 Conservative and Non-Conservative Forces 356 8.3 Conservation of Energy 359 8.4 Potential Energy Diagrams and Stability 365 8.5 Sources of Energy 369 Chapter Review 373

CHAPTER 9
Linear Momentum and Collisions 385
Introduction 385 9.1 Linear Momentum 386 9.2 Impulse and Collisions 388 9.3 Conservation of Linear Momentum 401 9.4 Types of Collisions 411

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9.5 Collisions in Multiple Dimensions 419 9.6 Center of Mass 426 9.7 Rocket Propulsion 441 Chapter Review 447
CHAPTER 10
Fixed-Axis Rotation 461
Introduction 461 10.1 Rotational Variables 462 10.2 Rotation with Constant Angular Acceleration 471 10.3 Relating Angular and Translational Quantities 476 10.4 Moment of Inertia and Rotational Kinetic Energy 481 10.5 Calculating Moments of Inertia 488 10.6 Torque 496 10.7 Newton’s Second Law for Rotation 501 10.8 Work and Power for Rotational Motion 505 Chapter Review 511
CHAPTER 11
Angular Momentum 527
Introduction 527 11.1 Rolling Motion 528 11.2 Angular Momentum 535 11.3 Conservation of Angular Momentum 544 11.4 Precession of a Gyroscope 550 Chapter Review 554
CHAPTER 12
Static Equilibrium and Elasticity 565
Introduction 565 12.1 Conditions for Static Equilibrium 566 12.2 Examples of Static Equilibrium 574 12.3 Stress, Strain, and Elastic Modulus 587 12.4 Elasticity and Plasticity 597 Chapter Review 599
CHAPTER 13
Gravitation 611
Introduction 611 13.1 Newton's Law of Universal Gravitation 612 13.2 Gravitation Near Earth's Surface 617 13.3 Gravitational Potential Energy and Total Energy 624 13.4 Satellite Orbits and Energy 629 13.5 Kepler's Laws of Planetary Motion 636 13.6 Tidal Forces 642 13.7 Einstein's Theory of Gravity 648 Chapter Review 656

CHAPTER 14
Fluid Mechanics 665
Introduction 665 14.1 Fluids, Density, and Pressure 666 14.2 Measuring Pressure 677 14.3 Pascal's Principle and Hydraulics 682 14.4 Archimedes’ Principle and Buoyancy 687 14.5 Fluid Dynamics 691 14.6 Bernoulli’s Equation 696 14.7 Viscosity and Turbulence 702 Chapter Review 711
UNIT 2 WAVES AND ACOUSTICS
CHAPTER 15
Oscillations 723
Introduction 723 15.1 Simple Harmonic Motion 724 15.2 Energy in Simple Harmonic Motion 733 15.3 Comparing Simple Harmonic Motion and Circular Motion 740 15.4 Pendulums 744 15.5 Damped Oscillations 750 15.6 Forced Oscillations 753 Chapter Review 758
CHAPTER 16
Waves 767
Introduction 767 16.1 Traveling Waves 768 16.2 Mathematics of Waves 774 16.3 Wave Speed on a Stretched String 782 16.4 Energy and Power of a Wave 786 16.5 Interference of Waves 790 16.6 Standing Waves and Resonance 799 Chapter Review 808
CHAPTER 17
Sound 823
Introduction 823 17.1 Sound Waves 824 17.2 Speed of Sound 826 17.3 Sound Intensity 835 17.4 Normal Modes of a Standing Sound Wave 843 17.5 Sources of Musical Sound 851 17.6 Beats 855 17.7 The Doppler Effect 857 17.8 Shock Waves 863 Chapter Review 868
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Appendix A Units 883 Appendix B Conversion Factors 887 Appendix C Fundamental Constants 891 Appendix D Astronomical Data 893 Appendix E Mathematical Formulas 895 Appendix F Chemistry 899 Appendix G The Greek Alphabet 901 Answer Key 902
Index 963

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Preface

1

PREFACE

Welcome to University Physics, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.
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You can access this textbook for free in web view or PDF through OpenStax.org, and for a low cost in print.
About University Physics
University Physics is designed for the two- or threesemester calculus-based physics course. The text has been developed to meet the scope and sequence of most university physics courses and provides a foundation for a career in mathematics, science, or engineering. The book provides an important opportunity for students to learn the core concepts of physics and understand how those concepts apply to their lives and to the world around them.
Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency.
Coverage and scope
Our University Physics textbook adheres to the scope and sequence of most two- and threesemester physics courses nationwide. We have worked to make physics interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of this textbook has been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but to work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from science educators dedicated to the project.
VOLUME I
Unit 1: Mechanics

2

Preface

• Chapter 1: Units and Measurement • Chapter 2: Vectors • Chapter 3: Motion Along a Straight Line • Chapter 4: Motion in Two and Three Dimensions • Chapter 5: Newton’s Laws of Motion • Chapter 6: Applications of Newton’s Laws • Chapter 7: Work and Kinetic Energy • Chapter 8: Potential Energy and Conservation of
Energy • Chapter 9: Linear Momentum and Collisions • Chapter 10: Fixed-Axis Rotation • Chapter 11: Angular Momentum • Chapter 12: Static Equilibrium and Elasticity • Chapter 13: Gravitation • Chapter 14: Fluid Mechanics
Unit 2: Waves and Acoustics
• Chapter 15: Oscillations • Chapter 16: Waves • Chapter 17: Sound
VOLUME II
Unit 1: Thermodynamics
• Chapter 1: Temperature and Heat • Chapter 2: The Kinetic Theory of Gases • Chapter 3: The First Law of Thermodynamics • Chapter 4: The Second Law of Thermodynamics
Unit 2: Electricity and Magnetism
• Chapter 5: Electric Charges and Fields • Chapter 6: Gauss’s Law • Chapter 7: Electric Potential • Chapter 8: Capacitance • Chapter 9: Current and Resistance • Chapter 10: Direct-Current Circuits • Chapter 11: Magnetic Forces and Fields • Chapter 12: Sources of Magnetic Fields • Chapter 13: Electromagnetic Induction • Chapter 14: Inductance • Chapter 15: Alternating-Current Circuits • Chapter 16: Electromagnetic Waves
VOLUME III
Unit 1: Optics
• Chapter 1: The Nature of Light • Chapter 2: Geometric Optics and Image
Formation • Chapter 3: Interference • Chapter 4: Diffraction
Unit 2: Modern Physics
• Chapter 5: Relativity

• Chapter 6: Photons and Matter Waves • Chapter 7: Quantum Mechanics • Chapter 8: Atomic Structure • Chapter 9: Condensed Matter Physics • Chapter 10: Nuclear Physics • Chapter 11: Particle Physics and Cosmology
Pedagogical foundation
Throughout University Physics you will find derivations of concepts that present classical ideas and techniques, as well as modern applications and methods. Most chapters start with observations or experiments that place the material in a context of physical experience. Presentations and explanations rely on years of classroom experience on the part of long-time physics professors, striving for a balance of clarity and rigor that has proven successful with their students. Throughout the text, links enable students to review earlier material and then return to the present discussion, reinforcing connections between topics. Key historical figures and experiments are discussed in the main text (rather than in boxes or sidebars), maintaining a focus on the development of physical intuition. Key ideas, definitions, and equations are highlighted in the text and listed in summary form at the end of each chapter. Examples and chapter-opening images often include contemporary applications from daily life or modern science and engineering that students can relate to, from smart phones to the internet to GPS devices.
Assessments that reinforce key concepts
In-chapter Examples generally follow a three-part format of Strategy, Solution, and Significance to emphasize how to approach a problem, how to work with the equations, and how to check and generalize the result. Examples are often followed by Check Your Understanding questions and answers to help reinforce for students the important ideas of the examples. Problem-Solving Strategies in each chapter break down methods of approaching various types of problems into steps students can follow for guidance. The book also includes exercises at the end of each chapter so students can practice what they’ve learned.
• Conceptual questions do not require calculation but test student learning of the key concepts.
• Problems categorized by section test student problem-solving skills and the ability to apply ideas to practical situations.
• Additional Problems apply knowledge across the chapter, forcing students to identify what

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Preface

3

concepts and equations are appropriate for solving given problems. Randomly located throughout the problems are Unreasonable Results exercises that ask students to evaluate the answer to a problem and explain why it is not reasonable and what assumptions made might not be correct. • Challenge Problems extend text ideas to interesting but difficult situations.
Answers for selected exercises are available in an Answer Key at the end of the book.
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About the authors
Senior contributing authors
Samuel J. Ling, Truman State University

Dr. Samuel Ling has taught introductory and advanced physics for over 25 years at Truman State University, where he is currently Professor of Physics and the Department Chair. Dr. Ling has two PhDs from Boston University, one in Chemistry and the other in Physics, and he was a Research Fellow at the Indian Institute of Science, Bangalore, before joining Truman. Dr. Ling is also an author of A First Course in Vibrations and Waves, published by Oxford University Press. Dr. Ling has considerable experience with research in Physics Education and has published research on collaborative learning methods in physics teaching. He was awarded a Truman Fellow and a Jepson fellow in recognition of his innovative teaching methods. Dr. Ling’s research publications have spanned Cosmology, Solid State Physics, and Nonlinear Optics.
Jeff Sanny, Loyola Marymount University Dr. Jeff Sanny earned a BS in Physics from Harvey Mudd College in 1974 and a PhD in Solid State Physics from the University of California–Los Angeles in 1980. He joined the faculty at Loyola Marymount University in the fall of 1980. During his tenure, he has served as department Chair as well as Associate Dean. Dr. Sanny enjoys teaching introductory physics in particular. He is also passionate about providing students with research experience and has directed an active undergraduate student research group in space physics for many years.
William Moebs, Formerly of Loyola Marymount University Dr. William Moebs earned a BS and PhD (1959 and 1965) from the University of Michigan. He then joined their staff as a Research Associate for one year, where he continued his doctoral research in particle physics. In 1966, he accepted an appointment to the Physics Department of Indiana Purdue Fort Wayne (IPFW), where he served as Department Chair from 1971 to 1979. In 1979, he moved to Loyola Marymount University (LMU), where he served as Chair of the Physics Department from 1979 to 1986. He retired from LMU in 2000. He has published research in particle physics, chemical kinetics, cell division, atomic physics, and physics teaching.
Contributing authors
Stephen D. Druger, Northwestern University Alice Kolakowska, University of Memphis David Anderson, Albion College Daniel Bowman, Ferrum College Dedra Demaree, Georgetown University

4

Preface

Edw. S. Ginsberg, University of Massachusetts Joseph Trout, Richard Stockton College Kevin Wheelock, Bellevue College David Smith, University of the Virgin Islands Takashi Sato, Kwantlen Polytechnic University Gerald Friedman, Santa Fe Community College Lev Gasparov, University of North Florida Lee LaRue, Paris Junior College Mark Lattery, University of Wisconsin Richard Ludlow, Daniel Webster College Patrick Motl, Indiana University Kokomo Tao Pang, University of Nevada, Las Vegas Kenneth Podolak, Plattsburgh State University
Reviewers
Salameh Ahmad, Rochester Institute of Technology–Dubai John Aiken, University of Colorado–Boulder Raymond Benge, Terrant County College Gavin Buxton, Robert Morris University Erik Christensen, South Florida State College Clifton Clark, Fort Hays State University Nelson Coates, California Maritime Academy Herve Collin, Kapi’olani Community College Carl Covatto, Arizona State University Alejandro Cozzani, Imperial Valley College Danielle Dalafave, The College of New Jersey Nicholas Darnton, Georgia Institute of Technology Ethan Deneault, University of Tampa Kenneth DeNisco, Harrisburg Area Community College Robert Edmonds, Tarrant County College William Falls, Erie Community College Stanley Forrester, Broward College Umesh Garg, University of Notre Dame Maurizio Giannotti, Barry University

Bryan Gibbs, Dallas County Community College Lynn Gillette, Pima Community College–West Campus Mark Giroux, East Tennessee State University Matthew Griffiths, University of New Haven Alfonso Hinojosa, University of Texas–Arlington Steuard Jensen, Alma College David Kagan, University of Massachusetts Sergei Katsev, University of Minnesota–Duluth Gregory Lapicki, East Carolina University Jill Leggett, Florida State College–Jacksonville Alfredo Louro, University of Calgary James Maclaren, Tulane University Ponn Maheswaranathan, Winthrop University Seth Major, Hamilton College Oleg Maksimov, Excelsior College Aristides Marcano, Delaware State University James McDonald, University of Hartford Ralph McGrew, SUNY–Broome Community College Paul Miller, West Virginia University Tamar More, University of Portland Farzaneh Najmabadi, University of Phoenix Richard Olenick, The University of Dallas Christopher Porter, Ohio State University Liza Pujji, Manakau Institute of Technology Baishali Ray, Young Harris University Andrew Robinson, Carleton University Aruvana Roy, Young Harris University Gajendra Tulsian, Daytona State College Adria Updike, Roger Williams University Clark Vangilder, Central Arizona University Steven Wolf, Texas State University Alexander Wurm, Western New England University Lei Zhang, Winston Salem State University Ulrich Zurcher, Cleveland State University

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CHAPTER 1
Units and Measurement

Figure 1.1 This image might be showing any number of things. It might be a whirlpool in a tank of water or perhaps

a collage of paint and shiny beads done for art class. Without knowing the size of the object in units we all recognize,

such as meters or inches, it is difficult to know what we’re looking at. In fact, this image shows the Whirlpool Galaxy

(and its companion galaxy), which is about 60,000 light-years in diameter (about

across). (credit:

modification of work by S. Beckwith (STScI) Hubble Heritage Team, (STScI/AURA), ESA, NASA)

Chapter Outline
1.1 The Scope and Scale of Physics 1.2 Units and Standards 1.3 Unit Conversion 1.4 Dimensional Analysis 1.5 Estimates and Fermi Calculations 1.6 Significant Figures 1.7 Solving Problems in Physics

6

1 • Units and Measurement

INTRODUCTION As noted in the figure caption, the chapter-opening image is of the Whirlpool Galaxy, which we examine in the first section of this chapter. Galaxies are as immense as atoms are small, yet the same laws of physics describe both, along with all the rest of nature—an indication of the underlying unity in the universe. The laws of physics are surprisingly few, implying an underlying simplicity to nature’s apparent complexity. In this text, you learn about the laws of physics. Galaxies and atoms may seem far removed from your daily life, but as you begin to explore this broad-ranging subject, you may soon come to realize that physics plays a much larger role in your life than you first thought, no matter your life goals or career choice.
1.1 The Scope and Scale of Physics
Learning Objectives By the end of this section, you will be able to:
• Describe the scope of physics. • Calculate the order of magnitude of a quantity. • Compare measurable length, mass, and timescales quantitatively. • Describe the relationships among models, theories, and laws.

Physics is devoted to the understanding of all natural phenomena. In physics, we try to understand physical phenomena at all scales—from the world of subatomic particles to the entire universe. Despite the breadth of the subject, the various subfields of physics share a common core. The same basic training in physics will prepare you to work in any area of physics and the related areas of science and engineering. In this section, we investigate the scope of physics; the scales of length, mass, and time over which the laws of physics have been shown to be applicable; and the process by which science in general, and physics in particular, operates.

The Scope of Physics

Take another look at the chapter-opening image. The Whirlpool Galaxy contains billions of individual stars as

well as huge clouds of gas and dust. Its companion galaxy is also visible to the right. This pair of galaxies lies a

staggering billion trillion miles

from our own galaxy (which is called the Milky Way). The stars

and planets that make up the Whirlpool Galaxy might seem to be the furthest thing from most people’s

everyday lives, but the Whirlpool is a great starting point to think about the forces that hold the universe

together. The forces that cause the Whirlpool Galaxy to act as it does are thought to be the same forces we

contend with here on Earth, whether we are planning to send a rocket into space or simply planning to raise

the walls for a new home. The gravity that causes the stars of the Whirlpool Galaxy to rotate and revolve is

thought to be the same as what causes water to flow over hydroelectric dams here on Earth. When you look up

at the stars, realize the forces out there are the same as the ones here on Earth. Through a study of physics, you

may gain a greater understanding of the interconnectedness of everything we can see and know in this

universe.

Think, now, about all the technological devices you use on a regular basis. Computers, smartphones, global positioning systems (GPSs), MP3 players, and satellite radio might come to mind. Then, think about the most exciting modern technologies you have heard about in the news, such as trains that levitate above tracks, “invisibility cloaks” that bend light around them, and microscopic robots that fight cancer cells in our bodies. All these groundbreaking advances, commonplace or unbelievable, rely on the principles of physics. Aside from playing a significant role in technology, professionals such as engineers, pilots, physicians, physical therapists, electricians, and computer programmers apply physics concepts in their daily work. For example, a pilot must understand how wind forces affect a flight path; a physical therapist must understand how the muscles in the body experience forces as they move and bend. As you will learn in this text, the principles of physics are propelling new, exciting technologies, and these principles are applied in a wide range of careers.

The underlying order of nature makes science in general, and physics in particular, interesting and enjoyable to study. For example, what do a bag of chips and a car battery have in common? Both contain energy that can be converted to other forms. The law of conservation of energy (which says that energy can change form but is never lost) ties together such topics as food calories, batteries, heat, light, and watch springs. Understanding this law makes it easier to learn about the various forms energy takes and how they relate to one another. Apparently unrelated topics are connected through broadly applicable physical laws, permitting an

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1.1 • The Scope and Scale of Physics

7

understanding beyond just the memorization of lists of facts.
Science consists of theories and laws that are the general truths of nature, as well as the body of knowledge they encompass. Scientists are continuously trying to expand this body of knowledge and to perfect the expression of the laws that describe it. Physics, which comes from the Greek phúsis, meaning “nature,” is concerned with describing the interactions of energy, matter, space, and time to uncover the fundamental mechanisms that underlie every phenomenon. This concern for describing the basic phenomena in nature essentially defines the scope of physics.
Physics aims to understand the world around us at the most basic level. It emphasizes the use of a small number of quantitative laws to do this, which can be useful to other fields pushing the performance boundaries of existing technologies. Consider a smartphone (Figure 1.2). Physics describes how electricity interacts with the various circuits inside the device. This knowledge helps engineers select the appropriate materials and circuit layout when building a smartphone. Knowledge of the physics underlying these devices is required to shrink their size or increase their processing speed. Or, think about a GPS. Physics describes the relationship between the speed of an object, the distance over which it travels, and the time it takes to travel that distance. When you use a GPS in a vehicle, it relies on physics equations to determine the travel time from one location to another.

Figure 1.2 The Apple iPhone is a common smartphone with a GPS function. Physics describes the way that electricity flows through the circuits of this device. Engineers use their knowledge of physics to construct an iPhone with features that consumers will enjoy. One specific feature of an iPhone is the GPS function. A GPS uses physics equations to determine the drive time between two locations on a map. (credit: Jane Whitney)
Knowledge of physics is useful in everyday situations as well as in nonscientific professions. It can help you understand how microwave ovens work, why metals should not be put into them, and why they might affect pacemakers. Physics allows you to understand the hazards of radiation and to evaluate these hazards rationally and more easily. Physics also explains the reason why a black car radiator helps remove heat in a car engine, and it explains why a white roof helps keep the inside of a house cool. Similarly, the operation of a car’s ignition system as well as the transmission of electrical signals throughout our body’s nervous system are much easier to understand when you think about them in terms of basic physics.
Physics is a key element of many important disciplines and contributes directly to others. Chemistry, for example—since it deals with the interactions of atoms and molecules—has close ties to atomic and molecular physics. Most branches of engineering are concerned with designing new technologies, processes, or structures within the constraints set by the laws of physics. In architecture, physics is at the heart of structural

8

1 • Units and Measurement

stability and is involved in the acoustics, heating, lighting, and cooling of buildings. Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, and heat transfer within Earth. Some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines.
Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties of cells and their environments. On the macroscopic level, it explains the heat, work, and power associated with the human body and its various organ systems. Physics is involved in medical diagnostics, such as radiographs, magnetic resonance imaging, and ultrasonic blood flow measurements. Medical therapy sometimes involves physics directly; for example, cancer radiotherapy uses ionizing radiation. Physics also explains sensory phenomena, such as how musical instruments make sound, how the eye detects color, and how lasers transmit information.
It is not necessary to study all applications of physics formally. What is most useful is knowing the basic laws of physics and developing skills in the analytical methods for applying them. The study of physics also can improve your problem-solving skills. Furthermore, physics retains the most basic aspects of science, so it is used by all the sciences, and the study of physics makes other sciences easier to understand.
The Scale of Physics
From the discussion so far, it should be clear that to accomplish your goals in any of the various fields within the natural sciences and engineering, a thorough grounding in the laws of physics is necessary. The reason for this is simply that the laws of physics govern everything in the observable universe at all measurable scales of length, mass, and time. Now, that is easy enough to say, but to come to grips with what it really means, we need to get a little bit quantitative. So, before surveying the various scales that physics allows us to explore, let’s first look at the concept of “order of magnitude,” which we use to come to terms with the vast ranges of length, mass, and time that we consider in this text (Figure 1.3).

