Force & Motion - Chpt4,5 PDF

4 Force and Motion
A drag race is a
memorable
example of the
connection
between force
and motion.
Looking Ahead This drag racer can cover a quarter mile from a standing start in less than 5 sec-
The goal of Chapter 4 is to onds. Not bad! We could use kinematics to describe the car’s motion with pictures,
establish a connection between graphs, and equations. By defining position, velocity, and acceleration and dress-
force and motion. In this chapter ing them in mathematical clothing, kinematics provides a language to describe
you will learn to: how something moves. But kinematics would tell us nothing about why the car
accelerates so quickly. For the more fundamental task of understanding the cause
■ Recognize what a force is and of motion, we turn our attention to dynamics. Dynamics joins with kinematics to
is not. form mechanics, the general science of motion. We study dynamics qualitatively
■ Identify the specific forces in this chapter, then develop it quantitatively in the next four chapters.
acting on an object. The theory of mechanics originated in the mid-1600s when Sir Isaac Newton
■ Draw free-body diagrams. formulated his laws of motion. These fundamental principles of mechanics
■ Understand the connection explain how motion occurs as a consequence of forces. Newton’s laws are more
between force and motion. than 300 years old, but they still form the basis for our contemporary understand-
ing of motion.
Looking Back A challenge in learning physics is that a textbook is not an experiment. The
To master the material introduced book can assert that an experiment will have a certain outcome, but you may not
in this chapter, you must be convinced unless you see or do the experiment yourself. Newton’s laws are
understand how acceleration is frequently contrary to our intuition, and a lack of familiarity with the evidence for
determined and how vectors are Newton’s laws is a source of difficulty for many people. You will have an oppor-
used. Please review: tunity through lecture demonstrations and in the laboratory to see for yourself the
■ Section 1.5 Acceleration. evidence supporting Newton’s laws. Physics is not an arbitrary collection of defi-
■ Section 3.2 Properties of nitions and formulas, but a consistent theory as to how the universe really works.
vectors. It is only with experience and evidence that we learn to separate physical fact
from fantasy.
97
98 CHAPTER 4 . Force and Motion
4.1 Force
If you kick a ball, it rolls across the floor. If you pull on a door handle, the door
opens. You know, from many years of experience, that some sort of force is
required to move these objects. Our goal is to understand why motion occurs, and
the observation that force and motion are related is a good place to start.
The two major issues that this chapter will examine are:
■ What is a force?
■ What is the connection between force and motion?
We begin with the first of these questions in the table below.
What is a force?
A force is a push or a pull.

Our commonsense idea of a force is that it is a push or a pull. We will refine this idea as we go
along, but it is an adequate starting point. Notice our careful choice of words: We refer to “a force,”
rather than simply “force.” We want to think of a force as a very specific action, so that we can talk
about a single force or perhaps about two or three individual forces that we can clearly distinguish.
Hence the concrete idea of “a force” acting on an object.
A force acts on an object.

Object
Implicit in our concept of force is that a force acts on an object. In other words, pushes and pulls are
applied to something—an object. From the object’s perspective, it has a force exerted on it. Forces
do not exist in isolation from the object that experiences them.
A force requires an agent.

Every force has an agent, something that acts or exerts power. That is, a force has a specific,
identifiable cause. As you throw a ball, it is your hand, while in contact with the ball, that is the
agent or the cause of the force exerted on the ball. If a force is being exerted on an object, you must
Agent be able to identify a specific cause (i.e., the agent) of that force. Conversely, a force is not exerted
on an object unless you can identify a specific cause or agent. Although this idea may seem to be
stating the obvious, you will find it to be a powerful tool for avoiding some common misconceptions
about what is and is not a force.
A force is a vector.
If you push an object, you can push either gently or very hard. Similarly, you can push either left or
right, up or down. To quantify a push, we need to specify both a magnitude and a direction. It should
thus
r
come as no surprise that a force is a vector quantity. The symbol for a force is the vector symbol
F . The size or strength of a force is its magnitude F.
A force can be either a contact force . . .

There are two basic classes of forces, depending on whether the agent touches the object or not.
Contact forces are forces that act on an object by touching it at a point of contact. The bat must
touch the ball to hit it. A string must be tied to an object to pull it. The majority of forces that we
will examine are contact forces.
. . . or a long-range force.
Long-range forces are forces that act on an object without physical contact. Magnetism is an
example of a long-range force. You have undoubtedly held a magnet over a paper clip and seen
the paper clip leap up to the magnet. A coffee cup released from your hand is pulled to the earth
by the long-range force of gravity.
4.1 . Force 99
Let’s summarize these ideas as our definition of force:

■ A force is a push or a pull on an object.
■ A force is a vector. It has both a magnitude and a direction.
■ A force requires an agent. Something does the pushing or pulling.
■ A force is either a contact force or a long-range force. Gravity is the only long-
range force we will deal with until much later in the book.
There’s one more important aspect of forces. If you push against a door (the
object) to close it, the door pushes back against your hand (the agent). If a tow
rope pulls on a car (the object), the car pulls back on the rope (the agent). Thus, in
general, if an agent exerts a force on an object, the object exerts a force on the
agent. We really need to think of a force as an interaction between two objects.
Although the interaction perspective is a more exact way to view forces, it adds
complications that we would like to avoid for now. Our approach will be to start
by focusing on how a single object responds to forces exerted on it. Then, in
Chapter 8, we’ll return to the larger issue of how two or more objects interact
with each other.
NOTE In the particle model, objects cannot exert forces on themselves. A
force on an object will always have an agent or cause external to the object.
Now, there are certainly objects that have internal forces (think of all the
forces inside the engine of your car!), but the particle model is not valid if you
need to consider those internal forces. If you are going to treat your car as a
particle and look only at the overall motion of the car as a whole, that motion
will be a consequence of external forces acting on the car.
Force Vectors
We can use a simple diagram to visualize how forces are exerted on objects.
Because we are using the particle model, in which objects are treated as points,
the process of drawing a force vector is straightforward. Here is how it goes:
TACTICS BOX 4.1 Drawing force vectors
1 Represent the object as a particle.
2 Place the tail of the force vector

on the particle.
3 Draw the force vector as an arrow pointing

in the proper direction and with a length
proportional to the size of the force.
r
F
4 Give the vector an appropriate label.
Step 2 may seem contrary to what a “push” should do, but recall that moving a
vector does not change it as long as the length and angle do not change. The vec-
r
tor F is the same regardless of whether the tail or the tip is placed on the particle.
Our reason for using the tail will become clear when we consider how to combine
several forces.
Figure 4.1 on the next page shows three examples of force vectors. One is a
push, one a pull, and one a long-range force, but in all three the tail of the force
vector is placed on the particle representing the object.
The rope is the agent. The spring is the agent.
Box
Long-range
force of
gravity
Box Box
Pulling force of rope Pushing force of spring

Earth is the agent.
FIGURE 4.1 Three force vectors.
(a) Combining Forces

Figure 4.2a shows a box being pulled by two ropes, each exerting a force on the
box. How will the box respond? Experimentally, we find that when several indi-
r r r
vidual forces F1, F2, F3, . . . are exerted on an object, they combine to form a net
force given by the vector sum of the individual forces:
Top view
N
of box r r r r r
Fnet ; a Fi 5 F1 1 F2 1 c1 FN (4.1)
i51
Recall that ; is the symbol meaning “is defined as.” Mathematically, this sum-
mation is called a superposition of forces. The net force is sometimes called the
resultant force. Figure 4.2b shows the net force on the box.
STOP TO THINK 4.1 Two forces are exerted on an object. What third force
(b) r would make the net force point to the left?
F1
Pulling force
r r
of the ropes F1 F3
r r
Find the net F2 F3
force on the box. r
F3
Two forces r
F3
r r r exerted on
Fnet 5 F1 1 F2
an object
(a) (b) (c) (d)
r
F2
FIGURE 4.2 Two forces applied to a box.
4.2 A Short Catalog of Forces

