Ch4Forces and Newton's Laws of Motion

F O R C E S A N D N E W T O N ’ S L AW S O F M O T I O N

&

1

As this windsurfer is propelled through the air, his motion is determined by forces due to the wind and his weight.The relationship between the forces acting on an object and the resulting motion is discussed in this chapter. (Tom

King/The Image Bank/Getty Images)

C | H | A | P | T | E | R 4

THE CONCEPTS OF FORCE AND MASS

In common usage, a force is a push or a pull, as the examples in Figure 4.1 illus-trate. In basketball, a player launches a shot by pushing on the ball. The tow bar attachedto a speeding boat pulls a water skier. Forces such as those that launch the basketball orpull the skier are called contact forces, because they arise from the physical contact be-tween two objects. There are circumstances, however, in which two objects exert forceson one another even though they are not touching. Such forces are referred to as noncon-tact forces or action-at-a-distance forces. One example of such a noncontact force occurswhen a diver is pulled toward the earth because of the force of gravity. The earth exertsthis force even when it is not in direct contact with the diver. In Figure 4.1, arrows areused to represent the forces. It is appropriate to use arrows, because a force is a vectorquantity and has both a magnitude and a direction. The direction of the arrow gives thedirection of the force, and the length is proportional to its strength or magnitude.

4.1

Cutnell

Johnson

PHYSICS, 7e

coming

January 2006

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2 | CHAPTER 4 | FORCES AND NEW TON’S LAWS OF MOTION

The word mass is just as familiar as the word force. A massive supertanker, for in-stance, is one that contains an enormous amount of mass. As we will see in the next sec-tion, it is difficult to set such a massive object into motion and difficult to bring it to a halt once it is moving. In comparison, a penny does not contain much mass. The empha-sis here is on the amount of mass, and the idea of direction is of no concern. Therefore,mass is a scalar quantity.

During the seventeenth century, Isaac Newton, starting with the work of Galileo, de-veloped three important laws that deal with force and mass. Collectively they are called“Newton’s laws of motion” and provide the basis for understanding the effect that forceshave on an object. Because of the importance of these laws, a separate section will be de-voted to each one.

NEWTON’S FIRST LAW OF MOTION

THE FIRST LAW

To gain some insight into Newton’s first law, think about the game of ice hockey(Figure 4.2). If a player does not hit a stationary puck, it will remain at rest on the ice. Af-ter the puck is struck, however, it coasts on its own across the ice, slowing down onlyslightly because of friction. Since ice is very slippery, there is only a relatively smallamount of friction to slow down the puck. In fact, if it were possible to remove all frictionand wind resistance, and if the rink were infinitely large, the puck would coast forever in astraight line at a constant speed. Left on its own, the puck would lose none of the velocityimparted to it at the time it was struck. This is the essence of Newton’s first law of motion:

(c)

F

(b)

F

Figure 4.1 The arrow labeled represents the force that acts on (a) thebasketball, (b) the water skier, and (c) the cliff diver. (a, © Nathaniel S.Butler/NBAE/Getty Images; b, © P.BeavisMasterfile; c, © Amy and ChuckWiley/Wales/Index Stock)

FB

NEWTON’S FIRST LAW OF MOTION

An object continues in a state of rest or in a state of motion at a constant speed along astraight line, unless compelled to change that state by a net force.

In the first law the phrase “net force” is crucial. Often, several forces act simultane-ously on a body, and the net force is the vector sum of all of them. Individual forces mat-ter only to the extent that they contribute to the total. For instance, if friction and other op-posing forces were absent, a car could travel forever at 30 m/s in a straight line, withoutusing any gas after it has come up to speed. In reality gas is needed, but only so that theengine can produce the necessary force to cancel opposing forces such as friction. Thiscancellation ensures that there is no net force to change the state of motion of the car.

When an object moves at a constant speed along a straight line, its velocity is con-stant. Newton’s first law indicates that a state of rest (zero velocity) and a state of constant velocity are completely equivalent, in the sense that neither one requires the ap-plication of a net force to sustain it. The purpose served when a net force acts on an ob-ject is not to sustain the object’s velocity, but, rather, to change it.

4.2(a)

F

Figure 4.2 The game of ice hockey cangive some insight into Newton’s lawsof motion. (© Royalty-Free/Corbis Images)

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4.1 | NEW TON’S F IRST LAW OF MOTION | 3

The SI unit for mass is the kilogram (kg), whereas the units in the CGS system andthe BE system are the gram (g) and the slug (sl), respectively. Conversion factors be-tween these units are given on the page facing the inside of the front cover. Figure 4.3gives the masses of various objects, ranging from a penny to a supertanker. The larger themass, the greater is the inertia. Often the words “mass” and “weight” are used inter-changeably, but this is incorrect. Mass and weight are different concepts, and Section 4.7will discuss the distinction between them.

Figure 4.4 shows a useful application of inertia. Automobile seat belts unwind freelywhen pulled gently, so they can be buckled. But in an accident, they hold you safely inplace. One seat-belt mechanism consists of a ratchet wheel, a locking bar, and a pendu-lum. The belt is wound around a spool mounted on the ratchet wheel. While the car is atrest or moving at a constant velocity, the pendulum hangs straight down, and the lockingbar rests horizontally, as the gray part of the drawing shows. Consequently, nothing pre-vents the ratchet wheel from turning, and the seat belt can be pulled out easily. When thecar suddenly slows down in an accident, however, the relatively massive lower part of thependulum keeps moving forward because of its inertia. The pendulum swings on its pivotinto the position shown in color and causes the locking bar to block the rotation of theratchet wheel, thus preventing the seat belt from unwinding.

AN INERTIAL REFERENCE FRAME

Newton’s first law (and also the second law) can appear to be invalid to certainobservers. Suppose, for instance, that you are a passenger riding in a friend’s car. Whilethe car moves at a constant speed along a straight line, you do not feel the seat pushingagainst your back to any unusual extent. This experience is consistent with the first law,which indicates that in the absence of a net force you should move with a constant veloc-ity. Suddenly the driver floors the gas pedal. Immediately you feel the seat pressingagainst your back as the car accelerates. Therefore, you sense that a force is being appliedto you. The first law leads you to believe that your motion should change, and, relative tothe ground outside, your motion does change. But relative to the car, you can see thatyour motion does not change, because you remain stationary with respect to the car.Clearly, Newton’s first law does not hold for observers who use the accelerating car as aframe of reference. As a result, such a reference frame is said to be noninertial. All accel-erating reference frames are noninertial. In contrast, observers for whom the law of iner-tia is valid are said to be using inertial reference frames for their observations, as definedbelow:

DEFINITION OF AN INERTIAL REFERENCE FRAME

An inertial reference frame is one in which Newton’s law of inertia is valid.

INERTIA AND MASS

A greater net force is required to change the velocity of some objects than of others. For instance, a net force that is just enough to cause a bicycle to pick up speedwill cause only an imperceptible change in the motion of a freight train. In comparison tothe bicycle, the train has a much greater tendency to remain at rest. Accordingly, we saythat the train has more inertia than the bicycle. Quantitatively, the inertia of an object ismeasured by its mass. The following definition of inertia and mass indicates why New-ton’s first law is sometimes called the law of inertia:

Figure 4.4 Inertia plays a central rolein one seat-belt mechanism. The graypart of the drawing applies when thecar is at rest or moving at a constantvelocity. The colored parts show whathappens when the car suddenly slowsdown, as in an accident.

Figure 4.3 The masses of variousobjects.

Seatbelt

Motion of car

Ratchetwheel

Lockingbar

Pivots

Pendulum

The physics ofseat belts.

DEFINITION OF INERTIA AND MASS

Inertia is the natural tendency of an object to remain at rest or in motion at a constantspeed along a straight line. The mass of an object is a quantitative measure of inertia.

SI Unit of Inertia and Mass: kilogram (kg)

INGOD WE TRUST

LIBERTY1996

Penny(0.003 kg)

Book(2 kg)

Bicycle(15 kg)

Car(2000 kg)

Jetliner(1.2 x 105 kg)

Supertanker(1.5 x 108 kg)

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The acceleration of an inertial reference frame is zero, so it moves with a constant veloc-ity. All of Newton’s laws of motion are valid in inertial reference frames, and when weapply these laws, we will be assuming such a reference frame. In particular, the earth it-self is a good approximation of an inertial reference frame.

NEWTON’S SECOND LAW OF MOTION

Newton’s first law indicates that if no net force acts on an object, then the velocityof the object remains unchanged. The second law deals with what happens when a netforce does act. Consider a hockey puck once again. When a player strikes a stationarypuck, he causes the velocity of the puck to change. In other words, he makes the puck ac-celerate. The cause of the acceleration is the force that the hockey stick applies. As longas this force acts, the velocity increases, and the puck accelerates. Now, suppose anotherplayer strikes the puck and applies twice as much force as the first player does. Thegreater force produces a greater acceleration. In fact, if the friction between the puck andthe ice is negligible, and if there is no wind resistance, the acceleration of the puck is di-rectly proportional to the force. Twice the force produces twice the acceleration. More-over, the acceleration is a vector quantity, just as the force is, and points in the same di-rection as the force.

Often, several forces act on an object simultaneously. Friction and wind resistance,for instance, do have some effect on a hockey puck. In such cases, it is the net force, orthe vector sum of all the forces acting, that is important. Mathematically, the net force iswritten as � , where the Greek capital letter � (sigma) denotes the vector sum. Newton’ssecond law states that the acceleration is proportional to the net force acting on the object.

In Newton’s second law, the net force is only one of two factors that determine theacceleration. The other is the inertia or mass of the object. After all, the same net forcethat imparts an appreciable acceleration to a hockey puck (small mass) will impart verylittle acceleration to a semitrailer truck (large mass). Newton’s second law states that for agiven net force, the magnitude of the acceleration is inversely proportional to the mass.Twice the mass means one-half the acceleration, if the same net force acts on both ob-jects. Thus, the second law shows how the acceleration depends on both the net force andthe mass, as given in Equation 4.1.

FB

NEWTON’S SECOND LAW OF MOTION

When a net external force � acts on an object of mass m, the acceleration that re-sults is directly proportional to the net force and has a magnitude that is inversely pro-portional to the mass. The direction of the acceleration is the same as the direction ofthe net force.

(4.1)

SI Unit of Force: (N)kg �m/s2 � newton

aB ��FB

m or �F

B� maB

aBFB

Note that the net force in Equation 4.1 includes only the forces that the environmentexerts on the object of interest. Such forces are called external forces. In contrast, inter-nal forces are forces that one part of an object exerts on another part of the object and arenot included in Equation 4.1.

According to Equation 4.1, the SI unit for force is the unit for mass (kg) times theunit for acceleration (m/s2), or

The combination of kg � m/s2 is called a newton (N) and is a derived SI unit, not a baseunit; 1 newton � 1 N � 1 kg � m/s2.

SI unit for force � (kg) � m

s2 � �kg �m

s2

4.3

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4.3 | NEW TON’S SECOND LAW OF MOTION | 5

In the CGS system, the procedure for establishing the unit of force is the same aswith SI units, except that mass is expressed in grams (g) and acceleration in cm/s2. Theresulting unit for force is the dyne; 1 dyne � 1 g � cm/s2.

In the BE system, the unit for force is defined to be the pound (lb),* and the unit foracceleration is ft /s2. With this procedure, Newton’s second law can then be used to obtainthe unit for mass:

The combination of lb � s2/ft is the unit for mass in the BE system and is called the slug(sl); 1 slug � 1 sl � 1 lb � s2/ft.

Table 4.1 summarizes the various units for mass, acceleration, and force. Conversionfactors between force units from different systems are provided on the page facing the in-side of the front cover.

When using the second law to calculate the acceleration, it is necessary to determine thenet force that acts on the object. In this determination a free-body diagram helps enormously.A free-body diagram is a diagram that represents the object and the forces that act on it. Onlythe forces that act on the object appear in a free-body diagram. Forces that the object exertson its environment are not included. Example 1 illustrates the use of a free-body diagram.

▼Example 1 | Pushing a Stalled Car

Two people are pushing a stalled car, as Figure 4.5a indicates. The mass of the car is 1850 kg.One person applies a force of 275 N to the car, while the other applies a force of 395 N. Bothforces act in the same direction. A third force of 560 N also acts on the car, but in a directionopposite to that in which the people are pushing. This force arises because of friction and theextent to which the pavement opposes the motion of the tires. Find the acceleration of the car.

Reasoning According to Newton’s second law, the acceleration is the net force divided by themass of the car. To determine the net force, we use the free-body diagram in Figure 4.5b. Inthis diagram, the car is represented as a dot, and its motion is along the �x axis. The diagrammakes it clear that the forces all act along one direction. Therefore, they can be added as colin-ear vectors to obtain the net force.

Solution From Equation 4.1, the acceleration is a � (�F)/m. The net force is

�F � �275 N � 395 N � 560 N � �110 N

Unit for mass �lb �s2

ft

BE unit for force � lb � (unit for mass) � ft

s2 �

Table 4.1 Units for Mass, Acceleration, and Force

System Mass Acceleration Force

SI kilogram (kg) meter /second2 (m/s2) newton (N)CGS gram (g) centimeter /second2 (cm/s2) dyne (dyn)BE slug (sl) foot/second2 (ft/s2) pound (lb)

* We refer here to the gravitational version of the BE system, in which a force of one pound is defined to be thepull of the earth on a certain standard body at a location where the acceleration due to gravity is 32.174 ft/s2.

+x

+y

560 N 395 N

275 N

(b) Free-body diagram of the car(a)

Opposing force = 560 N

275 N

395 N Figure 4.5 (a) Two people push astalled car, in opposition to a forcecreated by friction and the pavement.(b) A free-body diagram that shows thehorizontal forces acting on the car.

Problem solving insight

A free-body diagram is very helpful whenapplying Newton’s second law. Alwaysstart a problem by drawing the free-bodydiagram.

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The acceleration can now be obtained:

(4.1)

The plus sign indicates that the acceleration points along the �x axis, in the same direction asthe net force.

▲

THE VECTOR NATURE OF NEWTON’S SECOND LAW OF MOTION

When a football player throws a pass, the direction of the force he applies to theball is important. Both the force and the resulting acceleration of the ball are vector quan-tities, as are all forces and accelerations. The directions of these vectors can be taken intoaccount in two dimensions by using x and y components. The net force � in Newton’ssecond law has components �Fx and �Fy , while the acceleration has components ax

and ay . Consequently, Newton’s second law, as expressed in Equation 4.1, can be writtenin an equivalent form as two equations, one for the x components and one for the y com-ponents:

(4.2a)

(4.2b)

This procedure is similar to that employed in Chapter 3 for the equations of two-dimen-sional kinematics (see Table 3.1). The components in Equations 4.2a and 4.2b are scalarcomponents and will be either positive or negative numbers, depending on whether theypoint along the positive or negative x or y axis. The remainder of this section deals withexamples that show how these equations are used.

▼Example 2 | Applying Newton’s Second Law Using Components

A man is stranded on a raft (mass of man and raft � 1300 kg), as shown in Figure 4.6a. Bypaddling, he causes an average force of 17 N to be applied to the raft in a direction due eastP

B

�Fy � may

�Fx � max

aBFB

�0.059 m/s2a ��F

m�

�110 N

1850 kg�


The direction of the acceleration is alwaysthe same as the direction of the net force.

+x

+y

+x

+y

A sin 67°

A = 15 N

P = 17 N

x = 48 m

y = 23 m

A cos 67°

67°

A

Free-body diagram of the raft

+x

ax

ay

N

S

W E

+y

+x

+y

(b)(a)

(d )(c)

B

Figure 4.6 (a) A man is paddling a raft,as in Examples 2 and 3. (b) The free-body diagram shows the forces and

that act on the raft. Forces acting onthe raft in a direction perpendicular tothe surface of the water play no role inthe examples and are omitted forclarity. (c) The raft’s accelerationcomponents ax and ay . (d) In 65 s, thecomponents of the raft’s displacementare x � 48 m and y � 23 m.

AB

PB

4.4

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4.4 | THE VECTOR NATURE OF NEW TON’S SECOND LAW OF MOTION | 7

(the �x direction). The wind also exerts a force on the raft. This force has a magnitude of15 N and points 67° north of east. Ignoring any resistance from the water, find the x and ycomponents of the raft’s acceleration.

Reasoning Since the mass of the man and the raft is known, Newton’s second law can beused to determine the acceleration components from the given forces. According to the formof the second law in Equations 4.2a and 4.2b, the acceleration component in a given directionis the component of the net force in that direction divided by the mass. As an aid in determin-ing the components �Fx and �Fy of the net force, we use the free-body diagram in Figure4.6b. In this diagram, the directions due east and due north are the �x and �y directions, re-spectively.

Solution Figure 4.6b shows the force components:

AB

The plus signs indicate that �Fx points in the direction of the �x axis and �Fy points in the di-rection of the �y axis. The x and y components of the acceleration point in the directions of�Fx and �Fy , respectively, and can now be calculated:

(4.2a)

(4.2b)

These acceleration components are shown in Figure 4.6c.▲

▼Example 3 | The Displacement of a Raft

At the moment the forces and begin acting on the raft in Example 2, the velocity of theraft is 0.15 m/s, in a direction due east (the �x direction). Assuming that the forces are main-tained for 65 s, find the x and y components of the raft’s displacement during this time interval.

Reasoning Once the net force acting on an object and the object’s mass have been used inNewton’s second law to determine the acceleration, it becomes possible to use the equationsof kinematics to describe the resulting motion. We know from Example 2 that the accelera-tion components are ax � �0.018 m/s2 and ay � �0.011 m/s2, and it is given here that theinitial velocity components are v0x � �0.15 m/s and v0y � 0 m/s. Thus, Equation 3.5a

and Equation 3.5b (y � can be used with t � 65 s to de-termine the x and y components of the raft’s displacement.

Solution According to Equations 3.5a and 3.5b, the x and y components of the displacementare

Figure 4.6d shows the final location of the raft.▲

23 my � v0yt � 12 ayt 2 � (0 m/s)(65 s) � 1

2 (0.011 m/s2)(65 s)2 �

48 mx � v0xt � 12 axt 2 � (0.15 m/s)(65 s) � 1

2 (0.018 m/s2)(65 s)2 �

� 0y t � 12 ayt 2)(x � � 0x t � 1

2 axt 2)

AB

PB

�0.011 m/s2ay ��Fy

m�

�14 N

1300 kg�

�0.018 m/s2ax ��Fx

m�

�23 N

1300 kg�

Force x Component y Component

�17 N 0 N�(15 N) cos 67° � �6 N �(15 N) sin 67° � �14 N

�Fy � �14 N�Fx � �17 N � 6 N � �23 N

ABPB

Need more practice?

Interactive LearningWare 4.1 Acatapult on an aircraft carrier is ca-pable of accelerating a 13 300-kgplane from 0 to 56.0 m/s in a dis-tance of 80.0 m. Find the net force,assumed constant, that the jet’sengine and the catapult exert onthe plane.

Related Homework: Problems 4, 6

For an interactive solution, go towww.wiley.com/college/cutnell

✔C H E C K Y O U R U N D E R S T A N D I N G 1

All of the following, except one, cause the acceleration of an object to double. Which one isit? (a) All forces acting on the object double. (b) The net force acting on the object doubles.(c) Both the net force acting on the object and the mass of the object double. (d) The massof the object is reduced by a factor of two. (The answer is given at the end of the book.)

Background: This problem depends on the concepts of force, net force, mass, andacceleration, because Newton’s second law of motion deals with them.

For similar questions (including calculational counterparts), consult Self-Assessment Test

4.1. This test is described at the end of Section 4.5.


Applications of Newton’s second lawalways involve the net external force,which is the vector sum of all the externalforces that act on an object. Eachcomponent of the net force leads to acorresponding component of theacceleration.

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NEWTON’S THIRD LAW OF MOTION

Imagine you are in a football game. You line up facing your opponent, the ball issnapped, and the two of you crash together. No doubt, you feel a force. But think aboutyour opponent. He too feels something, for while he is applying a force to you, you areapplying a force to him. In other words, there isn’t just one force on the line of scrim-mage; there is a pair of forces. Newton was the first to realize that all forces occur in pairsand there is no such thing as an isolated force, existing all by itself. His third law of mo-tion deals with this fundamental characteristic of forces.

