FORCES AND NEWTON S LAWS OF MOTION

(1)

F O R C E S A N D N E W T O N ’ S L A W 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 attached

to a speeding boat pulls a water skier. Forces such as those that launch the basketball or

pull 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 forces

on 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 occurs

when a diver is pulled toward the earth because of the force of gravity. The earth exerts

this force even when it is not in direct contact with the diver. In Figure 4.1, arrows are

used to represent the forces. It is appropriate to use arrows, because a force is a vector

quantity and has both a magnitude and a direction. The direction of the arrow gives the

direction of the force, and the length is proportional to its strength or magnitude.

4.

1 Cutnell

Johnson

PHY

SICS,

7e

c

_oming

January

2006

45807_04_p1-48 6/17/05 3:31 PM Page 1

(2)

2 | C H A P T E R 4 | F O RC E S A N D N E W TO N ’S L AW S O F M OT I O N

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 forces

have 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 only

slightly because of friction. Since ice is very slippery, there is only a relatively small

amount of friction to slow down the puck. In fact, if it were possible to remove all friction

and wind resistance, and if the rink were infinitely large, the puck would coast forever in a

straight line at a constant speed. Left on its own, the puck would lose none of the velocity

imparted 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) the basketball, (b) the water skier, and (c) the cliff diver. (a,© Nathaniel S. Butler/NBAE/Getty Images; b,© P. BeavisMasterfile; c,© Amy and Chuck Wiley/Wales/Index Stock)

F B

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 a

straight 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, without

using any gas after it has come up to speed. In reality gas is needed, but only so that the

engine can produce the necessary force to cancel opposing forces such as friction. This

cancellation 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 obob-ject’s velocity, but, rather, to change it.

4.2

(a)

F

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

(3)

4 .1 | N E W TO N ’S F I R ST L AW O F M OT I O N | 3

The SI unit for mass is the kilogram (kg), whereas the units in the CGS system and

the 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 4.3

gives the masses of various objects, ranging from a penny to a supertanker. The larger the

mass, 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 4.7

will discuss the distinction between them.

Figure 4.4 shows a useful application of inertia. Automobile seat belts unwind freely

when pulled gently, so they can be buckled. But in an accident, they hold you safely in

place. 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 at

rest or moving at a constant velocity, the pendulum hangs straight down, and the locking

bar 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 the

car suddenly slows down in an accident, however, the relatively massive lower part of the

pendulum keeps moving forward because of its inertia. The pendulum swings on its pivot

into the position shown in color and causes the locking bar to block the rotation of the

ratchet 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 certain

observers. Suppose, for instance, that you are a passenger riding in a friend’s car. While

the car moves at a constant speed along a straight line, you do not feel the seat pushing

against 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 pressing

against your back as the car accelerates. Therefore, you sense that a force is being applied

to you. The first law leads you to believe that your motion should change, and, relative to

the ground outside, your motion does change. But

relative to the car,

you can see that

your 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 a

frame 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 defined

below:

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 speed

will cause only an imperceptible change in the motion of a freight train. In comparison to

the bicycle, the train has a much greater tendency to remain at rest. Accordingly, we say

that the train has more

inertia

than the bicycle. Quantitatively, the inertia of an object is

measured 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 role in one seat-belt mechanism. The gray part of the drawing applies when the car is at rest or moving at a constant velocity. The colored parts show what happens when the car suddenly slows down, as in an accident.

Figure 4.3 The masses of various objects. Seat belt Motion of car Ratchet wheel Locking bar Pivots Pendulum The physics of seat belts.

DEFINITION OF INERTIA AND MASS

Inertia is the natural tendency of an object to remain at rest or in motion at a constant

speed 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 LIBERTY 1996 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) 45807_04_p1-48 6/17/05 3:31 PM Page 3

(4)

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 we

apply 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 velocity

of the object remains unchanged. The second law deals with what happens when a net

force does act. Consider a hockey puck once again. When a player strikes a stationary

puck, 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 long

as this force acts, the velocity increases, and the puck accelerates. Now, suppose another

player strikes the puck and applies twice as much force as the first player does. The

greater force produces a greater acceleration. In fact, if the friction between the puck and

the 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, or

the vector sum of all the forces acting, that is important. Mathematically, the net force is

written as

, where the Greek capital letter

(sigma) denotes the vector sum. Newton’s

second 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 the

acceleration. The other is the inertia or mass of the object. After all, the same net force

that imparts an appreciable acceleration to a hockey puck (small mass) will impart very

little acceleration to a semitrailer truck (large mass). Newton’s second law states that for a

given 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 and

the mass, as given in Equation 4.1.