Figure 1.3 (a) Using a scanning tunneling microscope, scientists can see the individual atoms (diameters around 10–10 m) that compose this sheet of gold. (b) Tiny phytoplankton swim among crystals of ice in the Antarctic Sea. They range from a few micrometers (1 μm is 10–6 m) to as much as 2 mm (1 mm is 10–3 m) in length. (c) These two colliding galaxies, known as NGC 4676A (right) and NGC 4676B (left), are nicknamed “The Mice” because of the tail of gas emanating from each one. They are located 300 million light-years from Earth in the constellation Coma Berenices. Eventually, these two galaxies will merge into one. (credit a: modification of work by "Erwinrossen"/Wikimedia Commons; credit b: modification of work by Prof. Gordon T. Taylor, Stony Brook University; NOAA Corps Collections; credit c: modification of work by NASA, H. Ford (JHU), G. Illingworth (UCSC/LO), M. Clampin (STScI), G. Hartig (STScI), the ACS Science Team, and ESA)

Order of magnitude

The order of magnitude of a number is the power of 10 that most closely approximates it. Thus, the order of

magnitude refers to the scale (or size) of a value. Each power of 10 represents a different order of magnitude.

For example,

and so forth, are all different orders of magnitude, as are

and

To find the order of magnitude of a number, take the base-10 logarithm of the number and round it to

the nearest integer, then the order of magnitude of the number is simply the resulting power of 10. For

example, the order of magnitude of 800 is 103 because

which rounds to 3. Similarly, the

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1.1 • The Scope and Scale of Physics

9

order of magnitude of 450 is 103 because

which rounds to 3 as well. Thus, we say the

numbers 800 and 450 are of the same order of magnitude: 103. However, the order of magnitude of 250 is 102

because

which rounds to 2.

An equivalent but quicker way to find the order of magnitude of a number is first to write it in scientific

notation and then check to see whether the first factor is greater than or less than

The idea

is that

is halfway between

and

on a log base-10 scale. Thus, if the first factor is

less than

then we round it down to 1 and the order of magnitude is simply whatever power of 10 is

required to write the number in scientific notation. On the other hand, if the first factor is greater than

then we round it up to 10 and the order of magnitude is one power of 10 higher than the power needed to write the number in scientific notation. For example, the number 800 can be written in scientific notation as

Because 8 is bigger than

we say the order of magnitude of 800 is

The number

450 can be written as

so its order of magnitude is also 103 because 4.5 is greater than 3. However,

250 written in scientific notation is

and 2.5 is less than 3, so its order of magnitude is

The order of magnitude of a number is designed to be a ballpark estimate for the scale (or size) of its value. It is

simply a way of rounding numbers consistently to the nearest power of 10. This makes doing rough mental

math with very big and very small numbers easier. For example, the diameter of a hydrogen atom is on the order of 10−10 m, whereas the diameter of the Sun is on the order of 109 m, so it would take roughly

hydrogen atoms to stretch across the diameter of the Sun. This is much easier to do in your

head than using the more precise values of

for a hydrogen atom diameter and

for the Sun’s diameter, to find that it would take

hydrogen atoms to stretch across the Sun’s

diameter. In addition to being easier, the rough estimate is also nearly as informative as the precise

calculation.

Known ranges of length, mass, and time
The vastness of the universe and the breadth over which physics applies are illustrated by the wide range of examples of known lengths, masses, and times (given as orders of magnitude) in Figure 1.4. Examining this table will give you a feeling for the range of possible topics in physics and numerical values. A good way to appreciate the vastness of the ranges of values in Figure 1.4 is to try to answer some simple comparative questions, such as the following:

• How many hydrogen atoms does it take to stretch across the diameter of the Sun? (Answer: 109 m/10–10 m = 1019 hydrogen atoms)
• How many protons are there in a bacterium? (Answer: 10–15 kg/10–27 kg = 1012 protons)
• How many floating-point operations can a supercomputer do in 1 day? (Answer: 105 s/10–17 s = 1022 floating-point operations)

In studying Figure 1.4, take some time to come up with similar questions that interest you and then try answering them. Doing this can breathe some life into almost any table of numbers.

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1 • Units and Measurement

Figure 1.4 This table shows the orders of magnitude of length, mass, and time.
INTERACTIVE
Visit this site (https://openstax.org/l/21scaleuniv) to explore interactively the vast range of length scales in our universe. Scroll down and up the scale to view hundreds of organisms and objects, and click on the individual objects to learn more about each one.
Building Models
How did we come to know the laws governing natural phenomena? What we refer to as the laws of nature are concise descriptions of the universe around us. They are human statements of the underlying laws or rules that all natural processes follow. Such laws are intrinsic to the universe; humans did not create them and cannot change them. We can only discover and understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle, triumph, and disappointment inherent in any creative effort (Figure 1.5). The cornerstone of discovering natural laws is observation; scientists must describe the universe as it is, not as we imagine it to be.
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1.1 • The Scope and Scale of Physics

11

Figure 1.5 (a) Enrico Fermi (1901–1954) was born in Italy. On accepting the Nobel Prize in Stockholm in 1938 for his work on artificial radioactivity produced by neutrons, he took his family to America rather than return home to the government in power at the time. He became an American citizen and was a leading participant in the Manhattan Project. (b) Marie Curie (1867–1934) sacrificed monetary assets to help finance her early research and damaged her physical well-being with radiation exposure. She is the only person to win Nobel prizes in both physics and chemistry. One of her daughters also won a Nobel Prize. (credit a: modification of work by United States Department of Energy)
A model is a representation of something that is often too difficult (or impossible) to display directly. Although a model is justified by experimental tests, it is only accurate in describing certain aspects of a physical system. An example is the Bohr model of single-electron atoms, in which the electron is pictured as orbiting the nucleus, analogous to the way planets orbit the Sun (Figure 1.6). We cannot observe electron orbits directly, but the mental image helps explain some of the observations we can make, such as the emission of light from hot gases (atomic spectra). However, other observations show that the picture in the Bohr model is not really what atoms look like. The model is “wrong,” but is still useful for some purposes. Physicists use models for a variety of purposes. For example, models can help physicists analyze a scenario and perform a calculation or models can be used to represent a situation in the form of a computer simulation. Ultimately, however, the results of these calculations and simulations need to be double-checked by other means—namely, observation and experimentation.
Figure 1.6 What is a model? The Bohr model of a single-electron atom shows the electron orbiting the nucleus in one of several possible circular orbits. Like all models, it captures some, but not all, aspects of the physical system.
The word theory means something different to scientists than what is often meant when the word is used in

12

1 • Units and Measurement

everyday conversation. In particular, to a scientist a theory is not the same as a “guess” or an “idea” or even a “hypothesis.” The phrase “it’s just a theory” seems meaningless and silly to scientists because science is founded on the notion of theories. To a scientist, a theory is a testable explanation for patterns in nature supported by scientific evidence and verified multiple times by various groups of researchers. Some theories include models to help visualize phenomena whereas others do not. Newton’s theory of gravity, for example, does not require a model or mental image, because we can observe the objects directly with our own senses. The kinetic theory of gases, on the other hand, is a model in which a gas is viewed as being composed of atoms and molecules. Atoms and molecules are too small to be observed directly with our senses—thus, we picture them mentally to understand what the instruments tell us about the behavior of gases. Although models are meant only to describe certain aspects of a physical system accurately, a theory should describe all aspects of any system that falls within its domain of applicability. In particular, any experimentally testable implication of a theory should be verified. If an experiment ever shows an implication of a theory to be false, then the theory is either thrown out or modified suitably (for example, by limiting its domain of applicability).

A law uses concise language to describe a generalized pattern in nature supported by scientific evidence and

repeated experiments. Often, a law can be expressed in the form of a single mathematical equation. Laws and

theories are similar in that they are both scientific statements that result from a tested hypothesis and are

supported by scientific evidence. However, the designation law is usually reserved for a concise and very

general statement that describes phenomena in nature, such as the law that energy is conserved during any

process, or Newton’s second law of motion, which relates force (F), mass (m), and acceleration (a) by the simple

equation

A theory, in contrast, is a less concise statement of observed behavior. For example, the

theory of evolution and the theory of relativity cannot be expressed concisely enough to be considered laws.

The biggest difference between a law and a theory is that a theory is much more complex and dynamic. A law

describes a single action whereas a theory explains an entire group of related phenomena. Less broadly

applicable statements are usually called principles (such as Pascal’s principle, which is applicable only in

fluids), but the distinction between laws and principles often is not made carefully.

The models, theories, and laws we devise sometimes imply the existence of objects or phenomena that are as yet unobserved. These predictions are remarkable triumphs and tributes to the power of science. It is the underlying order in the universe that enables scientists to make such spectacular predictions. However, if experimentation does not verify our predictions, then the theory or law is wrong, no matter how elegant or convenient it is. Laws can never be known with absolute certainty because it is impossible to perform every imaginable experiment to confirm a law for every possible scenario. Physicists operate under the assumption that all scientific laws and theories are valid until a counterexample is observed. If a good-quality, verifiable experiment contradicts a well-established law or theory, then the law or theory must be modified or overthrown completely.

The study of science in general, and physics in particular, is an adventure much like the exploration of an uncharted ocean. Discoveries are made; models, theories, and laws are formulated; and the beauty of the physical universe is made more sublime for the insights gained.
1.2 Units and Standards
Learning Objectives By the end of this section, you will be able to:
• Describe how SI base units are defined. • Describe how derived units are created from base units. • Express quantities given in SI units using metric prefixes.

As we saw previously, the range of objects and phenomena studied in physics is immense. From the incredibly short lifetime of a nucleus to the age of Earth, from the tiny sizes of subnuclear particles to the vast distance to the edges of the known universe, from the force exerted by a jumping flea to the force between Earth and the Sun, there are enough factors of 10 to challenge the imagination of even the most experienced scientist. Giving numerical values for physical quantities and equations for physical principles allows us to understand nature much more deeply than qualitative descriptions alone. To comprehend these vast ranges, we must also have accepted units in which to express them. We shall find that even in the potentially mundane discussion of

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1.2 • Units and Standards

13

meters, kilograms, and seconds, a profound simplicity of nature appears: all physical quantities can be expressed as combinations of only seven base physical quantities.
We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other measurements. For example, we might define distance and time by specifying methods for measuring them, such as using a meter stick and a stopwatch. Then, we could define average speed by stating that it is calculated as the total distance traveled divided by time of travel.
Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length of a race, which is a physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners). Without standardized units, it would be extremely difficult for scientists to express and compare measured values in a meaningful way (Figure 1.7).

Figure 1.7 Distances given in unknown units are maddeningly useless.
Two major systems of units are used in the world: SI units (for the French Système International d’Unités), also known as the metric system, and English units (also known as the customary or imperial system). English units were historically used in nations once ruled by the British Empire and are still widely used in the United States. English units may also be referred to as the foot–pound–second (fps) system, as opposed to the centimeter–gram–second (cgs) system. You may also encounter the term SAE units, named after the Society of Automotive Engineers. Products such as fasteners and automotive tools (for example, wrenches) that are measured in inches rather than metric units are referred to as SAE fasteners or SAE wrenches.
Virtually every other country in the world (except the United States) now uses SI units as the standard. The metric system is also the standard system agreed on by scientists and mathematicians.
SI Units: Base and Derived Units
In any system of units, the units for some physical quantities must be defined through a measurement process. These are called the base quantities for that system and their units are the system’s base units. All other physical quantities can then be expressed as algebraic combinations of the base quantities. Each of these physical quantities is then known as a derived quantity and each unit is called a derived unit. The choice of base quantities is somewhat arbitrary, as long as they are independent of each other and all other quantities can be derived from them. Typically, the goal is to choose physical quantities that can be measured accurately to a high precision as the base quantities. The reason for this is simple. Since the derived units can be expressed as algebraic combinations of the base units, they can only be as accurate and precise as the base units from which they are derived.
Based on such considerations, the International Standards Organization recommends using seven base quantities, which form the International System of Quantities (ISQ). These are the base quantities used to define the SI base units. Table 1.1 lists these seven ISQ base quantities and the corresponding SI base units.

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1 • Units and Measurement

ISQ Base Quantity

SI Base Unit

Length

meter (m)

Mass

kilogram (kg)

Time

second (s)

Electrical current

ampere (A)

Thermodynamic temperature kelvin (K)

Amount of substance

mole (mol)

Luminous intensity

candela (cd)

Table 1.1 ISQ Base Quantities and Their SI Units

You are probably already familiar with some derived quantities that can be formed from the base quantities in

Table 1.1. For example, the geometric concept of area is always calculated as the product of two lengths. Thus,

area is a derived quantity that can be expressed in terms of SI base units using square meters

Similarly, volume is a derived quantity that can be expressed in cubic meters

Speed is length per time; so

in terms of SI base units, we could measure it in meters per second (m/s). Volume mass density (or just

density) is mass per volume, which is expressed in terms of SI base units such as kilograms per cubic meter (kg/m3). Angles can also be thought of as derived quantities because they can be defined as the ratio of the arc

length subtended by two radii of a circle to the radius of the circle. This is how the radian is defined. Depending

on your background and interests, you may be able to come up with other derived quantities, such as the mass

flow rate (kg/s) or volume flow rate (m3/s) of a fluid, electric charge

mass flux density

and

so on. We will see many more examples throughout this text. For now, the point is that every physical quantity

can be derived from the seven base quantities in Table 1.1, and the units of every physical quantity can be

derived from the seven SI base units.

For the most part, we use SI units in this text. Non-SI units are used in a few applications in which they are in

very common use, such as the measurement of temperature in degrees Celsius

the measurement of fluid

volume in liters (L), and the measurement of energies of elementary particles in electron-volts (eV). Whenever

non-SI units are discussed, they are tied to SI units through conversions. For example, 1 L is

INTERACTIVE

Check out a comprehensive source of information on SI units (https://openstax.org/l/21SIUnits) at the National Institute of Standards and Technology (NIST) Reference on Constants, Units, and Uncertainty.

Units of Time, Length, and Mass: The Second, Meter, and Kilogram
The initial chapters in this textbook are concerned with mechanics, fluids, and waves. In these subjects all pertinent physical quantities can be expressed in terms of the base units of length, mass, and time. Therefore, we now turn to a discussion of these three base units, leaving discussion of the others until they are needed later.
The second
The SI unit for time, the second (abbreviated s), has a long history. For many years it was defined as 1/86,400 of a mean solar day. More recently, a new standard was adopted to gain greater accuracy and to define the second in terms of a nonvarying or constant physical phenomenon (because the solar day is getting longer as a result of the very gradual slowing of Earth’s rotation). Cesium atoms can be made to vibrate in a very steady

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1.2 • Units and Standards

15

way, and these vibrations can be readily observed and counted. In 1967, the second was redefined as the time required for 9,192,631,770 of these vibrations to occur (Figure 1.8). Note that this may seem like more precision than you would ever need, but it isn’t—GPSs rely on the precision of atomic clocks to be able to give you turn-by-turn directions on the surface of Earth, far from the satellites broadcasting their location.

Figure 1.8 An atomic clock such as this one uses the vibrations of cesium atoms to keep time to a precision of better than a microsecond per year. The fundamental unit of time, the second, is based on such clocks. This image looks down from the top of an atomic fountain nearly 30 feet tall. (credit: Steve Jurvetson)
The meter
The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more precise. The meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement was improved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum–iridium bar now kept near Paris. By 1960, it had become possible to define the meter even more accurately in terms of the wavelength of light, so it was again redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meter was given its current definition (in part for greater accuracy) as the distance light travels in a vacuum in 1/299,792,458 of a second (Figure 1.9). This change came after knowing the speed of light to be exactly 299,792,458 m/s. The length of the meter will change if the speed of light is someday measured with greater accuracy.
Figure 1.9 The meter is defined to be the distance light travels in 1/299,792,458 of a second in a vacuum. Distance traveled is speed multiplied by time.
The kilogram
The SI unit for mass is the kilogram (abbreviated kg); From 1795–2018 it was defined to be the mass of a platinum–iridium cylinder kept with the old meter standard at the International Bureau of Weights and Measures near Paris. However, this cylinder has lost roughly 50 micrograms since it was created. Because this is the standard, this has shifted how we defined a kilogram. Therefore, a new definition was adopted in May 2019 based on the Planck constant and other constants which will never change in value. We will study Planck’s constant in quantum mechanics, which is an area of physics that describes how the smallest pieces of the universe work. The kilogram is measured on a Kibble balance (see Figure 1.10). When a weight is placed on a Kibble balance, an electrical current is produced that is proportional to Planck’s constant. Since Planck’s constant is defined, the exact current measurements in the balance define the kilogram.

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1 • Units and Measurement

Figure 1.10 Redefining the SI unit of mass. The U.S. National Institute of Standards and Technology’s Kibble balance is a machine that balances the weight of a test mass with the resulting electrical current needed for a force to balance it.
Metric Prefixes
SI units are part of the metric system, which is convenient for scientific and engineering calculations because the units are categorized by factors of 10. Table 1.2 lists the metric prefixes and symbols used to denote various factors of 10 in SI units. For example, a centimeter is one-hundredth of a meter (in symbols, 1 cm = 10–2 m) and a kilometer is a thousand meters (1 km = 103 m). Similarly, a megagram is a million grams (1 Mg = 106 g), a nanosecond is a billionth of a second (1 ns = 10–9 s), and a terameter is a trillion meters (1 Tm = 1012 m).

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1.2 • Units and Standards

17

Prefix Symbol Meaning Prefix Symbol Meaning

yotta- Y

1024

yocto- y

10–24

zetta- Z

1021

zepto- z

10–21

exa- E

1018

atto- a

10–18

peta- P

1015

femto- f

10–15

tera- T

1012

pico- p

10–12

giga- G

109

nano- n

10–9

mega- M

106

micro-

10–6

kilo- k

103

milli- m

10–3

hecto- h

102

centi- c

10–2

deka- da

101

deci- d

10–1

Table 1.2 Metric Prefixes for Powers of 10 and Their Symbols

The only rule when using metric prefixes is that you cannot “double them up.” For example, if you have measurements in petameters (1 Pm = 1015 m), it is not proper to talk about megagigameters, although
In practice, the only time this becomes a bit confusing is when discussing masses. As we have seen, the base SI unit of mass is the kilogram (kg), but metric prefixes need to be applied to the gram (g), because we are not allowed to “double-up” prefixes. Thus, a thousand kilograms (103 kg) is written as a megagram (1 Mg) since

Incidentally, 103 kg is also called a metric ton, abbreviated t. This is one of the units outside the SI system considered acceptable for use with SI units.
As we see in the next section, metric systems have the advantage that conversions of units involve only powers of 10. There are 100 cm in 1 m, 1000 m in 1 km, and so on. In nonmetric systems, such as the English system of units, the relationships are not as simple—there are 12 in. in 1 ft, 5280 ft in 1 mi, and so on.
Another advantage of metric systems is that the same unit can be used over extremely large ranges of values simply by scaling it with an appropriate metric prefix. The prefix is chosen by the order of magnitude of physical quantities commonly found in the task at hand. For example, distances in meters are suitable in construction, whereas distances in kilometers are appropriate for air travel, and nanometers are convenient in optical design. With the metric system there is no need to invent new units for particular applications. Instead, we rescale the units with which we are already familiar.

EXAMPLE 1.1
Using Metric Prefixes
Restate the mass one but less than 1000.

using a metric prefix such that the resulting numerical value is bigger than

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1 • Units and Measurement

Strategy Since we are not allowed to “double-up” prefixes, we first need to restate the mass in grams by replacing the prefix symbol k with a factor of 103 (see Table 1.2). Then, we should see which two prefixes in Table 1.2 are closest to the resulting power of 10 when the number is written in scientific notation. We use whichever of these two prefixes gives us a number between one and 1000.
Solution Replacing the k in kilogram with a factor of 103, we find that

From Table 1.2, we see that 1016 is between “peta-” (1015) and “exa-” (1018). If we use the “peta-” prefix, then

we find that

since

Alternatively, if we use the “exa-” prefix we find

that

since

Because the problem asks for the numerical value

between one and 1000, we use the “peta-” prefix and the answer is 19.3 Pg.

Significance

It is easy to make silly arithmetic errors when switching from one prefix to another, so it is always a good idea to check that our final answer matches the number we started with. An easy way to do this is to put both numbers in scientific notation and count powers of 10, including the ones hidden in prefixes. If we did not

make a mistake, the powers of 10 should match up. In this problem, we started with

so we have

13 + 3 = 16 powers of 10. Our final answer in scientific notation is powers of 10. So, everything checks out.

Pg, so we have 1 + 15 = 16

If this mass arose from a calculation, we would also want to check to determine whether a mass this large makes any sense in the context of the problem. For this, Figure 1.4 might be helpful.