There are many forces we will deal with over and over. This section will intro-
duce you to some of them. Many of these forces have special symbols. As you
The weight force
learn the major forces, be sure to learn the symbol for each.
pulls the box down.
Weight
A falling rock is pulled toward the earth by the long-range force of gravity. Grav-
ity is what keeps you in your chair, keeps the planets in their orbits around the
sun, and shapes the large-scale structure of the universe. We’ll have a thorough
r
w look at gravity in Chapter 12. For now we’ll concentrate on objects on or near the
surface of the earth (or other planet).
The gravitational pull of the earth on an object on or near the surface of the
Ground r
earth is called weight. The symbol for weight is w . Weight is the only long-range
FIGURE 4.3 Weight. force we will encounter in the next few chapters. The agent for the weight force is
4.2 . A Short Catalog of Forces 101
the entire earth pulling on an object. Weight acts on an object whether the object
is moving or at rest. The weight vector always points vertically downward, as
shown in Figure 4.3 on the previous page.
NOTE We often refer to “the weight” of an object. This is an informal
expression for w, the magnitude of the weight force exerted on the object. Note
that weight is not the same thing as mass. We will briefly examine mass
later in the chapter and explore the connection between weight and mass in
Chapter 5.
Spring Force
Springs exert one of the most common contact forces. A spring can either push
(when compressed) or pull (when stretched). Figure 4.4 shows the spring force.
In both cases, pushing and pulling, the tail of the force vector is placed on the
particle in the force diagram. There is no special symbol for a spring force, so we A stretched spring exerts a force on
r
simply use a subscript label: Fsp. an object.
A compressed spring exerts A stretched spring exerts

a pushing force on an object. a pulling force on an object.
r r
Fsp Fsp
(a) (b)
FIGURE 4.4 The spring force.
Although you may think of a spring as a metal coil that can be stretched or The rope exerts a tension
compressed, this is only one type of spring. Hold a ruler, or any other thin piece force on the sled.
of wood or metal, by the ends and bend it slightly. It flexes. When you let
r
go, it “springs” back to its original shape. This is just as much a spring as is a T
metal coil.
Tension Force FIGURE 4.5 Tension.

When a string or rope or wire pulls on an object, it exerts a contact force that we
r
call the tension force, represented by a capital T . The direction of the tension
force is always in the direction of the string or rope, as you can see in Figure 4.5.
The commonplace reference to “the tension” in a string is an informal expression
for T , the size or magnitude of the tension force.
r Rope r
If you were to use a very powerful microscope to look inside a rope, you T T
would “see” that it is made of atoms joined together by molecular bonds. Molecu-
lar bonds are not rigid connections between the atoms. They are more accurately
Atomic-level
thought of as tiny springs holding the atoms together, as in Figure 4.6. These are view of a rope,
very stiff springs, to be sure, but pulling on the ends of a string or rope stretches with molecular
the molecular springs ever so slightly. The tension within a rope and the tension Molecular bonds as springs
experienced by an object at the end of the rope are really the net spring force bonds
being exerted by billions and billions of microscopic springs. Atoms
This atomic-level view of tension introduces a new idea: a microscopic
atomic model for understanding the behavior and properties of macroscopic
objects. We will frequently use an atomic model to obtain a deeper understanding FIGURE 4.6 An atomic-level view
of our observations. of tension.
Normal Force
If you sit on a bed, the springs in the mattress compress and, as a consequence
of the compression, exert an upward force on you. Stiffer springs would show
less compression but still exert an upward force. The compression of extremely
stiff springs might be measurable only by sensitive instruments. Nonetheless,
the springs would compress ever so slightly and exert an upward spring force
The compressed
molecular springs push on you.
upward on the object. Figure 4.7 shows an object resting on top of a sturdy table. The table may not
visibly flex or sag, but—just as you do to the bed—the object compresses the
Atoms
molecular springs in the table. The size of the compression is very small because
Molecular bonds molecular springs are so stiff, but it is not zero. As a consequence, the com-
pressed molecular springs push upward on the object. We say that “the table”
exerts the upward force, but it is important to understand that the pushing is really
done by molecular springs. Similarly, an object resting on the ground compresses
FIGURE 4.7 Atomic-level view of the force the molecular springs holding the ground together and, as a consequence, the
exerted by a table. ground pushes up on the object.
We can extend this idea. Suppose you place your hand on a wall and lean
The compressed against it, as shown in Figure 4.8. Does the wall exert a force on your hand? As
molecular springs you lean, you compress the molecular springs in the wall and, as a consequence,
in the wall press
outward against
they push outward against your hand. So the answer is “yes,” the wall does exert a
her hand. force on you.
The force the table surface exerts is vertical, the force the wall exerts is hori-
zontal. But in all cases, the force exerted on an object that is pressing against a
surface is in a direction perpendicular to the surface. Mathematicians refer to a
line that is perpendicular to a surface as being normal to the surface. In keeping
with this terminology, we define the normal force as the force exerted by a sur-
face (the agent) against an object that is pressing against the surface. The symbol
for the normal force is nr.
FIGURE 4.8 The wall pushes outward We’re not using the word normal to imply that the force is an “ordinary” force
against your hand. or to distinguish it from an “abnormal force.” A surface exerts a force perpen-
dicular (i.e., normal) to itself as the molecular springs press outward. Figure 4.9
shows an object on an inclined surface, a common situation. Notice how the nor-
nr
mal force nr is perpendicular to the surface.
We have spent a lot of time describing the normal force because many people
have a difficult time understanding it. The normal force is a very real force arising
from the very real compression of molecular bonds. It is in essence just a spring
force, but one exerted by a vast number of microscopic springs acting at once.
The normal force is responsible for the “solidness” of solids. It is what prevents
you from passing right through the chair you are sitting in and what causes the
pain and the lump if you bang your head into a door. Your head can then tell you
The surface pushes outward that the force exerted on it by the door was very real!
against the bottom of the frog.
The push is perpendicular
to the surface. Friction
FIGURE 4.9 The normal force. You’ve certainly observed that a rolling or sliding object, if not pushed or pro-
pelled, slows down and eventually stops. You’ve probably discovered that you
can slide better across a sheet of ice than across asphalt. And you also know that
most objects stay in place on a table without sliding off even if the table isn’t
absolutely level. The force responsible for these sorts of behavior is friction. The
r
symbol for friction is a lower case f .
Friction, like the normal force, is exerted by a surface. On a microscopic level,
friction arises as atoms from the object and atoms on the surface run into each
other. The rougher the surface is, the more these atoms are forced into close prox-
imity and, as a result, the larger the friction force. We will develop a simple model
4.2 . A Short Catalog of Forces 103
of friction in the next chapter that will be sufficient for our needs. For now, it is
useful to distinguish between two kinds of friction:
r
■ Kinetic friction, denoted f k, appears as an object slides across a surface. This is
r
a force that “opposes the motion,” meaning that the friction force vector f k
r
points in a direction opposite the velocity vector v (i.e., “the motion”).
r
■ Static friction, denoted f s, is the force that keeps an object “stuck” on a surface
r
and prevents its motion. Finding the direction of f s is a little trickier than find-
r
ing it for f k. Static friction points opposite the direction in which the object
would move if there were no friction. That is, it points in the direction neces-
sary to prevent motion.
Figure 4.10 shows examples of kinetic and static friction.
vr
r
fs
Static friction acts