F

B

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 of

the net force.

(4.1)

SI Unit of Force:

kg

m /s

2

_newton

(N)

a

B

F

B

m

or

F

B

ma

B

a

B

F

B

Note that the net force in Equation 4.1 includes only the forces that the environment

exerts 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 are

not included in Equation 4.1.

According to Equation 4.1, the SI unit for force is the unit for mass (kg) times the

unit for acceleration (m /s

2

_{), or}

The combination of kg

m /s

2

_{is called a}

_newton

_{(N) and is a derived SI unit, not a base}

unit; 1 newton

1 N

1 kg

m /s

2

_.

SI unit for force

(kg)

m

s

2

kg

m

s

2

4.3

(5)

4 . 3 | N E W TO N ’S S E CO N D L AW O F M OT I O N | 5

In the CGS system, the procedure for establishing the unit of force is the same as

with SI units, except that mass is expressed in grams (g) and acceleration in cm /s

2

_{. The}

resulting unit for force is the

dyne

; 1 dyne

1 g

cm /s

2

_.

In the BE system, the unit for force is defined to be the pound (lb),* and the unit for

acceleration is ft /s

2

_{. With this procedure, Newton’s second law can then be used to obtain}

the unit for mass:

The combination of lb

s

2

_{/ft is the unit for mass in the BE system and is called the}

_slug

(sl); 1 slug

1 sl

1 lb

s

2

_/ft.

Table 4.1 summarizes the various units for mass, acceleration, and force. Conversion

factors 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 the

net 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. Only

the forces that

act on the object

appear in a free-body diagram. Forces that the object exerts

on 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 4.5aindicates. 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. Both forces act in the same direction. A third force of 560 N also acts on the car, but in a direction opposite to that in which the people are pushing. This force arises because of friction and the extent 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 the mass of the car. To determine the net force, we use the free-body diagram in Figure 4.5b. In this diagram, the car is represented as a dot, and its motion is along the xaxis. The diagram makes 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 N395 N560 N 110 N

Unit for mass

lb

s

2

ft

BE unit for force

lb

(unit for mass)

ft

s

2

Table 4.1 Units for Mass, Acceleration, and Force

System Mass Acceleration Force

SI kilogram (kg) meter /second2_{(m /s}2₎ _{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 the pull 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 a

stalled car, in opposition to a force created by friction and the pavement. (b) A free-body diagram that shows the horizontal forces acting on the car.

Problem solving insight

A free-body diagram is very helpful when applying Newton’s second law. Always start a problem by drawing the free-body diagram.

(6)

6 | C H A P T E R 4 | F O RC E S A N D N E W TO N ’S L AW S O F M OT I O N The acceleration can now be obtained:

(4.1) The plus sign indicates that the acceleration points along the xaxis, in the same direction as the 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 the

ball 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 into

account in two dimensions by using

x

and

y

components. The net force

in Newton’s

second law has components

F

x

and

F

y

, while the acceleration

has components

a

x

and

a

y

. Consequently, Newton’s second law, as expressed in Equation 4.1, can be written

in 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 scalar

components and will be either positive or negative numbers, depending on whether they

point along the positive or negative

x

or

y

axis. The remainder of this section deals with

examples 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 raft1300 kg), as shown in Figure 4.6a. By paddling, he causes an average force B

P

of 17 N to be applied to the raft in a direction due east

F

y

ma

y

F

x

ma

x

a

B

F

B 0.059 m /s2 a F m 110 N 1850 kg

The direction of the acceleration is always the 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 a_x a_y 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 on the raft in a direction perpendicular to the surface of the water play no role in the examples and are omitted for clarity. (c) The raft’s acceleration components axand ay. (d) In 65 s, the

components of the raft’s displacement are x48 m and y23 m. AB P B

4.