CHECK YOUR UNDERSTANDING 1.1

Restate 1000.

using a metric prefix such that the resulting number is bigger than one but less than

1.3 Unit Conversion
Learning Objectives By the end of this section, you will be able to:
• Use conversion factors to express the value of a given quantity in different units.
It is often necessary to convert from one unit to another. For example, if you are reading a European cookbook, some quantities may be expressed in units of liters and you need to convert them to cups. Or perhaps you are reading walking directions from one location to another and you are interested in how many miles you will be walking. In this case, you may need to convert units of feet or meters to miles.
Let’s consider a simple example of how to convert units. Suppose we want to convert 80 m to kilometers. The first thing to do is to list the units you have and the units to which you want to convert. In this case, we have units in meters and we want to convert to kilometers. Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio that expresses how many of one unit are equal to another unit. For example, there are 12 in. in 1 ft, 1609 m in 1 mi, 100 cm in 1 m, 60 s in 1 min, and so on. Refer to Appendix B for a more complete list of conversion factors. In this case, we know that there are 1000 m in 1 km. Now we can set up our unit conversion. We write the units we have and then multiply them by the conversion factor so the units cancel out, as shown:

Note that the unwanted meter unit cancels, leaving only the desired kilometer unit. You can use this method to
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1.3 • Unit Conversion

19

convert between any type of unit. Now, the conversion of 80 m to kilometers is simply the use of a metric prefix, as we saw in the preceding section, so we can get the same answer just as easily by noting that

since “kilo-” means 103 (see Table 1.2) and

However, using conversion factors is handy when

converting between units that are not metric or when converting between derived units, as the following

examples illustrate.

EXAMPLE 1.2
Converting Nonmetric Units to Metric
The distance from the university to home is 10 mi and it usually takes 20 min to drive this distance. Calculate the average speed in meters per second (m/s). (Note: Average speed is distance traveled divided by time of travel.)
Strategy First we calculate the average speed using the given units, then we can get the average speed into the desired units by picking the correct conversion factors and multiplying by them. The correct conversion factors are those that cancel the unwanted units and leave the desired units in their place. In this case, we want to convert miles to meters, so we need to know the fact that there are 1609 m in 1 mi. We also want to convert minutes to seconds, so we use the conversion of 60 s in 1 min.
Solution
1. Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now. Average speed and other motion concepts are covered in later chapters.) In equation form,

2. Substitute the given values for distance and time:

3. Convert miles per minute to meters per second by multiplying by the conversion factor that cancels miles and leave meters, and also by the conversion factor that cancels minutes and leave seconds:

Significance Check the answer in the following ways:
1. Be sure the units in the unit conversion cancel correctly. If the unit conversion factor was written upside down, the units do not cancel correctly in the equation. We see the “miles” in the numerator in 0.50 mi/ min cancels the “mile” in the denominator in the first conversion factor. Also, the “min” in the denominator in 0.50 mi/min cancels the “min” in the numerator in the second conversion factor.
2. Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of meters per second and, after the cancellations, the only units left are a meter (m) in the numerator and a second (s) in the denominator, so we have indeed obtained these units.

CHECK YOUR UNDERSTANDING 1.2
Light travels about 9 Pm in a year. Given that a year is about second?

what is the speed of light in meters per

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1 • Units and Measurement

EXAMPLE 1.3

Converting between Metric Units

The density of iron is

under standard conditions. Convert this to kg/m3.

Strategy

We need to convert grams to kilograms and cubic centimeters to cubic meters. The conversion factors we need

are

and

However, we are dealing with cubic centimeters

so we have to use the second conversion factor three times (that is, we need to cube it). The idea is still to multiply by the conversion factors in such a way that they cancel the units we want to get rid of and introduce the units we want to keep.

Solution

Significance Remember, it’s always important to check the answer.
1. Be sure to cancel the units in the unit conversion correctly. We see that the gram (“g”) in the numerator in 7.86 g/cm3 cancels the “g” in the denominator in the first conversion factor. Also, the three factors of “cm” in the denominator in 7.86 g/cm3 cancel with the three factors of “cm” in the numerator that we get by cubing the second conversion factor.
2. Check that the units of the final answer are the desired units. The problem asked for us to convert to kilograms per cubic meter. After the cancellations just described, we see the only units we have left are “kg” in the numerator and three factors of “m” in the denominator (that is, one factor of “m” cubed, or “m3”). Therefore, the units on the final answer are correct.
CHECK YOUR UNDERSTANDING 1.3
We know from Figure 1.4 that the diameter of Earth is on the order of 107 m, so the order of magnitude of its surface area is 1014 m2. What is that in square kilometers (that is, km2)? (Try doing this both by converting 107 m to km and then squaring it and then by converting 1014 m2 directly to square kilometers. You should get the same answer both ways.)
Unit conversions may not seem very interesting, but not doing them can be costly. One famous example of this situation was seen with the Mars Climate Orbiter. This probe was launched by NASA on December 11, 1998. On September 23, 1999, while attempting to guide the probe into its planned orbit around Mars, NASA lost contact with it. Subsequent investigations showed a piece of software called SM_FORCES (or “small forces”) was recording thruster performance data in the English units of pound-seconds (lb-s). However, other pieces of software that used these values for course corrections expected them to be recorded in the SI units of newton-seconds (N-s), as dictated in the software interface protocols. This error caused the probe to follow a very different trajectory from what NASA thought it was following, which most likely caused the probe either to burn up in the Martian atmosphere or to shoot out into space. This failure to pay attention to unit conversions cost hundreds of millions of dollars, not to mention all the time invested by the scientists and engineers who worked on the project.
CHECK YOUR UNDERSTANDING 1.4
Given that 1 lb (pound) is 4.45 N, were the numbers being output by SM_FORCES too big or too small?

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1.4 • Dimensional Analysis

21

1.4 Dimensional Analysis
Learning Objectives By the end of this section, you will be able to:
• Find the dimensions of a mathematical expression involving physical quantities. • Determine whether an equation involving physical quantities is dimensionally consistent.

The dimension of any physical quantity expresses its dependence on the base quantities as a product of

symbols (or powers of symbols) representing the base quantities. Table 1.3 lists the base quantities and the symbols used for their dimension. For example, a measurement of length is said to have dimension L or L1, a measurement of mass has dimension M or M1, and a measurement of time has dimension T or T1. Like units, dimensions obey the rules of algebra. Thus, area is the product of two lengths and so has dimension L2, or length squared. Similarly, volume is the product of three lengths and has dimension L3, or length cubed. Speed has dimension length over time, L/T or LT–1. Volumetric mass density has dimension M/L3 or ML–3, or mass

over length cubed. In general, the dimension of any physical quantity can be written as

for

some powers

and g. We can write the dimensions of a length in this form with

and the

remaining six powers all set equal to zero:

Any quantity with a dimension that can

be written so that all seven powers are zero (that is, its dimension is

) is called

dimensionless (or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists

often call dimensionless quantities pure numbers.

Base Quantity

Symbol for Dimension

Length

L

Mass

M

Time

T

Current

I

Thermodynamic temperature Θ

Amount of substance

N

Luminous intensity

J

Table 1.3 Base Quantities and Their Dimensions

Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of

that quantity. For example, if is the radius of a cylinder and is its height, then we write

and

to indicate the dimensions of the radius and height are both those of length, or L. Similarly, if we use the symbol for the surface area of a cylinder and for its volume, then [A] = L2 and [V] = L3. If we use the symbol

for the mass of the cylinder and for the density of the material from which the cylinder is made, then

and

The importance of the concept of dimension arises from the fact that any mathematical equation relating physical quantities must be dimensionally consistent, which means the equation must obey the following rules:

• Every term in an expression must have the same dimensions; it does not make sense to add or subtract quantities of differing dimension (think of the old saying: “You can’t add apples and oranges”). In particular, the expressions on each side of the equality in an equation must have the same dimensions.
• The arguments of any of the standard mathematical functions such as trigonometric functions (such as

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1 • Units and Measurement

sine and cosine), logarithms, or exponential functions that appear in the equation must be dimensionless. These functions require pure numbers as inputs and give pure numbers as outputs.
If either of these rules is violated, an equation is not dimensionally consistent and cannot possibly be a correct statement of physical law. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of physics, and even to suggest the form that new laws of physics might take. This last use of dimensions is beyond the scope of this text, but is something you will undoubtedly learn later in your academic career.
EXAMPLE 1.4
Using Dimensions to Remember an Equation
Suppose we need the formula for the area of a circle for some computation. Like many people who learned geometry too long ago to recall with any certainty, two expressions may pop into our mind when we think of circles: and One expression is the circumference of a circle of radius r and the other is its area. But which is which?
Strategy One natural strategy is to look it up, but this could take time to find information from a reputable source. Besides, even if we think the source is reputable, we shouldn’t trust everything we read. It is nice to have a way to double-check just by thinking about it. Also, we might be in a situation in which we cannot look things up (such as during a test). Thus, the strategy is to find the dimensions of both expressions by making use of the fact that dimensions follow the rules of algebra. If either expression does not have the same dimensions as area, then it cannot possibly be the correct equation for the area of a circle.
Solution We know the dimension of area is L2. Now, the dimension of the expression is

since the constant is a pure number and the radius is a length. Therefore, Similarly, the dimension of the expression is

has the dimension of area.

since the constants and are both dimensionless and the radius is a length. We see that dimension of length, which means it cannot possibly be an area.

has the

We rule out because it is not dimensionally consistent with being an area. We see that is dimensionally consistent with being an area, so if we have to choose between these two expressions, is the one to choose.

Significance

This may seem like kind of a silly example, but the ideas are very general. As long as we know the dimensions of the individual physical quantities that appear in an equation, we can check to see whether the equation is dimensionally consistent. On the other hand, knowing that true equations are dimensionally consistent, we can match expressions from our imperfect memories to the quantities for which they might be expressions. Doing this will not help us remember dimensionless factors that appear in the equations (for example, if you

had accidentally conflated the two expressions from the example into help), but it does help us remember the correct basic form of equations.

then dimensional analysis is no

CHECK YOUR UNDERSTANDING 1.5

Suppose we want the formula for the volume of a sphere. The two expressions commonly mentioned in

elementary discussions of spheres are

and

One is the volume of a sphere of radius r and the other

is its surface area. Which one is the volume?

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1.4 • Dimensional Analysis

23

EXAMPLE 1.5

Checking Equations for Dimensional Consistency

Consider the physical quantities

and with dimensions

Determine whether each of the following equations is dimensionally consistent: (a)

and (c)

and (b)

Strategy
By the definition of dimensional consistency, we need to check that each term in a given equation has the same dimensions as the other terms in that equation and that the arguments of any standard mathematical functions are dimensionless.

Solution

a. There are no trigonometric, logarithmic, or exponential functions to worry about in this equation, so we need only look at the dimensions of each term appearing in the equation. There are three terms, one in the left expression and two in the expression on the right, so we look at each in turn:

All three terms have the same dimension, so this equation is dimensionally consistent. b. Again, there are no trigonometric, exponential, or logarithmic functions, so we only need to look at the
dimensions of each of the three terms appearing in the equation:

None of the three terms has the same dimension as any other, so this is about as far from being dimensionally consistent as you can get. The technical term for an equation like this is nonsense. c. This equation has a trigonometric function in it, so first we should check that the argument of the sine function is dimensionless:
The argument is dimensionless. So far, so good. Now we need to check the dimensions of each of the two terms (that is, the left expression and the right expression) in the equation:
The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation is another example of “nonsense.” Significance If we are trusting people, these types of dimensional checks might seem unnecessary. But, rest assured, any textbook on a quantitative subject such as physics (including this one) almost certainly contains some equations with typos. Checking equations routinely by dimensional analysis save us the embarrassment of using an incorrect equation. Also, checking the dimensions of an equation we obtain through algebraic manipulation is a great way to make sure we did not make a mistake (or to spot a mistake, if we made one).
CHECK YOUR UNDERSTANDING 1.6
Is the equation v = at dimensionally consistent?

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1 • Units and Measurement

One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t, we have that the dimension of the derivative of v with respect to t is just the ratio of the dimension of v over that of t:

Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t:

By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.
1.5 Estimates and Fermi Calculations
Learning Objectives By the end of this section, you will be able to:
• Estimate the values of physical quantities.
On many occasions, physicists, other scientists, and engineers need to make estimates for a particular quantity. Other terms sometimes used are guesstimates, order-of-magnitude approximations, back-of-theenvelope calculations, or Fermi calculations. (The physicist Enrico Fermi mentioned earlier was famous for his ability to estimate various kinds of data with surprising precision.) Will that piece of equipment fit in the back of the car or do we need to rent a truck? How long will this download take? About how large a current will there be in this circuit when it is turned on? How many houses could a proposed power plant actually power if it is built? Note that estimating does not mean guessing a number or a formula at random. Rather, estimation means using prior experience and sound physical reasoning to arrive at a rough idea of a quantity’s value. Because the process of determining a reliable approximation usually involves the identification of correct physical principles and a good guess about the relevant variables, estimating is very useful in developing physical intuition. Estimates also allow us to perform “sanity checks” on calculations or policy proposals by helping us rule out certain scenarios or unrealistic numbers. They allow us to challenge others (as well as ourselves) in our efforts to learn truths about the world.
Many estimates are based on formulas in which the input quantities are known only to a limited precision. As you develop physics problem-solving skills (which are applicable to a wide variety of fields), you also will develop skills at estimating. You develop these skills by thinking more quantitatively and by being willing to take risks. As with any skill, experience helps. Familiarity with dimensions (see Table 1.3) and units (see Table 1.1 and Table 1.2), and the scales of base quantities (see Figure 1.4) also helps.
To make some progress in estimating, you need to have some definite ideas about how variables may be related. The following strategies may help you in practicing the art of estimation:
• Get big lengths from smaller lengths. When estimating lengths, remember that anything can be a ruler. Thus, imagine breaking a big thing into smaller things, estimate the length of one of the smaller things, and multiply to get the length of the big thing. For example, to estimate the height of a building, first count how many floors it has. Then, estimate how big a single floor is by imagining how many people would have to stand on each other’s shoulders to reach the ceiling. Last, estimate the height of a person. The product of these three estimates is your estimate of the height of the building. It helps to have memorized a few length scales relevant to the sorts of problems you find yourself solving. For example, knowing some of the length scales in Figure 1.4 might come in handy. Sometimes it also helps to do this in reverse—that is, to estimate the length of a small thing, imagine a bunch of them making up a bigger thing. For example, to estimate the thickness of a sheet of paper, estimate the thickness of a stack of paper and then divide by the
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1.5 • Estimates and Fermi Calculations

25

number of pages in the stack. These same strategies of breaking big things into smaller things or aggregating smaller things into a bigger thing can sometimes be used to estimate other physical quantities, such as masses and times. • Get areas and volumes from lengths. When dealing with an area or a volume of a complex object, introduce a simple model of the object such as a sphere or a box. Then, estimate the linear dimensions (such as the radius of the sphere or the length, width, and height of the box) first, and use your estimates to obtain the volume or area from standard geometric formulas. If you happen to have an estimate of an object’s area or volume, you can also do the reverse; that is, use standard geometric formulas to get an estimate of its linear dimensions. • Get masses from volumes and densities. When estimating masses of objects, it can help first to estimate its volume and then to estimate its mass from a rough estimate of its average density (recall, density has dimension mass over length cubed, so mass is density times volume). For this, it helps to remember that the density of air is around 1 kg/m3, the density of water is 103 kg/m3, and the densest everyday solids max out at around 104 kg/m3. Asking yourself whether an object floats or sinks in either air or water gets you a ballpark estimate of its density. You can also do this the other way around; if you have an estimate of an object’s mass and its density, you can use them to get an estimate of its volume. • If all else fails, bound it. For physical quantities for which you do not have a lot of intuition, sometimes the best you can do is think something like: Well, it must be bigger than this and smaller than that. For example, suppose you need to estimate the mass of a moose. Maybe you have a lot of experience with moose and know their average mass offhand. If so, great. But for most people, the best they can do is to think something like: It must be bigger than a person (of order 102 kg) and less than a car (of order 103 kg). If you need a single number for a subsequent calculation, you can take the geometric mean of the upper and lower bound—that is, you multiply them together and then take the square root. For the moose mass example, this would be

The tighter the bounds, the better. Also, no rules are unbreakable when it comes to estimation. If you think the value of the quantity is likely to be closer to the upper bound than the lower bound, then you may want to bump up your estimate from the geometric mean by an order or two of magnitude. • One “sig. fig.” is fine. There is no need to go beyond one significant figure when doing calculations to obtain an estimate. In most cases, the order of magnitude is good enough. The goal is just to get in the ballpark figure, so keep the arithmetic as simple as possible. • Ask yourself: Does this make any sense? Last, check to see whether your answer is reasonable. How does it compare with the values of other quantities with the same dimensions that you already know or can look up easily? If you get some wacky answer (for example, if you estimate the mass of the Atlantic Ocean to be bigger than the mass of Earth, or some time span to be longer than the age of the universe), first check to see whether your units are correct. Then, check for arithmetic errors. Then, rethink the logic you used to arrive at your answer. If everything checks out, you may have just proved that some slick new idea is actually bogus.

EXAMPLE 1.6

Mass of Earth’s Oceans
Estimate the total mass of the oceans on Earth.

Strategy

We know the density of water is about 103 kg/m3, so we start with the advice to “get masses from densities and

volumes.” Thus, we need to estimate the volume of the planet’s oceans. Using the advice to “get areas and

volumes from lengths,” we can estimate the volume of the oceans as surface area times average depth, or V =

AD. We know the diameter of Earth from Figure 1.4 and we know that most of Earth’s surface is covered in

water, so we can estimate the surface area of the oceans as being roughly equal to the surface area of the

planet. By following the advice to “get areas and volumes from lengths” again, we can approximate Earth as a

sphere and use the formula for the surface area of a sphere of diameter d—that is,

to estimate the

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1 • Units and Measurement

surface area of the oceans. Now we just need to estimate the average depth of the oceans. For this, we use the advice: “If all else fails, bound it.” We happen to know the deepest points in the ocean are around 10 km and that it is not uncommon for the ocean to be deeper than 1 km, so we take the average depth to be around

fine.”

Now we just need to put it all together, heeding the advice that “one ‘sig. fig.’ is

Solution We estimate the surface area of Earth (and hence the surface area of Earth’s oceans) to be roughly

Next, using our average depth estimate of volume of Earth’s oceans to be

which was obtained by bounding, we estimate the

Last, we estimate the mass of the world’s oceans to be

Thus, we estimate that the order of magnitude of the mass of the planet’s oceans is 1021 kg.

Significance

To verify our answer to the best of our ability, we first need to answer the question: Does this make any sense? From Figure 1.4, we see the mass of Earth’s atmosphere is on the order of 1019 kg and the mass of Earth is on the order of 1025 kg. It is reassuring that our estimate of 1021 kg for the mass of Earth’s oceans falls somewhere

between these two. So, yes, it does seem to make sense. It just so happens that we did a search on the Web for

“mass of oceans” and the top search results all said

which is the same order of magnitude as

our estimate. Now, rather than having to trust blindly whoever first put that number up on a website (most of

the other sites probably just copied it from them, after all), we can have a little more confidence in it.

CHECK YOUR UNDERSTANDING 1.7
Figure 1.4 says the mass of the atmosphere is 1019 kg. Assuming the density of the atmosphere is 1 kg/m3, estimate the height of Earth’s atmosphere. Do you think your answer is an underestimate or an overestimate? Explain why.

How many piano tuners are there in New York City? How many leaves are on that tree? If you are studying photosynthesis or thinking of writing a smartphone app for piano tuners, then the answers to these questions might be of great interest to you. Otherwise, you probably couldn’t care less what the answers are. However, these are exactly the sorts of estimation problems that people in various tech industries have been asking potential employees to evaluate their quantitative reasoning skills. If building physical intuition and evaluating quantitative claims do not seem like sufficient reasons for you to practice estimation problems, how about the fact that being good at them just might land you a high-paying job?
INTERACTIVE
For practice estimating relative lengths, areas, and volumes, check out this PhET (https://openstax.org/l/ 21lengthgame) simulation, titled “Estimation.”

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1.6 • Significant Figures

27

1.6 Significant Figures
Learning Objectives By the end of this section, you will be able to:
• Determine the correct number of significant figures for the result of a computation. • Describe the relationship between the concepts of accuracy, precision, uncertainty, and discrepancy. • Calculate the percent uncertainty of a measurement, given its value and its uncertainty. • Determine the uncertainty of the result of a computation involving quantities with given uncertainties.
Figure 1.11 shows two instruments used to measure the mass of an object. The digital scale has mostly replaced the double-pan balance in physics labs because it gives more accurate and precise measurements. But what exactly do we mean by accurate and precise? Aren’t they the same thing? In this section we examine in detail the process of making and reporting a measurement.