r Kinetic friction in the direction that
fk opposes the prevents slipping.
motion.
FIGURE 4.10 Kinetic and static friction.
NOTE A surface exerts a kinetic friction force when an object moves rela-
tive to the surface. A package on a conveyor belt is in motion, but it does not
experience a kinetic friction force because it is not moving relative to the belt.
So to be precise, we should say that the kinetic friction force points opposite to
an object’s motion relative to a surface.
Air resistance is a significant force
on falling leaves. It points opposite
Drag the direction of motion.
Friction at a surface is one example of a resistive force, a force that opposes or
resists motion. Resistive forces are also experienced by objects moving through
r
fluids—gases and liquids. The resistive force of a fluid is called drag and is sym- D
r r
bolized as D. Drag, like kinetic friction, points opposite the direction of motion. D
r
Figure 4.11 shows an example of drag. D
Drag can be a large force for objects moving at high speeds or in dense fluids.
Hold your arm out the window as you ride in a car and feel how the air resistance
against it increases rapidly as the car’s speed increases. Drop a lightweight object
into a beaker of water and watch how slowly it settles to the bottom. In both cases FIGURE 4.11 Air resistance is an example
the drag force is very significant. of drag.
For objects that are heavy and compact, that move in air, and whose speed is
not too great, the drag force of air resistance is fairly small. To keep things as sim-
ple as possible, you can neglect air resistance in all problems unless a problem
explicitly asks you to include it. The error introduced into calculations by this r
Fthrust
approximation is generally pretty small. This textbook will not consider objects Thrust force is exerted
on a rocket by exhaust
moving in liquids. gases.
Thrust
A jet airplane obviously has a force that propels it forward during takeoff. Like-
wise for the rocket being launched in Figure 4.12. This force, called thrust,
occurs when a jet or rocket engine expels gas molecules at high speed. Thrust is a FIGURE 4.12 Thrust force on a rocket.
contact force, with the exhaust gas being the agent that pushes on the engine. The
process by which thrust is generated is rather subtle, and we will postpone a full
discussion until we introduce Newton’s third law in Chapter 8. For now, we will
treat thrust as a force opposite the direction in which the exhaust gas is expelled.
r
There’s no special symbol for thrust, so we will call it Fthrust.
Force Notation Electric and Magnetic Forces

r
General force F Electricity and magnetism, like gravity, exert long-range forces. The forces of
r
Weight w electricity and magnetism act on charged particles. We will study electric and
r
Spring force Fsp magnetic forces in detail in Part VI of this textbook. For now, it is worth noting that
r
Tension T the forces holding molecules together—the molecular bonds—are not actually
Normal force nr tiny springs. Atoms and molecules are made of charged particles—electrons and
r
Static friction fs protons—and what we call a molecular bond is really an attractive electric force
r
Kinetic friction fk between these particles. So when we say that the normal force and the tension
r
Drag D force are due to “molecular springs,” or that friction is due to atoms running into
r each other, what we’re really saying is that these forces, at the most fundamental
Thrust Fthrust
level, are actually electric forces between the charged particles in the atoms.
4.3 Identifying Forces

Force and motion problems generally have two basic steps:
1. Identify all of the forces acting on an object.
2. Use Newton’s laws and kinematics to determine the motion.
Understanding the first step is the primary goal of this chapter. We’ll turn our
attention to step 2 in the next chapter.
A typical physics problem describes an object that is being pushed and pulled
in various directions. Some forces are given explicitly, others are only implied. In
order to proceed, it is necessary to determine all the forces that act on the object.
It is also necessary to avoid including forces that do not really exist. Now that you
have learned the properties of forces and seen a catalog of typical forces, we can
develop a step-by-step method for identifying each force in a problem. This
procedure for identifying forces is part of the physical representation of the
problem.
TACTICS BOX 4.2 Identifying forces

●
1 Identify “the system” and “the environment.” The system is the object
whose motion you wish to study; the environment is everything else.
●
2 Draw a picture of the situation. Show the object—the system—and
everything in the environment that touches the system. Ropes, springs,
and surfaces are all parts of the environment.
●
3 Draw a closed curve around the system. Only the object is inside the
curve; everything else is outside.
●
4 Locate every point on the boundary of this curve where the environ-
ment touches the system. These are the points where the environment
exerts contact forces on the object.
●
5 Name and label each contact force acting on the object. There is at
least one force at each point of contact; there may be more than one.
When necessary, use subscripts to distinguish forces of the same type.
●
6 Name and label each long-range force acting on the object. For now,
the only long-range force is weight.
4.3 . Identifying Forces 105
EXAMPLE 4.1 Forces on a bungee jumper

A bungee jumper has leapt off a bridge and is nearing the bottom of her fall. What forces are being exerted on the bungee
jumper?
r
Tension T
4 Locate the points where the environment touches

the system. Here the only point of contact is where
the cord attaches to her ankles.
1 Identify the system. Here the system is
the bungee jumper.
5 Name and label each contact force. The force
exerted by the cord is a tension force.
2 Draw a picture of the situation.
3 Draw a closed curve around the system.
6 Name and label long-range forces. Weight

r is the only one.
Weight w
FIGURE 4.13 Forces on a bungee jumper.
EXAMPLE 4.2 Forces on a skier

A skier is being towed up a snow-covered hill by a tow rope. What forces are being exerted on the skier?
r
Tension T
1 Identify the system. Here

the system is the skier.
4 Locate the points where the environment
2 Draw a picture of touches the system. Here the rope and the
the situation. ground touch the skier.
5 Name and label each contact force. The rope

3 Draw a closed curve
r exerts a tension force and the ground exerts
around the system. Normal force n r both a normal and a kinetic friction force.
r Kinetic friction fk
Weight w
6 Name and label long-range forces. Weight

is the only one.
FIGURE 4.14 Forces on a skier.
NOTE You might have expected two friction forces and two normal
forces in Example 4.2, one on each ski. Keep in mind, however, that we’re
working within the particle model, which represents the skier by a single
point. A particle has only one contact with the ground, so there is a single nor-
mal force and a single friction force. The particle model is valid if we want to
analyze the translational motion of the skier as a whole, but we would have to
go beyond the particle model to find out what happens to each ski.
Now that you’re getting the hang of this, the next example is meant to look
much more like a sketch you should make when asked to identify forces in a
homework problem.
EXAMPLE 4.3 Forces on a rocket Air r

Drag D
A rocket is being launched to place a new satellite
in orbit. Air resistance is not negligible. What forces
are being exerted on the rocket?
r
Weight w
r
Thrust Fthrust
FIGURE 4.15 Forces on a rocket. Exhaust
STOP TO THINK 4.2 You’ve just kicked a rock, and it is now sliding across the
ground about 2 meters in front of you. Which of these forces act on the rock? List
all that apply.
a. Gravity, acting downward.

b. The normal force, acting upward.
c. The force of the kick, acting in the direction of motion.
d. Friction, acting opposite the direction of motion.
e. Air resistance, acting opposite the direction of motion.
4.4 What Do Forces Do? A Virtual Experiment