These two wapiti (elk) exert action andreaction forces on each other. (FirstLight/Corbis Images)

–P+PB

B

Figure 4.7 The astronaut pushes on thespacecraft with a force � . Accordingto Newton’s third law, the spacecraftsimultaneously pushes back on theastronaut with a force � .P

B

PB

NEWTON’S THIRD LAW OF MOTION

Whenever one body exerts a force on a second body, the second body exerts an oppo-sitely directed force of equal magnitude on the first body.

The third law is often called the “action–reaction” law, because it is sometimes quoted asfollows: “For every action (force) there is an equal, but opposite, reaction.”

Figure 4.7 illustrates how the third law applies to an astronaut who is drifting justoutside a spacecraft and who pushes on the spacecraft with a force . According to thethird law, the spacecraft pushes back on the astronaut with a force � that is equal inmagnitude but opposite in direction. In Example 4, we examine the accelerations pro-duced by each of these forces.

▼Example 4 | The Accelerations Produced by Action and Reaction Forces

Suppose that the mass of the spacecraft in Figure 4.7 is mS � 11 000 kg and that the mass ofthe astronaut is mA � 92 kg. In addition, assume that the astronaut exerts a force of ��36 N on the spacecraft. Find the accelerations of the spacecraft and the astronaut.

Reasoning According to Newton’s third law, when the astronaut applies the force ��36 N to the spacecraft, the spacecraft applies a reaction force � � �36 N to the astronaut.As a result, the spacecraft and the astronaut accelerate in opposite directions. Although the ac-tion and reaction forces have the same magnitude, they do not create accelerations of the samemagnitude, because the spacecraft and the astronaut have different masses. According to New-ton’s second law, the astronaut, having a much smaller mass, will experience a much larger ac-celeration. In applying the second law, we note that the net force acting on the spacecraft is� � , while the net force acting on the astronaut is � � � .

Solution Using the second law, we find that the acceleration of the spacecraft is

The acceleration of the astronaut is

▲

�0.39 m/s2aBA ��P

B

mA�

�36 N

92 kg�

�0.0033 m/s2aBS �PB

mS�

�36 N

11 000 kg�

PB

FB

PB

FB

PB

PB

PB

PB

PB

4.5


Even though the magnitudes of the actionand reaction forces are always equal,these forces do not necessarily produceaccelerations that have equal magnitudes,since each force acts on a different objectthat may have a different mass.

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4.6 | TYPES OF FORCES: AN OVERVIEW | 9

There is a clever application of Newton’s third law in some rental trailers. As Figure4.8 illustrates, the tow bar connecting the trailer to the rear bumper of a car contains amechanism that can automatically actuate brakes on the trailer wheels. This mechanismworks without the need for electrical connections between the car and the trailer. Whenthe driver applies the car brakes, the car slows down. Because of inertia, however, thetrailer continues to roll forward and begins pushing against the bumper. In reaction,the bumper pushes back on the tow bar. The reaction force is used by the mechanism in the tow bar to “push the brake pedal” for the trailer.

Mechanism for actuatingtrailer brakes

Figure 4.8 Some rental trailers includean automatic brake-actuatingmechanism.

Newton’s Second Law

ΣF = ma

External Forces

1. Gravitational Force (Section 4.7)2. Normal Force (Section 4.8)3. Frictional Forces (Section 4.9)4. Tension Force (Section 4.10)

CONCEPTS AT A GLANCE

B

Figure 4.9 CONCEPTS AT AGLANCE When any of the fourexternal forces listed here act on anobject, they are included as part of thenet force � in any application ofNewton's second law. Each of the fourexternal forces acts on this (very)reluctant bull as the farmers join effortsto pull it aboard the boat. (DietherEndlicher/©AP/Wide World Photos)

FB

TYPES OF FORCES: AN OVERVIEW

� CONCEPTS AT A GLANCE Newton’s three laws of motion make it clear that forcesplay a central role in determining the motion of an object. In the next four sections somecommon forces will be discussed: the gravitational force (Section 4.7), the normal force(Section 4.8), frictional forces (Section 4.9), and the tension force (Section 4.10). Inlater chapters, we will encounter still others, such as electric and magnetic forces. It isimportant to realize that Newton’s second law is always valid, regardless of which ofthese forces may act on an object. One does not have a different law for every type ofcommon force. Thus, we need only to determine what forces are acting on an object, addthem together to form the net force, and then use Newton’s second law to determine theobject’s acceleration. The Concepts-at-a-Glance chart in Figure 4.9 illustrates this im-portant idea. �

The physics ofautomatic trailer brakes.

S E L F - A S S E S S M E N T T E S T 4 . 1 www.wiley.com/college/cutnell

Test your understanding of the material in Sections 4.1–4.5:

• Newton’s First Law • Newton’s Second Law • Newton’s Third Law

4.6

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In nature there are two general types of forces, fundamental and nonfundamental.Fundamental forces are the ones that are truly unique, in the sense that all other forcescan be explained in terms of them. Only three fundamental forces have been discovered:

1. Gravitational force

2. Strong nuclear force

3. Electroweak force

The gravitational force is discussed in the next section. The strong nuclear force plays aprimary role in the stability of the nucleus of the atom (see Section 31.2). The elec-troweak force is a single force that manifests itself in two ways (see Section 32.6). Onemanifestation is the electromagnetic force that electrically charged particles exert on oneanother (see Sections 18.5, 21.2, and 21.8). The other manifestation is the so-called weaknuclear force that plays a role in the radioactive disintegration of certain nuclei (see Sec-tion 31.5).

Except for the gravitational force, all of the forces discussed in this chapter are non-fundamental, because they are related to the electromagnetic force. They arise from theinteractions between the electrically charged particles that comprise atoms and molecules.Our understanding of which forces are fundamental, however, is continually evolving. Forinstance, in the 1860s and 1870s James Clerk Maxwell showed that the electric force andthe magnetic force could be explained as manifestations of a single electromagnetic force.Then, in the 1970s, Sheldon Glashow (1932– ), Abdus Salam (1926–1996), and StevenWeinberg (1933– ) presented the theory that explains how the electromagnetic force andthe weak nuclear force are related to the electroweak force. They received a Nobel prizein 1979 for their achievement. Today, efforts continue that have the goal of further reduc-ing the number of fundamental forces.

THE GRAVITATIONAL FORCE

NEWTON’S LAW OF UNIVERSAL GRAVITATION

Objects fall downward because of gravity, and Chapters 2 and 3 discuss how todescribe the effects of gravity by using a value of g � 9.80 m/s2 for the downward accel-eration it causes. However, nothing has been said about why g is 9.80 m/s2. The reason isfascinating, as we will now see.

The acceleration due to gravity is like any other acceleration, and Newton’s secondlaw indicates that it must be caused by a net force. In addition to his famous three laws ofmotion, Newton also provided a coherent understanding of the gravitational force. His“law of universal gravitation” is stated as follows:

The constant G that appears in Equation 4.3 is called the universal gravitationalconstant, because it has the same value for all pairs of particles anywhere in the universe,no matter what their separation. The value for G was first measured in an experiment by

NEWTON’S LAW OF UNIVERSAL GRAVITATION

Every particle in the universe exerts an attractive force on every other particle. A parti-cle is a piece of matter, small enough in size to be regarded as a mathematical point.For two particles that have masses m1 and m 2 and are separated by a distance r, theforce that each exerts on the other is directed along the line joining the particles (seeFigure 4.10) and has a magnitude given by

(4.3)

The symbol G denotes the universal gravitational constant, whose value is found ex-perimentally to be

G � 6.673 � 10�11 N�m2/kg2

F � G m1m2

r2

r

m1 m2–F+FB B

Figure 4.10 The two particles, whosemasses are m1 and m 2 , are attracted bygravitational forces � and � .F

BFB

4.7

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4.7 | THE GRAVITATIONAL FORCE | 11

the English scientist Henry Cavendish (1731–1810), more than a century after Newtonproposed his law of universal gravitation.

To see the main features of Newton’s law of universal gravitation, look at the twoparticles in Figure 4.10. They have masses m1 and m 2 and are separated by a distance r.In the picture, it is assumed that a force pointing to the right is positive. The gravitationalforces point along the line joining the particles and are

� , the gravitational force exerted on particle 1 by particle 2

� , the gravitational force exerted on particle 2 by particle 1

These two forces have equal magnitudes and opposite directions. They act on differentbodies, causing them to be mutually attracted. In fact, these forces are an action–reactionpair, as required by Newton’s third law. Example 5 shows that the magnitude of the gravi-tational force is extremely small for ordinary values of the masses and the distance be-tween them.

▼Example 5 | Gravitational Attraction

What is the magnitude of the gravitational force that acts on each particle in Figure 4.10, as-suming m1 � 12 kg (approximately the mass of a bicycle), m2 � 25 kg, and r � 1.2 m?

Reasoning and Solution The magnitude of the gravitational force can be found using Equa-tion 4.3:

For comparison, you exert a force of about 1 N when pushing a doorbell, so that the gravita-tional force is exceedingly small in circumstances such as those here. This result is due to thefact that G itself is very small. However, if one of the bodies has a large mass, like that of theearth (5.98 � 1024 kg), the gravitational force can be large.

▲

As expressed by Equation 4.3, Newton’s law of gravitation applies only to particles.However, most familiar objects are too large to be considered particles. Nevertheless, thelaw of universal gravitation can be applied to such objects with the aid of calculus. New-ton was able to prove that an object of finite size can be considered to be a particle forpurposes of using the gravitation law, provided the mass of the object is distributed withspherical symmetry about its center. Thus, Equation 4.3 can be applied when each objectis a sphere whose mass is spread uniformly over its entire volume. Figure 4.11 shows thiskind of application, assuming that the earth and the moon are such uniform spheres ofmatter. In this case, r is the distance between the centers of the spheres and not the dis-tance between the outer surfaces. The gravitational forces that the spheres exert on eachother are the same as if the entire mass of each were concentrated at its center. Even if theobjects are not uniform spheres, Equation 4.3 can be used to a good degree of approxima-tion if the sizes of the objects are small relative to the distance of separation r.

WEIGHT

The weight of an object arises because of the gravitational pull of the earth.

1.4 � 10�8 NF � G m1m 2

r 2 � (6.67 � 10�11 N �m2/kg2) (12 kg)(25 kg)

(1.2 m)2 �

FB

FB

Moon

r

Earth

ME

ME

MM

MM+F –FB B

+F –FB B

Figure 4.11 The gravitational force thateach uniform sphere of matter exerts onthe other is the same as if each spherewere a particle with its massconcentrated at its center. The earth(mass ME) and the moon (mass MM)approximate such uniform spheres.

DEFINITION OF WEIGHT

The weight of an object on or above the earth is the gravitational force that the earthexerts on the object. The weight always acts downward, toward the center of the earth.On or above another astronomical body, the weight is the gravitational force exertedon the object by that body.

SI Unit of Weight: newton (N)

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Using W for the magnitude of the weight,* m for the mass of the object, and ME forthe mass of the earth, it follows from Equation 4.3 that

(4.4)

Equation 4.4 and Figure 4.12 both emphasize that an object has weight whether or not itis resting on the earth’s surface, because the gravitational force is acting even when thedistance r is not equal to the radius RE of the earth. However, the gravitational force be-comes weaker as r increases, since r is in the denominator of Equation 4.4. Figure 4.13,for example, shows how the weight of the Hubble Space Telescope becomes smaller asthe distance r from the center of the earth increases. In Example 6 the telescope’s weightis determined when it is on earth and in orbit.

▼Example 6 | The Hubble Space Telescope

The mass of the Hubble Space Telescope is 11 600 kg. Determine the weight of the telescope(a) when it was resting on the earth and (b) as it is in its orbit 598 km above the earth’s surface.

Reasoning The weight of the Hubble Space Telescope is the gravitational force exerted on itby the earth. According to Equation 4.4, the weight varies inversely as the square of the radialdistance r. Thus, we expect the telescope’s weight on the earth’s surface (r smaller) to begreater than its weight in orbit (r larger).

Solution (a) On the earth’s surface, the weight is given by Equation 4.4 with r � 6.38 �106 m (the earth’s radius):

(b) When the telescope is 598 km above the surface, its distance from the center of the earth is

The weight now can be calculated as in part (a), except that the new value of r must be used:

. As expected, the weight is less in orbit.

▲

The space age has forced us to broaden our ideas about weight. For instance, an as-tronaut weighs only about one-sixth as much on the moon as on the earth. To obtain hisweight on the moon from Equation 4.4, it is only necessary to replace ME by MM (themass of the moon) and let r � RM (the radius of the moon).

RELATION BETWEEN MASS AND WEIGHT

Although massive objects weigh a lot on the earth, mass and weight are not thesame quantity. As Section 4.2 discusses, mass is a quantitative measure of inertia. Assuch, mass is an intrinsic property of matter and does not change as an object is movedfrom one location to another. Weight, in contrast, is the gravitational force acting on theobject and can vary, depending on how far the object is above the earth’s surface orwhether it is located near another body such as the moon.

The relation between weight W and mass m can be written in two ways:

(4.4)

(4.5)gW � m

mG

ME

r 2W �

W � 0.950 � 105 N

r � 6.38 � 106 m � 598 � 103 m � 6.98 � 106 m

W � 1.14 � 105 N

W � G ME m

r 2 �(6.67 � 10�11 N �m2/kg2)(5.98 � 10 24 kg)(11 600 kg)

(6.38 � 106 m)2

W � G MEm

r2

r

W

Object of mass m

RE

Mass of earth = ME

5

0.2

0

0.4

0.6

0.8

1.0

10 15 20

r (× 106 m)

RE = 6.38 x 106 m

W (×

10

5 N

)

Figure 4.12 On or above the earth, theweight of an object is thegravitational force exerted on the objectby the earth.

WoB

Figure 4.13 The weight of the HubbleSpace Telescope decreases as thetelescope gets farther from the earth.The distance from the center of theearth to the telescope is r.

* Often, the word “weight” and the phrase “magnitude of the weight” are used interchangeably, even thoughweight is a vector. Generally, the context makes it clear when the direction of the weight vector must be taken into account.

p


When applying Newton’s gravitation lawto uniform spheres of matter, rememberthat the distance r is between the centersof the spheres, not between the surfaces.

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4.7 | THE GRAVITATIONAL FORCE | 13

Equation 4.4 is Newton’s law of universal gravitation, and Equation 4.5 is Newton’s sec-ond law (net force equals mass times acceleration) incorporating the acceleration g due togravity. These expressions make the distinction between mass and weight stand out. Theweight of an object whose mass is m depends on the values for the universal gravitationalconstant G, the mass ME of the earth, and the distance r. These three parameters togetherdetermine the acceleration g due to gravity. The specific value of g � 9.80 m/s2 appliesonly when r equals the radius RE of the earth. For larger values of r, as would be the caseon top of a mountain, the effective value of g is less than 9.80 m/s2. The fact that g de-creases as the distance r increases means that the weight likewise decreases. The mass ofthe object, however, does not depend on these effects and does not change. ConceptualExample 7 further explores the difference between mass and weight.

▼Conceptual Example 7 | Mass Versus Weight

A vehicle is being designed for use in exploring the moon’s surface and is being tested onearth, where it weighs roughly six times more than it will on the moon. In one test, the accel-eration of the vehicle along the ground is measured. To achieve the same acceleration on themoon, will the net force acting on the vehicle be greater than, less than, or the same as that re-quired on earth?

Reasoning and Solution The net force � required to accelerate the vehicle is specified byNewton’s second law as � � m , where m is the vehicle’s mass and is the accelerationalong the ground. For a given acceleration, the net force depends only on the mass. But themass is an intrinsic property of the vehicle and is the same on the moon as it is on the earth.Therefore, the same net force would be required for a given acceleration on the moon as onthe earth. Do not be misled by the fact that the vehicle weighs more on earth. The greaterweight occurs only because the earth’s mass and radius are different than the moon’s. In anyevent, in Newton’s second law, the net force is proportional to the vehicle’s mass, not itsweight.

Related Homework: Problems 22, 87▲

aBaBFB

FB

The Lunar Roving Vehicle thatastronaut Eugene Cernan is driving onthe moon and the Lunar ExcursionModule (behind the Roving Vehicle)have the same mass that they have onthe earth. However, their weight isdifferent on the moon than on the earth,as Conceptual Example 7 discusses.(NASA/Johnson Space Center)


One object has a mass m1 , and a second object has a mass m2 , which is greater than m1 .The two are separated by a distance 2d. A third object has a mass m3 . All three objects arelocated on the same straight line. The net gravitational force acting on the third object iszero. Which of the drawings correctly represents the locations of the objects? The answer isgiven at the end of the book.)

m2m1 m3

dd

m2m1 m3

dd

m2m1 m3

dd

(a) (b)

m2m1 m3

dd

(d)

(c)

Background: The gravitational force and Newton’s law of universal gravitation are the focusof this problem.




Mass and weight are different quantities.They cannot be interchanged when solvingproblems.

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THE NORMAL FORCE

THE DEFINITION AND INTERPRETATION OF THE NORMAL FORCE

In many situations, an object is in contact with a surface, such as a tabletop. Be-cause of the contact, there is a force acting on the object. The present section discussesonly one component of this force, the component that acts perpendicular to the surface.The next section discusses the component that acts parallel to the surface. The perpendic-ular component is called the normal force.

Figure 4.14 shows a block resting on a horizontal table and identifies the two forcesthat act on the block, the weight and the normal force N . To understand how an inani-mate object, such as a tabletop, can exert a normal force, think about what happens whenyou sit on a mattress. Your weight causes the springs in the mattress to compress. As a re-sult, the compressed springs exert an upward force (the normal force) on you. In a similarmanner, the weight of the block causes invisible “atomic springs” in the surface of thetable to compress, thus producing a normal force on the block.

Newton’s third law plays an important role in connection with the normal force. InFigure 4.14, for instance, the block exerts a force on the table by pressing down on it.Consistent with the third law, the table exerts an oppositely directed force of equal magni-tude on the block. This reaction force is the normal force. The magnitude of the normalforce indicates how hard the two objects press against each other.

If an object is resting on a horizontal surface and there are no vertically acting forcesexcept the object’s weight and the normal force, the magnitudes of these two forces areequal; that is, FN � W. This is the situation in Figure 4.14. The weight must be balancedby the normal force for the object to remain at rest on the table. If the magnitudes of theseforces were not equal, there would be a net force acting on the block, and the block wouldaccelerate either upward or downward, in accord with Newton’s second law.

If other forces in addition to and N act in the vertical direction, the magnitudes of the normal force and the weight are no longer equal. In Figure 4.15a, forinstance, a box whose weight is 15 N is being pushed downward against a table. Thepushing force has a magnitude of 11 N. Thus, the total downward force exerted on thebox is 26 N, and this must be balanced by the upward-acting normal force if the box is toremain at rest. In this situation, then, the normal force is 26 N, which is considerablylarger than the weight of the box.

Figure 4.15b illustrates a different situation. Here, the box is being pulled upward bya rope that applies a force of 11 N. The net force acting on the box due to its weight andthe rope is only 4 N, downward. To balance this force, the normal force needs to be only4 N. It is not hard to imagine what would happen if the force applied by the rope were in-creased to 15 N—exactly equal to the weight of the box. In this situation, the normalforce would become zero. In fact, the table could be removed, since the block would besupported entirely by the rope. The situations in Figure 4.15 are consistent with the ideathat the magnitude of the normal force indicates how hard two objects press against eachother. Clearly, the box and the table press against each other harder in part a of the picturethan in part b.

Like the box and the table in Figure 4.15, various parts of the human body pressagainst one another and exert normal forces. Example 8 illustrates the remarkable abilityof the human skeleton to withstand a wide range of normal forces.

FB

WoB

FB

WoB

DEFINITION OF THE NORMAL FORCE

The normal force N is one component of the force that a surface exerts on an objectwith which it is in contact—namely, the component that is perpendicular to the surface.