4

45807_04_p1-48 6/17/05 3:31 PM Page 6

(7)

4 .4 | T H E V E C TO R N AT U R E O F N E W TO N ’S S E CO N D L AW O F M OT I O N | 7 (the xdirection). The wind also exerts a force on the raft. This force has a magnitude of

15 N and points 67° north of east. Ignoring any resistance from the water, find the xand y components of the raft’s acceleration.

Reasoning

Since the mass of the man and the raft is known, Newton’s second law can be used to determine the acceleration components from the given forces. According to the form of the second law in Equations 4.2a and 4.2b, the acceleration component in a given direction is the component of the net force in that direction divided by the mass. As an aid in determin-ing the components

F

xand

F

y of the net force, we use the free-body diagram in Figure

4.6

b

. In this diagram, the directions due east and due north are the

x

and

y

directions, re-spectively.

Solution

Figure 4.6

b

shows the force components:

AB

The plus signs indicate that Fxpoints in the direction of the xaxis and Fypoints in the

di-rection of the yaxis. The xand ycomponents 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 the raft is 0.15 m /s, in a direction due east (the xdirection). Assuming that the forces are main-tained for 65 s, find the xand ycomponents 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 in Newton’s second law to determine the acceleration, it becomes possible to use the equations of kinematics to describe the resulting motion. We know from Example 2 that the accelera-tion components are

a

x 0.018 m/s2and

a

y 0.011 m/s2, and it is given here that the

initial velocity components are

v

0x 0.15 m/s and

v

0y0 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 displacement are

Figure 4.6dshows the final location of the raft.

▲

23 m yv0yt 1 2ayt2(0 m /s)(65 s) 1 2(0.011 m /s 2_{)(65 s)}2 48 m xv0xt 1 2axt2(0.15 m /s)(65 s) 1 2(0.018 m /s 2_{)(65 s)}2 0yt 1 2ayt2) (x0xt 1 2axt2)

A

B

P

B 0.011 m /s2 ay Fy m 14 N 1300 kg 0.018 m /s2 ax Fx m 23 N 1300 kg

Force xComponent yComponent

17 N 0 N (15 N) cos 67° 6 N (15 N) sin 67° 14 N Fy 14 N Fx 17 N6 N 23 N AB P B

Need more practice?

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

Related Homework:Problems 4, 6

For an interactive solution, go to www.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 is it? (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 mass of 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, and acceleration, 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 law always involve the net external force, which is the vector sum of all the external forces that act on an object. Each component of the net force leads to a corresponding component of the acceleration.

(8)

NEWTON’S THIRD LAW OF MOTION

Imagine you are in a football game. You line up facing your opponent, the ball is

snapped, and the two of you crash together. No doubt, you feel a force. But think about

your opponent. He too feels something, for while he is applying a force to you, you are

applying 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 pairs

and 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 and reaction forces on each other. (First Light/Corbis Images)

–P

+PB

B

Figure 4.7 The astronaut pushes on the spacecraft with a force . According to Newton’s third law, the spacecraft simultaneously pushes back on the astronaut with a force PB.

P B

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 as

follows: “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 just

outside a spacecraft and who pushes on the spacecraft with a force . According to the

third law, the spacecraft pushes back on the astronaut with a force

that is equal in

magnitude 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 mS11 000 kg and that the mass of

the astronaut is mA92 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 reacac-tion forces have the same magnitude, they do not create acceleraac-tions of the same magnitude, 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 /s2 a B A PB mA _{92 kg}36 N 0.0033 m /s2 a B S P B mS _{11 000 kg}36 N P B F B P B FB P B P B P B P B P B

4.5

Even though the magnitudes of the action and reaction forces are always equal, these forces do not necessarily produce accelerations that have equal magnitudes, since each force acts on a different object that may have a different mass.