Slack rubber band The fundamental question is: How does an object move when a force is exerted on
it? The only way to answer this question is to do experiments. To do experiments,
however, we need a way to reproduce the same amount of force again and again.
Let’s conduct a “virtual experiment,” one you can easily visualize. Imagine
using your fingers to stretch a rubber band to a certain length—say 10 centimeters—
that you can measure with a ruler. We’ll call this the standard length. Figure 4.16
shows the idea. You know that a stretched rubber band exerts a force because your
fingers feel the pull. Furthermore, this is a reproducible force. The rubber band
Standard length r
exerts the same force every time you stretch it to the standard length. We’ll call this
F the standard force F.
What happens to the force if you put two identical rubber bands around your
fingers and stretch both to the standard length? If you are not sure, find a few rub-
ber bands and try this. You will discover that two rubber bands exert a larger
One rubber band pulling force than one rubber band. This is not surprising. If two rubber bands are
stretched the standard
length exerts the each pulling equally hard, the net pull is twice that of one rubber band: Fnet 5 2F.
standard force F. N side-by-side rubber bands, each pulled to the standard length, will exert N times
the standard force: Fnet 5 NF.
Standard length
Now we’re ready to start the virtual experiment. Imagine an object to which
r
2F you can attach rubber bands, such as a block of wood with a hook. If you attach a
rubber band and stretch it to the standard length, the object experiences the same
force F as did your finger. N rubber bands attached to the object will exert N times
the force of one rubber band. The rubber bands give us a way of applying a
known and reproducible force to an object.
Two rubber bands
stretched the standard Our next task is to measure the object’s motion in response to these forces.
length exert twice Imagine using the rubber bands to pull the object across a horizontal table. Fric-
the standard force. tion between the object and the surface might affect our results, so let’s just elimi-
FIGURE 4.16 A reproducible force. nate friction. This is, after all, a virtual experiment! (In practice you could nearly
4.4 . What Do Forces Do? A Virtual Experiment 107
eliminate friction by pulling a smooth block over a smooth sheet of ice or by sup-
porting the object on a cushion of air.) To make measurements, lay a meter stick
along the edge of the table and hang a movie camera over the table to record
the motion.
Our experiment will be easiest to interpret if the force is constant throughout
the object’s motion. If you stretch the rubber band and then release the object, it
moves toward your hand. But as it does so, the rubber band gets shorter and the
pulling force decreases. To keep the pulling force constant, you must move your
hand at just the right speed to keep the length of the rubber band from changing!
Figure 4.17 shows the experiment being carried out. Once the motion is complete,
you can use motion diagrams (made from the movie frames) and kinematics to
analyze the object’s motion.
Camera
Maintain constant stretch.
Frictionless
surface Rubber band Pull
Motion diagram vr
ar
FIGURE 4.17 Measuring the motion of an object that is pulled with a constant force.
The first important finding of this experiment is that an object pulled with a TABLE 4.1 Acceleration due to an
constant force moves with a constant acceleration. This finding could not have increasing force
been anticipated in advance. It’s conceivable that the object would speed up for a Rubber bands Force Acceleration
while, then move with a steady speed. Or that it would continue to speed up,
1 F a1
but that the rate of increase, the acceleration, would steadily decline. These
2 2F a2 5 2a1
are conceivable motions, but they’re not what happens. Instead, the object con-
3 3F a3 5 3a1
tinues to accelerate with a constant acceleration for as long as you pull it with a
constant force. ( ( (
The next question is: What happens if you increase the force by using several N NF aN 5 Na1
rubber bands? To find out, use 2 rubber bands. Stretch both to the standard
length to double the force, then measure the acceleration. Then measure the
acceleration due to 3 rubber bands, then 4, and so on. Table 4.1 shows the results
of this experiment. You can see that doubling the force causes twice the accelera-
tion, tripling the force causes three times the acceleration, and so on. Acceleration is directly
Figure 4.18 is a graph of the data. Force is the independent variable, the one proportional to force.
The slope of the line is c.
you can control, so we’ve placed force on the horizontal axis to make an accelera-
Acceleration (in multiples of a1)
tion-versus-force graph. The graph shows that the acceleration is directly pro- 5a1
portional to the force. This is our second important finding. Recall that 4a1
proportionality indicates a linear relationship whose graph passes through the
3a1
origin (y-intercept of zero). This result can be written
2a1
a 5 cF (4.2)
where c is called the proportionality constant. The proportionality constant c is 1a1
the slope of the graph. 0
The final question for our virtual experiment is: How does the acceleration 0 1 2 3 4 5
depend on the size of the object? (The “size” of an object is somewhat ambigu- Force (number of rubber bands)
ous. We’ll be more precise below.) To find out, glue the original object and an FIGURE 4.18 Graph of acceleration versus
identical copy together, and then, applying the same force as you applied to the force.
TABLE 4.2Acceleration with different original, single object, measure the acceleration of this new object. Three objects
numbers of objects glued together make an object three times the size of the original. Doing several
Number of objects Acceleration such experiments, applying the same force to each object, would give you the
results shown in Table 4.2. An object twice the size of the original has only half
1 a1
the acceleration of the original object when both are subjected to the same force.
2 a2 5 12 a1
An object three times the size of the original has one-third the acceleration.
3 a3 5 13 a1 Figure 4.19 shows these results added to the graph of Figure 4.18. You can see
( ( that the proportionality constant c between acceleration and force —the slope of
N aN 5 N1 a1 the line— changes with the size of the object. The graph for an object twice the
size of the original is a line with half the slope. It may seem surprising that larger
objects have smaller slopes, so you’ll want to think about this carefully.
Mass
The slope is inversely 1 object Now, “twice the size” is a little vague; we could mean the object’s external
proportional to the Slope 5 1
number of objects dimensions or some other measure. Although mass is a common word, we’ve
being accelerated. avoided the term so far because we first need to define what mass is. Because we
Acceleration
made the larger objects in our experiment from the same material as the original
2 objects
Slope 5 12 object, an object twice the size has twice as many atoms—twice the amount of
3 objects matter—as the original. Thus it should come as no surprise that it has twice the
Slope 5 13 mass as the original. Loosely speaking, an object’s mass is a measure of the
4 objects amount of matter it contains. This is certainly our everyday meaning of mass,
Slope 5 14 but it is not yet a precise definition.
Force Figure 4.19 showed that an object with twice the amount of matter as the origi-
FIGURE 4.19 Acceleration-versus-force nal accelerates only half as quickly if both experience the same force. An object
graphs for objects of different size. with N times as much matter has only N1 of the original acceleration. The more
matter an object has, the more it resists accelerating in response to a force. You’re
familiar with this idea: Your car is much harder to push than your bicycle. The
tendency of an object to resist a change in its velocity (i.e., to resist acceleration)
is called inertia. Figure 4.19 tells us that larger objects have more inertia than
smaller objects of the same material.
We can make this idea precise by defining the inertial mass m of an object to be
1 F
m; 5
slope of the acceleration-versus-force graph a
We usually refer to the inertial mass as simply “the mass.” Mass is an intrinsic
property of an object. It is the property that determines how an object accelerates
in response to an applied force.
STOP TO THINK 4.3 Two rubber bands stretched to the standard length cause an
object to accelerate at 2 m/s2. Suppose another object with twice the mass is
pulled by four rubber bands stretched to the standard length. The acceleration of
this second object is
a. 1 m/s2. b. 2 m/s2. c. 4 m/s2. d. 8 m/s2. e. 16 m/s2.
4.5 Newton’s Second Law

We can now summarize the results of our experiment. Figure 4.18 showed that the
acceleration is directly proportional to the force, a conclusion that we wrote in
Equation 4.2 with the unspecified proportionality constant c. Now we see that c, the
slope of the acceleration-versus-force graph, is the inverse of the inertial mass m.
4.5 . Newton’s Second Law 109
Thus we’ve found that a force of magnitude F causes an object of mass m to