FB

FN

W

B

11 N

FN = 26 N

W = 15 N(a)

11 N

W = 15 N(b)

FN = 4 N

B

B

Figure 4.14 Two forces act on theblock, its weight and the normalforce N exerted by the surface of thetable.

FB

WoB

Figure 4.15 (a) The normal force N isgreater than the weight of the box,because the box is being presseddownward with an 11-N force. (b) Thenormal force is smaller than the weight,because the rope supplies an upwardforce of 11 N that partially supports thebox.

FB

4.8

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4.8 | THE NORMAL FORCE | 15

▼Example 8 | A Balancing Act

In a circus balancing act, a woman performs a headstand on top of a standing performer’shead, as Figure 4.16a illustrates. The woman weighs 490 N, and the standing performer’s headand neck weigh 50 N. It is primarily the seventh cervical vertebra in the spine that supports allthe weight above the shoulders. What is the normal force that this vertebra exerts on the neckand head of the standing performer (a) before the act and (b) during the act?

Reasoning To begin, we draw a free-body diagram for the neck and head of the standing per-former. Before the act, there are only two forces, the weight of the standing performer’s headand neck, and the normal force. During the act, an additional force is present due to thewoman’s weight. In both cases, the upward and downward forces must balance for the headand neck to remain at rest. This condition of balance will lead us to values for the normalforce.

Solution (a) Figure 4.16b shows the free-body diagram for the standing performer’s headand neck before the act. The only forces acting are the normal force N and the 50-N weight.These two forces must balance for the standing performer’s head and neck to remain at rest.Therefore, the seventh cervical vertebra exerts a normal force of .

(b) Figure 4.16c shows the free-body diagram that applies during the act. Now, the totaldownward force exerted on the standing performer’s head and neck is 50 N � 490 N � 540N, which must be balanced by the upward normal force, so that .

▲

In summary, the normal force does not necessarily have the same magnitude as theweight of the object. The value of the normal force depends on what other forces are present. It also depends on whether the objects in contact are accelerating. In one situa-tion that involves accelerating objects, the magnitude of the normal force can be regardedas a kind of “apparent weight,” as we will now see.

APPARENT WEIGHT

Usually, the weight of an object can be determined with the aid of a scale. How-ever, even though a scale is working properly, there are situations in which it does notgive the correct weight. In such situations, the reading on the scale gives only the “appar-ent” weight, rather than the gravitational force or “true” weight. The apparent weight isthe force that the object exerts on the scale with which it is in contact.

FN � 540 N

FN � 50 N

FB

+x

+y

+x

+y

FN

FN

FN

FN

50 N

50 N

50 N

490 N

490 N

50 N

(b)(a) (c)

Seventhcervicalvertebra

Free-body diagram Free-body diagram

Figure 4.16 (a) A young woman keepsher balance during a performance byChina’s Sincuan Acrobatic group. Afree-body diagram is shown (above theshoulders) for the standing performer(b) before the act and (c) during the act.For convenience, the scales used for thevectors in parts b and c are different. (©SUPRI/Reuters/Landov LLC)

The physics of the human skeleton.

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To see the discrepancies that can arise between true weight and apparent weight, con-sider the scale in the elevator in Figure 4.17. The reasons for the discrepancies will be ex-plained shortly. A person whose true weight is 700 N steps on the scale. If the elevator isat rest or moving with a constant velocity (either upward or downward), the scale regis-ters the true weight, as Figure 4.17a illustrates.

If the elevator is accelerating, the apparent weight and the true weight are not equal.When the elevator accelerates upward, the apparent weight is greater than the true weight,as Figure 4.17b shows. Conversely, if the elevator accelerates downward, as in part c, theapparent weight is less than the true weight. In fact, if the elevator falls freely, so its ac-celeration is equal to the acceleration due to gravity, the apparent weight becomes zero,as part d indicates. In a situation such as this, where the apparent weight is zero, the per-son is said to be “weightless.” The apparent weight, then, does not equal the true weight ifthe scale and the person on it are accelerating.

The discrepancies between true weight and apparent weight can be understood withthe aid of Newton’s second law. Figure 4.18 shows a free-body diagram of the person inthe elevator. The two forces that act on him are the true weight � m and the normalforce N exerted by the platform of the scale. Applying Newton’s second law in the verti-cal direction gives

where a is the acceleration of the elevator and person. In this result, the symbol g standsfor the magnitude of the acceleration due to gravity and can never be a negative quantity.However, the acceleration a may be either positive or negative, depending on whether theelevator is accelerating upward (�) or downward (�). Solving for the normal force FN

shows that

(4.6)

In Equation 4.6, FN is the magnitude of the normal force exerted on the person by thescale. But in accord with Newton’s third law, FN is also the magnitude of the downwardforce that the person exerts on the scale—namely, the apparent weight.

Equation 4.6 contains all the features shown in Figure 4.17. If the elevator is not ac-celerating, a � 0 m/s2, and the apparent weight equals the true weight. If the elevator ac-celerates upward, a is positive, and the equation shows that the apparent weight is greaterthan the true weight. If the elevator accelerates downward, a is negative, and the apparentweight is less than the true weight. If the elevator falls freely, a � �g, and the apparentweight is zero. The apparent weight is zero because when both the person and the scalefall freely, they cannot push against one another. In this text, when the weight is given, itis assumed to be the true weight, unless stated otherwise.

True123

weightApparent

123

weight

FN � mg � ma

�Fy � �FN � mg � ma

FB

gBWoB

(a) No acceleration (v = constant) (b) Upward acceleration (c) Downward acceleration (d) Free-fall

0

700

W = 700 N

0

W = 700 N

0

400

W = 700 N

0

W = 700 N

a a

a = g

1000

Figure 4.17 (a) When the elevator isnot accelerating, the scale registers thetrue weight (W � 700 N) of the person.(b) When the elevator acceleratesupward, the apparent weight (1000 N)exceeds the true weight. (c) When theelevator accelerates downward, theapparent weight (400 N) is less than thetrue weight. (d ) The apparent weight iszero if the elevator falls freely—that is,if it falls with the acceleration due togravity.

+x

W = mg

FN

+y

B

Figure 4.18 A free-body diagramshowing the forces acting on the personriding in the elevator of Figure 4.17. is the true weight, and N is the normalforce exerted on the person by theplatform of the scale.

FB

WoB

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4.9 | STATIC AND KINETIC FRICTIONAL FORCES | 17

STATIC AND KINETIC FRICTIONAL FORCES

When an object is in contact with a surface, there is a force acting on the object.The previous section discusses the component of this force that is perpendicular to thesurface, which is called the normal force. When the object moves or attempts to movealong the surface, there is also a component of the force that is parallel to the surface.This parallel force component is called the frictional force, or simply friction.

In many situations considerable engineering effort is expended trying to reduce fric-tion. For example, oil is used to reduce the friction that causes wear and tear in the pis-tons and cylinder walls of an automobile engine. Sometimes, however, friction is ab-solutely necessary. Without friction, car tires could not provide the traction needed tomove the car. In fact, the raised tread on a tire is designed to maintain friction. On a wetroad, the spaces in the tread pattern (see Figure 4.19) provide channels for the water tocollect and be diverted away. Thus, these channels largely prevent the water from comingbetween the tire surface and the road surface, where it would reduce friction and allowthe tire to skid.

Surfaces that appear to be highly polished can actually look quite rough when exam-ined under a microscope. Such an examination reveals that two surfaces in contact touchonly at relatively few spots, as Figure 4.20 illustrates. The microscopic area of contact forthese spots is substantially less than the apparent macroscopic area of contact between thesurfaces—perhaps thousands of times less. At these contact points the molecules of thedifferent bodies are close enough together to exert strong attractive intermolecular forceson one another, leading to what are known as “cold welds.” Frictional forces are associ-ated with these welded spots, but the exact details of how frictional forces arise are notwell understood. However, some empirical relations have been developed that make itpossible to account for the effects of friction.

Figure 4.21 helps to explain the main features of the type of friction known as staticfriction. The block in this drawing is initially at rest on a table, and as long as there is noattempt to move the block, there is no static frictional force. Then, a horizontal force isapplied to the block by means of a rope. If is small, as in part a, experience tells us thatthe block still does not move. Why? It does not move because the static frictional force s

exactly cancels the effect of the applied force. The direction of s is opposite to that of ,and the magnitude of s equals the magnitude of the applied force, fs � F. Increasing theapplied force in Figure 4.21 by a small amount still does not cause the block to move.There is no movement because the static frictional force also increases by an amount thatcancels out the increase in the applied force (see part b of the drawing). If the applied

fB

FB

fB

fB

FB

FB

Microscopiccontact points

Figure 4.19 This photo, shot fromunderneath a transparent surface, showsa tire rolling under wet conditions. Thechannels in the tire collect and divertwater away from the regions where thetire contacts the surface, thus providingbetter traction. (Courtesy Goodyear Tire &Rubber Co.)

Figure 4.20 Even when two highlypolished surfaces are in contact, theytouch only at relatively few points.

F

No movement(a)

No movement(b)

When movement just begins(c)

F

fsMAX

F

fs

fs

B

B

B

Figure 4.21 Applying a small force to the block, as in parts a and b,produces no movement, because thestatic frictional force s exactly balancesthe applied force. (c) The block justbegins to move when the applied forceis slightly greater than the maximumstatic frictional force s

MAX.fB

fB

FB

4.9

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force continues to increase, however, there comes a point when the block finally “breaksaway” and begins to slide. The force just before breakaway represents the maximum staticfrictional force that the table can exert on the block (see part c of the drawing). Anyapplied force that is greater than cannot be balanced by static friction, and the re-sulting net force accelerates the block to the right.

Experimental evidence shows that, to a good degree of approximation, the maxi-mum static frictional force between a pair of dry, unlubricated surfaces has two maincharacteristics. It is independent of the apparent macroscopic area of contact between the objects, provided that the surfaces are hard or nondeformable. For instance, in Figure 4.22 the maximum static frictional force that the surface of the table can exert on a block is the same, whether the block is resting on its largest or its smallest side. The other main characteristic of is that its magnitude is proportional to the mag-nitude of the normal force N . As Section 4.8 points out, the magnitude of the normalforce indicates how hard two surfaces are being pressed together. The harder they arepressed, the larger is presumably because the number of “cold-welded,” micro-scopic contact points is increased. Equation 4.7 expresses the proportionality between

and FN with the aid of a proportionality constant �s , which is called the coefficientof static friction.fs

MAX

fs

MAX,

FB

fB

s

MAX

fB

s

MAXfB

s

MAX

STATIC FRICTIONAL FORCE

The magnitude fs of the static frictional force can have any value from zero up to amaximum value of depending on the applied force. In other words, ,where the symbol “�” is read as “less than or equal to.” The equality holds only whenfs attains its maximum value, which is

(4.7)

In Equation 4.7, �s is the coefficient of static friction, and FN is the magnitude of thenormal force.

fs

MAX � �sFN

fs � fs

MAXfs

MAX,

Figure 4.22 The maximum staticfrictional force would be thesame, no matter which side of the blockis in contact with the table.

fB

s

MAX

It should be emphasized that Equation 4.7 relates only the magnitudes of and N ,not the vectors themselves. This equation does not imply that the directions of the vectorsare the same. In fact, is parallel to the surface, while N is perpendicular to it.

The coefficient of static friction, being the ratio of the magnitudes of two forceshas no units. Also, it depends on the type of material from which each

surface is made (steel on wood, rubber on concrete, etc.), the condition of the surfaces(polished, rough, etc.), and other variables such as temperature. Table 4.2 gives some typ-ical values of �s for various surfaces. Example 9 illustrates the use of Equation 4.7 for de-termining the maximum static frictional force.

(�s � fs

MAX/FN),

FB

fB

s

MAX

FB

fB

s

MAX

Table 4.2 Approximate Values of the Coefficients of Friction for Various Surfaces*

Coefficient of Static Coefficient of Materials Friction, �s Kinetic Friction, �k

Glass on glass (dry) 0.94 0.4Ice on ice (clean, 0 °C) 0.1 0.02Rubber on dry concrete 1.0 0.8Rubber on wet concrete 0.7 0.5Steel on ice 0.1 0.05Steel on steel (dry hard steel) 0.78 0.42Teflon on Teflon 0.04 0.04Wood on wood 0.35 0.3

*The last column gives the coefficients of kinetic friction, a concept that will be discussedshortly.

Need more practice?

Interactive LearningWare 4.2 Asofa rests on the horizontal bed ofa moving van. The coefficient ofstatic friction between the sofaand van bed is 0.30. The van startsfrom rest and accelerates for atime of 5.1 s. What is the maxi-mum distance that the van cantravel in this time period withouthaving the sofa slide?

Related Homework: Problem 82


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Example 9 | The Force Needed To Start a Skier Moving

A skier is standing motionless on a horizontal patch of snow. She is holding onto a hori-zontal tow rope, which is about to pull her forward (see Figure 4.23a). The skier’s mass is59 kg, and the coefficient of static friction between the skis and snow is 0.14. What is themagnitude of the maximum force that the tow rope can apply to the skier without causingher to move?

Reasoning When the rope applies a relatively small force, the skier does not accelerate.The reason is that the static frictional force opposes the applied force and the two forceshave the same magnitude. We can apply Newton’s second law in the horizontal directionto this situation. In order for the rope to pull the skier forward, it must exert a force largeenough to overcome the maximum static frictional force acting on the skis. The magnitudeof the maximum static frictional force depends on the coefficient of static friction (whichis known) and on the magnitude of the normal force. We can determine the magnitude ofthe normal force by using Newton's second law, along with the fact that the skier does notaccelerate in the vertical direction.

Knowns and Unknowns The data for this problem are:

Description Symbol Value

Mass of skier m 59 kgCoefficient of static friction �s 0.14

Unknown VariableMagnitude of maximum horizontal

force that tow rope can apply F ?

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

(a)

+y

+x

(b)

FN

mg

fsffMAX

F

Figure 4.23 (a) Two horizontal forcesact on the skier in the horizontaldirection just before she begins to move.(b) Two vertical forces act on the skier.Modeling the Problem

Newton’s Second Law (Horizontal Direction) Figure 4.23a shows the two

horizontal forces that act on the skier just before she begins to move: the force appliedby the tow rope and the maximum static frictional force . Since the skier is standingmotionless, she is not accelerating in the horizontal or x direction, so ax � 0 m/s2. Apply-ing Newton’s second law (Equation 4.2a) to this situation, we have

�Fx � max � 0

Since the net force �Fx in the x direction is �Fx � �F � Newton’s second lawcan be written as �F � � 0 . Thus,

We do not know , but its value will be determined in Steps 2 and 3.

The Maximum Static Frictional Force The magnitude of the maxi-mum static frictional force is related to the coefficient of static friction �s and the magni-tude FN of the normal force by Equation 4.7:

(4.7)

We now substitute this result into Equation 1, as indicated in the right column. The coeffi-cient of static friction is known, but FN is not. An expression for FN will be obtained inthe next step.

fs

MAX � �sFN

fs

MAXS T E P 2

fs

MAX

F � fs

MAX

f s

MAXf s

MAX,

fB

s

MAXFB

S T E P 1

(1)

(1)

(4.7)fs

MAX � �sFN

F � fs

MAX

F � fs

MAX

q

45807_04_p1-48 6/20/05 7:39 AM


The physics of rock climbing.

Newton’s Second Law (Vertical Direction) We can find the magnitude FN ofthe normal force by noting that the skier does not accelerate in the vertical or y direction,so ay � 0 m/s2. Figure 4.23b shows the two vertical forces that act on the skier: the nor-mal force and her weight . Applying Newton’s second law (Equation 4.2b) to thevertical direction gives

�Fy � may � 0

The net force in the y direction is �Fy � �FN � mg, so Newton’s second law becomes�FN � mg � 0. Thus,

We now substitute this result into Equation 4.7, as shown at the right.

Solution Algebraically combining the results of the three steps, we have

The magnitude F of the maximum force is

F � �smg � (0.14)(59 kg)(9.80 m/s2) �

If the force exerted by the tow rope exceeds this value, the skier will begin to accelerateforward.

Related Homework: Problems 100, 110, 113

81 N

F � fsMAX � msFN � msmg

S T E P 1 S T E P 2 S T E P 3

FN � mg

mgBFB

N

S T E P 3

* The word “kinetic” is derived from the Greek word kinetikos, meaning “of motion.”

Figure 4.24 In maneuvering her way up El Matador at Devil’s Tower National Monument inWyoming, this rock climber uses the static frictional forces between her hands and feet and thevertical rock walls to support her weight. (© Corey Rich/Coreyography)

F � fsMAX

fsMAX � msFN

FN � mg

(1)

(4.7)

Static friction is often essential, as it is to the rock climber in Figure 4.24, for in-stance. She presses outward against the walls of the rock formation with her hands andfeet to create sufficiently large normal forces, so that the static frictional forces can sup-port her weight.

Once two surfaces begin sliding over one another, the static frictional force is nolonger of any concern. Instead, a type of friction known as kinetic* friction comes intoplay. The kinetic frictional force opposes the relative sliding motion. If you have everpushed an object across a floor, you may have noticed that it takes less force to keep theobject sliding than it takes to get it going in the first place. In other words, the kinetic fric-tional force is usually less than the static frictional force.

Experimental evidence indicates that the kinetic frictional force k has three maincharacteristics, to a good degree of approximation. It is independent of the apparent areaof contact between the surfaces (see Figure 4.22). It is independent of the speed of thesliding motion, if the speed is small. And lastly, the magnitude of the kinetic frictionalforce is proportional to the magnitude of the normal force. Equation 4.8 expresses thisproportionality with the aid of a proportionality constant �k , which is called the coeffi-cient of kinetic friction.

fB

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Example 10 | Sled RidingA sled and its rider are moving at a speed of4.0 m/s along a horizontal stretch of snow, asFigure 4.25a illustrates. The snow exerts a ki-netic frictional force on the runners of thesled, so the sled slows down and eventuallycomes to a stop. The coefficient of kinetic fric-tion is 0.050. What is the displacement x of thesled?

Reasoning As the sled slows down, its ve-locity is decreasing. As our discussions inChapters 2 and 3 indicate, the changing veloc-ity is described by an acceleration (which inthis case is a deceleration since the sled isslowing down). Assuming that the accelerationis constant, we can use one of the equations ofkinematics from Chapter 3 to relate the dis-placement to the initial and final velocities and to the acceleration. The acceleration of the sled is not given directly. However,we can determine it by using Newton's second law of motion, which relates the acceleration to the net force (which is the kineticfrictional force in this case) acting on the sled and its mass.

Knowns and Unknowns The data for this problem are listed in the table:

Description Symbol Value Comment

Explicit DataInitial velocity v0x �4.0 m/s Positive, because the velocity points in the �x direction.

See drawing.Coefficient of kinetic friction �k 0.050

Implicit DataFinal velocity vx 0 m/s The sled comes to a stop.

Unknown VariableDisplacement x ?

Equation 4.8, like Equation 4.7, is a relationship between only the magnitudes of thefrictional and normal forces. The directions of these forces are perpendicular to eachother. Moreover, like the coefficient of static friction, the coefficient of kinetic friction is anumber without units and depends on the type and condition of the two surfaces that arein contact. As indicated in Table 4.2, values for �k are typically less than those for �s , re-flecting the fact that kinetic friction is generally less than static friction. The next exampleillustrates the effect of kinetic friction.

0 = 4.0 m/s = 0 m/s

+x

+y

FN

fk

W = mg

(b) Free-body diagram for the sled and rider

(a)

x

Figure 4.25 (a) The moving sled decelerates because of the kinetic frictional force. (b)Three forces act on the moving sled, the weight of the sled and its rider, the normalforce N, and the kinetic frictional force k. The free-body diagram for the sled showsthese forces.

fB

FB

WoB

KINETIC FRICTIONAL FORCE

The magnitude fk of the kinetic frictional force is given by

(4.8)

In Equation 4.8, �k is the coefficient of kinetic friction, and FN is the magnitude of thenormal force.

fk � �kFN


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Modeling the Problem

Displacement To obtain the displacement x of the sled we will use Equation3.6a from the equations of kinematics:

Solving for the displacement x gives the result shown at the right. This equation is useful because two of the variables, v0x and vx, are known and the acceleration ax can be found by applying Newton’s second law to the accelerating sled (see Step 2).