(9)

4 . 6 | T Y P E S O F F O RC E S : A N OV E RV I E W | 9

There is a clever application of Newton’s third law in some rental trailers. As Figure

4.8 illustrates, the tow bar connecting the trailer to the rear bumper of a car contains a

mechanism that can automatically actuate brakes on the trailer wheels. This mechanism

works without the need for electrical connections between the car and the trailer. When

the driver applies the car brakes, the car slows down. Because of inertia, however, the

trailer 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 actuating trailer brakes

Figure 4.8 Some rental trailers include an automatic brake-actuating

mechanism.

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 A GLANCE When any of the four external forces listed here act on an object, they are included as part of the net force in any application of Newton's second law. Each of the four external forces acts on this (very) reluctant bull as the farmers join efforts to pull it aboard the boat. (Diether Endlicher/©AP/Wide World Photos)

F B

TYPES OF FORCES: AN OVERVIEW

CONCEPTS AT A GLANCE

Newton’s three laws of motion make it clear that forces

play a central role in determining the motion of an object. In the next four sections some

common 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). In

later chapters, we will encounter still others, such as electric and magnetic forces. It is

important to realize that Newton’s second law is always valid, regardless of which of

these forces may act on an object. One does not have a different law for every type of

common force. Thus, we need only to determine what forces are acting on an object, add

them together to form the net force, and then use Newton’s second law to determine the

object’s acceleration. The Concepts-at-a-Glance chart in Figure 4.9 illustrates this

im-portant idea.

The physics of

automatic 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

(10)

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 forces

can 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 a

primary 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). One

manifestation is the electromagnetic force that electrically charged particles exert on one

another (see Sections 18.5, 21.2, and 21.8). The other manifestation is the so-called weak

nuclear 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 the

interactions between the electrically charged particles that comprise atoms and molecules.

Our understanding of which forces are fundamental, however, is continually evolving. For

instance, in the 1860s and 1870s James Clerk Maxwell showed that the electric force and

the magnetic force could be explained as manifestations of a single electromagnetic force.

Then, in the 1970s, Sheldon Glashow (1932 – ), Abdus Salam (1926 – 1996), and Steven

Weinberg (1933 – ) presented the theory that explains how the electromagnetic force and

the weak nuclear force are related to the electroweak force. They received a Nobel prize

in 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 to

describe the effects of gravity by using a value of

g

9.80 m /s

2

_{for the downward}

accel-eration it causes. However, nothing has been said about why

g

is 9.80 m /s

2

_{. The reason is}

fascinating, as we will now see.

The acceleration due to gravity is like any other acceleration, and Newton’s second

law indicates that it must be caused by a net force. In addition to his famous three laws of

motion, 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 gravitational

constant,

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

m

1

and

m

2

and are separated by a distance

r

, the

force that each exerts on the other is directed along the line joining the particles (see

Figure 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

_m

2

_/kg

2

F

G

m

1

m

2

r

2 r m₁ m₂ –F +FB B

Figure 4.10 The two particles, whose masses are m1and m2, are attracted by gravitational forces FBand FB.

4.

7

(11)

4 .7 | T H E G R AV I TAT I O N A L F O RC E | 1 1

the English scientist Henry Cavendish (1731 – 1810), more than a century after Newton

proposed his law of universal gravitation.

To see the main features of Newton’s law of universal gravitation, look at the two

particles in Figure 4.10. They have masses

m

1

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 gravitational

forces 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 different

bodies, causing them to be mutually attracted. In fact, these forces are an action – reaction

pair, 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 m112 kg (approximately the mass of a bicycle),m225 kg, and r1.2 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 the fact that Gitself is very small. However, if one of the bodies has a large mass, like that of the earth (5.981024_{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, the

law 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 for

purposes of using the gravitation law, provided the mass of the object is distributed with

spherical symmetry about its center. Thus, Equation 4.3 can be applied when each object

is a sphere whose mass is spread uniformly over its entire volume. Figure 4.11 shows this

kind of application, assuming that the earth and the moon are such uniform spheres of

matter. 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 each

other are the same as if the entire mass of each were concentrated at its center. Even if the

objects 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 separaapproxima-tion

r

.