accelerate with
F
a5
m
Equation 4.3 answers the question with which we started: How does an object
move when a force is exerted on it? A force causes an object to accelerate! Fur-
thermore, the size of the acceleration is directly proportional to the size of the
force and inversely proportional to the object’s mass.
This is an important finding, but our experiment was limited to looking at an
object’s response to a single applied force. Realistically, an object is likely to be
r r r
subjected to several distinct forces F1, F2, F3, . . . that may point in different
directions. What happens then? In that case, it is found experimentally that the
acceleration is determined by the net force.
Newton was the first to recognize the connection between force and motion.
This relationship is known today as Newton’s second law.
r r
Newton’s second law An object of mass m subjected to forces F1, F2,
r
F3, . . . will undergo an acceleration ar given by
r
Fnet
ar 5 (4.3)
m
r r r r
where the net force Fnet 5 F1 1 F2 1 F3 1 cis the vector sum of the indi-
vidual forces. The acceleration vector ar points in the same direction as the
r
net force vector Fnet.
It may seem puzzling that we’ve skipped over Newton’s first law. The reasons for
this will become clear as we continue our discussion of dynamics. For now, the
critical idea is that an object’s acceleration vector ar points in the same direc-
r
tion as the net force vector Fnet.
The significance of Newton’s second law cannot be overstated. There was no
reason to suspect that there should be any simple relationship between force and
acceleration. Yet there it is, a simple but exceedingly powerful equation relating
r
the two. Newton’s work, preceded to some extent by Galileo’s, marks the begin- T2
ning of a highly successful period in the history of science during which it was
learned that the behavior of physical objects can often be described and predicted
by mathematical relationships. While some relationships are found to apply only
in special circumstances, others seem to have universal applicability. Those equa- Top view
of box
tions that appear to apply at all times and under all conditions have come to be
called “laws of nature.” Newton’s second law is a law of nature; you will meet
others as we go through this book.
We can rewrite Newton’s second law in the form Two ropes exerting
tension forces on a box
r
Fnet 5 mar (4.4) r
T2
which is how you’ll see it presented in many textbooks. Equations 4.3 and 4.4 are
mathematically equivalent, but Equation 4.3 better describes the central idea of
r
Newtonian mechanics: A force applied to an object causes the object to accelerate. Fnet
The acceleration
Be careful not to think that one force “overcomes” the others to determine the is inr the direction ar
r r
motion. Forces are not in competition with each other! It is Fnet, the sum of all the T1 of Fnet.
forces, that determines the acceleration ar. r
T1
As an example, Figure 4.20a shows a box being pulled by two ropes. The
r r (a) (b)
ropes exert tension forces T 1 and T 2 on the box. Figure 4.20b represents the box
as a particle, shows the forces acting on the box, and adds them graphically to FIGURE 4.20 Acceleration of a pulled box.
r r
find the net force Fnet. The box will accelerate in the direction of Fnet with an accel-
eration of magnitude
r r r
r Fnet T1 1 T2
a5 5
m m
NOTE The acceleration is not ( T1 1 T2 ) /m. You must add the forces as vec-
tors, not merely add their magnitudes as scalars.
Units of Force
r
Because Fnet 5 mar, the units of force must be mass units multiplied by accelera-
tion units. We’ve previously specified the SI unit of mass as the kilogram. We can
now define the basic unit of force as “the force that causes a 1 kg mass to acceler-
ate at 1 m/s2.” From the second law, this force is
m kg m
1 basic unit of force ; 1 kg 3 1 2
51 2
s s
This basic unit of force is called a newton:
One newton is the force that causes a 1 kg mass to accelerate at 1 m/s2. The
abbreviation for newton is N. Mathematically, 1 N 5 1 kg m/s2.
The newton is a secondary unit, meaning that it is defined in terms of the pri-
mary units of kilograms, meters, and seconds. We will introduce other secondary
units as needed.
TABLE 4.3 Approximate magnitude It is important to develop a feeling for what the size of forces should be.
of some typical forces Table 4.3 shows some typical forces. As you can see, “typical” forces on “typical”
Approximate objects are likely to be in the range 0.01–10,000 N. Forces less than 0.01 N are
magnitude too small to consider unless you are dealing with very small objects. Forces
Force (newtons) greater than 10,000 N would make sense only if applied to very massive objects.
Weight of a U.S. quarter 0.05 The unit of force in the English system is the pound (abbreviated lb). Although
Weight of a 1-pound object 5 the definition of the pound has varied throughout history, it is now defined in
Weight of a 110-pound person 500 terms of the newton:
Propulsion force of a car 5,000 1 pound 5 1 lb ; 4.45 N
Thrust force of a
You very likely associate pounds with kilograms rather than with newtons. Every-
rocket motor 5,000,000
day language often confuses the ideas of mass and weight, but we’re going to
need to make a clear distinction between them. More on this in the next chapter.
STOP TO THINK 4.4 Three forces act on an object. In which direction does the
object accelerate?
r
F1
In which direction ar ar
r
F2 does the object ar ar
ar
accelerate?
r
F3 (a) (b) (c) (d) (e)
4.6 Newton’s First Law

As we remarked earlier, Aristotle and his contemporaries in the world of ancient
Greece were very interested in motion. One question they asked was: What is the
“natural state” of an object if left to itself? It does not take an expensive research
4.6 . Newton’s First Law 111
program to see that every moving object on earth, if left to itself, eventually
comes to rest. Aristotle concluded that the natural state of an earthly object is to
be at rest. An object at rest requires no explanation; it is doing precisely what
comes naturally to it. A moving object, though, is not in its natural state and thus
requires an explanation: Why is this object moving? What keeps it going and pre-
vents it from being in its natural state?
Galileo reopened the question of the “natural state” of objects. He suggested
focusing on the limiting case in which resistance to the motion (e.g., friction or air
resistance) is zero. This is an idealization that may not be realizable in practice,
but Galileo had asserted previously, with great success, that the idealized case can
establish a general principle. Many careful experiments in which he minimized
the influence of friction led Galileo to a conclusion that was in sharp contrast to
Aristotle’s belief that rest is an object’s natural state.
Galileo found that an external influence (i.e., a force) is needed to make an
object accelerate—to change its velocity. In particular, a force is needed to put an
object in motion. But, in the absence of friction or air resistance, a moving object
continues to move along a straight line forever with no loss of speed. In other
words, the natural state of an object—its behavior if free of external influences—
is uniform motion with constant velocity! This does not happen in practice
because friction or air resistance prevents the object from being left alone. “At
rest” has no special significance in Galileo’s view of motion; it is simply uniform
r
motion that happens to have rv 5 0.
Galileo’s experiments were limited to motion along horizontal surfaces. It was
left to Newton to generalize this result, and today we call it Newton’s first law
of motion.
Newton’s first law An object that is at rest will remain at rest, or an object
that is moving will continue to move in a straight line with constant velocity,
if and only if the net force acting on the object is zero.
Newton’s first law is also known as the law of inertia. If an object is at rest, it has
a tendency to stay at rest. If it is moving, it has a tendency to continue moving
An object at rest is
with the same velocity. in staticr equilibrium:
r
Fnet 5 0.
NOTE The first law refers to net force. An object can remain at rest, or can
move in a straight line with constant velocity, even though forces are exerted
on it as long as the net force is zero. r
vr 5 0 r
Notice the “if and only if” aspect of Newton’s first law. If an object is at rest or ar 5 0
moves with constant velocity, then we can conclude that there is no net force acting
on it. Conversely, if no net force is acting on it, we can conclude that the object will
have constant velocity, not just constant speed. The direction remains constant, too!
r r
An object on which the net force is zero, Fnet 5 0 is said to be in mechanical
equilibrium. According to Newton’s first law, there are two distinct forms of
mechanical equilibrium:
r
v
1. The object is at rest. This is static equilibrium. ar 5 0
r
2. The object is moving in a straight line with constant velocity. This is An object moving in a straight line at constant
r r
dynamic equilibrium. velocity is in dynamic equilibrium: Fnet 5 0.
Two examples of mechanical equilibrium are shown in Figure 4.21. Both share FIGURE 4.21 Two examples of mechanical
r
the common feature that the acceleration is zero: ar 5 0. equilibrium.
What Good Is Newton’s First Law?

The first law completes our definition of force. It answers the question: What is a
force? If an “influence” on an object causes the object’s velocity to change, the
influence is a force.
Newton’s first law changes the question the ancient Greeks were trying to
answer: What causes an object to move? Newton’s first law says no cause is
needed for an object to move! Uniform motion is the object’s natural state.
Nothing at all is required for it to remain in that state. The proper question,
according to Newton, is: What causes an object to change its velocity? Newton,
with Galileo’s help, also gave us the answer. A force is what causes an object to
change its velocity.
The preceding paragraph contains the essence of Newtonian mechanics. This
new perspective on motion, however, is often contrary to our common experi-
ence. We all know perfectly well that you must keep pushing an object— exerting
a force on it—to keep it moving. Newton is asking us to change our point of view
and to consider motion from the object’s perspective rather than from our personal
perspective. As far as the object is concerned, our push is just one of several
forces acting on it. Others might include friction, air resistance, or gravity. Only
by knowing the net force can we determine the object’s motion.
Newton’s first law may seem to be merely a special case of Newton’s sec-
r
ond law. After all, the equation Fnet 5 mar tells us that an object moving with
r r r
constant velocity (ar 5 0 ) has Fnet 5 0. The difficulty is that the second law
assumes that we already know what force is. The purpose of the first law is to
identify a force as something that disturbs a state of equilibrium. The second
law then describes how the object responds to this force. Thus from a logical
perspective, the first law really is a separate statement that must precede the
second law. But this is a rather formal distinction. From a pedagogical per-
spective it is better—as we have done—to use a commonsense understanding
of force and start with Newton’s second law.
Inertial Reference Frames