Newton’s Second Law Newton’s second law, as given in Equation 4.2a, statesthat the acceleration ax is equal to the net force �Fx divided by the mass m:

The free-body diagram in Figure 4.25b shows that the only force acting on the sled in the horizontal or x direction is the kinetic frictional force . We can write this force as �fk, where fk is the magnitude of the force and the minus sign indicates that it points in the �x direction. Since the net force is �Fx � �fk, Equation 4.2abecomes

This result can now be substituted into Equation 1, as shown at the right.

Kinetic Frictional Force We do not know the magnitude fk of the kineticfrictional force, but we do know the coefficient of kinetic friction �k. According to Equa-tion 4.8, the two are related by

(4.8)

where FN is the magnitude of the normal force. This relation can be substituted into Equation 2, as shown at the right. An expression for FN will be obtained in the next step.

Normal Force The magnitude FN of the normal force can be found by notingthat the sled does not accelerate in the vertical or y direction (ay � 0 m/s2). Thus,Newton’s second law, as given in Equation 4.2b, becomes

�Fy � may � 0

There are two forces acting on the sled in the y direction, the normal force and its weight [see part (b) of the drawing]. Therefore, the net force in the y direction is

�Fy � �FN � W

where W � mg (Equation 4.5). Thus, Newton’s second law becomes

�FN � mg � 0 or FN � mg

This result for FN can be substituted into Equation 4.8, as shown at the right.

WoB

FB

N

S T E P 4

fk � �kFN

S T E P 3

ax ��fk

m

fB

k

ax ��Fx

m

S T E P 2

vx2 � v0x

2 � 2axx

S T E P 1


(1)x �vx

2 � v0x2

2ax

x �vx

2 � v0x2

2ax

ax ��fk

m

x �vx

2 � v0x2

2ax

ax ��fk

m

fk � mkFN

x �vx

2 � v0x2

2ax

ax ��fk

m

fk � mkFN

FN � mg

(1)

(2)

(1)

(2)

(4.8)

(1)

(2)

(4.8)

45807_04_p1-48 6/21/05 9:39 AM

4.10 | THE TENSION FORCE | 23

Static friction opposes the impending relative motion between two objects, while ki-netic friction opposes the relative sliding motion that actually does occur. In either case,relative motion is opposed. However, this opposition to relative motion does not meanthat friction prevents or works against the motion of all objects. For instance, the foot of aperson walking exerts a force on the earth, and the earth exerts a reaction force on thefoot. This reaction force is a static frictional force, and it opposes the impending back-ward motion of the foot, propelling the person forward in the process. Kinetic friction canalso cause an object to move, all the while opposing relative motion, as it does in Exam-ple 10. In this example the kinetic frictional force acts on the sled and opposes the rela-tive motion of the sled and the earth. Newton’s third law indicates, however, that since theearth exerts the kinetic frictional force on the sled, the sled must exert a reaction force onthe earth. In response, the earth accelerates, but because of the earth’s huge mass, the mo-tion is too slight to be noticed.

T

(a)

T

(b)

T

–T

(c)

B

B

B

B

Figure 4.26 (a) A force is beingapplied to the right end of a rope. (b)The force is transmitted to the box. (c)Forces are applied to both ends of therope. These forces have equalmagnitudes and opposite directions.

TB


A box has a weight of 150 N and is being pulled across a horizontal floor by a force thathas a magnitude of 110 N. The pulling force can be applied to the box in two ways. It canpoint horizontally, or it can point above the horizontal at an angle �. When the pulling forceis applied horizontally, the kinetic frictional force acting on the box is twice as large aswhen it is applied at an angle �. What is the value of �? (The answer is given at the end ofthe book.)

Background: Vector components, the normal force, and the kinetic frictional force play rolesin this question. See Section 1.7 to review vector components.

For similar questions (including conceptual counterparts), consult Self-Assessment Test


Solution Algebraically combining the results of each step, we find that

Note that the mass m of the sled and rider is algebraically eliminated from the final result.Thus, the displacement of the sled is


�16 mx �vx

2 � v0x2

2(��kg)�

(0 m/s)2 � (�4.0 m/s)2

2[�(0.050)(9.80 m/s2)]�

x �vx

2 � v0x2

2ax

�vx

2 � v0x2

S T E P 1

2 ��fk

m �vx

2 � v0x2

�

2 ��mkFN

m �vx

2 � v0x2

2 ��mkmg

m ��

2 (�mkg)vx

2 � v0x2

�

S T E P 3 S T E P 4

S T E P 2

The physics of walking.

THE TENSION FORCE

Forces are often applied by means of cables or ropes that are used to pull an ob-ject. For instance, Figure 4.26a shows a force being applied to the right end of a ropeattached to a box. Each particle in the rope in turn applies a force to its neighbor. As a re-sult, the force is applied to the box, as part b of the drawing shows.

TB

4.10

45807_04_p1-48 6/21/05 9:39 AM



Test your understanding of the material in Sections 4.6–4.10:

• Gravitational Force • Normal Force • Frictional Forces • Tension Force

* If a rope is not accelerating, is zero in the second law, and regardless of the mass of therope. Then, the rope can be ignored, no matter what mass it has.† In this discussion of equilibrium we ignore rotational motion, which is discussed in Chapters 8 and 9. InSection 9.2 a more complete treatment of the equilibrium of a rigid object is presented and takes into accountthe concept of torque and the fact that objects can rotate.

�FB

� maB � 0,aB

T

–T

B

B

Figure 4.27 The force applied atone end of a massless rope istransmitted undiminished to the otherend, even when the rope bends arounda pulley, provided the pulley is alsomassless and friction is absent.

TB

In situations such as that in Figure 4.26, we say that the force is applied to the boxbecause of the tension in the rope, meaning that the tension and the force applied to thebox have the same magnitude. However, the word “tension” is commonly used to meanthe tendency of the rope to be pulled apart. To see the relationship between these two usesof the word “tension,” consider the left end of the rope, which applies the force to thebox. In accordance with Newton’s third law, the box applies a reaction force to the rope.The reaction force has the same magnitude as but is oppositely directed. In otherwords, a force � acts on the left end of the rope. Thus, forces of equal magnitude act onopposite ends of the rope, as in Figure 4.26c, and tend to pull it apart.

In the previous discussion, we have used the concept of a “massless” rope (m �0 kg) without saying so. In reality, a massless rope does not exist, but it is useful as anidealization when applying Newton’s second law. According to the second law, a netforce is required to accelerate an object that has mass. In contrast, no net force is neededto accelerate a massless rope, since � � m and m � 0 kg. Thus, when a force isapplied to one end of a massless rope, none of the force is needed to accelerate the rope.As a result, the force is also applied undiminished to the object attached at the otherend, as we assumed in Figure 4.26.* If the rope had mass, however, some of the force would have to be used to accelerate the rope. The force applied to the box would then beless than , and the tension would be different at different locations along the rope. Inthis text we will assume that a rope connecting one object to another is massless, unlessstated otherwise. The ability of a massless rope to transmit tension undiminished fromone end to the other is not affected when the rope passes around objects such as the pul-ley in Figure 4.27 (provided the pulley itself is massless and frictionless).

TB

TB

TB

TB

aBFB

TB

TB

TB

TB

DEFINITION OF EQUILIBRIUM†

An object is in equilibrium when it has zero acceleration.

EQUILIBRIUM APPLICATIONS OF NEWTON’S LAWS OF MOTION

Have you ever been so upset that it took days to recover your “equilibrium?” Inthis context, the word “equilibrium” refers to a balanced state of mind, one that is notchanging wildly. In physics, the word “equilibrium” also refers to a lack of change, but inthe sense that the velocity of an object isn’t changing. If its velocity doesn’t change, anobject is not accelerating. Our definition of equilibrium, then, is as follows:

� CONCEPTS AT A GLANCE The concept of equilibrium arises directly from Newton’ssecond law. The Concepts-at-a-Glance chart in Figure 4.28, which is an enhanced versionof the chart in Figure 4.9, illustrates this important point. When the acceleration of an ob-ject is zero ( � 0 m/s2), the object is in equilibrium, as the upper-right part of the chartindicates. This section presents several examples involving equilibrium situations. On theother hand, when the acceleration is not zero ( � 0 m/s2), we have a nonequilibriumaB

aB

4.11

45807_04_p1-48 6/17/05 3:32 PM

R E A S O N I N G S T R A T E G Y

Analyzing Equilibrium Situations1. Select the object (often called the “system”) to which Equations 4.9a and 4.9b

are to be applied. It may be that two or more objects are connected by means ofa rope or a cable. If so, it may be necessary to treat each object separately ac-cording to the following steps.

2. Draw a free-body diagram for each object chosen above. Be sure to include onlyforces that act on the object. Do not include forces that the object exerts on itsenvironment.

3. Choose a set of x, y axes for each object and resolve all forces in the free-bodydiagram into components that point along these axes. Select the axes so that asmany forces as possible point along the x axis or the y axis. Such a choice mini-mizes the calculations needed to determine the force components.

4. Apply Equations 4.9a and 4.9b by setting the sum of the x components and thesum of the y components of the forces each equal to zero.

5. Solve the two equations obtained in Step 4 for the desired unknown quantities, re-membering that two equations can yield answers for only two unknowns at most.

4.11 | EQUILIBRIUM APPLICATIONS OF NEW TON’S LAWS OF MOTION | 25

situation, as the lower-right part of the chart suggests. Section 4.12 deals with nonequilib-rium applications of Newton’s second law. �

Since the acceleration is zero for an object in equilibrium, all of the accelerationcomponents are also zero. In two dimensions, this means that ax � 0 m/s2 and ay �0 m/s2. Substituting these values into the second law (�Fx � max and �Fy � may) showsthat the x component and the y component of the net force must each be zero. In otherwords, the forces acting on an object in equilibrium must balance. Thus, in two dimen-sions, the equilibrium condition is expressed by two equations:

(4.9a)

(4.9b)

In using Equations 4.9a and 4.9b to solve equilibrium problems, we will use the fol-lowing five-step reasoning strategy:

�Fy � 0

�Fx � 0

Equilibrium(Section 4.11)

Nonequilibrium(Section 4.12)

Newton’s Second Law

ΣF = ma

External Forces

1. Gravitational Force (Section 4.7)2. Normal Force (Section 4.8)3. Frictional Forces (Section 4.9)4. Tension Force (Section 4.10)

Equilibrium Nonequilibrium

a = 0 m/s2

a ≠ 0 m/s2

CONCEPTS AT A GLANCE

B

Figure 4.28 CONCEPTS AT AGLANCE Both equilibrium andnonequilibrium problems can be solvedwith the aid of Newton’s second law.For equilibrium situations, such as thegymnast holding the “Iron Cross”position in the left photograph, theacceleration is zero Fornonequilibrium situations, such as thefreely falling gymnast in the rightphotograph, the acceleration is not zero

(© Both, MikePowell/Allsport/Getty Images)( aB � 0 m/s2).

( aB � 0 m/s2).

Example 11 illustrates how these steps are followed. It deals with a traction device inwhich three forces act together to bring about the equilibrium.

45807_04_p1-48 6/17/05 3:32 PM


▼Example 11 | Traction for the Foot

Figure 4.29a shows a traction device used with a foot injury. The weight of the 2.2-kg objectcreates a tension in the rope that passes around the pulleys. Therefore, tension forces 1 and

2 are applied to the pulley on the foot. It may seem surprising that the rope applies a force toeither side of the foot pulley. A similar effect occurs when you place a finger inside a rubberband and push downward. You can feel each side of the rubber band pulling upward on the fin-ger. The foot pulley is kept in equilibrium because the foot also applies a force to it. Thisforce arises in reaction (Newton’s third law) to the pulling effect of the forces 1 and 2 . Ig-noring the weight of the foot, find the magnitude of .

Reasoning The forces 1 , 2 , and keep the pulley on the foot at rest. The pulley, there-fore, has no acceleration and is in equilibrium. As a result, the sum of the x components andthe sum of the y components of the three forces must each be zero. Figure 4.29b shows thefree-body diagram of the pulley on the foot. The x axis is chosen to be along the direction offorce , and the components of the forces are indicated in the drawing. (See Section 1.7 for areview of vector components.)

Solution Since the sum of the y components of the forces is zero, it follows that

(4.9b)

or T1 � T2 . In other words, the magnitudes of the tension forces are equal. In addition, thesum of the x components of the forces is zero, so we have that

(4.9a)

Solving for F and letting T1 � T2 � T, we find that F � 2T cos 35°. However, the tension T inthe rope is determined by the weight of the 2.2-kg object: T � mg, where m is its mass and gis the acceleration due to gravity. Therefore, the magnitude of is

▲

Example 12 presents another situation in which three forces are responsible for theequilibrium of an object. However, in this example all the forces have different magnitudes.

▼Example 12 | Replacing an Engine

An automobile engine has a weight , whose magnitude is W � 3150 N. This engine is be-ing positioned above an engine compartment, as Figure 4.30a illustrates. To position the en-gine, a worker is using a rope. Find the tension 1 in the supporting cable and the tension 2

in the positioning rope.

Reasoning Under the influence of the forces , 1 , and 2 the ring in Figure 4.30a is at restand, therefore, in equilibrium. Consequently, the sum of the x components and the sum of they components of these forces must each be zero; �Fx � 0 and �Fy � 0. By using these rela-tions, we can find T1 and T2 . Figure 4.30b shows the free-body diagram of the ring and theforce components for a suitable x, y axis system.

TB

TB

WoB

TB

TB

WoB

35 NF � 2T cos 35 � 2mg cos 35 � 2(2.2 kg)(9.80 m/s2) cos 35 �

FB

�Fx � �T1 cos 35 � T2 cos 35 � F � 0

�Fy � �T1 sin 35 � T2 sin 35 � 0

FB

FB

TB

TB

FB

TB

TB

FB

TB

TB

F

T135°

35°

2.2 kg

T2

F

T1

T1 sin 35°

T2 sin 35°

T2 cos 35°

T1 cos 35°

T1

T2

35° +x

+y

35°

T2

(a) (b) Free-body diagram for the foot pulley

B

B

B

B

B

B

Figure 4.29 (a) A traction device forthe foot. (b) The free-body diagram forthe pulley on the foot.


Choose the orientation of the x, y axes forconvenience. In Example 11, the axes havebeen rotated so the force points alongthe x axis. Since does not have acomponent along the y axis, the analysis issimplified.

FB

FB

The physics of traction for a foot injury.

45807_04_p1-48 6/17/05 3:32 PM

4.11 | EQUILIBRIUM APPLICATIONS OF NEW TON’S LAWS OF MOTION | 27

Solution The free-body diagram shows the components for each of the three forces, and thecomponents are listed in the following table:


1 �T1 sin 10.0° �T1 cos 10.0°

2 �T2 sin 80.0° �T2 cos 80.0°0 �WW

oBTBTB

The plus signs in the table denote components that point along the positive axes, and the mi-nus signs denote components that point along the negative axes. Setting the sum of the x com-ponents and the sum of the y components equal to zero leads to the following equations:

(4.9a)

(4.9b)

Solving the first of these equations for T1 shows that

Substituting this expression for T1 into the second equation gives

Setting W � 3150 N in this result yields .

Since it follows that .

▲

An object can be moving and still be in equilibrium, provided there is no accelera-tion. Example 13 illustrates such a case, and the solution is again obtained using the five-step reasoning strategy summarized at the beginning of the section.

T1 � 3.30 � 103 NT1 � � sin 80.0

sin 10.0 � T2 and T2 � 582 N,

T2 � 582 N

T2 �W

� sin 80.0

sin 10.0 � cos 10.0 � cos 80.0

� sin 80.0

sin 10.0 � T2 cos 10.0 � T2 cos 80.0 � W � 0

T1 � � sin 80.0

sin 10.0 � T2

�Fy � �T1 cos 10.0 � T2 cos 80.0 � W � 0

�Fx � �T1 sin 10.0 � T2 sin 80.0 � 0

10.0°

(a) (b) Free-body diagram for the ring

10.0°

80.0°

80.0°

Ring

T1

T1

WW

T2

+y

+x

T1

T1 cos 10.0°

T1 sin 10.0°

T2

T2 cos 80.0°

T2 sin 80.0°

T2

B

B

B

B

Figure 4.30 (a) The ring is inequilibrium because of the three forces

1 (the tension force in the supportingcable), 2 (the tension force in thepositioning rope), and (the weight ofthe engine). (b) The free-body diagramfor the ring.

WoB

TB

TB


When an object is in equilibrium, as herein Example 12, the net force is zero,

This does not mean that eachindividual force is zero. It means that thevector sum of all the forces is zero.

�FB

� 0.

45807_04_p1-48 6/17/05 3:32 PM


▼Example 13 | Equilibrium at Constant Velocity

A jet plane is flying with a constant speed along a straight line, at an angle of 30.0° above the horizontal, as Figure 4.31a indicates. The plane has a weight whose magnitude is W � 86 500 N, and its engines provide a forward thrust of magnitude T � 103 000 N.In addition, the lift force (directed perpendicular to the wings) and the force of air resistance (directed opposite to the motion) act on the plane. Find and .

Reasoning Figure 4.31b shows the free-body diagram of the plane, including the forces ,, , and . Since the plane is not accelerating, it is in equilibrium, and the sum of the x

components and the sum of the y components of these forces must be zero. The lift force and the force of air resistance can be obtained from these equilibrium conditions. To calcu-late the components, we have chosen axes in the free-body diagram that are rotated by 30.0°from their usual horizontal–vertical positions. This has been done purely for convenience,since the weight is then the only force that does not lie along either axis.

Solution When determining the components of the weight, it is necessary to realize that theangle � in Figure 4.31a is 30.0°. Part c of the drawing focuses attention on the geometry thatis responsible for this fact. There it can be seen that � � � � 90° and � � 30° � 90°, withthe result that � � 30°. The table below lists the components of the forces that act on the jet.

WoB

RB

LB

RB

TB

LB

WoB

RB

LB

RB

LB

TB

WoB


�W sin 30.0° �W cos 30.0°0 �L

�T 0�R 0R

BTBLBWoB

TL

R

+y

+x+y

+x

W

(a) (c)(b) Free-body diagram

30°90°

90°

30°βα

αβ

W sin 30°W cos 30°30°

BB

B

TL

R

B B

B

W


A moving object is in equilibrium if itmoves with a constant velocity; then itsacceleration is zero. A zero acceleration isthe fundamental characteristic of anobject in equilibrium.

Figure 4.31 (a) A plane moves with a constant velocity at an angle of 30.0° above the horizontaldue to the action of four forces, the weight , the lift , the engine thrust , and the air resistance

. (b) The free-body diagram for the plane. (c) This geometry occurs often in physics.RB

TB

LB

WoB

Need more practice?

Interactive LearningWare 4.3 A 0.600-kg kite is being flown at the end of a string. Thestring is straight and makes an angle of 55.0° above the horizontal. The kite is stationary,and the tension in the string is 35.0 N. Determine the force (both magnitude and direction)that the wind exerts on the kite. Specify the angle relative to the horizontal.

Related Homework: Problems 46, 56

For an interactive solution, go to www.wiley.com/college/cutnell

Setting the sum of the x component of the forces to zero gives

(4.9a)

Setting the sum of the y component of the forces to zero gives

(4.9b)

▲74 900 NL � W cos 30.0� � (86 500 N) cos 30.0� �

�Fy � �W cos 30.0� � L � 0

59 800 NR � T � W sin 30.0� � 103 000 N � (86 500 N) sin 30.0� �

�Fx � �W sin 30.0� � T � R � 0

45807_04_p1-48 6/21/05 4:29 PM

4.12 | NONEQUILIBRIUM APPLICATIONS OF NEW TON’S LAWS OF MOTION | 29

NONEQUILIBRIUM APPLICATIONS OF NEWTON’S LAWS OF MOTION

When an object is accelerating, it is not in equilibrium, as indicated in Figure4.28. The forces acting on it are not balanced, so the net force is not zero in Newton’ssecond law. However, with one exception, the reasoning strategy followed in solving non-equilibrium problems is identical to that used in equilibrium situations. The exception oc-curs in Step 4 of the five steps outlined at the beginning of the last section. Since the ob-ject is now accelerating, the representation of Newton’s second law in Equations 4.2a and4.2b applies instead of Equations 4.9a and 4.9b:

(4.2b)

Example 14 uses these equations in a situation where the forces are applied in directionssimilar to those in Example 11, except that now an acceleration is present.