WEIGHT

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

1.4108_N FG m1m2 r2 (6.6710 11_N_m2_{/ kg}2₎(12 kg)(25 kg) (1.2 m)2

F

B

F

B Moon r Earth M_E M_E M_M M_M +FB –FB +FB –BF

Figure 4.11 The gravitational force that each uniform sphere of matter exerts on the other is the same as if each sphere were a particle with its mass

concentrated 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 earth

exerts 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 exerted

on the object by that body.

SI Unit of Weight:

newton (N)

45807_04_p1-48 6/17/05 3:31 PM Page 11

(12)

Using

W

for the magnitude of the weight,*

m

for the mass of the object, and

M

E

for

the 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 it

is resting on the earth’s surface, because the gravitational force is acting even when the

distance

r

is not equal to the radius

R

E

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 as

the distance

r

from the center of the earth increases. In Example 6 the telescope’s weight

is 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 it by the earth. According to Equation 4.4, the weight varies inversely as the square of the radial distance

r

. Thus, we expect the telescope’s weight on the earth’s surface (rsmaller) to be greater than its weight in orbit (rlarger).

Solution

(a) On the earth’s surface, the weight is given by Equation 4.4 with r6.38 106_{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 rmust 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 his

weight on the moon from Equation 4.4, it is only necessary to replace

M

E

by

M

(the

mass of the moon) and let

r

R

M

(the radius of the moon).

RELATION BETWEEN MASS AND WEIGHT

Although massive objects weigh a lot on the earth, mass and weight are not the

same quantity. As Section 4.2 discusses, mass is a quantitative measure of inertia. As

such, mass is an intrinsic property of matter and does not change as an object is moved

from one location to another. Weight, in contrast, is the gravitational force acting on the

object and can vary, depending on how far the object is above the earth’s surface or

whether 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)

g

W

m

G

M

E

r

2

W

W0.950105_N r6.38106_m₅₉₈₁₀3_m_6.98₁₀6_m W1.14105_N WG MEm r2 (6.671011_N_m2_{/ kg}2_)(5.98₁₀24_{kg)(11 600 kg)} (6.38106_m)2

W

G

M

E

m

r

2 r W Object of mass m R_E Mass of earth = M_E 5 0.2 0 0.4 0.6 0.8 1.0 10 15 20 r (× 106 m) R_E = 6.38 x 106 m W ( × 10 5 N)

Figure 4.12 On or above the earth, the weight of an object is the

gravitational force exerted on the object by the earth.

W oB

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

* Often, the word “weight” and the phrase “magnitude of the weight” are used interchangeably, even though weight 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 law to uniform spheres of matter, remember that the distance ris between the centers of the spheres, not between the surfaces. 45807_04_p1-48 6/20/05 7:39 AM Page 12

(13)

4 .7 | T H E G R AV I TAT I O N A L F O RC E | 1 3

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 to

gravity. These expressions make the distinction between mass and weight stand out. The

weight of an object whose mass is

m

depends on the values for the universal gravitational

constant

G

, the mass

M

E

of the earth, and the distance

r

. These three parameters together

determine the acceleration

g

due to gravity. The specific value of

g

9.80 m /s

2

_applies

only when

r

equals the radius

R

E

of the earth. For larger values of

r

, as would be the case

on top of a mountain, the effective value of

g

is less than 9.80 m /s

2

_{. The fact that}

_g

de-creases as the distance

r

increases means that the weight likewise decreases. The mass of

the object, however, does not depend on these effects and does not change. Conceptual

Example 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 on earth, 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 accelaccel-eration on the moon, 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 by Newton’s second law as m , where mis the vehicle’s mass and is the acceleration along the ground. For a given acceleration, the net force depends only on the mass. But the mass 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 on the earth. Do not be misled by the fact that the vehicle weighs more on earth. The greater weight occurs only because the earth’s mass and radius are different than the moon’s. In any event,in Newton’s second law, the net force is proportional to the vehicle’s mass, not its weight.

Related Homework:

Problems 22, 87

▲

a B a B F B F B

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

✔

C H E C K Y O U R U N D E R S T A N D I N G 2

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 are

located on the same straight line. The net gravitational force acting on the third object is zero. Which of the drawings correctly represents the locations of the objects? The answer is given at the end of the book.)

m₂ m₁ _m 3 d d m₂ m₁ _m 3 d d m₂ m₁ _m 3 d d (a) (b) m₂ m₁ _m 3 d d (d) (c)

Background: The gravitational force and Newton’s law of universal gravitation are the focus

of this problem.