If a car stops suddenly, you may be “thrown” into the windshield if you’re not
wearing your seat belt. You have a very real forward acceleration relative to the
car, but is there a force pushing you forward? A force is a push or a pull caused by
This guy thinks there’s a force hurling him an identifiable agent in contact with the object. Although you seem to be pushed
into the windshield. What a dummy! forward, there’s no agent to do the pushing.
The difficulty—an acceleration without an apparent force—comes from using
an inappropriate coordinate system. Your acceleration measured in a coordinate
(a) system attached to the car is not the same as your acceleration measured in a coor-
r
dinate system attached to the ground. Newton’s second law says Fnet 5 mar. But
r r
ar 5 0 which a? Measured in which coordinate system?
We define an inertial reference frame as a coordinate system in which New-
ton’s laws are valid. The first law provides a convenient way to test whether a
r
The ball stays in place. coordinate system is an inertial reference frame. If ar 5 0 (an object is at rest or
r r
A ball with no horizontal forces stays at rest
moving with constant velocity) only when Fnet 5 0, then the coordinate system in
in an airplane cruising at constant velocity. which ar is measured is an inertial reference frame.
The airplane is an inertial reference frame. Not all coordinate systems are inertial reference frames. Figure 4.22a shows a
physics student cruising at constant velocity in an airplane. If the student places a
(b) ball on the floor, it stays there. There are no horizontal forces, and the ball
r
remains at rest relative to the airplane. That is, ar 5 0 in the airplane’s coordinate
r r
Accelerating system when Fnet 5 0. Newton’s first law is satisfied, so this airplane is an inertial
reference frame.
The physics student in Figure 4.22b conducts the same experiment during
takeoff. She carefully places the ball on the floor just as the airplane starts to
The ball rolls to the back. accelerate down the runway. You can imagine what happens. The ball rolls to the
The ball rolls to the back of the plane during back of the plane as the passengers are being pressed back into their seats. Noth-
takeoff. An accelerating plane is not an ing exerts a horizontal contact force on the ball, yet the ball accelerates in the
inertial reference frame. plane’s coordinate system. This violates Newton’s first law, so the plane is not an
FIGURE 4.22 Reference frames. inertial reference frame during takeoff.
4.6 . Newton’s First Law 113
In the first example, the plane is traveling with constant velocity. In the sec-
ond, the plane is accelerating. Accelerating reference frames are not inertial
reference frames. Consequently, Newton’s laws are not valid in a coordinate sys-
tem attached to an accelerating object.
But accelerating with respect to what? The plane accelerated with respect
to the earth. But the earth is accelerating as it rotates on its axis and revolves
around the sun. The entire solar system is accelerating as our Milky Way galaxy
rotates.
This is a subtle and difficult question, one that Einstein grappled with as he
developed his theory of relativity. We cannot give a complete answer here, but
suffice it to say that an inertial reference frame is a reference frame that is not
accelerating with respect to the distant stars. Thus the earth is not exactly an iner-
tial reference frame. However, the earth’s acceleration with respect to the distant
stars is so small that violations of Newton’s laws can be measured only in
extremely high-precision experiments. We will treat the earth and laboratories
attached to the earth as inertial reference frames, an approximation that is exceed-
ingly well justified.
We will prove in Chapter 6 that a coordinate system moving with constant
velocity relative to an inertial reference frame is also an inertial reference frame, a
reference frame in which Newton’s laws are valid. Because the earth is an inertial
reference frame, the airplane of Figure 4.22a is also an inertial reference frame.
But a car braking to a stop is not, so you cannot use Newton’s laws in the car’s
reference frame.
To understand the motion of objects in the car, such as the passengers, you need
to measure velocities and accelerations relative to the ground. From the perspec-
tive of an observer on the ground, the body of a passenger in a braking car tries to
continue moving forward with constant velocity, exactly as we would expect on
the basis of Newton’s first law, while his immediate surroundings are decelerat-
ing. The passenger is not “thrown” into the windshield. Instead, the windshield
runs into the passenger!
Common Misconceptions About Force

It is important to identify correctly all the forces acting on an object. It is equally
important not to include forces that do not really exist. We have established a
number of criteria for identifying forces; the two critical ones are:
■ A force has an agent. Something tangible and identifiable causes the force.
■ Forces exist at the point of contact between the agent and the object experi-
encing the force (except for the few special cases of long-range forces).
We all have had many experiences suggesting that a force is necessary to keep
something moving. Consider a bowling ball rolling along on a smooth floor. It is
very tempting to think that a horizontal “force of motion” keeps it moving in the
forward direction. But if we draw a closed curve around the ball, nothing contacts
it except the floor. No agent is giving the ball a forward push. According to our
definition, then, there is no forward “force of motion” acting on the ball. So what
keeps it going? Recall our discussion of the first law: no cause is needed to keep an
object moving at constant velocity. It continues to move forward simply because
of its inertia.
One reason for wanting to include a “force of motion” is that we tend to view
the problem from our perspective as one of the agents of force. You certainly have
to keep pushing to shove a box across the floor at constant velocity. If you stop, it
stops. Newton’s laws, though, require that we adopt the object’s perspective. The
box experiences your pushing force in one direction and a friction force in the
opposite direction. The box moves at constant velocity if the net force is zero.
This will be true as long as your pushing force exactly balances the friction force.
When you stop pushing, the friction force causes an acceleration that slows and
stops the box.
A related problem occurs if you throw a ball. A pushing force was indeed
required to accelerate the ball as it was thrown. But that force disappears the
instant the ball loses contact with your hand. The force does not stick with the ball
as the ball travels through the air. Once the ball has acquired a velocity, nothing is
needed to keep it moving with that velocity.
A final difficulty worth noting is the force due to air pressure. You may have
learned in an earlier science class that air, like any fluid, exerts forces on objects.
Perhaps you learned this idea as “the air presses down with a weight of 15 pounds
on every square inch.” There is only one error here, but it is a serious one: the word
down. Air pressure, at sea level, does indeed exert a force of 15 pounds per square
inch, but in all directions. It presses down on the top of an object, but also inward
on the sides and upward on the bottom. For most purposes, the net force due to air
pressure is zero! The only way to experience an air pressure force is to form a seal
around one side of the object and then remove the air, creating a vacuum. When
you press a suction cup against the wall, you press the air out and the rubber forms
a seal that prevents the air from returning. Now the air pressure does hold the suc-
tion cup in place! We do not need to be concerned with air pressure until Part III of
this book.
4.7 Free-Body Diagrams

Having discussed at length what is and is not a force, we are ready to assemble
our knowledge about force and motion into a single diagram called a free-body
diagram. You will learn in the next chapter how to write the equations of motion
directly from the free-body diagram. Solution of the equations is a mathematical
exercise—possibly a difficult one, but nonetheless an exercise that could be done
by a computer. The physics of the problem, as distinct from the purely calcula-
tional aspects, are the steps that lead to the free-body diagram.
A free-body diagram represents the object as a particle and shows all of the
forces acting on the object. The free-body diagram joins with the motion diagram
and force identification to form the full physical representation of a problem.
TACTICS BOX 4.3 Drawing a free-body diagram

●
1 Identify all forces acting on the object. This step was described in Tac-
tics Box 4.2.
●
2 Draw a coordinate system. Use the axes defined in your pictorial repre-
sentation. If those axes are tilted, for motion along an incline, then the
axes of the free-body diagram should be similarly tilted.
●
3 Represent the object as a dot at the origin of the coordinate axes.
This is the particle model.
●
4 Draw vectors representing each of the identified forces. This was
described in Tactics Box 4.1. Be sure to label each force vector.
r
●
5 Draw and label the net force vector Fnet. Draw this vector beside the
r r
diagram, not on the particle. Or, if appropriate, write Fnet 5 0. Then
r
check that Fnet points in the same direction as the acceleration vector ar
on your motion diagram.
4.7 . Free-Body Diagrams 115
EXAMPLE 4.4 An elevator accelerates upward