▼Example 14 | Towing a Supertanker

A supertanker of mass m � 1.50 � 108 kg is being towed by two tugboats, as in Figure 4.32a.The tensions in the towing cables apply the forces 1 and 2 at equal angles of 30.0° with re-spect to the tanker’s axis. In addition, the tanker’s engines produce a forward drive force ,whose magnitude is D � 75.0 � 103 N. Moreover, the water applies an opposing force ,whose magnitude is R � 40.0 � 103 N. The tanker moves forward with an acceleration thatpoints along the tanker’s axis and has a magnitude of m/s2. Find the magnitudesof the tensions 1 and 2 .

Reasoning The unknown forces 1 and 2 contribute to the net force that accelerates thetanker. To determine 1 and 2 , therefore, we analyze the net force, which we will do usingcomponents. The various force components can be found by referring to the free-body dia-

TB

TB

TB

TB

TB

TB

2.00 � 10�3

RBDB

TB

TB

�Fx � max (4.2a) and �Fy � may

30.0°30.0°

30.0° 30.0°

(b) Free-body diagram for the tanker(a)

T1

T2T2

T1

+y

+xR D

T2 sin 30.0°

T2 cos 30.0°

T1 cos 30.0°

T1 sin 30.0° T1

T2

D

a

RB

B

B

B

B

B

Figure 4.32 (a) Four forces act on asupertanker: 1 and 2 are the tensionforces due to the towing cables, isthe forward drive force produced by thetanker’s engines, and is the forcewith which the water opposes thetanker’s motion. (b) The free-bodydiagram for the tanker.

RB

DB

TB

TB


In which one of the following situations could an object possibly be in equilibrium? (a)Three forces act on the object. The forces all point along the same line but may have differ-ent directions. (b) Two perpendicular forces act on the object. (c) A single force acts on theobject. (d) In none of the situations described in (a), (b), and (c) could the object possibly bein equilibrium. (The answer is given at the end of the book.)

Background: A number of concepts enter the picture here: force, net force, and equilibrium.The addition of vectors also plays a role, as does Newton’s second law of motion.



4.12

45807_04_p1-48 6/17/05 3:32 PM


gram for the tanker in Figure 4.32b, where the ship’s axis is chosen as the x axis. We will thenuse Newton’s second law in its component form, �Fx � max and �Fy � may , to obtain themagnitudes of 1 and 2 .

Solution The individual force components are summarized as follows:

TB

TB


1 �T1 cos 30.0° �T1 sin 30.0°

2 �T2 cos 30.0° �T2 sin 30.0°�D 0�R 0R

BDBTBTB

Trailer Truck

(b) Free-body diagrams

+x +xT D

T

Drawbar

(a)

m1 = 8500 kgm2 = 27 000 kga = 0.78 m/s2

DT´

DT´

B

B B

B

Figure 4.33 (a) The force acts on thetruck and propels it forward. Thedrawbar exerts the tension force � onthe truck and the tension force on thetrailer. (b) The free-body diagrams forthe trailer and the truck, ignoring thevertical forces.

TB

TB

DB

Since the acceleration points along the x axis, there is no y component of the acceleration (ay � 0 m/s2). Consequently, the sum of the y components of the forces must be zero:

This result shows that the magnitudes of the tensions in the cables are equal, T1 � T2 . Sincethe ship accelerates along the x direction, the sum of the x components of the forces is notzero. The second law indicates that

Since T1 � T2, we can replace the two separate tension symbols by a single symbol T, themagnitude of the tension. Solving for T gives

▲

It often happens that two objects are connected somehow, perhaps by a drawbar likethat used when a truck pulls a trailer. If the tension in the connecting device is of no inter-est, the objects can be treated as a single composite object when applying Newton’s sec-ond law. However, if it is necessary to find the tension, as in the next example, then thesecond law must be applied separately to at least one of the objects.

▼Example 15 | Hauling a Trailer

A truck is hauling a trailer along a level road, as Figure 4.33a illustrates. The mass of the truckis m1 � 8500 kg and that of the trailer is m 2 � 27 000 kg. The two move along the x axis withan acceleration of ax � 0.78 m/s2. Ignoring the retarding forces of friction and air resistance,determine (a) the tension in the horizontal drawbar between the trailer and the truck and (b)the force that propels the truck forward.D

BTB

1.53 � 105 N �

�(1.50 � 108 kg)(2.00 � 10�3 m/s2) � 40.0 � 103 N � 75.0 � 103 N

2 cos 30.0

T �max � R � D

2 cos 30.0

�Fx � T1 cos 30.0 � T2 cos 30.0 � D � R � max

�Fy � �T1 sin 30.0 � T2 sin 30.0 � 0

45807_04_p1-48 6/17/05 3:32 PM


Reasoning Since the truck and the trailer accelerate along the horizontal direction and fric-tion is being ignored, only forces that have components in the horizontal direction are of inter-est. Therefore, Figure 4.33 omits the weight and the normal force, which act vertically. To de-termine the tension force in the drawbar, we draw the free-body diagram for the trailer andapply Newton’s second law, �Fx � max . Similarly, we can determine the propulsion force by drawing the free-body diagram for the truck and applying Newton’s second law.

Solution (a) The free-body diagram for the trailer is shown in Figure 4.33b. There is onlyone horizontal force acting on the trailer, the tension force due to the drawbar. Therefore, itis straightforward to obtain the tension from �Fx � m 2ax , since the mass of the trailer and theacceleration are known:

(b) Two horizontal forces act on the truck, as the free-body diagram in Figure 4.33b shows.One is the desired force . The other is the force �. According to Newton’s third law, � isthe force with which the trailer pulls back on the truck, in reaction to the truck pulling for-ward. If the drawbar has negligible mass, the magnitude of � is equal to the magnitude of

—namely, 21 000 N. Since the magnitude of �, the mass of the truck, and the accelerationare known, �Fx � m1ax can be used to determine the drive force:

▲

In Section 4.11 we examined situations where the net force acting on an object iszero, and in this section we have considered two examples where the net force is not zero.Conceptual Example 16 illustrates a common situation where the net force is zero at cer-tain times but is not zero at other times.

▼Conceptual Example 16 | The Motion of a Water Skier

Figure 4.34 shows a water skier at four different moments:

(a) The skier is floating motionless in the water.

(b) The skier is being pulled out of the water and up onto the skis.

(c) The skier is moving at a constant speed along a straight line.

(d) The skier has let go of the tow rope and is slowing down.

For each moment, explain whether the net force acting on the skier is zero.

Reasoning and Solution According to Newton’s second law, if an object has zero accelera-tion, the net force acting on it is zero. In such a case, the object is in equilibrium. In contrast,if the object has an acceleration, the net force acting on it is not zero. Such an object is not inequilibrium. We will apply this criterion to each of the four phases of the motion to decidewhether the net force is zero.

(a) The skier is floating motionless in the water, so her velocity and acceleration are bothzero. Therefore, the net force acting on her is zero, and she is in equilibrium.

28 000 N D � m1ax � T � � (8500 kg)(0.78 m /s2) � 21 000 N �

�Fx � �D � T � � m1ax

TB

TB

TB

TB

TB

DB

21 000 N�Fx � T � m2ax � (27 000 kg)(0.78 m/s2) �

TB

DB

TB

Net forceNet force

(a) (c)(b) (d )

Figure 4.34 A water skier (a) floatingin water, (b) being pulled up by theboat, (c) moving at a constant velocity,and (d ) slowing down.

45807_04_p1-48 6/17/05 3:32 PM


Example 17 | Hauling a CrateA flatbed truck is carrying a crate up a 10.0° hill, asFigure 4.35a illustrates. The coefficient of staticfriction between the truck bed and the crate is0.350. Find the maximum acceleration that thetruck can attain before the crate begins to slip back-ward relative to the truck.

Reasoning The crate will not slip as long as it hasthe same acceleration as the truck. Therefore, a netforce must act on the crate to accelerate it, and thestatic frictional force contributes to this net force.Since the crate tends to slip backward, the staticfrictional force is directed forward, up the hill.

As the acceleration of the truck increases, mustalso increase to produce a corresponding increasein the acceleration of the crate. However, the staticfrictional force can increase only until its maximummagnitude is reached, at which point thecrate and truck have the maximum acceleration . If the acceleration increases even more, the crate will slip.

To find , we will employ Newton's second law, the definition of weight, and the relationship between the maximum staticfrictional force and the normal force.

Knowns and Unknowns The data for this problem are:

Description Symbol Value

Angle of hill � 10.0°Coefficient of static friction �s 0.350

Unknown VariableMaximum acceleration before

crate slips aMAX ?

aB

MAX

aB

MAXfB

s

MAX

fB

s

fB

s

(b) As the skier is being pulled up and out of the water, her velocity is increasing. Thus, sheis accelerating, and the net force acting on her is not zero. The skier is not in equilibrium. Thedirection of the net force is shown in Figure 4.34b.

(c) The skier is now moving at a constant speed along a straight line, so her velocity is con-stant. Since her velocity is constant, her acceleration is zero. Thus, the net force acting on heris zero, and she is again in equilibrium, even though she is moving.

(d) After the skier lets go of the tow rope, her speed decreases, so she is decelerating. Thus,the net force acting on her is not zero, and she is not in equilibrium. The direction of the netforce is shown in Figure 4.34d, and it is opposite to that in (b).

Related Homework: Problem 68▲

The force of gravity is often present among the forces that affect the acceleration ofan object. Examples 17–19 deal with typical situations.

Need more practice?

Interactive LearningWare 4.4 Apassenger is pulling on the strapof a 15-kg suitcase with a force of70.0 N. The strap makes an angleof 35.0° above the horizontal. A37.8-N friction force opposes thehorizontal motion of the suitcase.Determine the acceleration of thesuitcase.

Related Homework: Problem 90


= 10.0°

mg cos

FN FN

W = mg

fsMAX

f sMAX

= 10.0° = 10.0°θ

θ

θθ

(a)

W = mg

mg sin

(b) Free-body diagram of the crate

+x

+y

B B

θ

Figure 4.35 (a) A crate on a truck is kept from slipping by the static frictionalforce s

MAX. The other forces that act on the crate are its weight and thenormal force N . (b) The free-body diagram of the crate.F

BWoB

fB


45807_04_p1-48 6/21/05 4:30 PM


Modeling the Problem

Newton’s Second Law (x direction) Taking the x direction to be parallel tothe acceleration of the truck, Newton’s second law for this direction can be written as (seeEquation 4.2a) ax

MAX � (�Fx)/m, where �Fx is the net force acting on the crate in the x di-rection and m is the crate’s mass. Using the x components of the forces shown in Figure4.35b, the net force is �Fx � �mg sin � � fs

MAX. Substituting this expression into New-ton’s second law gives

The acceleration due to gravity g and the angle � are known, but m and fsMAX are not. We

will now turn our attention to finding fsMAX.

The Maximum Static Frictional Force The magnitude fsMAX of the maxi-

mum static frictional force is related to the coefficient of static friction �s and the magni-tude FN of the normal force by Equation 4.7:

(4.7)

This result can be substituted into Equation 1, as shown at the right. Although �s isknown, FN is not known. An expression for FN will be found in Step 3, however.

Newton’s Second Law (y direction) We can determine the magnitude FN

of the normal force by noting that the crate does not accelerate in the y direction (ay � 0 m/s2). Thus, Newton’s second law as given in Equation 4.2b becomes

There are two forces acting on the crate in the y direction (see Figure 4.35b): the normalforce �FN and the y component of the weight �mg cos � (The minus sign is included because this component points along the negative y direction.) Thus, the net force is �Fy � �FN � mg cos �. Newton’s second law becomes

or

This result for FN can be substituted into Equation 4.7, as indicated at the right.

Solution Algebraically combining the results of the three steps, we find that

� �g sin � � �sg cos �

Note that the mass m of the crate is algebraically eliminated from the final result. Thus,the maximum acceleration is

axMAX � �g sin � � �sg cos �

� �(9.80 m/s2) sin 10.0° � (0.350)(9.80 m/s2) cos 10 0° �


1.68 m/s2

S T E P 1 S T E P 2 S T E P 3

axMAX � �

�mg sin u � fsMAX

m

�mg sin u �msFN

m�

�mg sin u �ms mg cos u

m

FN � mg cos ��FN � mg cos � � 0

�Fy � may � 0

S T E P 3

fsMAX � �sFN

S T E P 2

axMAX �

�Fx

m�

�mg sin � � fsMAX

m

S T E P 1

(1)axMAX �

�mg sin � � fsMAX

m

axMAX �


m

fsMAX � msFN

axMAX �


m

fsMAX � msFN

FN � mg cos u

(1)

(4.7)

(1)

(4.7)

45807_04_p1-48 6/17/05 3:32 PM


▼Example 18 | Accelerating Blocks

Block 1 (mass m1 � 8.00 kg) is moving on a frictionless 30.0° incline. This block is connectedto block 2 (mass m2 � 22.0 kg) by a massless cord that passes over a massless and frictionlesspulley (see Figure 4.36a). Find the acceleration of each block and the tension in the cord.

Reasoning Since both blocks accelerate, there must be a net force acting on each one. Thekey to this problem is to realize that Newton’s second law can be used separately for eachblock to relate the net force and the acceleration. Note also that both blocks have accelerationsof the same magnitude a, since they move as a unit. We assume that block 1 accelerates up theincline and choose this direction to be the �x axis. If block 1 in reality accelerates down theincline, then the value obtained for the acceleration will be a negative number.

Solution Three forces act on block 1: (1) 1 is its weight [W1 � m1g � (8.00 kg) �(9.80 m/s2) � 78.4 N], (2) is the force applied because of the tension in the cord, and (3)

N is the normal force that the incline exerts. Figure 4.36b shows the free-body diagram forblock 1. The weight is the only force that does not point along the x, y axes, and its x and ycomponents are given in the diagram. Applying Newton’s second law (�Fx � m1ax) to block1 shows that

where we have set ax � a. This equation cannot be solved as it stands, since both T and a areunknown quantities. To complete the solution, we next consider block 2.

Two forces act on block 2, as the free-body diagram in Figure 4.36b indicates: (1) 2 isits weight [W2 � m 2 g � (22.0 kg)(9.80 m/s2) � 216 N] and (2) � is exerted as a result ofblock 1 pulling back on the connecting cord. Since the cord and the frictionless pulley aremassless, the magnitudes of � and are the same: T� � T. Applying Newton’s second law(�Fy � m 2ay) to block 2 reveals that

The acceleration ay has been set equal to �a since block 2 moves downward along the �yaxis in the free-body diagram, consistent with the assumption that block 1 moves up the in-cline. Now there are two equations in two unknowns, and they may be solved simultaneously(see Appendix C) to give T and a:

and

▲

▼Example 19 | Hoisting a Scaffold

A window washer on a scaffold is hoisting the scaffold up the side of a building by pulling down-ward on a rope, as in Figure 4.37a. The magnitude of the pulling force is 540 N, and the com-bined mass of the worker and the scaffold is 155 kg. Find the upward acceleration of the unit.

a � 5.89 m/s2T � 86.3 N

�Fy � T � W2 � m2(�a)

TB

TB

TB

WoB

�Fx � �W1 sin 30.0 � T � m1a

FB

TB

WoB

(a) (b) Free–body diagrams

T´ T´

m1

a

a30.0°

30.0°

T

Block 1 Block 2

FN

FN

W1W2

W1

W2

T

W1 cos 30.0°

30.0°

W1 sin 30.0°

+x

+x

+y

+y

m2

B B

BB

BB

Figure 4.36 (a) Three forces act onblock 1: its weight 1 , the normalforce N , and the force due to thetension in the cord. Two forces act onblock 2: its weight 2 and the force �due to the tension. The acceleration islabeled according to its magnitude a.(b) Free-body diagrams.

TB

WoB

TB

FB

WoB

45807_04_p1-48 6/17/05 3:32 PM

4.13 | CONCEPTS & CALCULATIONS | 35

Reasoning The worker and the scaffold form a single unit, on which the rope exerts a forcein three places. The left end of the rope exerts an upward force on the worker’s hands. Thisforce arises because he pulls downward with a 540-N force, and the rope exerts an oppositelydirected force of equal magnitude on him, in accord with Newton’s third law. Thus, the magni-tude T of the upward force is T � 540 N and is the magnitude of the tension in the rope. If themasses of the rope and each pulley are negligible and if the pulleys are friction-free, the ten-sion is transmitted undiminished along the rope. Then, a 540-N tension force acts upwardon the left side of the scaffold pulley (see part a of the drawing). A tension force is also ap-plied to the point P, where the rope attaches to the roof. The roof pulls back on the rope in ac-cord with the third law, and this pull leads to the 540-N tension force that acts on the rightside of the scaffold pulley. In addition to the three upward forces, the weight of the unit mustbe taken into account [W � mg � (155 kg)(9.80 m/s2) � 1520 N]. Part b of the drawingshows the free-body diagram.

Solution Newton’s second law (�Fy � may) can be applied to calculate the acceleration ay :

▲

0.65 m/s2 ay �3T � W

m�

3(540 N) � 1520 N

155 kg�

�Fy � �T � T � T � W � may

TB

TB

TB


Two boxes have masses m1 and m2 , and m2 is greater than m1. The boxes are beingpushed across a frictionless horizontal surface. As the drawing shows, there are two possi-ble arrangements, and the pushing force is the same in each. In which arrangement doesthe force that the left box applies to the right box have a greater magnitude, or is the mag-nitude the same in both cases? (The answer is given at the end of the book.)


Test your understanding of the material in Sections 4.11 and 4.12:

• Equilibrium Applications of Newton’s Laws of Motion• Nonequilibrium Applications of Newton’s Laws of Motion

ay

P

T

T

T T

+x

+y

(a)

(b) Free-body diagram of the unit

T

T

W = mg

W = mg

B

B B

BBB

Figure 4.37 (a) A window washer pullsdown on the rope to hoist the scaffoldup the side of a building. The force results from the effort of the windowwasher and acts on him and the scaffoldin three places, as discussed inExample 19. (b) The free-body diagramof the unit comprising the man and thescaffold.

TB

CONCEPTS & CALCULATIONS

Newton’s three laws of motion provide the basis for understanding the effect offorces on the motion of an object, as we have seen. The second law is especially impor-tant, because it provides the quantitative relationship between force and acceleration. Theexamples in this section serve as a review of the essential features of this relationship.

m1 m1m2

(a) (b)

m2

Pushingforce

Pushingforce

Background: This question deals with net force, acceleration, and Newton’s second andthird laws of motion.


4.3. The test is described next.

4.13

45807_04_p1-48 6/17/05 3:32 PM


Concepts & Calculations Example 20 | Velocity, Acceleration, and

▼Newton’s Second Law of Motion

Figure 4.38 shows two forces, 1 � �3000 N and 2 � �5000 N, acting on a spacecraft,where the plus signs indicate that the forces are directed along the �x axis. A third force 3

also acts on the spacecraft but is not shown in the drawing. The craft is moving with a constantvelocity of �850 m/s. Find the magnitude and direction of 3 .

Concept Questions and Answers Suppose the spacecraft were stationary. What would bethe direction of 3 ?

Answer If the spacecraft were stationary, its acceleration would be zero. According toNewton’s second law, the acceleration of an object is proportional to the net force actingon it. Thus, the net force must also be zero. But the net force is the vector sum of thethree forces in this case. Therefore, the force 3 must have a direction such that it balances to zero the forces 1 and 2 . Since 1 and 2 point along the �x axis in Figure4.38, 3 must then point along the �x axis.