Figure 1.11 (a) A double-pan mechanical balance is used to compare different masses. Usually an object with unknown mass is placed in one pan and objects of known mass are placed in the other pan. When the bar that connects the two pans is horizontal, then the masses in both pans are equal. The “known masses” are typically metal cylinders of standard mass such as 1 g, 10 g, and 100 g. (b) Many mechanical balances, such as double-pan balances, have been replaced by digital scales, which can typically measure the mass of an object more precisely. A mechanical balance may read only the mass of an object to the nearest tenth of a gram, but many digital scales can measure the mass of an object up to the nearest thousandth of a gram. (credit a: modification of work by Serge Melki; credit b: modification of work by Karel Jakubec)
Accuracy and Precision of a Measurement
Science is based on observation and experiment—that is, on measurements. Accuracy is how close a measurement is to the accepted reference value for that measurement. For example, let’s say we want to measure the length of standard printer paper. The packaging in which we purchased the paper states that it is 11.0 in. long. We then measure the length of the paper three times and obtain the following measurements: 11.1 in., 11.2 in., and 10.9 in. These measurements are quite accurate because they are very close to the reference value of 11.0 in. In contrast, if we had obtained a measurement of 12 in., our measurement would not be very accurate. Notice that the concept of accuracy requires that an accepted reference value be given.
The precision of measurements refers to how close the agreement is between repeated independent measurements (which are repeated under the same conditions). Consider the example of the paper measurements. The precision of the measurements refers to the spread of the measured values. One way to analyze the precision of the measurements is to determine the range, or difference, between the lowest and the highest measured values. In this case, the lowest value was 10.9 in. and the highest value was 11.2 in. Thus, the measured values deviated from each other by, at most, 0.3 in. These measurements were relatively precise because they did not vary too much in value. However, if the measured values had been 10.9 in., 11.1 in., and 11.9 in., then the measurements would not be very precise because there would be significant variation from one measurement to another. Notice that the concept of precision depends only on the actual measurements

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1 • Units and Measurement

acquired and does not depend on an accepted reference value.
The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let’s consider an example of a GPS attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull’seye target and think of each GPS attempt to locate the restaurant as a black dot. In Figure 1.12(a), we see the GPS measurements are spread out far apart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target. This indicates a low-precision, high-accuracy measuring system. However, in Figure 1.12(b), the GPS measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high-precision, low-accuracy measuring system.

Figure 1.12 A GPS attempts to locate a restaurant at the center of the bull’s-eye. The black dots represent each attempt to pinpoint the location of the restaurant. (a) The dots are spread out quite far apart from one another, indicating low precision, but they are each rather close to the actual location of the restaurant, indicating high accuracy. (b) The dots are concentrated rather closely to one another, indicating high precision, but they are rather far away from the actual location of the restaurant, indicating low accuracy. (credit a and credit b: modification of works by "DarkEvil"/Wikimedia Commons)
Accuracy, Precision, Uncertainty, and Discrepancy
The precision of a measuring system is related to the uncertainty in the measurements whereas the accuracy is related to the discrepancy from the accepted reference value. Uncertainty is a quantitative measure of how much your measured values deviate from one another. There are many different methods of calculating uncertainty, each of which is appropriate to different situations. Some examples include taking the range (that is, the biggest less the smallest) or finding the standard deviation of the measurements. Discrepancy (or “measurement error”) is the difference between the measured value and a given standard or expected value. If the measurements are not very precise, then the uncertainty of the values is high. If the measurements are not very accurate, then the discrepancy of the values is high.
Recall our example of measuring paper length; we obtained measurements of 11.1 in., 11.2 in., and 10.9 in., and the accepted value was 11.0 in. We might average the three measurements to say our best guess is 11.1 in.; in this case, our discrepancy is 11.1 – 11.0 = 0.1 in., which provides a quantitative measure of accuracy. We might calculate the uncertainty in our best guess by using half of the range of our measured values: 0.15 in. Then we would say the length of the paper is 11.1 in. plus or minus 0.15 in. The uncertainty in a measurement, A, is often denoted as δA (read “delta A”), so the measurement result would be recorded as A ± δA. Returning to our paper example, the measured length of the paper could be expressed as 11.1 ± 0.15 in. Since the discrepancy of 0.1 in. is less than the uncertainty of 0.15 in., we might say the measured value agrees with the accepted reference value to within experimental uncertainty.
Some factors that contribute to uncertainty in a measurement include the following:
• Limitations of the measuring device • The skill of the person taking the measurement • Irregularities in the object being measured • Any other factors that affect the outcome (highly dependent on the situation)
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1.6 • Significant Figures

29

In our example, such factors contributing to the uncertainty could be the smallest division on the ruler is 1/16 in., the person using the ruler has bad eyesight, the ruler is worn down on one end, or one side of the paper is slightly longer than the other. At any rate, the uncertainty in a measurement must be calculated to quantify its precision. If a reference value is known, it makes sense to calculate the discrepancy as well to quantify its accuracy.
Percent uncertainty
Another method of expressing uncertainty is as a percent of the measured value. If a measurement A is expressed with uncertainty δA, the percent uncertainty is defined as

EXAMPLE 1.7

Calculating Percent Uncertainty: A Bag of Apples
A grocery store sells 5-lb bags of apples. Let’s say we purchase four bags during the course of a month and weigh the bags each time. We obtain the following measurements:

• Week 1 weight: 4.8 lb • Week 2 weight: 5.3 lb • Week 3 weight: 4.9 lb • Week 4 weight: 5.4 lb

We then determine the average weight of the 5-lb bag of apples is 5.1 ± 0.3 lb from using half of the range. What is the percent uncertainty of the bag’s weight?

Strategy First, observe that the average value of the bag’s weight, A, is 5.1 lb. The uncertainty in this value, We can use the following equation to determine the percent uncertainty of the weight:

is 0.3 lb.

1.1

Solution Substitute the values into the equation:

Significance

We can conclude the average weight of a bag of apples from this store is 5.1 lb ± 6%. Notice the percent

uncertainty is dimensionless because the units of weight in

lb canceled those in A = 5.1 lb when we

took the ratio.

CHECK YOUR UNDERSTANDING 1.8
A high school track coach has just purchased a new stopwatch. The stopwatch manual states the stopwatch has an uncertainty of ±0.05 s. Runners on the track coach’s team regularly clock 100-m sprints of 11.49 s to 15.01 s. At the school’s last track meet, the first-place sprinter came in at 12.04 s and the second-place sprinter came in at 12.07 s. Will the coach’s new stopwatch be helpful in timing the sprint team? Why or why not?

Uncertainties in calculations
Uncertainty exists in anything calculated from measured quantities. For example, the area of a floor calculated from measurements of its length and width has an uncertainty because the length and width have

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1 • Units and Measurement

uncertainties. How big is the uncertainty in something you calculate by multiplication or division? If the

measurements going into the calculation have small uncertainties (a few percent or less), then the method of

adding percents can be used for multiplication or division. This method states the percent uncertainty in a

quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to

make the calculation. For example, if a floor has a length of 4.00 m and a width of 3.00 m, with uncertainties of

2% and 1%, respectively, then the area of the floor is 12.0 m2 and has an uncertainty of 3%. (Expressed as an

area, this is 0.36 m2 [

], which we round to 0.4 m2 since the area of the floor is given to a tenth of

a square meter.)

Precision of Measuring Tools and Significant Figures

An important factor in the precision of measurements involves the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, a standard ruler can measure length to the nearest millimeter whereas a caliper can measure length to the nearest 0.01 mm. The caliper is a more precise measuring tool because it can measure extremely small differences in length. The more precise the measuring tool, the more precise the measurements.

When we express measured values, we can only list as many digits as we measured initially with our measuring tool. For example, if we use a standard ruler to measure the length of a stick, we may measure it to be 36.7 cm. We can’t express this value as 36.71 cm because our measuring tool is not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement. For example, the person measuring the length of a stick with a ruler notices the stick length seems to be somewhere in between 36.6 cm and 36.7 cm, and he or she must estimate the value of the last digit. Using the method of significant figures, the rule is that the last digit written down in a measurement is the first digit with some uncertainty. To determine the number of significant digits in a value, start with the first measured value at the left and count the number of digits through the last digit written on the right. For example, the measured value 36.7 cm has three digits, or three significant figures. Significant figures indicate the precision of the measuring tool used to measure a value.

Zeros

Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant

because they are placeholders that locate the decimal point. There are two significant figures in 0.053. The

zeros in 10.053 are not placeholders; they are significant. This number has five significant figures. The zeros in

1300 may or may not be significant, depending on the style of writing numbers. They could mean the number

is known to the last digit or they could be placeholders. So 1300 could have two, three, or four significant

figures. To avoid this ambiguity, we should write 1300 in scientific notation as

or

depending on whether it has two, three, or four significant figures. Zeros are significant except

when they serve only as placeholders.

Significant figures in calculations
When combining measurements with different degrees of precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least-precise measured value. There are two different rules, one for multiplication and division and the other for addition and subtraction.

1. For multiplication and division, the result should have the same number of significant figures as the quantity with the least number of significant figures entering into the calculation. For example, the area of a circle can be calculated from its radius using A = πr2. Let’s see how many significant figures the area has if the radius has only two—say, r = 1.2 m. Using a calculator with an eight-digit output, we would calculate

But because the radius has only two significant figures, it limits the calculated quantity to two significant figures, or

although π is good to at least eight digits. 2. For addition and subtraction, the answer can contain no more decimal places than the least-precise
measurement. Suppose we buy 7.56 kg of potatoes in a grocery store as measured with a scale with

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1.7 • Solving Problems in Physics

31

precision 0.01 kg, then we drop off 6.052 kg of potatoes at your laboratory as measured by a scale with precision 0.001 kg. Then, we go home and add 13.7 kg of potatoes as measured by a bathroom scale with precision 0.1 kg. How many kilograms of potatoes do we now have and how many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction:

Next, we identify the least-precise measurement: 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer is rounded to the tenths place, giving us 15.2 kg.
Significant figures in this text
In this text, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significant figures are used in all worked examples. An answer given to three digits is based on input good to at least three digits, for example. If the input has fewer significant figures, the answer will also have fewer significant figures. Care is also taken that the number of significant figures is reasonable for the situation posed. In some topics, particularly in optics, more accurate numbers are needed and we use more than three significant figures. Finally, if a number is exact, such as the two in the formula for the circumference of a circle, C = 2πr, it does not affect the number of significant figures in a calculation. Likewise, conversion factors such as 100 cm/1 m are considered exact and do not affect the number of significant figures in a calculation.
1.7 Solving Problems in Physics
Learning Objectives By the end of this section, you will be able to:
• Describe the process for developing a problem-solving strategy. • Explain how to find the numerical solution to a problem. • Summarize the process for assessing the significance of the numerical solution to a problem.

Figure 1.13 Problem-solving skills are essential to your success in physics. (credit: “scui3asteveo”/Flickr)

32

1 • Units and Measurement

Problem-solving skills are clearly essential to success in a quantitative course in physics. More important, the ability to apply broad physical principles—usually represented by equations—to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday life.
As you are probably well aware, a certain amount of creativity and insight is required to solve problems. No rigid procedure works every time. Creativity and insight grow with experience. With practice, the basics of problem solving become almost automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and then progressing to the more difficult. After you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.
Although there is no simple step-by-step method that works for every problem, the following three-stage process facilitates problem solving and makes it more meaningful. The three stages are strategy, solution, and significance. This process is used in examples throughout the book. Here, we look at each stage of the process in turn.
Strategy
Strategy is the beginning stage of solving a problem. The idea is to figure out exactly what the problem is and then develop a strategy for solving it. Some general advice for this stage is as follows:
• Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch at the outset. You often need to decide which direction is positive and note that on your sketch. When you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless.
• Make a list of what is given or can be inferred from the problem as stated (identify the “knowns”). Many problems are stated very succinctly and require some inspection to determine what is known. Drawing a sketch can be very useful at this point as well. Formally identifying the knowns is of particular importance in applying physics to real-world situations. For example, the word stopped means the velocity is zero at that instant. Also, we can often take initial time and position as zero by the appropriate choice of coordinate system.
• Identify exactly what needs to be determined in the problem (identify the unknowns). In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help identify the unknowns.
• Determine which physical principles can help you solve the problem. Since physical principles tend to be expressed in the form of mathematical equations, a list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown—that is, all the other variables are known—so you can solve for the unknown easily. If the equation contains more than one unknown, then additional equations are needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer.
Solution
The solution stage is when you do the math. Substitute the knowns (along with their units) into the appropriate equation and obtain numerical solutions complete with units. That is, do the algebra, calculus, geometry, or arithmetic necessary to find the unknown from the knowns, being sure to carry the units through the calculations. This step is clearly important because it produces the numerical answer, along with its units. Notice, however, that this stage is only one-third of the overall problem-solving process.

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1.7 • Solving Problems in Physics

33

Significance
After having done the math in the solution stage of problem solving, it is tempting to think you are done. But, always remember that physics is not math. Rather, in doing physics, we use mathematics as a tool to help us understand nature. So, after you obtain a numerical answer, you should always assess its significance:
• Check your units. If the units of the answer are incorrect, then an error has been made and you should go back over your previous steps to find it. One way to find the mistake is to check all the equations you derived for dimensional consistency. However, be warned that correct units do not guarantee the numerical part of the answer is also correct.
• Check the answer to see whether it is reasonable. Does it make sense? This step is extremely important: –the goal of physics is to describe nature accurately. To determine whether the answer is reasonable, check both its magnitude and its sign, in addition to its units. The magnitude should be consistent with a rough estimate of what it should be. It should also compare reasonably with magnitudes of other quantities of the same type. The sign usually tells you about direction and should be consistent with your prior expectations. Your judgment will improve as you solve more physics problems, and it will become possible for you to make finer judgments regarding whether nature is described adequately by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to solve a problem mechanically.
• Check to see whether the answer tells you something interesting. What does it mean? This is the flip side of the question: Does it make sense? Ultimately, physics is about understanding nature, and we solve physics problems to learn a little something about how nature operates. Therefore, assuming the answer does make sense, you should always take a moment to see if it tells you something about the world that you find interesting. Even if the answer to this particular problem is not very interesting to you, what about the method you used to solve it? Could the method be adapted to answer a question that you do find interesting? In many ways, it is in answering questions such as these that science progresses.

34 1 • Chapter Review

CHAPTER REVIEW

Key Terms

accuracy the degree to which a measured value

agrees with an accepted reference value for that

measurement

base quantity physical quantity chosen by

convention and practical considerations such

that all other physical quantities can be

expressed as algebraic combinations of them

base unit standard for expressing the

measurement of a base quantity within a

particular system of units; defined by a particular

procedure used to measure the corresponding

base quantity

conversion factor a ratio that expresses how many

of one unit are equal to another unit

derived quantity physical quantity defined using

algebraic combinations of base quantities

derived units units that can be calculated using

algebraic combinations of the fundamental units

dimension expression of the dependence of a

physical quantity on the base quantities as a

product of powers of symbols representing the

base quantities; in general, the dimension of a

quantity has the form

for some

powers a, b, c, d, e, f, and g.

dimensionally consistent equation in which every

term has the same dimensions and the

arguments of any mathematical functions

appearing in the equation are dimensionless

dimensionless quantity with a dimension of

also called quantity of

dimension 1 or a pure number

discrepancy the difference between the measured

value and a given standard or expected value

English units system of measurement used in the

United States; includes units of measure such as

feet, gallons, and pounds

estimation using prior experience and sound

physical reasoning to arrive at a rough idea of a

quantity’s value; sometimes called an “order-of-

magnitude approximation,” a “guesstimate,” a

“back-of-the-envelope calculation”, or a “Fermi

calculation”

Key Equations

Percent uncertainty

kilogram SI unit for mass, abbreviated kg law description, using concise language or a
mathematical formula, of a generalized pattern in nature supported by scientific evidence and repeated experiments meter SI unit for length, abbreviated m method of adding percents the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to make the calculation. metric system system in which values can be calculated in factors of 10 model representation of something often too difficult (or impossible) to display directly order of magnitude the size of a quantity as it relates to a power of 10 percent uncertainty the ratio of the uncertainty of a measurement to the measured value, expressed as a percentage physical quantity characteristic or property of an object that can be measured or calculated from other measurements physics science concerned with describing the interactions of energy, matter, space, and time; especially interested in what fundamental mechanisms underlie every phenomenon precision the degree to which repeated measurements agree with each other second the SI unit for time, abbreviated s SI units the international system of units that scientists in most countries have agreed to use; includes units such as meters, liters, and grams significant figures used to express the precision of a measuring tool used to measure a value theory testable explanation for patterns in nature supported by scientific evidence and verified multiple times by various groups of researchers uncertainty a quantitative measure of how much measured values deviate from one another units standards used for expressing and comparing measurements

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1 • Chapter Review 35

Summary
1.1 The Scope and Scale of Physics
• Physics is about trying to find the simple laws that describe all natural phenomena.
• Physics operates on a vast range of scales of length, mass, and time. Scientists use the concept of the order of magnitude of a number to track which phenomena occur on which scales. They also use orders of magnitude to compare the various scales.
• Scientists attempt to describe the world by formulating models, theories, and laws.
1.2 Units and Standards
• Systems of units are built up from a small number of base units, which are defined by accurate and precise measurements of conventionally chosen base quantities. Other units are then derived as algebraic combinations of the base units.
• Two commonly used systems of units are English units and SI units. All scientists and most of the other people in the world use SI, whereas nonscientists in the United States still tend to use English units.
• The SI base units of length, mass, and time are the meter (m), kilogram (kg), and second (s), respectively.
• SI units are a metric system of units, meaning values can be calculated by factors of 10. Metric prefixes may be used with metric units to scale the base units to sizes appropriate for almost any application.
1.3 Unit Conversion
• To convert a quantity from one unit to another, multiply by conversions factors in such a way that you cancel the units you want to get rid of and introduce the units you want to end up with.
• Be careful with areas and volumes. Units obey the rules of algebra so, for example, if a unit is squared we need two factors to cancel it.
1.4 Dimensional Analysis
• The dimension of a physical quantity is just an expression of the base quantities from which it is derived.
• All equations expressing physical laws or principles must be dimensionally consistent. This fact can be used as an aid in remembering physical laws, as a way to check whether claimed relationships between physical

quantities are possible, and even to derive new physical laws.
1.5 Estimates and Fermi Calculations
• An estimate is a rough educated guess at the value of a physical quantity based on prior experience and sound physical reasoning. Some strategies that may help when making an estimate are as follows: ◦ Get big lengths from smaller lengths. ◦ Get areas and volumes from lengths. ◦ Get masses from volumes and densities. ◦ If all else fails, bound it. ◦ One “sig. fig.” is fine. ◦ Ask yourself: Does this make any sense?
1.6 Significant Figures
• Accuracy of a measured value refers to how close a measurement is to an accepted reference value. The discrepancy in a measurement is the amount by which the measurement result differs from this value.
• Precision of measured values refers to how close the agreement is between repeated measurements. The uncertainty of a measurement is a quantification of this.
• The precision of a measuring tool is related to the size of its measurement increments. The smaller the measurement increment, the more precise the tool.
• Significant figures express the precision of a measuring tool.
• When multiplying or dividing measured values, the final answer can contain only as many significant figures as the value with the least number of significant figures.
• When adding or subtracting measured values, the final answer cannot contain more decimal places than the least-precise value.
1.7 Solving Problems in Physics
The three stages of the process for solving physics problems used in this book are as follows:
• Strategy: Determine which physical principles are involved and develop a strategy for using them to solve the problem.
• Solution: Do the math necessary to obtain a numerical solution complete with units.
• Significance: Check the solution to make sure it makes sense (correct units, reasonable magnitude and sign) and assess its significance.

36 1 • Chapter Review

Conceptual Questions
1.1 The Scope and Scale of Physics
1. What is physics? 2. Some have described physics as a “search for
simplicity.” Explain why this might be an appropriate description. 3. If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assuming both use accepted rules of logic)? 4. What determines the validity of a theory? 5. Certain criteria must be satisfied if a measurement or observation is to be believed. Will the criteria necessarily be as strict for an expected result as for an unexpected result? 6. Can the validity of a model be limited or must it be universally valid? How does this compare with the required validity of a theory or a law?

1.2 Units and Standards
7. Identify some advantages of metric units. 8. What are the SI base units of length, mass, and
time? 9. What is the difference between a base unit and a
derived unit? (b) What is the difference between a base quantity and a derived quantity? (c) What is the difference between a base quantity and a base unit?