An elevator, suspended by a cable, speeds up as it moves upward from the ground
floor. Draw a free-body diagram of the elevator.
MODEL Treat the elevator as a particle.
VISUALIZE
Force identification Free-body diagram 2 Draw a coordinate

r y system.
Tension T
r
T 3 Represent the
object as a dot
x at the origin.
r
w r
Fnet
1 Identify all r
Weight w
forces acting 5 Draw and label
r
on the object. 4 Draw vectors for Fnet beside the
the identified forces. diagram.
FIGURE 4.23 Free-body diagram of an elevator accelerating upward.
ASSESS The coordinate axes, with a vertical y-axis, are the ones we would use in a
r
pictorial representation of the motion. The elevator is accelerating upward, so F net
r
must point upward. For this to be true, the magnitude of T must be larger than the
r
magnitude of w . The diagram has been drawn accordingly.
EXAMPLE 4.5 An ice block shoots across a frozen lake MODEL Treat the block of ice as a particle. The full physical
Bobby straps a small model rocket to a block of ice and shoots representation consists of a motion diagram to determine ar , a
it across the smooth surface of a frozen lake. Friction is negligi- force identification picture, and a free-body diagram. The state-
ble. Draw a full physical representation of the block of ice. ment of the situation implies that friction is negligible.
VISUALIZE
Motion diagram Force identification Free-body diagram

r
Thrust force Fthrust
Start
vr
nr r
Fthrust
x
ar r
r
Weight w w
r
Normal force nr Fnet
r r
Check that Fnet points in the same direction as a.
FIGURE 4.24 Physical representation for a block of ice shooting across a frictionless frozen lake.
ASSESS The motion diagram tells us that the acceleration is in ( wy 5 2ny ) . The vectors have been drawn accordingly, and
the 1x-direction. According to the rules of vector addition, this this leaves the net force vector pointing toward the right, in
can be true only if the upward-pointing nr and the downward- agreement with ar from the motion diagram.
r
pointing w are equal in magnitude and thus cancel each other
EXAMPLE 4.6 A skier is pulled up a hill treated as a particle in dynamic equilibrium. If we were doing
A tow rope pulls a skier up a snow-covered hill at a constant a kinematics problem, the pictorial representation would use
speed. Draw a full physical representation of the skier. a tilted coordinate system with the x-axis parallel to the slope,
MODEL This is Example 4.2 again with the additional informa- so we use these same tilted coordinate axes for the free-body
tion that the skier is moving at constant speed. The skier will be diagram.
VISUALIZE
Motion diagram Force identification Free-body diagram
r y
Tension T
nr
x
r
T
r
fk
Notice that the angle between
r
the w and the negative y-axis
r
vr ar 5 0 is the same as the angle of
Normal force
r
nr r
w
the incline.
r Friction fk r r
Weight w Fnet 5 0
r
Check that Fnet matches ar.
FIGURE 4.25 Physical representation for a skier being towed at a constant speed.
r r
ASSESS We have shown T pulling parallel to the slope and f k, the incline above the horizontal. The skier moves in a straight
r
which opposes the direction of motion, pointing down the slope. line with constant speed, so ar 5 0 and, from Newton’s first
r r
nr is perpendicular to the surface and thus along the y-axis. law, F net 5 0. Thus we have drawn the vectors such that the
r r r
Finally, and this is important, the weight w is vertically down- y-component of w is equal in magnitude to nr. Similarly, T must
r
ward, not along the negative y-axis. In fact, you should convince be large enough to match the negative x-components of both f k
r r
yourself from the geometry that the angle u between the w and w.
vector and the negative y-axis is the same as the angle u of
Free-body diagrams will be our major tool for the next several chapters. Care-
ful practice with the workbook exercises and homework in this chapter will pay
immediate benefits in the next chapter. Indeed, it is not too much to assert that a
problem is half solved, or even more, when you complete the free-body diagram.
STOP TO THINK 4.5 An elevator suspended by a cable is moving upward and

slowing to a stop. Which free-body diagram is correct?
y y y y
r
T r
T r r r
T T Felevator
r
x x x x Fnet 5 0
r
w
r
w
r r
w w
(a) (b) (c) (d) (e)

Summary 117
SUMMARY
The goal of Chapter 4 has been to learn how force and motion are connected.
GENERAL PRINCIPLES
Newton’s First Law Newton’s Second Law
An object at rest will remain at rest, or an An object with mass m will undergo acceleration
Newton’s laws are
object that is moving will continue to move 1 r
valid only in inertial
in a straight line with constant velocity, if ar 5 F
and only if the net force on the object is
reference frames. m net
r r r r
zero. where F net 5 F 1 1 F 2 1 F 3 1 c is the vector
sum of all the individual forces acting on the object.
r r
Fnet 5 0 r
F
vr vr vr vr vr vr vr vr vr vr
r
ar 5 0
ar
The first law tells us that no “cause” is The second law tells us that a net force causes
needed for motion. Uniform motion is the an object to accelerate. This is the connection
“natural state” of an object. between force and motion that we are seeking.
IMPORTANT CONCEPTS
Acceleration is the link to kinematics. Mass is the resistance of an Force is a push or a pull on an object.
object to acceleration. It is
From a, find v and x. • Force is a vector, with a magnitude and
an intrinsic property of an
From v and x, find a. a direction.
r
object.
ar 5 0 is the condition for equilibrium. • Force requires an agent.
r r
Static equilibrium if v 5 0. • Force is either a contact force or a
Dynamic equilibrium if rv 5 constant. long-range force.
r r
Equilibrium occurs if and only if F net 5 0.
KEY SKILLS
y
Identifying Forces r
Thrust force Fthrust Free-Body Diagrams
Forces are identified by locating A free-body diagram represents the
the points where the environment object as a particle at the origin of nr r
Fthrust
touches the system. These are points a coordinate system. Force vectors x
where contact forces are exerted. In are drawn with their tails on the wr
addition, objects with mass feel a particle. The net force vector is r
Fnet
r
long-range weight force. Weight w Normal force nr drawn beside the diagram.
TERMS AND NOTATION

r
dynamics superposition of forces thrust, F thrust static equilibrium
r
mechanics weight, w inertia dynamic equilibrium
r r
force, F tension force, T inertial mass, m inertial reference frame
agent atomic model Newton’s second law free-body diagram
contact force normal force, nr newton, N
r r
long-range force friction, f k or f s Newton’s first law
r r
net force, F net drag, D mechanical equilibrium
EXERCISES AND PROBLEMS

Exercises a (m/s2)
?
Section 4.1 Force
?
1. Write a one-paragraph essay on the topic “What Is a Force?”
Explain in your own words how you recognize a force and
?
what the properties of forces are.
?
Section 4.3 Identifying Forces 0 F (N)

FIGURE EX4.8 0 1 2
2. A mountain climber is hanging from a rope in the middle of a
crevasse. The rope is vertical. Identify the forces on the moun- 9. Figure Ex4.9 shows an object’s acceleration-versus-force
tain climber. graph. What is the object’s mass?
3. A baseball player is sliding into second base. Identify the
a (m/s2)
forces on the baseball player. 4
4. A jet plane is speeding down the runway during takeoff. Air
resistance is not negligible. Identify the forces on the jet. 3
2
Section 4.4 What Do Forces Do?
1
5. Figure Ex4.5 shows an acceleration-versus-force graph for
three objects pulled by rubber bands. The mass of object 2 is 0 F (N)
0.20 kg. What are the masses of objects 1 and 3? Explain your FIGURE EX4.9 0.0 0.5 1.0
reasoning.
10. Based on the information in Table 4.3, estimate
a. The weight of this textbook.
Acceleration (multiples of a1)
1 b. The propulsion force of a bicycle.