When the spacecraft is moving at a constant velocity of �850 m/s, what is the direction of 3 ?

Answer Since the velocity is constant, the acceleration is still zero. As a result, every-thing we said in the stationary case applies again here. The net force is zero, and the force

3 must point along the �x axis in Figure 4.38.

Solution Since the velocity is constant, the acceleration is zero. The net force must also bezero, so that

Solving for F3 yields

The minus sign in the answer means that 3 points opposite to the sum of 1 and 2 , or alongthe �x axis in Figure 4.38. The force 3 has a magnitude of 8000 N, which is the magnitudeof the sum of the forces 1 and 2 . The answer is independent of the velocity of the space-craft, as long as that velocity remains constant.

▲

▼Concepts & Calculations Example 21 | The Importance of Mass

On earth a block has a weight of 88 N. This block is sliding on a horizontal surface on themoon, where the acceleration due to gravity is 1.60 m/s2. As Figure 4.39a shows, the block isbeing pulled by a horizontal rope in which the tension is T � 24 N. The coefficient of kineticfriction between the block and the surface is �k � 0.20. Determine the acceleration of theblock.

Concept Questions and Answers Which of Newton’s laws of motion provides a way to de-termine the acceleration of the block?

Answer Newton’s second law allows us to calculate the acceleration as ax � �Fx /m,where �Fx is the net force acting in the horizontal direction and m is the mass of theblock.

This problem deals with a situation on the moon, but the block’s mass on the moon is notgiven. Instead, the block’s earth-weight is given. Why can the earth-weight be used to obtain avalue for the block’s mass that applies on the moon?

Answer Since the block’s earth-weight Wearth is related to the block’s mass according toWearth � mgearth , we can use Wearth � 88 N and gearth � 9.80 m/s2 to obtain m. But mass isan intrinsic property of the block and does not depend on whether it is on the earth or onthe moon. Therefore, the value obtained for m applies on the moon as well as on theearth.

Does the net force �Fx equal the tension T?

Answer No. The net force �Fx is the vector sum of all the external forces acting in thehorizontal direction. It includes the kinetic frictional force fk as well as the tension T.

FB

FB

FB

FB

FB

FB

�8000 NF3 � �(F1 � F2) � �(3000 N � 5000 N) �

�Fx � F1 � F2 � F3 � 0

FB

FB

FB

FB

FB

FB

FB

FB

FB

FB

FB

FB

FB

F2

F1

+x

B

B

Figure 4.38 Two horizontal forces, 1

and 2 , act on the spacecraft. A thirdforce 3 also acts but is not shown.F

BFB

FB

T

ax

(a)

(b) Free-body diagram for the block

+x

+y

FN

fk T

mgmoon

B

B

Figure 4.39 (a) A block is sliding on ahorizontal surface on the moon. Thetension in the rope is . (b) The free-body diagram for the block, including akinetic frictional force k.f

B

TB

45807_04_p1-48 6/21/05 9:40 AM

CONCEPT SUMMARY | 37

Solution Figure 4.39b shows the free-body diagram for the block. The net force along the xaxis is �Fx � �T � fk , where T is the magnitude of the tension in the rope and fk is the mag-nitude of the kinetic frictional force. According to Equation 4.8, fk is related to the magnitudeFN of the normal force by fk � �kFN , where � k is the coefficient of kinetic friction. The ac-celeration ax of the block is given by Newton’s second law as

We can obtain an expression for FN by noting that the block does not move in the y direction,so ay � 0 m/s2. Therefore, the net force �Fy along the y direction must also be zero. An ex-amination of the free-body diagram reveals that �Fy � �FN � mgmoon � 0, so that FN �mgmoon . The acceleration in the x direction becomes

Using the earth-weight of the block to determine its mass, we find

The acceleration of the block is, then,

▲

At the end of the problem set for this chapter, you will find homework problems thatcontain both conceptual and quantitative parts. These problems are grouped under theheading Concepts & Calculations.

�2.3 m/s2ax ��T � �kmgmoon

m�

24 N � (0.20)(9.0 kg)(1.60 m/s2)

9.0 kg�

Wearth � mgearth or m �Wearth

gearth�

88 N

9.80 m/s2 � 9.0 kg

ax ��T � �kmgmoon

m

ax ��Fx

m�

�T � �kFN

m

If you need more help with a concept, use the text examples and media noted next to the discussion or equation. Go towww.wiley.com/college/cutnell for:

Interactive LearningWare (ILW) —examples solved in five-step interactive format.

Concept Simulations (CS) —animated text figures or animations of important concepts.

Self-Assessment Tests —qualitative and quantitative questions with extensive feedback.

Topic Discussion Learning Aids

CONCEPT SUMMARY

Contact and noncontact forces

Mass

4.1 THE CONCEPTS OF FORCE AND MASS

A force is a push or a pull and is a vector quantity. Contact forces arise from the physical contactbetween two objects. Noncontact forces are also called action-at-a-distance forces, because theyarise without physical contact between two objects.

Mass is a property of matter that determines how difficult it is to accelerate or decelerate an ob-ject. Mass is a scalar quantity.

4.2 NEWTON’S FIRST LAW OF MOTION

Newton’s first law of motion, sometimes called the law of inertia, states that an object continuesin a state of rest or in a state of motion at a constant speed along a straight line unless compelledto change that state by a net force.

Inertia is the natural tendency of an object to remain at rest or in motion at a constant speed alonga straight line. The mass of a body is a quantitative measure of inertia and is measured in an SIunit called the kilogram (kg). An inertial reference frame is one in which Newton’s law of inertiais valid.

Inertia

Mass

Newton’s first law

Inertial referenceframe

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Newton’s third law of motion

Fundamental forces

Ex. 4

Use Self-Assessment Test 4.1 to evaluate your understanding of Sections 4.1 – 4.5.

4.6 TYPES OF FORCES: AN OVERVIEW

Only three fundamental forces have been discovered: the gravitational force, the strong nuclearforce, and the electroweak force. The electroweak force manifests itself as either the electromag-netic force or the weak nuclear force.

4.7 THE GRAVITATIONAL FORCE

Newton’s law of universal gravitation states that every particle in the universe exerts an attractiveforce on every other particle. For two particles that are separated by a distance r and have massesm1 and m2, the law states that the magnitude of this attractive force is

(4.3)

The direction of this force lies along the line between the particles. The constant G has a value ofand is called the universal gravitational constant.

The weight W of an object on or above the earth is the gravitational force that the earth exerts onthe object and can be calculated from the mass m of the object and the acceleration g due to theearth’s gravity according to

(4.5)

4.8 THE NORMAL FORCE

The normal force N is one component of the force that a surface exerts on an object with which itis in contact—namely, the component that is perpendicular to the surface.

The apparent weight is the force that an object exerts on the platform of a scale and may belarger or smaller than the true weight mg if the object and the scale have an acceleration a (� ifupward, � if downward). The apparent weight is

(4.6)

4.9 STATIC AND KINETIC FRICTIONAL FORCES

A surface exerts a force on an object with which it is in contact. The component of the force perpendic-ular to the surface is called the normal force. The component parallel to the surface is called friction.

The force of static friction between two surfaces opposes any impending relative motion of thesurfaces. The magnitude of the static frictional force depends on the magnitude of the appliedforce and can assume any value up to a maximum of

(4.7)fs

MAX � � s FN

Apparent weight � mg � ma

FB

W � mg

G � 6.673 � 10�11 N�m2 /kg2

F � Gm 1m 2

r 2


4.3 NEWTON’S SECOND LAW OF MOTION

4.4 THE VECTOR NATURE OF NEWTON’S SECOND LAW OF MOTION

Newton’s second law of motion states that when a net force � acts on an object of mass m, theacceleration of the object can be obtained from the following equation:

(4.1)

This is a vector equation and, for motion in two dimensions, is equivalent to the following twoequations:

(4.2a)

(4.2b)

In these equations the x and y subscripts refer to the scalar components of the force and accelera-tion vectors. The SI unit of force is the Newton (N).

When determining the net force, a free-body diagram is helpful. A free-body diagram is a diagramthat represents the object and the forces acting on it.

4.5 NEWTON’S THIRD LAW OF MOTION

Newton’s third law of motion, often called the action–reaction law, states that whenever one ob-ject exerts a force on a second object, the second object exerts an oppositely directed force ofequal magnitude on the first object.

�Fy � may

�Fx � ma x

�FB

� maBaB

FB

Newton’s second law (vector form)

Newton’s second law (component form)

Free-body diagram

CS 4.1

Ex. 1 – 3, 20, 21

ILW 4.1

Ex. 5, 6Newton’s law of universalgravitation

Weight and mass

Normal force Ex. 8

Ex. 7

Friction

Apparent weight

Ex. 9CS 4.2, 4.4

Maximum static frictionalforce

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CONCEPTUAL QUESTIONS | 39

Kinetic frictional force

ILW 4.2

Ex. 10CS 4.3

where �s is the coefficient of static friction and FN is the magnitude of the normal force.

The force of kinetic friction between two surfaces sliding against one another opposes the relativemotion of the surfaces. This force has a magnitude given by

(4.8)

where �k is the coefficient of kinetic friction.

4.10 THE TENSION FORCE

The word “tension” is commonly used to mean the tendency of a rope to be pulled apart due toforces that are applied at each end. Because of tension, a rope transmits a force from one end tothe other. When a rope is accelerating, the force is transmitted undiminished only if the rope ismassless.

fk � � k FN


Use Self-Assessment Test 4.2 to evaluate your understanding of Sections 4.6 – 4.10.

4.11 EQUILIBRIUM APPLICATIONS OF NEWTON’S LAWS OF MOTION

An object is in equilibrium when the object has zero acceleration, or, in other words, when itmoves at a constant velocity (which may be zero). The sum of the forces that act on an object inequilibrium is zero. Under equilibrium conditions in two dimensions, the separate sums of theforce components in the x direction and in the y direction must each be zero:

(4.9a)

(4.9b)

4.12 NONEQUILIBRIUM APPLICATIONS OF NEWTON’S LAWS OF MOTION

If an object is not in equilibrium, then Newton’s second law must be used to account for the accel-eration:

(4.2a)

(4.2b)�Fy � may

�Fx � max

�Fy � 0

�Fx � 0

Definition of equilibrium

The equilibrium condition Ex. 11, 12, 13

ILW 4.3

Use Self-Assessment Test 4.3 to evaluate your understanding of Sections 4.11 and 4.12.

1. Why do you lunge forward when your car suddenly comes to ahalt? Why are you pressed backward against the seat when your carrapidly accelerates? In your explanation, refer to the most appropri-ate one of Newton’s three laws of motion.

2. A bird feeder of large mass ishung from a tree limb, as thedrawing shows. A cord attached tothe bottom of the feeder has beenleft dangling free. Curiosity getsthe best of a child, who pulls onthe dangling cord in an attempt tosee what’s in the feeder. The dan-gling cord is cut from the samesource as the cord attached to thelimb. Is the cord between thefeeder and the limb more likely tosnap with a slow continuous pull or a sudden downward pull? Giveyour reasoning.

3. ssm The net external force acting on an object is zero. Is it possi-ble for the object to be traveling with a velocity that is not zero? Ifyour answer is yes, state whether any conditions must be placed on

the magnitude and direction of the velocity. If your answer is no,provide a reason for your answer.

4. Is a net force being applied to an object when the object is mov-ing downward (a) with a constant acceleration of 9.80 m/s2 and (b)with a constant velocity of 9.80 m/s? Explain.

5. Newton’s second law indicates that when a net force acts on anobject, it must accelerate. Does this mean that when two or moreforces are applied to an object simultaneously, it must accelerate?Explain.

6. A father and his seven-year-old daughter are facing each other onice skates. With their hands, they push off against one another. (a)Compare the magnitudes of the pushing forces that they experience.(b) Which one, if either, experiences the larger acceleration? Ac-count for your answers.

7. A gymnast is bouncing on a trampoline. After a high bounce thegymnast comes down and hits the elastic surface of the trampoline.In so doing the gymnast applies a force to the trampoline. (a) De-scribe the effect this force has on the elastic surface. (b) The surfaceapplies a reaction force to the gymnast. Describe the effect that thisreaction force has on the gymnast.

CONCEPTUAL QUESTIONSssm Solution is in the Student Solutions Manual.

Ex. 14 – 19ILW 4.4

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8. According to Newton’s third law, when you push on an object,the object pushes back on you with an oppositely directed force ofequal magnitude. If the object is a massive crate resting on the floor,it will probably not move. Some people think that the reason thecrate does not move is that the two oppositely directed pushingforces cancel. Explain why this logic is faulty and why the cratedoes not move.

9. Three particles have identical masses. Each particle experiencesonly the gravitational forces due to the other two particles. Howshould the particles be arranged so each one experiences a net gravi-tational force that has the same magnitude? Give your reasoning.

10. When a body is moved from sea level to the top of a mountain,what changes—the body’s mass, its weight, or both? Explain.

11. The force of air resistance acts to oppose the motion of an ob-ject moving through the air. A ball is thrown upward and eventuallyreturns to the ground. (a) As the ball moves upward, is the net forcethat acts on the ball greater than, less than, or equal to its weight?Justify your answer. (b) Repeat part (a) for the downward motion ofthe ball.

12. Object A weighs twice as much as object B at the same spot onthe earth. Would the same be true at a given spot on Mars? Accountfor your answer.

13. Does the acceleration of a freely falling object depend to anyextent on the location—that is, whether the object is on top of Mt.Everest or in Death Valley, California? Explain.

14. A “bottle rocket” is a type of fireworks that has a long thin tailthat you insert into an empty bottle, to provide a launch platform.One of these rockets is fired with the bottle pointing vertically up-ward. An identical rocket is fired with the bottle lying on its side,pointing horizontally. In which case does the rocket leave the bottlewith the greater acceleration? Explain, ignoring air resistance andfriction.

15. A 10-kg suitcase is placed on a scale that is in an elevator. Is theelevator accelerating up or down when the scale reads (a) 75 N and(b) 120 N? Justify your answers.

16. ssm A stack of books whose true weight is 165 N is placed on ascale in an elevator. The scale reads 165 N. Can you tell from thisinformation whether the elevator is moving with a constant velocityof 2 m/s upward or 2 m/s downward or whether the elevator is atrest? Explain.

17. Suppose you are in an elevator that is moving upward with aconstant velocity. A scale inside the elevator shows your weight tobe 600 N. (a) Does the scale register a value that is greater than, lessthan, or equal to 600 N during the time when the elevator slowsdown as it comes to a stop? (b) What is the reading when the eleva-tor is stopped? (c) How does the value registered on the scale com-pare to 600 N during the time when the elevator picks up speedagain on its way back down? Give your reasoning in each case.

18. A person has a choice of either pushing or pulling a sled at aconstant velocity, as the drawing illustrates. Friction is present. If theangle � is the same in both cases, does it require less force to pushor to pull? Account for your answer.

θθ

19. Suppose that the coefficients of static and kinetic friction havevalues such that �s � 1.4�k for a crate in contact with a cementfloor. Does this mean that the magnitude of the static frictional forceacting on the crate at rest would always be 1.4 times the magnitudeof the kinetic frictional force acting on the moving crate? Give yourreasoning.

20. A box rests on the floor of an elevator. Because of static friction,a force is required to start the box sliding across the floor when theelevator is (a) stationary, (b) accelerating upward, and (c) accelerat-ing downward. Rank the forces required in these three situations inascending order—that is, smallest first. Explain.

21. A rope is used in a tug-of-war between two teams of five peopleeach. Both teams are equally strong, so neither team wins. An iden-tical rope is tied to a tree, and the same ten people pull just as hardon the loose end as they did in the contest. In both cases, the peoplepull steadily with no jerking. Which rope, if either, is more likely tobreak? Justify your answer.

22. A stone is thrown from the top of a cliff. As the stone falls, is itin equilibrium? Explain, ignoring air resistance.

23. ssm Can an object ever be in equilibrium if the object is actedon by only (a) a single nonzero force, (b) two forces that point inmutually perpendicular directions, and (c) two forces that point indirections that are not perpendicular? Account for your answers.

24. A circus performer hangs stationary from a rope. She then be-gins to climb upward by pulling herself up, hand over hand. Whenshe starts climbing, is the tension in the rope less than, equal to, orgreater than it is when she hangs stationary? Explain.

25. During the final stages of descent, a sky diver with an openparachute approaches the ground with a constant velocity. The winddoes not blow him from side to side. Is the sky diver in equilibriumand, if so, what forces are responsible for the equilibrium?

26. A weight hangs from a ring at the middle of a rope, as the draw-ing illustrates. Can the person who is pulling on the right end of therope ever make the rope perfectly horizontal? Explain your answerin terms of the forces that act on the ring.

27. A freight train is accelerating on a level track. Other things be-ing equal, would the tension in the coupling between the engine andthe first car change if some of the cargo in the last car were trans-ferred to any one of the other cars? Account for your answer.

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ssm Solution is in the Student Solutions Manual.

www Solution is available on the World Wide Web at www.wiley.com/college/cutnell.

PROBLEMS | 41

Section 4.3 Newton’s Second Law of Motion

1. An airplane has a mass of 3.1 � 104 kg and takes off under theinfluence of a constant net force of 3.7 � 104 N. What is the netforce that acts on the plane’s 78-kg pilot?

2. Concept Simulation 4.1 at www.wiley.com/college/cutnell re-views the central idea in this problem. A boat has a mass of 6800 kg.Its engines generate a drive force of 4100 N, due west, while thewind exerts a force of 800 N, due east, and the water exerts a resis-tive force of 1200 N due east. What is the magnitude and directionof the boat’s acceleration?

3. In the amusement park ride known as Magic Mountain Super-man, powerful magnets accelerate a car and its riders from rest to 45 m/s (about 100 mi/h) in a time of 7.0 s. The mass of the car andriders is 5.5 � 103 kg. Find the average net force exerted on the carand riders by the magnets.

4. Review Interactive LearningWare 4.1 at www.wiley.com/college/cutnell in preparation for this problem. During a circus per-formance, a 72-kg human cannonball is shot out of an 18-m-longcannon. If the human cannonball spends 0.95 s in the cannon, deter-mine the average net force exerted on him in the barrel of the cannon.

5. ssm A 15-g bullet is fired from a rifle. It takes 2.50 � 10�3 sfor the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m /s. Assuming that the acceleration ofthe bullet is constant, find the average net force exerted on the bullet.

6. Interactive LearningWare 4.1 at www.wiley.com/college/cutnellreviews the approach taken in problems such as this one. A 1580-kgcar is traveling with a speed of 15.0 m/s. What is the magnitude ofthe horizontal net force that is required to bring the car to a halt in adistance of 50.0 m?

7. ssm A person with a black belt in karate has a fist that has amass of 0.70 kg. Starting from rest, this fist attains a velocity of 8.0 m/s in 0.15 s. What is the magnitude of the average net force ap-plied to the fist to achieve this level of performance?

* 8. An arrow, starting from rest, leaves the bow with a speed of 25.0 m/s. If the average force exerted on the arrow by the bow weredoubled, all else remaining the same, with what speed would the ar-row leave the bow?

* 9. ssm www Two forces A and B are applied to an object whosemass is 8.0 kg. The larger force is A. When both forces point due east,the object’s acceleration has a magnitude of 0.50 m/s2. However, when

A points due east and B points due west, the acceleration is 0.40 m/s2, due east. Find (a) the magnitude of A and (b) the magni-tude of B .

Section 4.4 The Vector Nature of Newton’s Second Law of Motion, Section 4.5 Newton’s Third Law of Motion

10. A 350-kg sailboat has an acceleration of 0.62 m/s2 at an angleof 64° north of east. Find the magnitude and direction of the netforce that acts on the sailboat.

FB

FB

FB

FB

FBFB

FB

11. Two forces, 1 and 2, acton the 7.00-kg block

shown in the drawing. The magni-tudes of the forces are F1 � 59.0 Nand F2 � 33.0 N. What is the hori-zontal acceleration (magnitude anddirection) of the block?