Problems

1.1 The Scope and Scale of Physics

14. Find the order of magnitude of the following

physical quantities. (a) The mass of Earth’s

atmosphere:

(b) The mass of the

Moon’s atmosphere: 25,000 kg; (c) The mass of

Earth’s hydrosphere:

(d) The

mass of Earth:

(e) The mass of

the Moon:

(f) The Earth–Moon

distance (semimajor axis):

(g)

The mean Earth–Sun distance:

(h) The equatorial radius of Earth:

(i) The mass of an electron:

(j) The mass of a proton:

(k) The mass of the Sun:

15. Use the orders of magnitude you found in the previous problem to answer the following questions to within an order of magnitude. (a) How many electrons would it take to equal the mass of a proton? (b) How many Earths would it

10. For each of the following scenarios, refer to Figure 1.4 and Table 1.2 to determine which metric prefix on the meter is most appropriate for each of the following scenarios. (a) You want to tabulate the mean distance from the Sun for each planet in the solar system. (b) You want to compare the sizes of some common viruses to design a mechanical filter capable of blocking the pathogenic ones. (c) You want to list the diameters of all the elements on the periodic table. (d) You want to list the distances to all the stars that have now received any radio broadcasts sent from Earth 10 years ago.
1.6 Significant Figures
11. (a) What is the relationship between the precision and the uncertainty of a measurement? (b) What is the relationship between the accuracy and the discrepancy of a measurement?
1.7 Solving Problems in Physics
12. What information do you need to choose which equation or equations to use to solve a problem?
13. What should you do after obtaining a numerical answer when solving a problem?
take to equal the mass of the Sun? (c) How many Earth–Moon distances would it take to cover the distance from Earth to the Sun? (d) How many Moon atmospheres would it take to equal the mass of Earth’s atmosphere? (e) How many moons would it take to equal the mass of Earth? (f) How many protons would it take to equal the mass of the Sun? For the remaining questions, you need to use Figure 1.4 to obtain the necessary orders of magnitude of lengths, masses, and times.
16. Roughly how many heartbeats are there in a lifetime?
17. A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0?
18. Roughly how many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human?
19. Calculate the approximate number of atoms in a

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1 • Chapter Review 37

bacterium. Assume the average mass of an atom in the bacterium is 10 times the mass of a proton. 20. (a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is 10 times the mass of a bacterium. (b) Making the same assumption, how many cells are there in a human? 21. Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second? 22. About how many floating-point operations can a supercomputer perform each year? 23. Roughly how many floating-point operations can a supercomputer perform in a human lifetime?

1.2 Units and Standards
24. The following times are given using metric prefixes on the base SI unit of time: the second. Rewrite them in scientific notation without the prefix. For example, 47 Ts would be rewritten as (a) 980 Ps; (b) 980 fs; (c) 17 ns; (d)

25. The following times are given in seconds. Use

metric prefixes to rewrite them so the

numerical value is greater than one but less

than 1000. For example,

could be

written as either 7.9 cs or 79 ms. (a)

(b) 0.045 s; (c)

(d)

26. The following lengths are given using metric

prefixes on the base SI unit of length: the meter.

Rewrite them in scientific notation without the

prefix. For example, 4.2 Pm would be rewritten

as

(a) 89 Tm; (b) 89 pm; (c) 711

mm; (d)

27. The following lengths are given in meters. Use

metric prefixes to rewrite them so the

numerical value is bigger than one but less than

1000. For example,

could be

written either as 7.9 cm or 79 mm. (a)

(b) 0.0074 m; (c)

(d)

28. The following masses are written using metric

prefixes on the gram. Rewrite them in scientific

notation in terms of the SI base unit of mass: the

kilogram. For example, 40 Mg would be written

as

(a) 23 mg; (b) 320 Tg; (c) 42 ng;

(d) 7 g; (e) 9 Pg.

29. The following masses are given in kilograms.

Use metric prefixes on the gram to rewrite them

so the numerical value is bigger than one but

less than 1000. For example,

could

be written as 70 cg or 700 mg. (a)

(b)

(c)

(d)

(e)

1.3 Unit Conversion

30. The volume of Earth is on the order of 1021 m3. (a) What is this in cubic kilometers (km3)? (b) What is it in cubic miles (mi3)? (c) What is it in cubic centimeters (cm3)?

31. The speed limit on some interstate highways is

roughly 100 km/h. (a) What is this in meters per

second? (b) How many miles per hour is this?

32. A car is traveling at a speed of 33 m/s. (a) What

is its speed in kilometers per hour? (b) Is it

exceeding the 90 km/h speed limit?

33. In SI units, speeds are measured in meters per

second (m/s). But, depending on where you live,

you’re probably more comfortable of thinking of

speeds in terms of either kilometers per hour

(km/h) or miles per hour (mi/h). In this

problem, you will see that 1 m/s is roughly 4

km/h or 2 mi/h, which is handy to use when

developing your physical intuition. More

precisely, show that (a)

and

(b)

34. American football is played on a 100-yd-long

field, excluding the end zones. How long is the

field in meters? (Assume that 1 m = 3.281 ft.)

35. Soccer fields vary in size. A large soccer field is

115 m long and 85.0 m wide. What is its area in

square feet? (Assume that 1 m = 3.281 ft.)

36. What is the height in meters of a person who is

6 ft 1.0 in. tall?

37. Mount Everest, at 29,028 ft, is the tallest

mountain on Earth. What is its height in

kilometers? (Assume that 1 m = 3.281 ft.)

38. The speed of sound is measured to be 342 m/s

on a certain day. What is this measurement in

kilometers per hour?

39. Tectonic plates are large segments of Earth’s

crust that move slowly. Suppose one such plate

has an average speed of 4.0 cm/yr. (a) What

distance does it move in 1.0 s at this speed? (b)

What is its speed in kilometers per million

years?

40. The average distance between Earth and the

Sun is

(a) Calculate the average

speed of Earth in its orbit (assumed to be

circular) in meters per second. (b) What is this

38 1 • Chapter Review

speed in miles per hour? 41. The density of nuclear matter is about 1018 kg/
m3. Given that 1 mL is equal in volume to cm3,

what is the density of nuclear matter in

megagrams per microliter (that is,

)?

42. The density of aluminum is 2.7 g/cm3. What is

the density in kilograms per cubic meter?

43. A commonly used unit of mass in the English

system is the pound-mass, abbreviated lbm,

where 1 lbm = 0.454 kg. What is the density of

water in pound-mass per cubic foot?

44. A furlong is 220 yd. A fortnight is 2 weeks.

Convert a speed of one furlong per fortnight to

millimeters per second.

45. It takes radians (rad) to get around a circle,

which is the same as 360°. How many radians

are in 1°?

46. Light travels a distance of about

A

light-minute is the distance light travels in 1

min. If the Sun is

from Earth, how

far away is it in light-minutes?

47. A light-nanosecond is the distance light travels

in 1 ns. Convert 1 ft to light-nanoseconds.

48. An electron has a mass of

A

proton has a mass of

What is

the mass of a proton in electron-masses?

49. A fluid ounce is about 30 mL. What is the

volume of a 12 fl-oz can of soda pop in cubic

meters?

1.4 Dimensional Analysis

50. A student is trying to remember some formulas

from geometry. In what follows, assume is

area, is volume, and all other variables are

lengths. Determine which formulas are

dimensionally consistent. (a)

(b)

(c)

(d)

(e)

51. Consider the physical quantities s, v, a, and t

with dimensions

and

Determine whether

each of the following equations is dimensionally

consistent. (a)

(b)

(c)

(d)

52. Consider the physical quantities

and

with dimensions [m] = M, [s] = L, [v] = LT–1, [a] =

LT–2, and [t] = T. Assuming each of the following

equations is dimensionally consistent, find the

dimension of the quantity on the left-hand side of the equation: (a) F = ma; (b) K = 0.5mv2; (c) p

= mv; (d) W = mas; (e) L = mvr.

53. Suppose quantity is a length and quantity is a

time. Suppose the quantities and are defined by v = ds/dt and a = dv/dt. (a) What is the dimension of v? (b) What is the dimension of the quantity a? What are the dimensions of (c)

(d)

and (e) da/dt?

54. Suppose [V] = L3,

and [t] = T. (a)

What is the dimension of

(b) What is

the dimension of dV/dt? (c) What is the

dimension of

55. The arc length formula says the length of arc
subtended by angle Ɵ in a circle of radius is

given by the equation

Ɵ What are the

dimensions of (a) s, (b) r, and (c) Ɵ

1.5 Estimates and Fermi Calculations
56. Assuming the human body is made primarily of water, estimate the volume of a person.
57. Assuming the human body is primarily made of water, estimate the number of molecules in it. (Note that water has a molecular mass of 18 g/ mol and there are roughly 1024 atoms in a mole.)
58. Estimate the mass of air in a classroom. 59. Estimate the number of molecules that make up
Earth, assuming an average molecular mass of 30 g/mol. (Note there are on the order of 1024 objects per mole.) 60. Estimate the surface area of a person. 61. Roughly how many solar systems would it take to tile the disk of the Milky Way? 62. (a) Estimate the density of the Moon. (b) Estimate the diameter of the Moon. (c) Given that the Moon subtends at an angle of about half a degree in the sky, estimate its distance from Earth. 63. The average density of the Sun is on the order 103 kg/m3. (a) Estimate the diameter of the Sun. (b) Given that the Sun subtends at an angle of about half a degree in the sky, estimate its distance from Earth. 64. Estimate the mass of a virus. 65. A floating-point operation is a single arithmetic operation such as addition, subtraction, multiplication, or division. (a) Estimate the maximum number of floating-point operations a human being could possibly perform in a lifetime. (b) How long would it take a supercomputer to perform that many floatingpoint operations?

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1 • Chapter Review 39

1.6 Significant Figures

66. Consider the equation 4000/400 = 10.0.

Assuming the number of significant figures in

the answer is correct, what can you say about

the number of significant figures in 4000 and

400?

67. Suppose your bathroom scale reads your mass

as 65 kg with a 3% uncertainty. What is the

uncertainty in your mass (in kilograms)?

68. A good-quality measuring tape can be off by

0.50 cm over a distance of 20 m. What is its

percent uncertainty?

69. An infant’s pulse rate is measured to be 130 ± 5

beats/min. What is the percent uncertainty in

this measurement?

70. (a) Suppose that a person has an average heart

rate of 72.0 beats/min. How many beats does he

or she have in 2.0 years? (b) In 2.00 years? (c) In

2.000 years?

71. A can contains 375 mL of soda. How much is left

after 308 mL is removed?

72. State how many significant figures are proper in

the results of the following calculations: (a)

(b)

(c)

73. (a) How many significant figures are in the numbers 99 and 100.? (b) If the uncertainty in each number is 1, what is the percent uncertainty in each? (c) Which is a more meaningful way to express the accuracy of these two numbers: significant figures or percent uncertainties?

74. (a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h, what is the percent uncertainty? (b) If it has the same percent uncertainty when it reads 60 km/h, what is the range of speeds you could be going?
75. (a) A person’s blood pressure is measured to be What is its percent
uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mm Hg? 76. A person measures his or her heart rate by counting the number of beats in 30 s. If 40 ± 1 beats are counted in 30.0 ± 0.5 s, what is the heart rate and its uncertainty in beats per minute? 77. What is the area of a circle 3.102 cm in diameter? 78. Determine the number of significant figures in the following measurements: (a) 0.0009, (b) 15,450.0, (c) 6×103, (d) 87.990, and (e) 30.42. 79. Perform the following calculations and express your answer using the correct number of significant digits. (a) A woman has two bags weighing 13.5 lb and one bag with a weight of 10.2 lb. What is the total weight of the bags? (b) The force F on an object is equal to its mass m multiplied by its acceleration a. If a wagon with mass 55 kg accelerates at a rate of 0.0255 m/s2, what is the force on the wagon? (The unit of force is called the newton and it is expressed with the symbol N.)

Additional Problems
80. Consider the equation y = mt +b, where the dimension of y is length and the dimension of t is time, and m and b are constants. What are the dimensions and SI units of (a) m and (b) b?
81. Consider the equation
where s is a length and t is a time. What are the dimensions and SI units of (a) (b) (c) (d)
(e) and (f) c? 82. (a) A car speedometer has a 5% uncertainty.
What is the range of possible speeds when it reads 90 km/h? (b) Convert this range to miles per hour. Note 1 km = 0.6214 mi.

83. A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the percent uncertainty in the elapsed time. (c) What is the average speed in meters per second? (d) What is the uncertainty in the average speed?
84. The sides of a small rectangular box are measured to be 1.80 ± 0.1 cm, 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm long. Calculate its volume and uncertainty in cubic centimeters.

40 1 • Chapter Review

85. When nonmetric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was used, where 1 lbm = 0.4539 kg. (a) If there is an uncertainty of 0.0001 kg in the pound-mass unit, what is its percent uncertainty? (b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?

86. The length and width of a rectangular room are measured to be 3.955 ± 0.005 m and 3.050 ± 0.005 m. Calculate the area of the room and its uncertainty in square meters.
87. A car engine moves a piston with a circular cross-section of 7.500 ± 0.002 cm in diameter a distance of 3.250 ± 0.001 cm to compress the gas in the cylinder. (a) By what amount is the gas decreased in volume in cubic centimeters? (b) Find the uncertainty in this volume.

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1 • Chapter Review 41

Challenge Problems

88. The first atomic bomb was detonated on July 16,

1945, at the Trinity test site about 200 mi south

of Los Alamos. In 1947, the U.S. government

declassified a film reel of the explosion. From

this film reel, British physicist G. I. Taylor was

able to determine the rate at which the radius of

the fireball from the blast grew. Using

dimensional analysis, he was then able to

deduce the amount of energy released in the

explosion, which was a closely guarded secret at

the time. Because of this, Taylor did not publish

his results until 1950. This problem challenges

you to recreate this famous calculation. (a)

Using keen physical insight developed from

years of experience, Taylor decided the radius r

of the fireball should depend only on time since

the explosion, t, the density of the air, and the

energy of the initial explosion, E. Thus, he made

the educated guess that

for some

dimensionless constant k and some unknown exponents a, b, and c. Given that [E] = ML2T–2,

determine the values of the exponents

necessary to make this equation dimensionally

consistent. (Hint: Notice the equation implies

that

and that

) (b) By

analyzing data from high-energy conventional

explosives, Taylor found the formula he derived

seemed to be valid as long as the constant k had

the value 1.03. From the film reel, he was able to

determine many values of r and the

corresponding values of t. For example, he

found that after 25.0 ms, the fireball had a

radius of 130.0 m. Use these values, along with an average air density of 1.25 kg/m3, to

calculate the initial energy release of the Trinity

detonation in joules (J). (Hint: To get energy in

joules, you need to make sure all the numbers

you substitute in are expressed in terms of SI

base units.) (c) The energy released in large

explosions is often cited in units of “tons of

TNT” (abbreviated “t TNT”), where 1 t TNT is

about 4.2 GJ. Convert your answer to (b) into

kilotons of TNT (that is, kt TNT). Compare your

answer with the quick-and-dirty estimate of 10

kt TNT made by physicist Enrico Fermi shortly

after witnessing the explosion from what was

thought to be a safe distance. (Reportedly, Fermi

made his estimate by dropping some shredded

bits of paper right before the remnants of the

shock wave hit him and looked to see how far

they were carried by it.)

89. The purpose of this problem is to show the entire concept of dimensional consistency can be summarized by the old saying “You can’t add apples and oranges.” If you have studied power series expansions in a calculus course, you know the standard mathematical functions such as trigonometric functions, logarithms, and exponential functions can be expressed as infinite sums of the form ∞
where the are dimensionless constants for all and x is the argument of the
function. (If you have not studied power series in calculus yet, just trust us.) Use this fact to explain why the requirement that all terms in an equation have the same dimensions is sufficient as a definition of dimensional consistency. That is, it actually implies the arguments of standard mathematical functions must be dimensionless, so it is not really necessary to make this latter condition a separate requirement of the definition of dimensional consistency as we have done in this section.

42 1 • Chapter Review Access for free at openstax.org.

CHAPTER 2
Vectors
Figure 2.1 A signpost gives information about distances and directions to towns or to other locations relative to the location of the signpost. Distance is a scalar quantity. Knowing the distance alone is not enough to get to the town; we must also know the direction from the signpost to the town. The direction, together with the distance, is a vector quantity commonly called the displacement vector. A signpost, therefore, gives information about displacement vectors from the signpost to towns. (credit: modification of work by "studio tdes"/Flickr, thedailyenglishshow.com)
Chapter Outline
2.1 Scalars and Vectors 2.2 Coordinate Systems and Components of a Vector 2.3 Algebra of Vectors 2.4 Products of Vectors INTRODUCTION Vectors are essential to physics and engineering. Many fundamental physical quantities are vectors, including displacement, velocity, force, and electric and magnetic vector fields. Scalar products of vectors define other fundamental scalar physical quantities, such as energy. Vector products of vectors define still other fundamental vector physical quantities, such as torque and angular momentum. In other words, vectors are a component part of physics in much the same way as sentences are a component part of literature.

44

2 • Vectors

In introductory physics, vectors are Euclidean quantities that have geometric representations as arrows in one dimension (in a line), in two dimensions (in a plane), or in three dimensions (in space). They can be added, subtracted, or multiplied. In this chapter, we explore elements of vector algebra for applications in mechanics and in electricity and magnetism. Vector operations also have numerous generalizations in other branches of physics.
2.1 Scalars and Vectors
Learning Objectives By the end of this section, you will be able to:
• Describe the difference between vector and scalar quantities. • Identify the magnitude and direction of a vector. • Explain the effect of multiplying a vector quantity by a scalar. • Describe how one-dimensional vector quantities are added or subtracted. • Explain the geometric construction for the addition or subtraction of vectors in a plane. • Distinguish between a vector equation and a scalar equation.

Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100 m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of “number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.

Scalar quantities that have the same physical units can be added or subtracted according to the usual rules of

algebra for numbers. For example, a class ending 10 min earlier than 50 min lasts

.

Similarly, a 60-cal serving of corn followed by a 200-cal serving of donuts gives

of

energy. When we multiply a scalar quantity by a number, we obtain the same scalar quantity but with a larger

(or smaller) value. For example, if yesterday’s breakfast had 200 cal of energy and today’s breakfast has four

times as much energy as it had yesterday, then today’s breakfast has

of energy. Two scalar

quantities can also be multiplied or divided by each other to form a derived scalar quantity. For example, if a

train covers a distance of 100 km in 1.0 h, its speed is 100.0 km/1.0 h = 27.8 m/s, where the speed is a derived

scalar quantity obtained by dividing distance by time.

Many physical quantities, however, cannot be described completely by just a single number of physical units. For example, when the U.S. Coast Guard dispatches a ship or a helicopter for a rescue mission, the rescue team must know not only the distance to the distress signal, but also the direction from which the signal is coming so they can get to its origin as quickly as possible. Physical quantities specified completely by giving a number of units (magnitude) and a direction are called vector quantities. Examples of vector quantities include displacement, velocity, position, force, and torque. In the language of mathematics, physical vector quantities are represented by mathematical objects called vectors (Figure 2.2). We can add or subtract two vectors, and we can multiply a vector by a scalar or by another vector, but we cannot divide by a vector. The operation of division by a vector is not defined.

Figure 2.2 We draw a vector from the initial point or origin (called the “tail” of a vector) to the end or terminal point (called the “head” of a vector), marked by an arrowhead. Magnitude is the length of a vector and is always a positive scalar quantity. (credit "photo": modification of work by Cate Sevilla)
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2.1 • Scalars and Vectors

45

Let’s examine vector algebra using a graphical method to be aware of basic terms and to develop a qualitative understanding. In practice, however, when it comes to solving physics problems, we use analytical methods, which we’ll see in the next section. Analytical methods are more simple computationally and more accurate than graphical methods. From now on, to distinguish between a vector and a scalar quantity, we adopt the common convention that a letter in bold type with an arrow above it denotes a vector, and a letter without an arrow denotes a scalar. For example, a distance of 2.0 km, which is a scalar quantity, is denoted by d = 2.0 km,
whereas a displacement of 2.0 km in some direction, which is a vector quantity, is denoted by .

Suppose you tell a friend on a camping trip that you have discovered a terrific fishing hole 6 km from your tent. It is unlikely your friend would be able to find the hole easily unless you also communicate the direction in which it can be found with respect to your campsite. You may say, for example, “Walk about 6 km northeast from my tent.” The key concept here is that you have to give not one but two pieces of information—namely, the distance or magnitude (6 km) and the direction (northeast).

Displacement is a general term used to describe a change in position, such as during a trip from the tent to the fishing hole. Displacement is an example of a vector quantity. If you walk from the tent (location A) to the hole

(location B), as shown in Figure 2.3, the vector , representing your displacement, is drawn as the arrow that originates at point A and ends at point B. The arrowhead marks the end of the vector. The direction of the

displacement vector is the direction of the arrow. The length of the arrow represents the magnitude D of

vector . Here, D = 6 km. Since the magnitude of a vector is its length, which is a positive number, the magnitude is also indicated by placing the absolute value notation around the symbol that denotes the vector;

so, we can write equivalently that

. To solve a vector problem graphically, we need to draw the vector

to scale. For example, if we assume 1 unit of distance (1 km) is represented in the drawing by a line segment of

length u = 2 cm, then the total displacement in this example is represented by a vector of length

, as shown in Figure 2.4. Notice that here, to avoid confusion, we used

to

denote the magnitude of the actual displacement and d = 12 cm to denote the length of its representation in

the drawing.