For similar questions (including calculational counterparts), consult Self-Assessment Test 4.2. This test is described at the end of Section 4.10.

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

(14)

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 discusses

only 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 forces

that 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 when

you 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 similar

manner, the weight of the block causes invisible “atomic springs” in the surface of the

table to compress, thus producing a normal force on the block.

Newton’s third law plays an important role in connection with the normal force. In

Figure 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 magnimagni-tude of the normal

force 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 forces

except the object’s weight and the normal force, the magnitudes of these two forces are

equal; that is,

F

N

W

. This is the situation in Figure 4.14. The weight must be balanced

by the normal force for the object to remain at rest on the table. If the magnitudes of these

forces were not equal, there would be a net force acting on the block, and the block would

accelerate 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 4.15

a

, for

instance, a box whose weight is 15 N is being pushed downward against a table. The

pushing force has a magnitude of 11 N. Thus, the total downward force exerted on the

box is 26 N, and this must be balanced by the upward-acting normal force if the box is to

remain at rest. In this situation, then, the normal force is 26 N, which is considerably

larger than the weight of the box.

Figure 4.15

b

illustrates a different situation. Here, the box is being pulled upward by

a rope that applies a force of 11 N. The net force acting on the box due to its weight and

the rope is only 4 N, downward. To balance this force, the normal force needs to be only

4 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 normal

force would become zero. In fact, the table could be removed, since the block would be

supported entirely by the rope. The situations in Figure 4.15 are consistent with the idea

that the magnitude of the normal force indicates how hard two objects press against each

other. Clearly, the box and the table press against each other harder in part

a

of the picture

than in part

b

.

Like the box and the table in Figure 4.15, various parts of the human body press

against one another and exert normal forces. Example 8 illustrates the remarkable ability

of the human skeleton to withstand a wide range of normal forces.

FB

W

oB F B

W

oB

DEFINITION OF THE NORMAL FORCE

The normal force

N

is one component of the force that a surface exerts on an object

with which it is in contact — namely, the component that is perpendicular to the

surface.

FB F_N W B 11 N F_N = 26 N W = 15 N (a) 11 N W = 15 N (b) F_N = 4 N B B

Figure 4.14 Two forces act on the block, its weight and the normal force Nexerted by the surface of the

table. FB

W oB

Figure 4.15 (a) The normal force Nis

greater than the weight of the box, because the box is being pressed downward with an 11-N force. (b) The normal force is smaller than the weight, because the rope supplies an upward force of 11 N that partially supports the box.

F B

4.8

(15)

4 . 8 | T H E N O R M A L F O RC E | 1 5

▼

Example 8

|

A Balancing Act

In a circus balancing act, a woman performs a headstand on top of a standing performer’s head, as Figure 4.16aillustrates. The woman weighs 490 N, and the standing performer’s head and neck weigh 50 N. It is primarily the seventh cervical vertebra in the spine that supports all the weight above the shoulders. What is the normal force that this vertebra exerts on the neck and 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 head and neck, and the normal force. During the act, an additional force is present due to the woman’s weight. In both cases, the upward and downward forces must balance for the head and neck to remain at rest. This condition of balance will lead us to values for the normal force.

Solution

(a) Figure 4.16bshows the free-body diagram for the standing performer’s head and neck before the act. The only forces acting are the normal force Nand 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 total downward force exerted on the standing performer’s head and neck is 50 N490 N540 N, which must be balanced by the upward normal force, so that .

▲

In summary, the normal force does not necessarily have the same magnitude as the

weight 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 regarded

as 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 not

give 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 is

the force that the object exerts on the scale with which it is in contact.

FN540 N FN50 N FB +x +y +x +y F_N F_N F_N F_N 50 N 50 N 50 N 490 N 490 N 50 N (b) (a) (c) Seventh cervical vertebra

Free-body diagram Free-body diagram

Figure 4.16 (a) A young woman keeps her balance during a performance by China’s Sincuan Acrobatic group. A free-body diagram is shown (above the shoulders) for the standing performer (b) before the act and (c) during the act. For convenience, the scales used for the vectors in parts band care different. (© SUPRI/Reuters/Landov LLC)

The physics of

the human skeleton.