12. When a parachute opens, the airexerts a large drag force on it. This upward force is initially greaterthan the weight of the sky diver and, thus, slows him down. Supposethe weight of the sky diver is 915 N and the drag force has a magni-tude of 1027 N. The mass of the sky diver is 93.4 kg. What are themagnitude and direction of his acceleration?

13. ssm Only two forces act on an object(mass � 3.00 kg), as in the drawing. Findthe magnitude and direction (relative to thex axis) of the acceleration of the object.

14. Airplane flight recorders must be ableto survive catastrophic crashes. Therefore,they are typically encased in crash-resistantsteel or titanium boxes that are subjected torigorous testing. One of the tests is an im-pact shock test, in which the box must sur-vive being thrown at high speeds against abarrier. A 41-kg box is thrown at a speed of220 m/s and is brought to a halt in a collision that lasts for a time of6.5 ms. What is the magnitude of the average net force that acts onthe box during the collision?

* 15. A duck has a mass of 2.5 kg. As the duck paddles, a force of0.10 N acts on it in a direction due east. In addition, the current ofthe water exerts a force of 0.20 N in a direction of 52° south of east.When these forces begin to act, the velocity of the duck is 0.11 m/sin a direction due east. Find the magnitude and direction (relative todue east) of the displacement that the duck undergoes in 3.0 s whilethe forces are acting.

** 16. At a time when mining asteroids has become feasible, astronautshave connected a line between their 3500-kg space tug and a 6200-kgasteroid. Using their ship’s engine, they pull on the asteroid with a forceof 490 N. Initially the tug and the asteroid are at rest, 450 m apart. Howmuch time does it take for the ship and the asteroid to meet?

** 17. ssm www A 325-kg boat is sailing 15.0° north of east at aspeed of 2.00 m/s. Thirty seconds later, it is sailing 35.0° north ofeast at a speed of 4.00 m/s. During this time, three forces act on theboat: a 31.0-N force directed 15.0° north of east (due to an auxiliaryengine), a 23.0-N force directed 15.0° south of west (resistance dueto the water), and W (due to the wind). Find the magnitude and di-rection of the force W. Express the direction as an angle with re-spect to due east.

Section 4.7 The Gravitational Force

18. On earth, two parts of a space probe weigh 11 000 N and 3400 N. These parts are separated by a center-to-center distance of

FB

FB

FB

FB

PROBLEMS

F1

F2

70.0°

7.00 kg

B

B

60.0 N

40.0 N

45.0°

+x

+y

Problem 13

Note to Instructors: Most of the homework problems in this chapter are available for assignment via an on-line homework management program suchas eGrade Plus or WebAssign, and those marked with the icon are presented in a guided tutorial format that provides enhanced interactivity. See Preface for additional details.

This icon represents a biomedical application.

45807_04_p1-48 6/20/05 7:40 AM


12 m and may be treated as uniform spherical objects. Find the mag-nitude of the gravitational force that each part exerts on the other outin space, far from any other objects.

19. A bowling ball (mass � 7.2 kg, radius � 0.11 m) and a billiardball (mass � 0.38 kg, radius � 0.028 m) may each be treated as uni-form spheres. What is the magnitude of the maximum gravitationalforce that each can exert on the other?

20. A rock of mass 45 kg accidentally breaks loose from the edge ofa cliff and falls straight down. The magnitude of the air resistancethat opposes its downward motion is 18 N. What is the magnitude ofthe acceleration of the rock?

21. ssm Saturn has an equatorial radius of 6.00 � 107 m and amass of 5.67 � 1026 kg. (a) Compute the acceleration of gravity atthe equator of Saturn. (b) What is the ratio of a person’s weight onSaturn to that on earth?

22. Review Conceptual Example 7 in preparation for this problem.In tests on earth a lunar surface exploration vehicle (mass �5.90 � 103 kg) achieves a forward acceleration of 0.220 m/s2. To achieve this same acceleration on the moon, the vehicle’s engines must produce a drive force of 1.43 � 103 N. What is themagnitude of the frictional force that acts on the vehicle on themoon?

23. ssm Synchronous communications satellites are placed in a cir-cular orbit that is 3.59 � 107 m above the surface of the earth. Whatis the magnitude of the acceleration due to gravity at this distance?

24. The drawing shows three particles far away from any other ob-jects and located on a straight line. The masses of these particles aremA � 363 kg, mB � 517 kg, and mC � 154 kg. Find the magnitudeand direction of the net gravitational force acting on (a) particle A,(b) particle B, and (c) particle C.

25. (a) Calculate the magnitude of the gravitational force exerted ona 425-kg satellite that is a distance of two earth radii from the centerof the earth. (b) What is the magnitude of the gravitational force ex-erted on the earth by the satellite? (c) Determine the magnitude ofthe satellite’s acceleration. (d) What is the magnitude of the earth’sacceleration?

26. A space traveler weighs 540 N on earth. What will the travelerweigh on another planet whose radius is three times that of earth andwhose mass is twice that of earth?

27. ssm Mars has a mass of 6.46 � 1023 kg and a radius of 3.39 � 106 m. (a) What is the acceleration due to gravity on Mars?(b) How much would a 65-kg person weigh on this planet?

* 28. Three uniform spheres are located at the corners of an equilat-eral triangle. Each side of the triangle has a length of 1.20 m. Two ofthe spheres have a mass of 2.80 kg each. The third sphere (mass un-known) is released from rest. Considering only the gravitationalforces that the spheres exert on each other, what is the magnitude ofthe initial acceleration of the third sphere?

* 29. ssm www Several people are riding in a hot-air balloon. Thecombined mass of the people and balloon is 310 kg. The balloon ismotionless in the air, because the downward-acting weight of thepeople and balloon is balanced by an upward-acting “buoyant”force. If the buoyant force remains constant, how much mass shouldbe dropped overboard so the balloon acquires an upward accelera-tion of 0.15 m/s2?

* 30. A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational forceexerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earthand the moon. The distance between the earth and the moon is 3.85 � 108 m , and the mass of the earth is 81.4 times as great asthat of the moon.

* 31. The sun is more massive than the moon, but the sun is fartherfrom the earth. Which one exerts a greater gravitational force on aperson standing on the earth? Give your answer by determining theratio Fsun /Fmoon of the magnitudes of the gravitational forces. Usethe data on the inside of the front cover.

* 32. A neutron star has a mass of 2.0 � 1030 kg (about the mass ofour sun) and a radius of 5.0 � 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surfaceof such a star. How fast would it be moving after it had fallen a dis-tance of 0.010 m? (Assume that the gravitational force is constantover the distance of the fall, and that the star is not rotating.)

** 33. Two particles are located on the x axis. Particle 1 has a mass mand is at the origin. Particle 2 has a mass 2m and is at x � �L. Athird particle is placed between particles 1 and 2. Where on the xaxis should the third particle be located so that the magnitude of thegravitational force on both particle 1 and particle 2 doubles? Expressyour answer in terms of L.

Section 4.8 The Normal Force, Section 4.9 Static and KineticFrictional Forces

34. A 35-kg crate rests on a horizontal floor, and a 65-kg person isstanding on the crate. Determine the magnitude of the normal force that(a) the floor exerts on the crate and (b) the crate exerts on the person.

35. ssm A student presses a book between hishands, as the drawing indicates. The forces that heexerts on the front and back covers of the book areperpendicular to the book and are horizontal. Thebook weighs 31 N. The coefficient of static frictionbetween his hands and the book is 0.40. To keepthe book from falling, what is the magnitude of theminimum pressing force that each hand must exert?

36. A 95.0-kg person stands on a scale in an elevator. What is theapparent weight when the elevator is (a) accelerating upward with anacceleration of 1.80 m/s2, (b) moving upward at a constant speed,and (c) accelerating downward with an acceleration of 1.30 m/s2?

37. A block whose weight is 45.0 N rests on a horizontal table. Ahorizontal force of 36.0 N is applied to the block. The coefficients ofstatic and kinetic friction are 0.650 and 0.420, respectively. Will theblock move under the influence of the force, and, if so, what will bethe block’s acceleration? Explain your reasoning.

38. A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road thatrises 15° above the horizontal. What are the magnitudes of (a) thenormal force and (b) the static frictional force that the ground exertson the tires?

39. ssm A 60.0-kg crate rests on a level floor at a shipping dock. Thecoefficients of static and kinetic friction are 0.760 and 0.410, respec-tively. What horizontal pushing force is required to (a) just start the cratemoving and (b) slide the crate across the dock at a constant speed?

40. A 6.00-kg box is sliding across the horizontal floor of an eleva-tor. The coefficient of kinetic friction between the box and the flooris 0.360. Determine the kinetic frictional force that acts on the boxwhen the elevator is (a) stationary, (b) accelerating upward with anacceleration whose magnitude is 1.20 m/s2, and (c) acceleratingdownward with an acceleration whose magnitude is 1.20 m/s2.

0.500 m

A B C

0.250 m

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PROBLEMS | 43

41. An 81-kg baseball player slides into second base. The co-efficient of kinetic friction between the player and the

ground is 0.49. (a) What is the magnitude of the frictional force? (b)If the player comes to rest after 1.6 s, what was his initial velocity?

* 42. Consult Multiple-Concept Example 10 in preparation for thisproblem. Traveling at a speed of 16.1 m/s, the driver of an automo-bile suddenly locks the wheels by slamming on the brakes. The co-efficient of kinetic friction between the tires and the road is 0.720.What is the speed of the automobile after 1.30 s have elapsed? Ig-nore the effects of air resistance.

* 43. ssm Refer to ConceptSimulation 4.4 at www.wi-ley.com/college/cutnell forbackground relating to thisproblem. The drawing shows alarge cube (mass � 25 kg) be-ing accelerated across a horizontal frictionless surface by a horizontalforce . A small cube (mass � 4.0 kg) is in contact with the frontsurface of the large cube and will slide downward unless is suffi-ciently large. The coefficient of static friction between the cubes is0.71. What is the smallest magnitude that can have in order to keepthe small cube from sliding downward?

* 44. Refer to Multiple-Concept Example 10 for help in solving prob-lems like this one. An ice skater is gliding horizontally across the icewith an initial velocity of �6.3 m/s. The coefficient of kinetic fric-tion between the ice and the skate blades is 0.081, and air resistanceis negligible. How much time elapses before her velocity is reducedto �2.8 m/s ?

** 45. The drawing shows a 25.0-kgcrate that is initially at rest. Notethat the view is one looking downon the top of the crate. Twoforces, 1 and 2 , are applied tothe crate, and it begins to move.The coefficient of kinetic frictionbetween the crate and the floor is� k � 0.350. Determine the mag-nitude and direction (relative to the x axis) of the acceleration of thecrate.

Section 4.10 The Tension Force, Section 4.11 EquilibriumApplications of Newton’s Laws of Motion

46. Review Interactive LearningWare 4.3 atwww.wiley.com/college/cutnell in prepara-tion for this problem. The helicopter in thedrawing is moving horizontally to the right ata constant velocity. The weight of the heli-copter is W � 53 800 N. The lift force generated by the rotating blade makes an an-gle of 21.0° with respect to the vertical. (a)What is the magnitude of the lift force? (b)Determine the magnitude of the air resistance

that opposes the motion.

47. A supertanker (mass � 1.70 � 108 kg) is moving with a con-stant velocity. Its engines generate a forward thrust of 7.40 � 105 N.Determine (a) the magnitude of the resistive force exerted on thetanker by the water and (b) the magnitude of the upward buoyantforce exerted on the tanker by the water.

48. Part a of the drawing shows a bucket of water suspended fromthe pulley of a well; the tension in the rope is 92.0 N. Part b shows

RB

LB

FB

FB

PB

PB

PB

the same bucket of water be-ing pulled up from the well ata constant velocity. What isthe tension in the rope in part b?

49. ssm Three forces act ona moving object. One forcehas a magnitude of 80.0 Nand is directed due north. Another has a magnitude of60.0 N and is directed duewest. What must be the mag-nitude and direction of thethird force, such that the object continues to move with a constant velocity?

50. The drawing shows a wire tooth brace used by orthodon-tists. The topmost tooth is protruding slightly, and thetension in the wire exerts two forces and � on this

tooth in order to bring it back into alignment. If the forces have thesame magnitude of 21.0 N, what is the magnitude of the net forceexerted on the tooth by these forces?

51. A stuntman is being pulled along a rough road at a constant ve-locity by a cable attached to a moving truck. The cable is parallel tothe ground. The mass of the stuntman is 109 kg, and the coefficientof kinetic friction between the road and him is 0.870. Find the ten-sion in the cable.

52. As preparation for this problem, review Example 13. Supposethat the pilot suddenly jettisons 2800 N of fuel. If the plane is tocontinue moving with the same velocity under the influence of thesame air resistance , by how much does the pilot have to reduce(a) the thrust and (b) the lift?

53. ssm A 1.40-kg bottle of vintagewine is lying horizontally in the rackshown in the drawing. The two surfaceson which the bottle rests are 90.0° apart,and the right surface makes an angle of45.0° with respect to the ground. Eachsurface exerts a force on the bottle thatis perpendicular to the surface. What isthe magnitude of each of these forces?

54. The drawing shows a circus clown whoweighs 890 N. The coefficient of static frictionbetween the clown’s feet and the ground is 0.53.He pulls vertically downward on a rope thatpasses around three pulleys and is tied aroundhis feet. What is the minimum pulling force thatthe clown must exert to yank his feet out fromunder himself?

RB

TB

TB

Frictionless

P

µ s = 0.71

B

(a) (b)

L

R

W

21.0°

v

B

B

90.0°

Problem 48

+x

+y

55.0˚

54.0 N

88.0

N

F1

F2

Top view

B

B

16.0°16.0°

Brace

T' T

45807_04_p1-48 6/20/05 7:40 AM


* 55. The drawing shows box 1 resting ona table, with box 2 resting on top of box1. A massless rope passes over a mass-less, frictionless pulley. One end of therope is connected to box 2 and the otherend is connected to box 3. The weights ofthe three boxes are W1 � 55 N, W2 �35 N, and W3 � 28 N. Determine themagnitude of the normal force that thetable exerts on box 1.

* 56. Interactive LearningWare 4.3 atwww.wiley.com/college/cutnell reviewsthe principles that play a role in thisproblem. During a storm, a tree limbbreaks off and comes to rest across a barbed wire fence at a pointthat is not in the middle between two fence posts. The limb exerts adownward force of 151 N on the wire. The left section of the wiremakes an angle of 14.0º relative to the horizontal and sustains a ten-sion of 447 N. Find the magnitude and direction of the tension thatthe right section of the wire sustains.

* 57. ssm A 44-kg chandelier is suspended 1.5 m below a ceiling bythree wires, each of which has the same tension and the same lengthof 2.0 m (see the drawing). Find the tension in each wire.

* 58. The person in the drawing is standing on crutches. As-sume that the force exerted on each crutch by the groundis directed along the crutch, as the force vectors in the

drawing indicate. If the coefficient of static friction between a crutchand the ground is 0.90, determine the largest angle �MAX that thecrutch can have just before it begins to slip on the floor.

* 59. A toboggan slides down a hill and has a constant veloc-ity. The angle of the hill is 8.00° with respect to the hori-

zontal. What is the coefficient of kinetic friction between the surfaceof the hill and the toboggan?

** 60. The weight of the block in the drawing is 88.9 N. The coeffi-cient of static friction between the block and the vertical wall is

0.560. (a) What minimum force is re-quired to prevent the block from slidingdown the wall? (Hint: The static frictionalforce exerted on the block is directed up-ward, parallel to the wall.) (b) What mini-mum force is required to start the blockmoving up the wall? (Hint: The static fric-tional force is now directed down thewall.)

** 61. ssm A bicyclist is coasting straightdown a hill at a constant speed. The mass of the rider and bicycle is80.0 kg, and the hill is inclined at 15.0° with respect to the horizon-tal. Air resistance opposes the motion of the cyclist. Later, the bicy-clist climbs the same hill at the same constant speed. How muchforce (directed parallel to the hill) must be applied to the bicycle inorder for the bicyclist to climb the hill?

** 62. A damp washcloth is hung over the edge of a table to dry. Thus,part (mass � mon ) of the washcloth rests on the table and part (mass � moff ) does not. The coefficient of static friction between thetable and the washcloth is 0.40. Determine the maximum fraction[moff /(mon � moff )] that can hang over the edge without causing thewhole washcloth to slide off the table.

Section 4.12 Nonequilibrium Applications of Newton’s Laws of Motion

63. Only two forces act on an object (mass �4.00 kg), as in the drawing. Find the magni-tude and direction (relative to the x axis) ofthe acceleration of the object.

64. A fisherman is fishing from a bridge andis using a “45-N test line.” In other words,the line will sustain a maximum force of 45 N without breaking. (a) What is theweight of the heaviest fish that can be pulledup vertically when the line is reeled in (a) ata constant speed and (b) with an accelerationwhose magnitude is 2.0 m/s2?

65. ssm A 1380-kg car is moving due east with an initial speed of27.0 m/s. After 8.00 s the car has slowed down to 17.0 m/s. Find themagnitude and direction of the net force that produces the decelera-tion.

66. A paraglider is flying horizontally at a constant speed. Assumethat only two forces act on it in the vertical direction, its weight anda vertical lift force exerted on its wings by the air. The lift force hasa magnitude of 1800 N. (a) What is the magnitude and direction ofthe force that the paraglider exerts on the earth? (b) If the lift forceshould suddenly decrease to 1200 N, what would be the vertical ac-celeration of the glider? For both questions, take the upward direc-tion to be the �y direction.

67. In the drawing, the weight of theblock on the table is 422 N and that ofthe hanging block is 185 N. Ignoring allfrictional effects and assuming the pul-ley to be massless, find (a) the accelera-tion of the two blocks and (b) the ten-sion in the cord.

68. Review Conceptual Example 16 asbackground for this problem. The water skier there has a mass of 73 kg. Find the magnitude of the net force acting on the skier when(a) she is accelerated from rest to a speed of 11 m/s in 8.0 s and (b)she lets go of the tow rope and glides to a halt in 21 s.

FB

Problem 55

2

3

1

185 N

422 N

60.0 N

40.0 N

+y

+x

Problem 63

F 40.0°

Block

Verticalwall

B

Problem 60

2.0 m1.5 m

F

θ

45807_04_p1-48 6/20/05 7:40 AM

PROBLEMS | 45

69. ssm www A student is skateboarding down a ramp that is 6.0 m long and inclined at 18° with respect to the horizontal. Theinitial speed of the skateboarder at the top of the ramp is 2.6 m/s.Neglect friction and find the speed at the bottom of the ramp.

70. A car is towing a boat on a trailer. The driver starts from rest andaccelerates to a velocity of �11 m/s in a time of 28 s. The combinedmass of the boat and trailer is 410 kg. The frictional force acting onthe trailer can be ignored. What is the tension in the hitch that con-nects the trailer to the car?

71. ssm In a supermarket parking lot, an employee is pushing tenempty shopping carts, lined up in a straight line. The acceleration ofthe carts is 0.050 m/s2. The ground is level, and each cart has a massof 26 kg. (a) What is the net force acting on any one of the carts? (b)Assuming friction is negligible, what is the force exerted by the fifthcart on the sixth cart?

72. The space probe Deep Space 1 was launched on October 24,1998. Its mass was 474 kg. The goal of the mission was to test a newkind of engine called an ion propulsion drive. This engine generatedonly a weak thrust, but it could do so over long periods of time withthe consumption of only small amounts of fuel. The mission wasspectacularly successful. At a thrust of 56 mN how many days wererequired for the probe to attain a velocity of 805 m/s (1800 mi /h),assuming that the probe started from rest and that the mass remainednearly constant?

73. A cable is lifting a constructionworker and a crate, as the drawingshows. The weights of the worker andcrate are 965 and 1510 N, respectively.The acceleration of the cable is 0.620 m/s2, upward. What is the tensionin the cable (a) below the worker and (b)above the worker?