Figure 2.3 The displacement vector from point A (the initial position at the campsite) to point B (the final position at the fishing hole) is indicated by an arrow with origin at point A and end at point B. The displacement is the same for any of the actual paths (dashed curves) that may be taken between points A and B.

46

2 • Vectors

Figure 2.4 A displacement of magnitude 6 km is drawn to scale as a vector of length 12 cm when the length of 2 cm represents 1 unit of displacement (which in this case is 1 km).

Suppose your friend walks from the campsite at A to the fishing pond at B and then walks back: from the

fishing pond at B to the campsite at A. The magnitude of the displacement vector

from A to B is the same

as the magnitude of the displacement vector

from B to A (it equals 6 km in both cases), so we can write

. However, vector

is not equal to vector

because these two vectors have different

directions:

. In Figure 2.3, vector

would be represented by a vector with an origin at point B

and an end at point A, indicating vector

points to the southwest, which is exactly

opposite to the

direction of vector . We say that vector

is antiparallel to vector

and write

, where

the minus sign indicates the antiparallel direction.

Two vectors that have identical directions are said to be parallel vectors—meaning, they are parallel to each

other. Two parallel vectors and are equal, denoted by

, if and only if they have equal magnitudes

. Two vectors with directions perpendicular to each other are said to be orthogonal vectors. These

relations between vectors are illustrated in Figure 2.5.

Figure 2.5 Various relations between two vectors and . (a)

because

. (c)

because they have different directions (even though

. (b) . (d)

because they are not parallel and because they are parallel and have

identical magnitudes A = B. (e) —meaning, they are orthogonal.

because they have different directions (are not parallel); here, their directions differ by

CHECK YOUR UNDERSTANDING 2.1

Two motorboats named Alice and Bob are moving on a lake. Given the information about their velocity vectors

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2.1 • Scalars and Vectors

47

in each of the following situations, indicate whether their velocity vectors are equal or otherwise. (a) Alice moves north at 6 knots and Bob moves west at 6 knots. (b) Alice moves west at 6 knots and Bob moves west at 3 knots. (c) Alice moves northeast at 6 knots and Bob moves south at 3 knots. (d) Alice moves northeast at 6 knots and Bob moves southwest at 6 knots. (e) Alice moves northeast at 2 knots and Bob moves closer to the shore northeast at 2 knots.
Algebra of Vectors in One Dimension
Vectors can be multiplied by scalars, added to other vectors, or subtracted from other vectors. We can illustrate these vector concepts using an example of the fishing trip seen in Figure 2.6.

Figure 2.6 Displacement vectors for a fishing trip. (a) Stopping to rest at point C while walking from camp (point A) to the pond (point B). (b) Going back for the dropped tackle box (point D). (c) Finishing up at the fishing pond.

Suppose your friend departs from point A (the campsite) and walks in the direction to point B (the fishing pond), but, along the way, stops to rest at some point C located three-quarters of the distance between A and B,

beginning from point A (Figure 2.6(a)). What is his displacement vector

when he reaches point C? We

know that if he walks all the way to B, his displacement vector relative to A is , which has magnitude

and a direction of northeast. If he walks only a 0.75 fraction of the total distance, maintaining the

northeasterly direction, at point C he must be

away from the campsite at A. So, his

displacement vector at the rest point C has magnitude

and is parallel to the

displacement vector . All of this can be stated succinctly in the form of the following vector equation:

In a vector equation, both sides of the equation are vectors. The previous equation is an example of a vector

multiplied by a positive scalar (number)

. The result, , of such a multiplication is a new vector

with a direction parallel to the direction of the original vector .

In general, when a vector is multiplied by a positive scalar , the result is a new vector that is parallel to :

2.1

The magnitude of this new vector is obtained by multiplying the magnitude expressed by the scalar equation:

of the original vector, as

2.2

In a scalar equation, both sides of the equation are numbers. Equation 2.2 is a scalar equation because the

48

2 • Vectors

magnitudes of vectors are scalar quantities (and positive numbers). If the scalar is negative in the vector equation Equation 2.1, then the magnitude of the new vector is still given by Equation 2.2, but the direction

of the new vector is antiparallel to the direction of . These principles are illustrated in Figure 2.7(a) by two

examples where the length of vector is 1.5 units. When

, the new vector

has length

(twice as long as the original vector) and is parallel to the original vector. When

, the

new vector

has length

antiparallel to the original vector.

(twice as long as the original vector) and is

Figure 2.7 Algebra of vectors in one dimension. (a) Multiplication by a scalar. (b) Addition of two vectors vectors and . (c) Subtraction of two vectors is the difference of vectors and .

is called the resultant of

Now suppose your fishing buddy departs from point A (the campsite), walking in the direction to point B (the fishing hole), but he realizes he lost his tackle box when he stopped to rest at point C (located three-quarters of the distance between A and B, beginning from point A). So, he turns back and retraces his steps in the direction toward the campsite and finds the box lying on the path at some point D only 1.2 km away from point

C (see Figure 2.6(b)). What is his displacement vector

when he finds the box at point D? What is his

displacement vector

from point D to the hole? We have already established that at rest point C his

displacement vector is

. Starting at point C, he walks southwest (toward the campsite), which

means his new displacement vector

from point C to point D is antiparallel to . Its magnitude

is

, so his second displacement vector is

. His total displacement

relative to the campsite is the vector sum of the two displacement vectors: vector

(from the

campsite to the rest point) and vector

(from the rest point to the point where he finds his box):

2.3

The vector sum of two (or more) vectors is called the resultant vector or, for short, the resultant. When the

vectors on the right-hand-side of Equation 2.3 are known, we can find the resultant

as follows:

2.4

When your friend finally reaches the pond at B, his displacement vector

from point A is the vector sum of

his displacement vector

from point A to point D and his displacement vector

from point D to the

fishing hole:

(see Figure 2.6(c)). This means his displacement vector

is the

difference of two vectors:

2.5

Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second

term in Equation 2.5 is vector

(which is antiparallel to

. When we substitute Equation 2.4 into

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2.1 • Scalars and Vectors

49

Equation 2.5, we obtain the second displacement vector:

2.6

This result means your friend walked finds his tackle box to the fishing hole.

from the point where he

When vectors and lie along a line (that is, in one dimension), such as in the camping example, their

resultant

and their difference

both lie along the same direction. We can illustrate the

addition or subtraction of vectors by drawing the corresponding vectors to scale in one dimension, as shown in

Figure 2.7.

To illustrate the resultant when and are two parallel vectors, we draw them along one line by placing the origin of one vector at the end of the other vector in head-to-tail fashion (see Figure 2.7(b)). The magnitude of this resultant is the sum of their magnitudes: R = A + B. The direction of the resultant is parallel to both vectors. When vector is antiparallel to vector , we draw them along one line in either head-to-head fashion (Figure 2.7(c)) or tail-to-tail fashion. The magnitude of the vector difference, then, is the absolute value
of the difference of their magnitudes. The direction of the difference vector is parallel to the direction of the longer vector.
In general, in one dimension—as well as in higher dimensions, such as in a plane or in space—we can add any number of vectors and we can do so in any order because the addition of vectors is commutative,

2.7

and associative,

2.8

Moreover, multiplication by a scalar is distributive:

2.9

We used the distributive property in Equation 2.4 and Equation 2.6.

When adding many vectors in one dimension, it is convenient to use the concept of a unit vector. A unit vector,

which is denoted by a letter symbol with a hat, such as , has a magnitude of one and does not have any

physical unit so that

. The only role of a unit vector is to specify direction. For example, instead of

saying vector

has a magnitude of 6.0 km and a direction of northeast, we can introduce a unit vector

that points to the northeast and say succinctly that

. Then the southwesterly direction is

simply given by the unit vector . In this way, the displacement of 6.0 km in the southwesterly direction is

expressed by the vector

EXAMPLE 2.1
A Ladybug Walker
A long measuring stick rests against a wall in a physics laboratory with its 200-cm end at the floor. A ladybug lands on the 100-cm mark and crawls randomly along the stick. It first walks 15 cm toward the floor, then it walks 56 cm toward the wall, then it walks 3 cm toward the floor again. Then, after a brief stop, it continues for 25 cm toward the floor and then, again, it crawls up 19 cm toward the wall before coming to a complete rest (Figure 2.8). Find the vector of its total displacement and its final resting position on the stick.

50

2 • Vectors

Strategy If we choose the direction along the stick toward the floor as the direction of unit vector , then the direction toward the floor is and the direction toward the wall is . The ladybug makes a total of five displacements:

The total displacement is the resultant of all its displacement vectors.

Figure 2.8 Five displacements of the ladybug. Note that in this schematic drawing, magnitudes of displacements are not drawn to scale. (credit "ladybug": modification of work by “Persian Poet Gal”/Wikimedia Commons)
Solution The resultant of all the displacement vectors is
In this calculation, we use the distributive law given by Equation 2.9. The result reads that the total displacement vector points away from the 100-cm mark (initial landing site) toward the end of the meter stick that touches the wall. The end that touches the wall is marked 0 cm, so the final position of the ladybug is at the (100 – 32)cm = 68-cm mark.
CHECK YOUR UNDERSTANDING 2.2
A cave diver enters a long underwater tunnel. When her displacement with respect to the entry point is 20 m, she accidentally drops her camera, but she doesn’t notice it missing until she is some 6 m farther into the tunnel. She swims back 10 m but cannot find the camera, so she decides to end the dive. How far from the
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2.1 • Scalars and Vectors

51

entry point is she? Taking the positive direction out of the tunnel, what is her displacement vector relative to the entry point?
Algebra of Vectors in Two Dimensions
When vectors lie in a plane—that is, when they are in two dimensions—they can be multiplied by scalars, added to other vectors, or subtracted from other vectors in accordance with the general laws expressed by Equation 2.1, Equation 2.2, Equation 2.7, and Equation 2.8. However, the addition rule for two vectors in a plane becomes more complicated than the rule for vector addition in one dimension. We have to use the laws of geometry to construct resultant vectors, followed by trigonometry to find vector magnitudes and directions. This geometric approach is commonly used in navigation (Figure 2.9). In this section, we need to have at hand two rulers, a triangle, a protractor, a pencil, and an eraser for drawing vectors to scale by geometric constructions.

Figure 2.9 In navigation, the laws of geometry are used to draw resultant displacements on nautical maps.

For a geometric construction of the sum of two vectors in a plane, we follow the parallelogram rule. Suppose

two vectors and are at the arbitrary positions shown in Figure 2.10. Translate either one of them in parallel to the beginning of the other vector, so that after the translation, both vectors have their origins at the

same point. Now, at the end of vector we draw a line parallel to vector and at the end of vector we draw a

line parallel to vector (the dashed lines in Figure 2.10). In this way, we obtain a parallelogram. From the

origin of the two vectors we draw a diagonal that is the resultant of the two vectors:

(Figure

2.10(a)). The other diagonal of this parallelogram is the vector difference of the two vectors

, as

shown in Figure 2.10(b). Notice that the end of the difference vector is placed at the end of vector .

Figure 2.10 The parallelogram rule for the addition of two vectors. Make the parallel translation of each vector to a point where their

52

2 • Vectors

origins (marked by the dot) coincide and construct a parallelogram with two sides on the vectors and the other two sides (indicated by

dashed lines) parallel to the vectors. (a) Draw the resultant vector along the diagonal of the parallelogram from the common point to the

opposite corner. Length R of the resultant vector is not equal to the sum of the magnitudes of the two vectors. (b) Draw the difference

vector

along the diagonal connecting the ends of the vectors. Place the origin of vector at the end of vector and the end

(arrowhead) of vector at the end of vector . Length D of the difference vector is not equal to the difference of magnitudes of the two

vectors.

It follows from the parallelogram rule that neither the magnitude of the resultant vector nor the magnitude of the difference vector can be expressed as a simple sum or difference of magnitudes A and B, because the length of a diagonal cannot be expressed as a simple sum of side lengths. When using a geometric construction
to find magnitudes and , we have to use trigonometry laws for triangles, which may lead to complicated
algebra. There are two ways to circumvent this algebraic complexity. One way is to use the method of components, which we examine in the next section. The other way is to draw the vectors to scale, as is done in navigation, and read approximate vector lengths and angles (directions) from the graphs. In this section we examine the second approach.

If we need to add three or more vectors, we repeat the parallelogram rule for the pairs of vectors until we find the resultant of all of the resultants. For three vectors, for example, we first find the resultant of vector 1 and vector 2, and then we find the resultant of this resultant and vector 3. The order in which we select the pairs of vectors does not matter because the operation of vector addition is commutative and associative (see Equation 2.7 and Equation 2.8). Before we state a general rule that follows from repetitive applications of the parallelogram rule, let’s look at the following example.

Suppose you plan a vacation trip in Florida. Departing from Tallahassee, the state capital, you plan to visit your uncle Joe in Jacksonville, see your cousin Vinny in Daytona Beach, stop for a little fun in Orlando, see a circus performance in Tampa, and visit the University of Florida in Gainesville. Your route may be represented by five

displacement vectors , , , and , which are indicated by the red vectors in Figure 2.11. What is your total displacement when you reach Gainesville? The total displacement is the vector sum of all five displacement vectors, which may be found by using the parallelogram rule four times. Alternatively, recall that the displacement vector has its beginning at the initial position (Tallahassee) and its end at the final position (Gainesville), so the total displacement vector can be drawn directly as an arrow connecting Tallahassee with Gainesville (see the green vector in Figure 2.11). When we use the parallelogram rule four times, the resultant

we obtain is exactly this green vector connecting Tallahassee with Gainesville:

.

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2.1 • Scalars and Vectors

53

Figure 2.11 When we use the parallelogram rule four times, we obtain the resultant vector vector connecting Tallahassee with Gainesville.

, which is the green

Drawing the resultant vector of many vectors can be generalized by using the following tail-to-head geometric
construction. Suppose we want to draw the resultant vector of four vectors , , , and (Figure 2.12(a)). We select any one of the vectors as the first vector and make a parallel translation of a second vector to a position where the origin (“tail”) of the second vector coincides with the end (“head”) of the first vector. Then, we select a third vector and make a parallel translation of the third vector to a position where the origin of the third vector coincides with the end of the second vector. We repeat this procedure until all the vectors are in a
head-to-tail arrangement like the one shown in Figure 2.12. We draw the resultant vector by connecting the origin (“tail”) of the first vector with the end (“head”) of the last vector. The end of the resultant vector is at the end of the last vector. Because the addition of vectors is associative and commutative, we obtain the same resultant vector regardless of which vector we choose to be first, second, third, or fourth in this construction.

Figure 2.12 Tail-to-head method for drawing the resultant vector

. (a) Four vectors of different magnitudes and

directions. (b) Vectors in (a) are translated to new positions where the origin (“tail”) of one vector is at the end (“head”) of another vector.

The resultant vector is drawn from the origin (“tail”) of the first vector to the end (“head”) of the last vector in this arrangement.

54

2 • Vectors

EXAMPLE 2.2

Geometric Construction of the Resultant

The three displacement vectors , , and in Figure 2.13 are specified by their magnitudes A = 10.0, B = 7.0,

and C = 8.0, respectively, and by their respective direction angles with the horizontal direction

,

, and

. The physical units of the magnitudes are centimeters. Choose a convenient scale and

use a ruler and a protractor to find the following vector sums: (a)

, (b)

(c)

.

Figure 2.13 Vectors used in Example 2.2 and in the Check Your Understanding feature that follows.

Strategy
In geometric construction, to find a vector means to find its magnitude and its direction angle with the horizontal direction. The strategy is to draw to scale the vectors that appear on the right-hand side of the equation and construct the resultant vector. Then, use a ruler and a protractor to read the magnitude of the resultant and the direction angle. For parts (a) and (b) we use the parallelogram rule. For (c) we use the tail-tohead method.

Solution

For parts (a) and (b), we attach the origin of vector to the origin of vector , as shown in Figure 2.14, and

construct a parallelogram. The shorter diagonal of this parallelogram is the sum

. The longer of the

diagonals is the difference

. We use a ruler to measure the lengths of the diagonals, and a protractor to

measure the angles with the horizontal. For the resultant , we obtain R = 5.8 cm and

. For the

difference , we obtain D = 16.2 cm and

, which are shown in Figure 2.14.

Figure 2.14 Using the parallelogram rule to solve (a) (finding the resultant, red) and (b) (finding the difference, blue). Access for free at openstax.org.

2.2 • Coordinate Systems and Components of a Vector

55

For (c), we can start with vector

and draw the remaining vectors tail-to-head as shown in Figure 2.15. In

vector addition, the order in which we draw the vectors is unimportant, but drawing the vectors to scale is very

important. Next, we draw vector from the origin of the first vector to the end of the last vector and place the

arrowhead at the end of . We use a ruler to measure the length of , and find that its magnitude is

S = 36.9 cm. We use a protractor and find that its direction angle is

. This solution is shown in

Figure 2.15.

Figure 2.15 Using the tail-to-head method to solve (c) (finding vector , green).

CHECK YOUR UNDERSTANDING 2.3

Using the three displacement vectors , , and in Figure 2.13, choose a convenient scale, and use a ruler

and a protractor to find vector given by the vector equation

.

INTERACTIVE
Observe the addition of vectors in a plane by visiting this vector calculator (https://openstax.org/l/ 21compveccalc) and this Phet simulation (https://openstax.org/l/21phetvecaddsim) .
2.2 Coordinate Systems and Components of a Vector
Learning Objectives By the end of this section, you will be able to:
• Describe vectors in two and three dimensions in terms of their components, using unit vectors along the axes.
• Distinguish between the vector components of a vector and the scalar components of a vector. • Explain how the magnitude of a vector is defined in terms of the components of a vector. • Identify the direction angle of a vector in a plane. • Explain the connection between polar coordinates and Cartesian coordinates in a plane.
Vectors are usually described in terms of their components in a coordinate system. Even in everyday life we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction north of east.

56

2 • Vectors

In a rectangular (Cartesian) xy-coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y). In a similar fashion, a vector in a plane is described by a pair of its vector coordinates. The x-coordinate of vector is called its x-component and the y-coordinate of vector is called its y-component. The vector x-component is a vector denoted by . The vector y-component is a vector denoted by . In the Cartesian system, the x and y vector components of a vector are the orthogonal projections of this vector onto the x- and y-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components:
2.10
As illustrated in Figure 2.16, vector is the diagonal of the rectangle where the x-component is the side parallel to the x-axis and the y-component is the side parallel to the y-axis. Vector component is orthogonal to vector component .

Figure 2.16 Vector in a plane in the Cartesian coordinate system is the vector sum of its vector x- and y-components. The x-vector component is the orthogonal projection of vector onto the x-axis. The y-vector component is the orthogonal projection of vector onto the y-axis. The numbers and that multiply the unit vectors are the scalar components of the vector.
It is customary to denote the positive direction on the x-axis by the unit vector and the positive direction on the y-axis by the unit vector . Unit vectors of the axes, and , define two orthogonal directions in the plane. As shown in Figure 2.16, the x- and y- components of a vector can now be written in terms of the unit vectors of the axes:
2.11

The vectors and defined by Equation 2.11 are the vector components of vector . The numbers and that define the vector components in Equation 2.11 are the scalar components of vector . Combining
Equation 2.10 with Equation 2.11, we obtain the component form of a vector:

2.12

If we know the coordinates

of the origin point of a vector (where b stands for “beginning”) and the

coordinates

of the end point of a vector (where e stands for “end”), we can obtain the scalar

components of a vector simply by subtracting the origin point coordinates from the end point coordinates:

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2.2 • Coordinate Systems and Components of a Vector

57

2.13

EXAMPLE 2.3

Displacement of a Mouse Pointer
A mouse pointer on the display monitor of a computer at its initial position is at point (6.0 cm, 1.6 cm) with respect to the lower left-side corner. If you move the pointer to an icon located at point (2.0 cm, 4.5 cm), what is the displacement vector of the pointer?

Strategy
The origin of the xy-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit vector on the x-axis points horizontally to the right and the unit vector on the y-axis points vertically upward. The origin of the displacement vector is located at point b(6.0, 1.6) and the end of the displacement vector is located at point e(2.0, 4.5). Substitute the coordinates of these points into Equation 2.13 to find the scalar components and of the displacement vector . Finally, substitute the coordinates into Equation 2.12 to write the displacement vector in the vector component form.

Solution

We identify

,

,

, and

y-components of the displacement vector are

, where the physical unit is 1 cm. The scalar x- and

The vector component form of the displacement vector is This solution is shown in Figure 2.17.