(16)

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 is

at rest or moving with a constant velocity (either upward or downward), the scale

regis-ters the true weight, as Figure 4.17

a

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 4.17

b

shows. Conversely, if the elevator accelerates downward, as in part

c

, the

apparent weight is less than the true weight. In fact, if the elevator falls freely, so its

ac-celeration is equal to the acac-celeration 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 if

the scale and the person on it are accelerating.

The discrepancies between true weight and apparent weight can be understood with

the aid of Newton’s second law. Figure 4.18 shows a free-body diagram of the person in

the elevator. The two forces that act on him are the true weight

m

and the normal

force

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

stands

for 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 the

elevator is accelerating upward (

) or downward (

). Solving for the normal force

F

N

shows that

(4.6)

In Equation 4.6,

F

N

is the magnitude of the normal force exerted on the person by the

scale. But in accord with Newton’s third law,

F

N

is also the magnitude of the downward

force 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 /s

2

_{, 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 greater

than the true weight. If the elevator accelerates downward,

a

is negative, and the apparent

weight is less than the true weight. If the elevator falls freely,

a

g

, and the apparent

weight is zero. The apparent weight is zero because when both the person and the scale

fall freely, they cannot push against one another. In this text, when the weight is given, it

is assumed to be the true weight, unless stated otherwise.

True 123 weight Apparent 123 weight

F

N

mg

ma

F

y

F

N

mg

ma

F

B

g

B

W

oB

(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 is not accelerating, the scale registers the true weight (W700 N) of the person. (b) When the elevator accelerates upward, the apparent weight (1000 N) exceeds the true weight. (c) When the elevator accelerates downward, the apparent weight (400 N) is less than the true weight. (d) The apparent weight is zero if the elevator falls freely — that is, if it falls with the acceleration due to gravity. +x W = mg F_N +y B

Figure 4.18 A free-body diagram showing the forces acting on the person riding in the elevator of Figure 4.17. is the true weight, and Nis the normal

force exerted on the person by the platform of the scale.

F

B W

oB 45807_04_p1-48 6/17/05 3:31 PM Page 16

(17)

4 .9 | STAT I C A N D K I N E T I C F R I C T I O N A L F O RC E S | 1 7

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 the

surface, which is called the normal force. When the object moves or attempts to move

along 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 to

move the car. In fact, the raised tread on a tire is designed to maintain friction. On a wet

road, the spaces in the tread pattern (see Figure 4.19) provide channels for the water to

collect and be diverted away. Thus, these channels largely prevent the water from coming

between the tire surface and the road surface, where it would reduce friction and allow

the 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 touch

only at relatively few spots, as Figure 4.20 illustrates. The microscopic area of contact for

these spots is substantially less than the apparent macroscopic area of contact between the

surfaces — perhaps thousands of times less. At these contact points the molecules of the

different bodies are close enough together to exert strong attractive intermolecular forces

on 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 not

well understood. However, some empirical relations have been developed that make it

possible to account for the effects of friction.

Figure 4.21 helps to explain the main features of the type of friction known as

static

friction.

The block in this drawing is initially at rest on a table, and as long as there is no

attempt to move the block, there is no static frictional force. Then, a horizontal force

is

applied to the block by means of a rope. If

is small, as in part

a

, experience tells us that

the 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,

f

s

F

. Increasing the

applied 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 that

cancels out the increase in the applied force (see part

b

of the drawing). If the applied

f

B

F

B

f

B

f

B

F

B

F

B Microscopic contact points

Figure 4.19 This photo, shot from underneath a transparent surface, shows a tire rolling under wet conditions. The channels in the tire collect and divert water away from the regions where the tire contacts the surface, thus providing better traction. (Courtesy Goodyear Tire & Rubber Co.)

Figure 4.20 Even when two highly polished surfaces are in contact, they touch only at relatively few points.

F

No movement (a)

No movement (b)

When movement just begins (c) F f_sMAX F f_s f_s B B B