* 74. This problem uses the same conceptsas Multiple-Concept Example 17, exceptthat kinetic, rather than static, friction isinvolved. A crate is siding down a rampthat is inclined at an angle of 38.0° above the horizontal. The coeffi-cient of kinetic friction between the crate and the ramp surface is0.600. Find the acceleration of the moving crate.

* 75. ssm A person whose weight is 5.20 � 102 N is being pulled upvertically by a rope from the bottom of a cave that is 35.1 m deep.The maximum tension that the rope can withstand without breakingis 569 N. What is the shortest time, starting from rest, in which theperson can be brought out of the cave?

* 76. The principles used to solve this problem are similar to those inMultiple-Concept Example 17. A 205-kg log is pulled up a ramp bymeans of a rope that is parallel tothe surface of the ramp. The rampis inclined at 30.0° with respect tothe horizontal. The coefficient ofkinetic friction between the logand the ramp is 0.900, and the loghas an acceleration of 0.800 m/s2.Find the tension in the rope.

* 77. The drawing showsRobin Hood (mass �

77.0 kg) about to escape from adangerous situation. With onehand, he is gripping the rope thatholds up a chandelier (mass �195 kg). When he cuts the rope

where it is tied to the floor, the chandelier will fall, and he will bepulled up toward a balcony above. Ignore the friction between therope and the beams over which it slides, and find (a) the accelerationwith which Robin is pulled upward and (b) the tension in the ropewhile Robin escapes.

* 78. A train consists of 50 cars, each of which has a mass of 6.8 � 103 kg. The train has an acceleration of �8.0 � 10�2 m/s2.Ignore friction and determine the tension in the coupling (a) be-tween the 30th and 31st cars and (b) between the 49th and 50th cars.

* 79. Consult Multiple-Concept Example 10 for insight into solvingthis type of problem. A box is sliding up an incline that makes anangle of 15.0° with respect to the horizontal. The coefficient of ki-netic friction between the box and the surface of the incline is 0.180.The initial speed of the box at the bottom of the incline is 1.50 m/s.How far does the box travel along the incline before coming to rest?

* 80. The alarm at a fire station rings and an 86-kg fireman, startingfrom rest, slides down a pole to the floor below (a distance of 4.0 m). Just before landing, his speed is 1.4 m/s . What is the magni-tude of the kinetic frictional force exerted on the fireman as he slidesdown the pole?

* 81. ssm At an airport, luggage is unloaded from a plane into thethree cars of a luggage carrier, as the drawing shows. The accelera-tion of the carrier is 0.12 m/s2, and friction is negligible. The cou-pling bars have negligible mass. By how much would the tension ineach of the coupling bars A, B, and C change if 39 kg of luggagewere removed from car 2 and placed in (a) car 1 and (b) car 3? If thetension changes, specify whether it increases or decreases.

* 82. Consult Interactive LearningWare 4.2 at www.wiley.com/college/cutnell before beginning this problem. A truck is traveling ata speed of 25.0 m/s along a level road. A crate is resting on the bedof the truck, and the coefficient of static friction between the crateand the truck bed is 0.650. Determine the shortest distance in whichthe truck can come to a halt without causing the crate to slip forwardrelative to the truck.

** 83. In the drawing, therope and the pulleys aremassless, and there is nofriction. Find (a) the ten-sion in the rope and (b)the acceleration of the10.0-kg block. (Hint:The larger mass movestwice as far as thesmaller mass.)

** 84. A small sphere ishung by a string fromthe ceiling of a van. When thevan is stationary, the spherehangs vertically. However, whenthe van accelerates, the sphere swings backward so that the stringmakes an angle of � with respect to the vertical. (a) Derive an ex-pression for the magnitude a of the acceleration of the van in termsof the angle � and the magnitude g of the acceleration due to gravity.(b) Find the acceleration of the van when � � 10.0°. (c) What is theangle � when the van moves with a constant velocity?

Couplingbar C

Car 3 Car 2 Car 1

Couplingbar B

Couplingbar A

10.0 kg

3.00 kg

Problem 83

a = 0.620 m/s2

Problem 73

Problem 77

45807_04_p1-48 6/17/05 3:32 PM


** 85. ssm A penguin slides at a constant velocity of 1.4 m/s down anicy incline. The incline slopes above the horizontal at an angle of6.9°. At the bottom of the incline, the penguin slides onto a horizon-tal patch of ice. The coefficient of kinetic friction between the pen-guin and the ice is the same for the incline as for the horizontalpatch. How much time is required for the penguin to slide to a haltafter entering the horizontal patch of ice?

ADDITIONAL PROBLEMS

87. ssm In preparation for this problem, review Conceptual Exam-ple 7. A space traveler whose mass is 115 kg leaves earth. What arehis weight and mass (a) on earth and (b) in interplanetary spacewhere there are no nearby planetary objects?

88. A person, sunbathing on a warm day, is lying horizontally onthe deck of a boat. Her mass is 59 kg, and the coefficient of staticfriction between the deck and her is 0.70. Assume that the person ismoving horizontally, and that the static frictional force is the onlyforce acting on her in this direction. (a) What is the magnitude of thestatic frictional force when the boat moves with a constant velocityof �8.0 m/s? (b) The boat speeds up with an acceleration of 1.6 m/s2, and she does not slip with respect to the deck. What is themagnitude of the static frictional force that acts on her? (c) What isthe magnitude of the maximum acceleration the boat can have be-fore she begins to slip relative to the deck?

89. ssm A rocket blasts off from rest and attains a speed of 45 m/sin 15 s. An astronaut has a mass of 57 kg. What is the astronaut’s ap-parent weight during takeoff ?

90. Interactive LearningWare 4.4 at www.wiley.com/college/cutnellprovides a review of the concepts that are important in this problem.A rocket of mass 4.50 � 105 kg is in flight. Its thrust is directed atan angle of 55.0º above the horizontal and has a magnitude of 7.50 � 106 N. Find the magnitude and direction of the rocket’s ac-celeration. Give the direction as an angle above the horizontal.

91. ssm A 20.0-kg sled is being pulled across a horizontal surfaceat a constant velocity. The pulling force has a magnitude of 80.0 Nand is directed at an angle of 30.0° above the horizontal. Determinethe coefficient of kinetic friction.

92. Concept Simulation 4.1 at www.wiley.com/college/cutnell re-views the concepts that are important in this problem. The speed of abobsled is increasing because it has an acceleration of 2.4 m/s2. At agiven instant in time, the forces resisting the motion, including ki-netic friction and air resistance, total 450 N. The mass of the bobsledand its riders is 270 kg. (a) What is the magnitude of the force pro-pelling the bobsled forward? (b) What is the magnitude of the netforce that acts on the bobsled?

93. A 1.14 � 104-kg lunar landing craft is about to touch down onthe surface of the moon, where the acceleration due to gravity is1.60 m/s2. At an altitude of 165 m the craft’s downward velocity is18.0 m/s. To slow down the craft, a retrorocket is firing to providean upward thrust. Assuming the descent is vertical, find the magni-tude of the thrust needed to reduce the velocity to zero at the instantwhen the craft touches the lunar surface.

94. A woman stands on a scale in a moving elevator. Her mass is60.0 kg, and the combined mass of the elevator and scale is an addi-tional 815 kg. Starting from rest, the elevator accelerates upward.During the acceleration, the hoisting cable applies a force of 9410 N. What does the scale read during the acceleration?

95. ssm When a 58-g tennis ball is served, it accelerates from restto a speed of 45 m/s. The impact with the racket gives the ball aconstant acceleration over a distance of 44 cm. What is the magni-tude of the net force acting on the ball?

96. A person in a kayak starts paddling, and it accelerates from 0 to0.60 m/s in a distance of 0.41 m. If the combined mass of the personand the kayak is 73 kg, what is the magnitude of the net force actingon the kayak?

97. The mass of a robot is 5450 kg. This robot weighs 3620 N moreon planet A than it does on planet B. Both planets have the same ra-dius of 1.33 � 107 m. What is the difference MA � MB in themasses of these planets?

98. The drawing (not to scale) shows one alignment of the sun,earth, and moon. The gravitational force SM that the sun exerts onthe moon is perpendicular to the force EM that the earth exerts onthe moon. The masses are: mass of sun � 1.99 � 1030 kg, mass ofearth � 5.98 � 1024 kg, mass of moon � 7.35 � 1022 kg. The dis-tances shown in the drawing are rSM � 1.50 � 1011 m and rEM �3.85 � 108 m. Determine the magnitude of the net gravitationalforce on the moon.

* 99. ssm Two objects (45.0 and 21.0 kg) are connected by a mass-less string that passes over a massless, frictionless pulley. The pulleyhangs from the ceiling. Find (a) the acceleration of the objects and(b) the tension in the string.

* 100. The central ideas in this problem are reviewed in Multiple-Concept Example 9. One block rests upon a horizontal surface. Asecond identical block rests upon the first one. The coefficient of sta-tic friction between the blocks is the same as the coefficient of staticfriction between the lower block and the horizontal surface. A hori-zontal force is applied to the upper block, and its magnitude isslowly increased. When the force reaches 47.0 N, the upper blockjust begins to slide. The force is then removed from the upper block,and the blocks are returned to their original configuration. What isthe magnitude of the horizontal force that should be applied to thelower block, so that it just begins to slide out from under the upperblock?

FB

FB

** 86. A 5.00-kg block is placed on top of a 12.0-kg block that rests ona frictionless table. The coefficient of static friction between the twoblocks is 0.600. What is the maximum horizontal force that can beapplied before the 5.00-kg block begins to slip relative to the 12.0-kg block, if the force is applied to (a) the more massive blockand (b) the less massive block?

Sun

Earth

Moon

rEM

rSM

FSM

FEM

B

B

45807_04_p1-48 6/17/05 3:32 PM

CONCEPTS & CALCULATIONS | 47

* 101. A skier is pulled up a slope at a constant velocity by a tow bar.The slope is inclined at 25.0° with respect to the horizontal. Theforce applied to the skier by the tow bar is parallel to the slope. Theskier’s mass is 55.0 kg, and the coefficient of kinetic friction be-tween the skis and the snow is 0.120. Find the magnitude of theforce that the tow bar exerts on the skier.

* 102. A mountain climber, in the process of crossing between twocliffs by a rope, pauses to rest. She weighs 535 N. As the drawingshows, she is closer to the left cliff than to the right cliff, with the re-sult that the tensions in the left and right sides of the rope are not thesame. Find the tensions in the rope to the left and to the right of themountain climber.

* 103. ssm To hoist himself into a tree, a 72.0-kg man ties one end ofa nylon rope around his waist and throws the other end over abranch of the tree. He then pulls downward on the free end of therope with a force of 358 N. Neglect any friction between the ropeand the branch, and determine the man’s upward acceleration.

* 104. Multiple-Concept Example 17 reviews the basic concepts in-volved in this problem. A sports car is accelerating up a hill thatrises 18° above the horizontal. The coefficient of static friction be-tween the wheels and the road is � s � 0.88. It is the static frictionalforce that propels the car forward. (a) What is the magnitude of themaximum acceleration that the car can have? (b) What is the magni-tude of the maximum acceleration if the car is being driven down thehill?

* 105. A space probe has two engines. Each generates the sameamount of force when fired, and the directions of these forces can beindependently adjusted. When the engines are fired simultaneouslyand each applies its force in the same direction, the probe, startingfrom rest, takes 28 s to travel a certain distance. How long does ittake to travel the same distance, again starting from rest, if the en-gines are fired simultaneously and the forces that they apply to theprobe are perpendicular?

* 106. Jupiter is the largest planet in our solar system, having a massand radius that are, respectively, 318 and 11.2 times that of earth.Suppose that an object falls from rest near the surface of eachplanet and that the acceleration due to gravity remains constant dur-ing the fall. Each object falls the same distance before striking theground. Determine the ratio of the time of fall on Jupiter to that onearth.

* 107. ssm A person is trying to judge whether a picture (mass �1.10 kg) is properly positioned by temporarily pressing it against awall. The pressing force is perpendicular to the wall. The coefficientof static friction between the picture and the wall is 0.660. What isthe minimum amount of pressing force that must be used?

** 108. As part a of the drawing shows, two blocks are connected by arope that passes over a set of pulleys. One block has a weight of 412 N, and the other has a weight of 908 N. The rope and the pulleysare massless and there is no friction. (a) What is the acceleration of thelighter block? (b) Suppose that the heavier block is removed, and adownward force of 908 N is provided by someone pulling on the rope,as part b of the drawing shows. Find the acceleration of the remainingblock. (c) Explain why the answers in (a) and (b) are different.

** 109. ssm The drawingshows three objects.They are connected bystrings that pass overmassless and friction-free pulleys. The ob-jects move, and the co-efficient of kineticfriction between themiddle object and thesurface of the table is 0.100. (a) What is the acceleration of thethree objects? (b) Find the tension in each of the two strings.

** 110. The basic concepts in this problem are presented in Multiple-Concept Example 9. A 225-kg crate rests on a surface that is in-clined above the horizontal at an angle of 20.0°. A horizontal force(magnitude � 535 N and parallel to the ground, not the incline) isrequired to start the crate moving down the incline. What is the coef-ficient of static friction between the crate and the incline?

** 111. While moving in, a new homeowner is pushing a box acrossthe floor at a constant velocity. The coefficient of kinetic friction be-tween the box and the floor is 0.41. The pushing force is directeddownward at an angle � below the horizontal. When � is greater thana certain value, it is not possible to move the box, no matter howlarge the pushing force is. Find that value of �.

CONCEPTS & CALCULATIONSNote: Each of these problems consists of Concept Questions

followed by a related quantitative Problem. The Concept Questions involvelittle or no mathematics. They focus on the concepts with which the problemsdeal. Recognizing the concepts is the essential initial step in any problem-solving technique.

112. Concept Questions Two skaters, a man and a woman,are standing on ice. Neglect any friction between the

skate blades and the ice. The woman pushes on the man with a cer-tain force that is parallel to the ground. (a) Must the man accelerateunder the action of this force? If so, what three factors determine the

magnitude and direction of his acceleration? (b) Is there a corre-sponding force exerted on the woman? If so, where does it origi-nate? Is this force related to the magnitude and direction of the forcethe woman exerts on the man? If so, how?

Problem The mass of the man is 82 kg and that of the woman is 48 kg. The woman pushes on the man with a force of 45 N due east.Determine the acceleration (magnitude and direction) of (a) the manand (b) the woman.

113. Concept Questions A person is attempting to push arefrigerator across a room. He exerts a horizontal force

25.0kg

80.0 kg

10.0kg

65.0° 80.0°

412 N 412 N908 N

(a) (b)

908 N

45807_04_p1-48 6/20/05 7:40 AM


on the refrigerator, but it does not move. (a) What other horizontalforce must be acting on the refrigerator? (b) How are the magnitudeand direction of this force related to the force that the person exerts?(c) Suppose that the person applies the force such that it is thelargest possible force before the refrigerator begins to move. Whatfactors determine the magnitude of this force?

Problem Consult Multiple-Concept Example 9 to explore a modelfor solving this problem. A person pushes on a 57-kg refrigeratorwith a horizontal force of �267 N; the minus sign indicates thatthe force is directed along the �x direction. The coefficient of sta-tic friction is 0.65. (a) If the refrigerator does not move, what isthe magnitude and direction of the static frictional force that thefloor exerts on the refrigerator? (b) What is the magnitude of thelargest pushing force that can be applied to the refrigerator beforeit just begins to move?

114. Concept Questions A large rock and a small pebbleare held at the same height above the ground. (a) Is the

gravitational force exerted on the rock greater than, less than, orequal to that exerted on the pebble? Justify your answer. (b) Whenthe rock and the pebble are released, is the downward accelerationof the rock greater than, less than, or equal to that of the pebble?Why?

Problem A 5.0-kg rock and a 3.0 � 10�4-kg pebble are held nearthe surface of the earth. (a) Determine the magnitude of the gravita-tional force exerted on each by the earth. (b) Calculate the accelera-tion of each object when released. Verify that your answers are con-sistent with your answers to the Concept Questions.

115. Concept Questions The earth exerts a gravitationalforce on a falling raindrop, pulling it down. Does the

raindrop exert a gravitational force on the earth, pulling it up? If so,is this force greater than, less than, or equal to the force that theearth exerts on the raindrop? Provide a reason for your answer.

Problem A raindrop has a mass of 5.2 � 10�7 kg, and is fallingnear the surface of the earth. Calculate the magnitude of the gravita-tional force exerted (a) on the raindrop by the earth and (b) on theearth by the raindrop.

116. Concept Questions Two horizontal forces, 1 and 2 ,are acting on a box, but only 1 is shown (see the draw-

ing). 2 can point either to the right or to the left. The box movesonly along the x axis. There is no friction between the box and thesurface. What is the direction of 2 and how does its magnitude com-pare to the magnitude of 1 when the acceleration of the box is (a)positive, (b) negative, and (c) zero?

Problem Suppose that 1 � �9.0 Nand the mass of the box is 3.0 kg.Find the magnitude and direction of

2 when the acceleration of the box is (a) �5.0 m/s2, (b) �5.0 m/s2,and (c) 0 m/s2. Be sure that your answers are consistent with your an-swers to the Concept Questions.

117. Concept Questions A car is driving up a hill. (a) Is themagnitude of the normal force exerted on the car equal

to the magnitude of its weight? Why or why not? (b) If the car drives up a steeper hill, does the normal force increase, decrease, orremain the same? Justify your answer. (c) Does the magnitude of thenormal force depend on whether the car is traveling up the hill ordown the hill? Give your reasoning.

FB

FB

FB

FB

FB

FB

FB

FB

Problem A car is traveling up a hill that is inclined at an angle of �above the horizontal. Determine the ratio of the magnitude of thenormal force to the weight of the car when (a) � � 15° and (b) � �35°. Check to see that your answers are consistent with your an-swers to the Concept Questions.

* 118. Concept Questions Two blocks are sliding to the rightacross a horizontal surface, as the drawing shows. In

Case A the masses of both blocks are 3.0 kg. In Case B the mass ofblock 1, the block behind, is 6.0 kg, and the mass of block 2 is 3.0 kg. No frictional force acts on block 1 in either Case A or CaseB. However, a kinetic frictional force does act on block 2 in bothcases and opposes the motion. (a) Identify the forces that contributeto the horizontal net force acting on block 1. (b) Identify the forcesthat contribute to the horizontal net force acting on block 2. (c) Inwhich case, if either, do the blocks push against each other withgreater forces? Explain. (d) Are the blocks accelerating or decelerat-ing, and in which case, if either, is the magnitude of the accelerationgreater?

Problem The magnitude of the kinetic frictional force acting onblock 2 in the drawing is 5.8 N. For both Case A and Case B deter-ine (a) the magnitude of the forces with which the blocks pushagainst each other and (b) the acceleration of the blocks. Check tosee that your answers are consistent with your answers to the Con-cept Questions.

* 119. Concept Questions A block is pressed against a verti-cal wall by a force , as the drawing shows. This force

can either push the block upward at a constant velocity or allow it toslide downward at a constant velocity, the magnitude of the force be-ing different in the two cases, while the direc-tional angle � is the same. Kinetic friction ex-ists between the block and the wall. (a) Is theblock in equilibrium in each case? Explain.(b) In each case what is the direction of thekinetic frictional force that acts on the block?Why? (c) In each case is the magnitude of thefrictional force the same or different? Justifyyour answer. (d) In which case is the magni-tude of the force greater? Provide a reasonfor your answer.

Problem The weight of the block is 39.0 N, and the coefficient ofkinetic friction between the block and the wall is 0.250. The direc-tion of the force is � � 30.0º. Determine the magnitude of when the block slides up the wall and when it slides down the wall.Check to see that your answers are consistent with your answers tothe Concept Questions.

PB

PB

PB

PB

+xF1B

Block 1 Block 2Block 1

Case A Case B

m1 = 3.0 kg m2 = 3.0 kg m1 = 6.0 kg m2 = 3.0 kg

Block 2

P θB

45807_04_p1-48 6/20/05 7:40 AM

Ch4Forces and Newton's Laws of Motion

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Transcript of Ch4Forces and Newton's Laws of Motion