2.14

Figure 2.17 The graph of the displacement vector. The vector points from the origin point at b to the end point at e.
Significance Notice that the physical unit—here, 1 cm—can be placed either with each component immediately before the unit vector or globally for both components, as in Equation 2.14. Often, the latter way is more convenient because it is simpler.

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2 • Vectors

The vector x-component

of the displacement vector has the magnitude

because the magnitude of the unit vector is

. Notice, too, that the direction of the

x-component is , which is antiparallel to the direction of the +x-axis; hence, the x-component vector

points to the left, as shown in Figure 2.17. The scalar x-component of vector is

.

Similarly, the vector y-component

of the displacement vector has magnitude

because the magnitude of the unit vector is

. The direction of the y-component is , which is parallel to

the direction of the +y-axis. Therefore, the y-component vector points up, as seen in Figure 2.17. The scalar

y-component of vector is components.

. The displacement vector is the resultant of its two vector

The vector component form of the displacement vector Equation 2.14 tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.

CHECK YOUR UNDERSTANDING 2.4
A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.

When we know the scalar components and of a vector , we can find its magnitude A and its direction angle . The direction angle—or direction, for short—is the angle the vector forms with the positive direction on the x-axis. The angle is measured in the counterclockwise direction from the +x-axis to the vector (Figure 2.18). Because the lengths A, , and form a right triangle, they are related by the Pythagorean theorem:
2.15
This equation works even if the scalar components of a vector are negative. The direction angle of a vector is defined via the tangent function of angle in the triangle shown in Figure 2.18:
2.16

Figure 2.18 When the vector lies either in the first quadrant or in the fourth quadrant, where component is positive (Figure 2.19), the Access for free at openstax.org.

2.2 • Coordinate Systems and Components of a Vector

59

direction angle in Equation 2.16) is identical to the angle .
When the vector lies either in the first quadrant or in the fourth quadrant, where component is positive (Figure 2.19), the angle in Equation 2.16 is identical to the direction angle . For vectors in the fourth quadrant, angle is negative, which means that for these vectors, direction angle is measured clockwise from the positive x-axis. Similarly, for vectors in the second quadrant, angle is negative. When the vector lies in either the second or third quadrant, where component is negative, the direction angle is (Figure 2.19).

Figure 2.19 Scalar components of a vector may be positive or negative. Vectors in the first quadrant (I) have both scalar components

positive and vectors in the third quadrant have both scalar components negative. For vectors in quadrants II and III, the direction angle of a

vector is

.

EXAMPLE 2.4

Magnitude and Direction of the Displacement Vector
You move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an icon located at point (2.0 cm, 4.5 cm). What are the magnitude and direction of the displacement vector of the pointer?

Strategy

In Example 2.3, we found the displacement vector of the mouse pointer (see Equation 2.14). We identify its

scalar components

and

and substitute into Equation 2.15 and Equation 2.16 to

find the magnitude D and direction , respectively.

Solution The magnitude of vector is

The direction angle is

Vector lies in the second quadrant, so its direction angle is

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2 • Vectors

CHECK YOUR UNDERSTANDING 2.5
If the displacement vector of a blue fly walking on a sheet of graph paper is magnitude and direction.

, find its

In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known.

Let us return to the right triangle in Figure 2.18. The quotient of the adjacent side to the hypotenuse A is

the cosine function of direction angle ,

, and the quotient of the opposite side to the

hypotenuse A is the sine function of ,

. When magnitude A and direction are known, we

can solve these relations for the scalar components:

2.17

When calculating vector components with Equation 2.17, care must be taken with the angle. The direction angle of a vector is the angle measured counterclockwise from the positive direction on the x-axis to the vector. The clockwise measurement gives a negative angle.

EXAMPLE 2.5

Components of Displacement Vectors
A rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes many trial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but his displacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, then he runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction
west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns west of south for 150.0 m. Find the scalar components of Trooper’s displacement vectors and his displacement vectors in vector component form for each leg.

Strategy
Let’s adopt a rectangular coordinate system with the positive x-axis in the direction of geographic east, with the positive y-direction pointed to geographic north. Explicitly, the unit vector of the x-axis points east and the unit vector of the y-axis points north. Trooper makes five legs, so there are five displacement vectors. We start by identifying their magnitudes and direction angles, then we use Equation 2.17 to find the scalar components of the displacements and Equation 2.12 for the displacement vectors.

Solution

On the first leg, the displacement magnitude is

and the direction is southeast. For direction

angle we can take either measured clockwise from the east direction or

measured

counterclockwise from the east direction. With the first choice,

. With the second choice,

. We can use either one of these two angles. The components are

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2.2 • Coordinate Systems and Components of a Vector

61

The displacement vector of the first leg is

On the second leg of Trooper’s wanderings, the magnitude of the displacement is

direction is north. The direction angle is

. We obtain the following results:

and the

On the third leg, the displacement magnitude is

and the direction is

direction angle measured counterclockwise from the eastern direction is

the following answers:

west of north. The . This gives

On the fourth leg of the excursion, the displacement magnitude is

direction angle can be taken as either

or

. We obtain

and the direction is south. The

On the last leg, the magnitude is which gives

and the angle is

west of south),

CHECK YOUR UNDERSTANDING 2.6
If Trooper runs 20 m west before taking a rest, what is his displacement vector?
Polar Coordinates
To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system these directions are given by unit vectors and along the x-axis and the y-axis, respectively. The Cartesian coordinate system is very convenient to use in describing displacements and velocities of objects and the forces acting on them. However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually work in the polar coordinate system.
In the polar coordinate system, the location of point P in a plane is given by two polar coordinates (Figure 2.20). The first polar coordinate is the radial coordinate r, which is the distance of point P from the origin. The second polar coordinate is an angle that the radial vector makes with some chosen direction, usually the positive x-direction. In polar coordinates, angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point P. This radial direction is described by a unit radial vector . The second unit vector is a vector orthogonal to the radial direction . The positive direction indicates how the angle changes in the counterclockwise direction. In this way, a point P that has

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2 • Vectors

coordinates (x, y) in the rectangular system can be described equivalently in the polar coordinate system by

the two polar coordinates

. Equation 2.17 is valid for any vector, so we can use it to express the x- and

y-coordinates of vector . In this way, we obtain the connection between the polar coordinates and rectangular

coordinates of point P:

2.18

Figure 2.20 Using polar coordinates, the unit vector defines the positive direction along the radius r (radial direction) and, orthogonal to it, the unit vector defines the positive direction of rotation by the angle .

EXAMPLE 2.6

Polar Coordinates
A treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction north of east and finds one gold coin at a location 10.0 m away from the well in the direction north of west. What are the polar and rectangular coordinates of these findings with respect to the well?

Strategy

The well marks the origin of the coordinate system and east is the +x-direction. We identify radial distances

from the locations to the origin, which are

(for the silver coin) and

(for the gold coin).

To find the angular coordinates, we convert to radians:

. We use Equation 2.18 to find

the x- and y-coordinates of the coins.

Solution

The angular coordinate of the silver coin is

, whereas the angular coordinate of the gold coin is

. Hence, the polar coordinates of the silver coin are

and those of

the gold coin are

. We substitute these coordinates into Equation 2.18 to obtain

rectangular coordinates. For the gold coin, the coordinates are

For the silver coin, the coordinates are

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63

Vectors in Three Dimensions
To specify the location of a point in space, we need three coordinates (x, y, z), where coordinates x and y specify locations in a plane, and coordinate z gives a vertical position above or below the plane. Threedimensional space has three orthogonal directions, so we need not two but three unit vectors to define a threedimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the x-axis and the unit vector of the y-axis . The third unit vector is the direction of the z-axis (Figure 2.21). The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order x-y-z, which is equivalent to the order - - , defines the standard right-handed coordinate system (positive orientation).

Figure 2.21 Three unit vectors define a Cartesian system in three-dimensional space. The order in which these unit vectors appear defines the orientation of the coordinate system. The order shown here defines the right-handed orientation.

In three-dimensional space, vector has three vector components: the x-component

, which is the

part of vector along the x-axis; the y-component

, which is the part of along the y-axis; and the

z-component

, which is the part of the vector along the z-axis. A vector in three-dimensional space is

the vector sum of its three vector components (Figure 2.22):

2.19

If we know the coordinates of its origin

and of its end

, its scalar components are

obtained by taking their differences: and are given by Equation 2.13 and the z-component is given by

2.20

Magnitude A is obtained by generalizing Equation 2.15 to three dimensions:

2.21

This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in

Figure 2.22, the diagonal in the xy-plane has length

and its square adds to the square to give

. Note that when the z-component is zero, the vector lies entirely in the xy-plane and its description is

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2 • Vectors

reduced to two dimensions.

Figure 2.22 A vector in three-dimensional space is the vector sum of its three vector components.
EXAMPLE 2.7 Takeoff of a Drone
During a takeoff of IAI Heron (Figure 2.23), its position with respect to a control tower is 100 m above the ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200 m to the east, and 2100 m to the north. What is the drone’s displacement vector with respect to the control tower? What is the magnitude of its displacement vector?
Figure 2.23 The drone IAI Heron in flight. (credit: SSgt Reynaldo Ramon, USAF)
Strategy We take the origin of the Cartesian coordinate system as the control tower. The direction of the +x-axis is given by unit vector to the east, the direction of the +y-axis is given by unit vector to the north, and the direction of the +z-axis is given by unit vector , which points up from the ground. The drone’s first position is the origin (or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement vector. Solution We identify b(300.0 m, 200.0 m, 100.0 m) and e(1200 m, 2100 m, 250 m), and use Equation 2.13 and Equation 2.20 to find the scalar components of the drone’s displacement vector:
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2.3 • Algebra of Vectors

65

We substitute these components into Equation 2.19 to find the displacement vector: We substitute into Equation 2.21 to find the magnitude of the displacement:

CHECK YOUR UNDERSTANDING 2.7
If the average velocity vector of the drone in the displacement in Example 2.7 is , what is the magnitude of the drone’s velocity vector?

2.3 Algebra of Vectors
Learning Objectives By the end of this section, you will be able to:
• Apply analytical methods of vector algebra to find resultant vectors and to solve vector equations for unknown vectors.
• Interpret physical situations in terms of vector expressions.
Vectors can be added together and multiplied by scalars. Vector addition is associative (Equation 2.8) and commutative (Equation 2.7), and vector multiplication by a sum of scalars is distributive (Equation 2.9). Also, scalar multiplication by a sum of vectors is distributive:
2.22

In this equation, is any number (a scalar). For example, a vector antiparallel to vector

can be expressed simply by multiplying by the scalar

:

2.23

EXAMPLE 2.8

Direction of Motion
In a Cartesian coordinate system where denotes geographic east, denotes geographic north, and denotes altitude above sea level, a military convoy advances its position through unknown territory with velocity
. If the convoy had to retreat, in what geographic direction would it be moving?

Solution

The velocity vector has the third component

, which says the convoy is climbing at a rate of

100 m/h through mountainous terrain. At the same time, its velocity is 4.0 km/h to the east and 3.0 km/h to the

north, so it moves on the ground in direction

north of east. If the convoy had to retreat, its

new velocity vector would have to be antiparallel to and be in the form

, where is a positive

number. Thus, the velocity of the retreat would be

. The negative sign of the

third component indicates the convoy would be descending. The direction angle of the retreat velocity is

south of west. Therefore, the convoy would be moving on the ground in direction

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2 • Vectors

south of west while descending on its way back.

The generalization of the number zero to vector algebra is called the null vector, denoted by . All components

of the null vector are zero,

, so the null vector has no length and no direction.

Two vectors and are equal vectors if and only if their difference is the null vector:

This vector equation means we must have simultaneously

,

, and

.

Hence, we can write

if and only if the corresponding components of vectors and are equal:

2.24

Two vectors are equal when their corresponding scalar components are equal.
Resolving vectors into their scalar components (i.e., finding their scalar components) and expressing them analytically in vector component form (given by Equation 2.19) allows us to use vector algebra to find sums or differences of many vectors analytically (i.e., without using graphical methods). For example, to find the resultant of two vectors and , we simply add them component by component, as follows:

In this way, using Equation 2.24, scalar components of the resultant vector sums of corresponding scalar components of vectors and :

are the

Analytical methods can be used to find components of a resultant of many vectors. For example, if we are to

sum up vectors

, where each vector is

, the resultant vector

is

Therefore, scalar components of the resultant vector are 2.25
Having found the scalar components, we can write the resultant in vector component form:
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2.3 • Algebra of Vectors

67

Analytical methods for finding the resultant and, in general, for solving vector equations are very important in physics because many physical quantities are vectors. For example, we use this method in kinematics to find resultant displacement vectors and resultant velocity vectors, in mechanics to find resultant force vectors and the resultants of many derived vector quantities, and in electricity and magnetism to find resultant electric or magnetic vector fields.

EXAMPLE 2.9

Analytical Computation of a Resultant

Three displacement vectors , , and in a plane (Figure 2.13) are specified by their magnitudes A = 10.0, B =

7.0, and C = 8.0, respectively, and by their respective direction angles with the horizontal direction

, and

. The physical units of the magnitudes are centimeters. Resolve the vectors to their

scalar components and find the following vector sums: (a)

, (b)

, and (c)

.

Strategy
First, we use Equation 2.17 to find the scalar components of each vector and then we express each vector in its vector component form given by Equation 2.12. Then, we use analytical methods of vector algebra to find the resultants.

Solution We resolve the given vectors to their scalar components:

For (a) we may substitute directly into Equation 2.25 to find the scalar components of the resultant:

Therefore, the resultant vector is

.

For (b), we may want to write the vector difference as

Then, the scalar components of the vector difference are

Hence, the difference vector is

.

For (c), we can write vector in the following explicit form:

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2 • Vectors

Then, the scalar components of are

The vector is

.

Significance
Having found the vector components, we can illustrate the vectors by graphing or we can compute magnitudes and direction angles, as shown in Figure 2.24. Results for the magnitudes in (b) and (c) can be compared with results for the same problems obtained with the graphical method, shown in Figure 2.14 and Figure 2.15. Notice that the analytical method produces exact results and its accuracy is not limited by the resolution of a ruler or a protractor, as it was with the graphical method used in Example 2.2 for finding this same resultant.

Figure 2.24 Graphical illustration of the solutions obtained analytically in Example 2.9.

CHECK YOUR UNDERSTANDING 2.8

Three displacement vectors , , and (Figure 2.13) are specified by their magnitudes A = 10.00, B = 7.00,

and F = 20.00, respectively, and by their respective direction angles with the horizontal direction

,

, and

. The physical units of the magnitudes are centimeters. Use the analytical method to

find vector G = 28.15 cm and that

. Verify that .

EXAMPLE 2.10

The Tug-of-War Game

Four dogs named Astro, Balto, Clifford, and Dug play a tug-of-war game with a toy (Figure 2.25). Astro pulls on

the toy in direction

south of east, Balto pulls in direction

east of north, and Clifford pulls in

direction

west of north. Astro pulls strongly with 160.0 units of force (N), which we abbreviate as A =

160.0 N. Balto pulls even stronger than Astro with a force of magnitude B = 200.0 N, and Clifford pulls with a

force of magnitude C = 140.0 N. When Dug pulls on the toy in such a way that his force balances out the

resultant of the other three forces, the toy does not move in any direction. With how big a force and in what

direction must Dug pull on the toy for this to happen?

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2.3 • Algebra of Vectors

69

Figure 2.25 Four dogs play a tug-of-war game with a toy.

Strategy

We assume that east is the direction of the positive x-axis and north is the direction of the positive y-axis. As in

Example 2.9, we have to resolve the three given forces— (the pull from Astro), (the pull from Balto), and (the pull from Clifford)—into their scalar components and then find the scalar components of the resultant

vector

. When the pulling force from Dug balances out this resultant, the sum of and

must give the null vector

. This means that

, so the pull from Dug must be antiparallel to .

Solution

The direction angles are

,

, and

them into Equation 2.17 gives the scalar components of the three given forces:

, and substituting

Now we compute scalar components of the resultant vector

:

The antiparallel vector to the resultant is The magnitude of Dug’s pulling force is The direction of Dug’s pulling force is

Dug pulls in the direction

south of west because both components are negative, which means the pull

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2 • Vectors

vector lies in the third quadrant (Figure 2.19).
CHECK YOUR UNDERSTANDING 2.9
Suppose that Balto in Example 2.10 leaves the game to attend to more important matters, but Astro, Clifford, and Dug continue playing. Astro and Clifford's pull on the toy does not change, but Dug runs around and bites on the toy in a different place. With how big a force and in what direction must Dug pull on the toy now to balance out the combined pulls from Clifford and Astro? Illustrate this situation by drawing a vector diagram indicating all forces involved.

EXAMPLE 2.11

Vector Algebra
Find the magnitude of the vector .

that satisfies the equation

, where

and

Strategy

We first solve the given equation for the unknown vector . Then we substitute each of the three directions , , and ; and identify the scalar components , substitute into Equation 2.21 to find magnitude C.

and ; group the terms along , and . Finally, we

Solution

The components are

,

, and

, and substituting into Equation 2.21 gives

EXAMPLE 2.12

Displacement of a Skier
Starting at a ski lodge, a cross-country skier goes 5.0 km north, then 3.0 km west, and finally 4.0 km southwest before taking a rest. Find his total displacement vector relative to the lodge when he is at the rest point. How far and in what direction must he ski from the rest point to return directly to the lodge?

Strategy

We assume a rectangular coordinate system with the origin at the ski lodge and with the unit vector pointing

east and the unit vector pointing north. There are three displacements: , , and . We identify their

magnitudes as

,

, and

. We identify their directions are the angles

,

, and

. We resolve each displacement vector to its scalar

components and substitute the components into Equation 2.25 to obtain the scalar components of the

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2.3 • Algebra of Vectors

71

resultant displacement from the lodge to the rest point. On the way back from the rest point to the lodge, the

displacement is

. Finally, we find the magnitude and direction of .

Solution Scalar components of the displacement vectors are

Scalar components of the net displacement vector are

Hence, the skier’s net displacement vector is

. On the way back to the

lodge, his displacement is

. Its magnitude is

and its direction angle is

.

Therefore, to return to the lodge, he must go 6.2 km in a direction about south of east.

Significance
Notice that no figure is needed to solve this problem by the analytical method. Figures are required when using a graphical method; however, we can check if our solution makes sense by sketching it, which is a useful final step in solving any vector problem.

EXAMPLE 2.13

Displacement of a Jogger

A jogger runs up a flight of 200 identical steps to the top of a hill and then runs along the top of the hill 50.0 m before he stops at a drinking fountain (Figure 2.26). His displacement vector from point A at the bottom of the

steps to point B at the fountain is

. What is the height and width of each step in the

flight? What is the actual distance the jogger covers? If he makes a loop and returns to point A, what is his net

displacement vector?

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2 • Vectors

Figure 2.26 A jogger runs up a flight of steps.

Strategy

The displacement vector

is the vector sum of the jogger’s displacement vector

along the stairs (from

point A at the bottom of the stairs to point T at the top of the stairs) and his displacement vector

on the top

of the hill (from point A at the top of the stairs to the fountain at point T). We must find the horizontal and the

vertical components of . If each step has width w and height h, the horizontal component of

must

have a length of 200w and the vertical component must have a length of 200h. The actual distance the jogger

covers is the sum of the distance he runs up the stairs and the distance of 50.0 m that he runs along the top of

the hill.

Solution In the coordinate system indicated in Figure 2.26, the jogger’s displacement vector on the top of the hill is
. His net displacement vector is

Therefore, his displacement vector

along the stairs is

Its scalar components are

and

. Therefore, we must have

Hence, the step width is w = 40.0 m/200 = 0.2 m = 20 cm, and the step height is h = 30.0 m/200 = 0.15 m = 15 cm. The distance that the jogger covers along the stairs is

Thus, the actual distance he runs is

. When he makes a loop and

comes back from the fountain to his initial position at point A, the total distance he covers is twice this

distance, or 200.0 m. However, his net displacement vector is zero, because when his final position is the same

as his initial position, the scalar components of his net displacement vector are zero (Equation 2.13).

In many physical situations, we often need to know the direction of a vector. For example, we may want to know the direction of a magnetic field vector at some point or the direction of motion of an object. We have already said direction is given by a unit vector, which is a dimensionless entity—that is, it has no physical units associated with it. When the vector in question lies along one of the axes in a Cartesian system of coordinates, the answer is simple, because then its unit vector of direction is either parallel or antiparallel to the direction

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2.3 • Algebra of Vectors

73

of the unit vector of an axis. For example, the direction of vector

is unit vector

. The general

rule of finding the unit vector of direction for any vector is to divide it by its magnitude V:

2.26

We see from this expression that the unit vector of direction is indeed dimensionless because the numerator and the denominator in Equation 2.26 have the same physical unit. In this way, Equation 2.26 allows us to express the unit vector of direction in terms of unit vectors of the axes. The following example illustrates this principle.

EXAMPLE 2.14

The Unit Vector of Direction
If the velocity vector of the military convoy in Example 2.8 is unit vector of its direction of motion?

, what is the

Strategy The unit vector of the convoy’s direction of motion is the unit vector that is parallel to the velocity vector. The unit vector is obtained by dividing a vector by its magnitude, in accordance with Equation 2.26.

Solution The magnitude of the vector is

To obtain the unit vector , divide by its magnitude:

Significance Note that when using the analytical method with a calculator, it is advisable to carry out your calculations to at least three decimal places and then round off the final answer to the required number of significant figures, which is the way we performed calculations in this example. If you round off your partial answer too early, you risk your final answer having a huge numerical error, and it may be far off from the exact answer or from a value measured in an experiment.
CHECK YOUR UNDERSTANDING 2.10
Verify that vector obtained in Example 2.14 is indeed a unit vector by computing its magnitude. If the convoy in Example 2.8 was moving across a desert flatland—that is, if the third component of its velocity was zero—what is the unit vector of its direction of motion? Which geographic direction does it represent?

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2 • Vectors

2.4 Products of Vectors
Learning Objectives By the end of this section, you will be able to:
• Explain the difference between the scalar product and the vector product of two vectors. • Determine the scalar product of two vectors. • Determine the vector product of two vectors. • Describe how the products of vectors are used in physics.
A vector can be multiplied by another vector but may not be divided by another vector. There are two kinds of products of vectors used broadly in physics and engineering. One kind of multiplication is a scalar multiplication of two vectors. Taking a scalar product of two vectors results in a number (a scalar), as its name indicates. Scalar products are used to define work and energy relations. For example, the work that a force (a vector) performs on an object while causing its displacement (a vector) is defined as a scalar product of the force vector with the displacement vector. A quite different kind of multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, as its name suggests. Vector products are used to define other derived vector quantities. For example, in describing rotations, a vector quantity called torque is defined as a vector product of an applied force (a vector) and its distance from pivot to force (a vector). It is important to distinguish between these two kinds of vector multiplications because the scalar product is a scalar quantity and a vector product is a vector quantity.
The Scalar Product of Two Vectors (the Dot Product)
Scalar multiplication of two vectors yields a scalar product.

Scalar Product (Dot Product)

The scalar product

of two vectors and is a number defined by the equation

2.27

where is the angle between the vectors (shown in Figure 2.27). The scalar product is also called the dot product because of the dot notation that indicates it.

In the definition of the dot product, the direction of angle does not matter, and can be measured from

either of the two vectors to the other because

. The dot product is a negative

number when

and is a positive number when

. Moreover, the dot product of two

parallel vectors is

, and the dot product of two antiparallel vectors is

. The scalar product of two orthogonal vectors vanishes:

.

The scalar product of a vector with itself is the square of its magnitude:

2.28

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2.4 • Products of Vectors

75

Figure 2.27 The scalar product of two vectors. (a) The angle between the two vectors. (b) The orthogonal projection

onto the direction of vector . (c) The orthogonal projection

of vector onto the direction of vector .

of vector

EXAMPLE 2.15

The Scalar Product

For the vectors shown in Figure 2.13, find the scalar product

.

Strategy

From Figure 2.13, the magnitudes of vectors and are A = 10.0 and F = 20.0. Angle , between them, is the

difference:

. Substituting these values into Equation 2.27 gives the scalar

product.

Solution A straightforward calculation gives us

CHECK YOUR UNDERSTANDING 2.11
For the vectors given in Figure 2.13, find the scalar products

and

.

In the Cartesian coordinate system, scalar products of the unit vector of an axis with other unit vectors of axes always vanish because these unit vectors are orthogonal:

2.29

In these equations, we use the fact that the magnitudes of all unit vectors are one: vectors of the axes, Equation 2.28 gives the following identities:

. For unit 2.30

The scalar product

can also be interpreted as either the product of B with the projection of vector

onto the direction of vector (Figure 2.27(b)) or the product of A with the projection of vector onto the

direction of vector (Figure 2.27(c)):

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2 • Vectors

For example, in the rectangular coordinate system in a plane, the scalar x-component of a vector is its dot product with the unit vector , and the scalar y-component of a vector is its dot product with the unit vector :

Scalar multiplication of vectors is commutative,

2.31

and obeys the distributive law:

2.32

We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot product of two vectors in terms of their scalar components.
CHECK YOUR UNDERSTANDING 2.12

For vector 2.32 to show that

in a rectangular coordinate system, use Equation 2.29 through Equation

and

.

When the vectors in Equation 2.27 are given in their vector component forms,

we can compute their scalar product as follows:

Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselves give one (see Equation 2.29 and Equation 2.30), there are only three nonzero terms in this expression. Thus, the scalar product simplifies to

2.33

We can use Equation 2.33 for the scalar product in terms of scalar components of vectors to find the angle

between two vectors. When we divide Equation 2.27 by AB, we obtain the equation for

, into which we

substitute Equation 2.33:

2.34

Angle between vectors and is obtained by taking the inverse cosine of the expression in Equation 2.34.

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2.4 • Products of Vectors

77

EXAMPLE 2.16

Angle between Two Forces

Three dogs are pulling on a stick in different directions, as shown in Figure 2.28. The first dog pulls with force

, the second dog pulls with force

, and the third dog pulls

with force

. What is the angle between forces and ?

Figure 2.28 Three dogs are playing with a stick.

Strategy

The components of force vector are

,

, and

, whereas those of force

vector are

,

, and

. Computing the scalar product of these vectors

and their magnitudes, and substituting into Equation 2.34 gives the angle of interest.

Solution The magnitudes of forces and are

and

Substituting the scalar components into Equation 2.33 yields the scalar product

Finally, substituting everything into Equation 2.34 gives the angle
Significance Notice that when vectors are given in terms of the unit vectors of axes, we can find the angle between them without knowing the specifics about the geographic directions the unit vectors represent. Here, for example, the +x-direction might be to the east and the +y-direction might be to the north. But, the angle between the forces in the problem is the same if the +x-direction is to the west and the +y-direction is to the south.

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2 • Vectors

CHECK YOUR UNDERSTANDING 2.13
Find the angle between forces and in Example 2.16.

EXAMPLE 2.17

The Work of a Force

When force pulls on an object and when it causes its displacement , we say the force performs work. The

amount of work the force does is the scalar product . If the stick in Example 2.16 moves momentarily and

gets displaced by vector

, how much work is done by the third dog in Example 2.16?

Strategy We compute the scalar product of displacement vector with force vector pull from the third dog. Let’s use to denote the work done by force on displacement .

, which is the

Solution Calculating the work is a straightforward application of the dot product:

Significance The SI unit of work is called the joule , where 1 J = 1
, so the answer can be expressed as

. The unit

can be written as .

CHECK YOUR UNDERSTANDING 2.14
How much work is done by the first dog and by the second dog in Example 2.16 on the displacement in Example 2.17?

The Vector Product of Two Vectors (the Cross Product)
Vector multiplication of two vectors yields a vector product.

Vector Product (Cross Product)

The vector product of two vectors and is denoted by

and is often referred to as a cross

product. The vector product is a vector that has its direction perpendicular to both vectors and . In

other words, vector

is perpendicular to the plane that contains vectors and , as shown in Figure

2.29. The magnitude of the vector product is defined as

2.35

where angle , between the two vectors, is measured from vector (first vector in the product) to vector (second vector in the product), as indicated in Figure 2.29, and is between and .

According to Equation 2.35, the vector product vanishes for pairs of vectors that are either parallel

or

antiparallel

because

.

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2.4 • Products of Vectors

79

Figure 2.29 The vector product of two vectors is drawn in three-dimensional space. (a) The vector product

is a vector

perpendicular to the plane that contains vectors and . Small squares drawn in perspective mark right angles between and , and

between and so that if and lie on the floor, vector points vertically upward to the ceiling. (b) The vector product

is a

vector antiparallel to vector

.

On the line perpendicular to the plane that contains vectors and there are two alternative directions—either up or down, as shown in Figure 2.29—and the direction of the vector product may be either one of them. In the standard right-handed orientation, where the angle between vectors is measured

counterclockwise from the first vector, vector

points upward, as seen in Figure 2.29(a). If we reverse the

order of multiplication, so that now comes first in the product, then vector

must point downward, as

seen in Figure 2.29(b). This means that vectors

and

are antiparallel to each other and that

vector multiplication is not commutative but anticommutative. The anticommutative property means the

vector product reverses the sign when the order of multiplication is reversed:

2.36

The corkscrew right-hand rule is a common mnemonic used to determine the direction of the vector product. As shown in Figure 2.30, a corkscrew is placed in a direction perpendicular to the plane that contains vectors
and , and its handle is turned in the direction from the first to the second vector in the product. The direction of the cross product is given by the progression of the corkscrew.

Figure 2.30 The corkscrew right-hand rule can be used to determine the direction of the cross product

. Place a corkscrew in the

direction perpendicular to the plane that contains vectors and , and turn it in the direction from the first to the second vector in the

80

2 • Vectors

product. The direction of the cross product is given by the progression of the corkscrew. (a) Upward movement means the cross-product vector points up. (b) Downward movement means the cross-product vector points downward.

EXAMPLE 2.18

The Torque of a Force

The mechanical advantage that a familiar tool called a wrench provides (Figure 2.31) depends on magnitude F of the applied force, on its direction with respect to the wrench handle, and on how far from the nut this force

is applied. The distance R from the nut to the point where force vector is attached is represented by the

radial vector . The physical vector quantity that makes the nut turn is called torque (denoted by , and it is

the vector product of the distance between the pivot to force with the force:

.

To loosen a rusty nut, a 20.00-N force is applied to the wrench handle at angle

and at a distance of

0.25 m from the nut, as shown in Figure 2.31(a). Find the magnitude and direction of the torque applied to the

nut. What would the magnitude and direction of the torque be if the force were applied at angle

, as

shown in Figure 2.31(b)? For what value of angle does the torque have the largest magnitude?

Figure 2.31 A wrench provides grip and mechanical advantage in applying torque to turn a nut. (a) Turn counterclockwise to loosen the nut. (b) Turn clockwise to tighten the nut.

Strategy

We adopt the frame of reference shown in Figure 2.31, where vectors and lie in the xy-plane and the origin

is at the position of the nut. The radial direction along vector (pointing away from the origin) is the reference

direction for measuring the angle because is the first vector in the vector product

. Vector

must lie along the z-axis because this is the axis that is perpendicular to the xy-plane, where both and lie.

To compute the magnitude , we use Equation 2.35. To find the direction of , we use the corkscrew right-hand

rule (Figure 2.30).

Solution

For the situation in (a), the corkscrew rule gives the direction of

in the positive direction of the z-axis.

Physically, it means the torque vector points out of the page, perpendicular to the wrench handle. We identify

F = 20.00 N and R = 0.25 m, and compute the magnitude using Equation 2.35:

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2.4 • Products of Vectors

81

For the situation in (b), the corkscrew rule gives the direction of

in the negative direction of the z-axis.

Physically, it means the vector points into the page, perpendicular to the wrench handle. The magnitude of

this torque is

The torque has the largest value when

, which happens when

. Physically, it means the

wrench is most effective—giving us the best mechanical advantage—when we apply the force perpendicular to

the wrench handle. For the situation in this example, this best-torque value is

.

Significance

When solving mechanics problems, we often do not need to use the corkscrew rule at all, as we’ll see now in

the following equivalent solution. Notice that once we have identified that vector we can write this vector in terms of the unit vector of the z-axis:

lies along the z-axis,

In this equation, the number that multiplies is the scalar z-component of the vector

. In the

computation of this component, care must be taken that the angle is measured counterclockwise from

(first vector) to (second vector). Following this principle for the angles, we obtain

for the situation in (a), and we obtain

for the situation in (b). In the latter case, the

angle is negative because the graph in Figure 2.31 indicates the angle is measured clockwise; but, the same

result is obtained when this angle is measured counterclockwise because

and

. In this way, we obtain the solution without reference to the corkscrew rule. For the

situation in (a), the solution is

; for the situation in (b), the solution is

.

CHECK YOUR UNDERSTANDING 2.15
For the vectors given in Figure 2.13, find the vector products

and

.

Similar to the dot product (Equation 2.32), the cross product has the following distributive property:

2.37

The distributive property is applied frequently when vectors are expressed in their component forms, in terms of unit vectors of Cartesian axes.

When we apply the definition of the cross product, Equation 2.35, to unit vectors , , and that define the positive x-, y-, and z-directions in space, we find that

2.38

All other cross products of these three unit vectors must be vectors of unit magnitudes because , , and are

orthogonal. For example, for the pair and , the magnitude is

. The

direction of the vector product

must be orthogonal to the xy-plane, which means it must be along the

z-axis. The only unit vectors along the z-axis are or . By the corkscrew rule, the direction of vector

must be parallel to the positive z-axis. Therefore, the result of the multiplication

is identical to . We

can repeat similar reasoning for the remaining pairs of unit vectors. The results of these multiplications are

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2 • Vectors

2.39
Notice that in Equation 2.39, the three unit vectors , , and appear in the cyclic order shown in a diagram in Figure 2.32(a). The cyclic order means that in the product formula, follows and comes before , or follows
and comes before , or follows and comes before . The cross product of two different unit vectors is always a third unit vector. When two unit vectors in the cross product appear in the cyclic order, the result of such a multiplication is the remaining unit vector, as illustrated in Figure 2.32(b). When unit vectors in the cross product appear in a different order, the result is a unit vector that is antiparallel to the remaining unit vector (i.e., the result is with the minus sign, as shown by the examples in Figure 2.32(c) and Figure 2.32(d). In practice, when the task is to find cross products of vectors that are given in vector component form, this rule for the cross-multiplication of unit vectors is very useful.

Figure 2.32 (a) The diagram of the cyclic order of the unit vectors of the axes. (b) The only cross products where the unit vectors appear in the cyclic order. These products have the positive sign. (c, d) Two examples of cross products where the unit vectors do not appear in the cyclic order. These products have the negative sign.

Suppose we want to find the cross product

for vectors

and

. We can use the distributive property (Equation 2.37), the anticommutative property (Equation 2.36), and the results in Equation 2.38 and Equation 2.39 for unit vectors to perform the following algebra:

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2.4 • Products of Vectors

83

When performing algebraic operations involving the cross product, be very careful about keeping the correct order of multiplication because the cross product is anticommutative. The last two steps that we still have to do to complete our task are, first, grouping the terms that contain a common unit vector and, second, factoring. In this way we obtain the following very useful expression for the computation of the cross product:
2.40
In this expression, the scalar components of the cross-product vector are

2.41

When finding the cross product, in practice, we can use either Equation 2.35 or Equation 2.40, depending on which one of them seems to be less complex computationally. They both lead to the same final result. One way to make sure if the final result is correct is to use them both.

EXAMPLE 2.19

A Particle in a Magnetic Field

When moving in a magnetic field, some particles may experience a magnetic force. Without going into details—a detailed study of magnetic phenomena comes in later chapters—let’s acknowledge that the magnetic

field is a vector, the magnetic force is a vector, and the velocity of the particle is a vector. The magnetic force vector is proportional to the vector product of the velocity vector with the magnetic field vector, which we

express as

. In this equation, a constant takes care of the consistency in physical units, so we can

omit physical units on vectors and . In this example, let’s assume the constant is positive.

A particle moving in space with velocity vector

enters a region with a magnetic field

and experiences a magnetic force. Find the magnetic force on this particle at the entry point to the region

where the magnetic field vector is (a)

and (b)

. In each case, find magnitude F of

the magnetic force and angle the force vector makes with the given magnetic field vector .

Strategy

First, we want to find the vector product

, because then we can determine the magnetic force using

. Magnitude F can be found either by using components,

, or by computing

the magnitude

directly using Equation 2.35. In the latter approach, we would have to find the angle

between vectors and . When we have , the general method for finding the direction angle involves the

computation of the scalar product

and substitution into Equation 2.34. To compute the vector product we

can either use Equation 2.40 or compute the product directly, whichever way is simpler.

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2 • Vectors

Solution

The components of the velocity vector are

,

(a) The components of the magnetic field vector are into Equation 2.41 gives the scalar components of vector

, and

.

,

, and

:

. Substituting them

Thus, the magnetic force is

and its magnitude is

To compute angle , we may need to find the magnitude of the magnetic field vector,

and the scalar product : Now, substituting into Equation 2.34 gives angle :

Hence, the magnetic force vector is perpendicular to the magnetic field vector. (We could have saved some time if we had computed the scalar product earlier.)

(b) Because vector product directly:

has only one component, we can perform the algebra quickly and find the vector

The magnitude of the magnetic force is

Because the scalar product is

the magnetic force vector is perpendicular to the magnetic field vector .

Significance

Even without actually computing the scalar product, we can predict that the magnetic force vector must always be perpendicular to the magnetic field vector because of the way this vector is constructed. Namely, the

magnetic force vector is the vector product

and, by the definition of the vector product (see Figure

2.29), vector must be perpendicular to both vectors and .

CHECK YOUR UNDERSTANDING 2.16

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2.4 • Products of Vectors

85

Given two vectors (d) the angle between

and

, find (a)

, (b)

and vector

.

, (c) the angle between and , and

In conclusion to this section, we want to stress that “dot product” and “cross product” are entirely different mathematical objects that have different meanings. The dot product is a scalar; the cross product is a vector. Later chapters use the terms dot product and scalar product interchangeably. Similarly, the terms cross product and vector product are used interchangeably.

86 2 • Chapter Review

CHAPTER REVIEW
Key Terms
anticommutative property change in the order of operation introduces the minus sign
antiparallel vectors two vectors with directions that differ by
associative terms can be grouped in any fashion commutative operations can be performed in any
order component form of a vector a vector written as the
vector sum of its components in terms of unit vectors corkscrew right-hand rule a rule used to determine the direction of the vector product cross product the result of the vector multiplication of vectors is a vector called a cross product; also called a vector product difference of two vectors vector sum of the first vector with the vector antiparallel to the second direction angle in a plane, an angle between the positive direction of the x-axis and the vector, measured counterclockwise from the axis to the vector displacement change in position distributive multiplication can be distributed over terms in summation dot product the result of the scalar multiplication of two vectors is a scalar called a dot product; also called a scalar product equal vectors two vectors are equal if and only if all their corresponding components are equal; alternately, two parallel vectors of equal magnitudes magnitude length of a vector null vector a vector with all its components equal to zero orthogonal vectors two vectors with directions that differ by exactly , synonymous with perpendicular vectors parallel vectors two vectors with exactly the same direction angles parallelogram rule geometric construction of the vector sum in a plane polar coordinate system an orthogonal coordinate
Key Equations
Multiplication by a scalar (vector equation)

system where location in a plane is given by polar coordinates polar coordinates a radial coordinate and an angle radial coordinate distance to the origin in a polar coordinate system resultant vector vector sum of two (or more) vectors scalar a number, synonymous with a scalar quantity in physics scalar component a number that multiplies a unit vector in a vector component of a vector scalar equation equation in which the left-hand and right-hand sides are numbers scalar product the result of the scalar multiplication of two vectors is a scalar called a scalar product; also called a dot product scalar quantity quantity that can be specified completely by a single number with an appropriate physical unit tail-to-head geometric construction geometric construction for drawing the resultant vector of many vectors unit vector vector of a unit magnitude that specifies direction; has no physical unit unit vectors of the axes unit vectors that define orthogonal directions in a plane or in space vector mathematical object with magnitude and direction vector components orthogonal components of a vector; a vector is the vector sum of its vector components. vector equation equation in which the left-hand and right-hand sides are vectors vector product the result of the vector multiplication of vectors is a vector called a vector product; also called a cross product vector quantity physical quantity described by a mathematical vector—that is, by specifying both its magnitude and its direction; synonymous with a vector in physics vector sum resultant of the combination of two (or more) vectors

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Multiplication by a scalar (scalar equation for magnitudes) Resultant of two vectors
Commutative law
Associative law
Distributive law The component form of a vector in two dimensions Scalar components of a vector in two dimensions
Magnitude of a vector in a plane
The direction angle of a vector in a plane Scalar components of a vector in a plane
Polar coordinates in a plane
The component form of a vector in three dimensions The scalar z-component of a vector in three dimensions Magnitude of a vector in three dimensions Distributive property
Antiparallel vector to
Equal vectors

2 • Chapter Review 87

88 2 • Chapter Review

Components of the resultant of N vectors

General unit vector
Definition of the scalar product
Commutative property of the scalar product
Distributive property of the scalar product
Scalar product in terms of scalar components of vectors
Cosine of the angle between two vectors
Dot products of unit vectors
Magnitude of the vector product (definition)
Anticommutative property of the vector product
Distributive property of the vector product

Cross products of unit vectors

The cross product in terms of scalar components of vectors
Summary
2.1 Scalars and Vectors
• A vector quantity is any quantity that has magnitude and direction, such as displacement
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or velocity. Vector quantities are represented by mathematical objects called vectors. • Geometrically, vectors are represented by