2
5a1
11. Based on the information in Table 4.3, estimate
4a1 a. The weight of a pencil.
3a1 b. The propulsion force of a sprinter.
2a1 3
1a1
Section 4.6 Newton’s First Law
0
0 1 2 3 4 5 6 Exercises 12 through 14 show two forces acting on an object.
FIGURE EX4.5 Force (number of rubber bands) Redraw the diagram, then add a third force that will cause the
r
object to be in equilibrium. Label the new force F 3.
6. Two rubber bands pulling on an object cause it to accelerate at
1.2 m/s2. 12. 13. r
F1
14.
r
a. What will be the object’s acceleration if it is pulled by four F1
r
rubber bands? F2
b. What will be the acceleration of two of these objects glued
together if they are pulled by two rubber bands? r
F1 r
F2
r
F2
Section 4.5 Newton’s Second Law FIGURE EX4.12 FIGURE EX4.13 FIGURE EX4.14
7. Write a one-paragraph essay on the topic “Force and Motion.”

Explain in your own words the connection between force
and motion. Where possible, cite evidence supporting your Section 4.7 Free-Body Diagrams
statements. Exercises 15 through 17 show a free-body diagram. For each:
8. Figure Ex4.8 shows an acceleration-versus-force graph for a a. Redraw the free-body diagram.
500 g object. Redraw this graph and add appropriate acceler- b. Write a short description of a real object for which this is the
ation values on the vertical scale. correct free-body diagram. Use Examples 4.4, 4.5, and 4.6 as
models of what a description should be like.
Exercises and Problems 119
15. y 16. y 24. A force with x-component Fx acts on a 2.0 kg object as it

moves along the x-axis. The object’s acceleration graph (ax
r
Fthrust versus t) is shown in Figure P4.24. Draw a graph of Fx versus t.
r
T r
Fnet
ax (m/s2)
x x
r r
D w 3
r
w
r r 2
Fnet 5 0
1
FIGURE EX4.15 FIGURE EX4.16
17. y 0 t (s)
1 2 3 4
FIGURE P4.24 21
nr 25. A force with x-component Fx acts on a 500 g object as it

r
fk moves along the x-axis. The object’s acceleration graph (ax
x versus t) is shown in Figure P4.25. Draw a graph of Fx versus t.
r
r
w ax (m/s2)
Fnet
1.5
FIGURE EX4.17
1.0
Exercises 18 through 21 describe a situation. For each, identify all
forces acting on the object and draw a free-body diagram of the 0.5
object.
0.0 t (s)
18. Your car is sitting in the parking lot. 1 2 3 4
19. An ice hockey puck glides across frictionless ice. FIGURE P4.25 20.5
20. An elevator, hanging from a cable, descends at steady speed.
21. Your physics textbook is sliding across the table. 26. A force with x-component Fx acts on a 2.0 kg object as it moves
along the x-axis. A graph of Fx versus t is shown in Figure P4.26.
Draw an acceleration graph (ax versus t) for this object.
Problems
Fx (N)
22. Redraw the two motion diagrams shown vr 3

in Figure P4.22, then draw a vector beside
2
each one to show the direction of the net
force acting on the object. Explain your 1
reasoning.
0 t (s)
1 2 3 4
FIGURE P4.26 21
27. A force with x-component Fx acts on a 500 g object as it

moves along the x-axis. A graph of Fx versus t is shown in Fig-
ure P4.27. Draw an acceleration graph (ax versus t) for this
vr object.
FIGURE P4.22 (a) (b)
Fx (N)
23. Redraw the two motion diagrams shown in Figure P4.23, then 1.5
draw a vector beside each one to show the direction of the net
force acting on the object. Explain your reasoning. 1.0
(a) r (b)
v 0.5
0.0 t (s)
1 2 3 4
FIGURE P4.27 20.5
28. A constant force is applied to an object, causing the object to

accelerate at 10 m/s2. What will the acceleration be if
a. The force is halved?
vr b. The object’s mass is halved?
FIGURE P4.23 c. The force and the object’s mass are both halved?
d. The force is halved and the object’s mass is doubled?
29. A constant force is applied to an object, causing the object to 35.

y
accelerate at 8.0 m/s2. What will the acceleration be if
a. The force is doubled? x
nr
b. The object’s mass is doubled?
r r
c. The force and the object’s mass are both doubled? T fk
r
d. The force is doubled and the object’s mass is halved? w r
Fnet
Problems 30 through 36 show a free-body diagram. For each: FIGURE P4.35
a. Redraw the diagram.
36.
b. Identify the direction of the acceleration vector ar and show it y
as a vector next to your diagram. Or, if appropriate, write
r
ar 5 0. r
c. If possible, identify the direction of the velocity vector vr and

Fthrust
x
show it as a labeled vector. wr
d. Write a short description of a real object for which this is the r
Fnet
correct free-body diagram. Use Examples 4.4, 4.5, and 4.6 as FIGURE P4.36
models of what a description should be like. Problems 37 through 46 describe a situation. For each, draw a full
30. physical representation. A full physical representation is a motion
y
diagram, force identification, and a free-body diagram.
37. An elevator, suspended by a single cable, has just left the
r
nr tenth floor and is speeding up as it descends toward the
fk r
T
x ground floor.
r
38. A rocket is being launched straight up. Air resistance is not
w
r negligible.
Fnet
39. A jet plane is speeding down the runway during takeoff. Air
FIGURE P4.30
resistance is not negligible.
31. 40. You’ve slammed on the brakes and your car is skidding to a
y stop while going down a 20° hill.
41. A skier is going down a 20° slope. A horizontal headwind is
blowing in the skier’s face. Friction is small, but not zero.
r
nr r
D Fthrust 42. You’ve just kicked a soccer ball and it is now rolling across
x
the grass.
r
w r r 43. A styrofoam ball has just been shot straight up. Air resistance
Fnet 5 0
is not negligible.
FIGURE P4.31
44. A spring-loaded gun shoots a plastic ball. The trigger has just
32. been pulled and the ball is starting to move down the barrel.
y
The barrel is horizontal.
45. A person on a bridge throws a rock straight down toward the
nr water. The rock has just been released.
r r
Fsp fk 46. A gymnast has just landed on a trampoline. She’s still moving
x downward as the trampoline stretches.
r
w
r
Fnet
FIGURE P4.32 Challenge Problems
33. y 47. A heavy box is in the back of a truck. The truck is accelerating
to the right. Draw a full physical representation of the box.
48. A bag of groceries is on the back seat of your car as you stop
for a stop light. The bag does not slide. Draw a full physical
x representation of the bag.
r
w 49. A rubber ball bounces. We’d like to understand how the ball
bounces.
FIGURE P4.33 a. A rubber ball has been dropped and is bouncing off the
floor. Draw a motion diagram of the ball during the brief
34. y time interval that it is in contact with the floor. Show 4 or 5
frames as the ball compresses, then another 4 or 5 frames
nr r
Fnet
as it expands. What is the direction of ar during each of
r
fk
these parts of the motion?
b. Draw a picture of the ball in contact with the floor and
r
x identify all forces acting on the ball.
w
FIGURE P4.34
c. Draw a free-body diagram of the ball during its contact b. Draw your free-body diagram. Is there a net force on you?
with the ground. Is there a net force acting on the ball? If If so, in which direction?
so, in which direction? c. Repeat parts a and b with the car slowing down.
d. Write a paragraph in which you describe what you learned d. Describe what happens to you as the car slows down.
from parts a to c and in which you answer the question: e. Use Newton’s laws to explain why you seem to be
How does a ball bounce? “thrown forward” as the car stops. Is there really a force
50. If a car stops suddenly, you feel “thrown forward.” We’d like pushing you forward?
to understand what happens to the passengers as a car stops. f. Suppose now that the bench is not slippery. As the car
Imagine yourself sitting on a very slippery bench inside a car. slows down, you stay on the bench and don’t slide off.
This bench has no friction, no seat back, and there’s nothing What force is responsible for your deceleration? In which
for you to hold to. direction does this force point? Include a free-body dia-
a. Draw a picture and identify all of the forces acting on you gram as part of your answer.
as the car travels at a perfectly steady speed on level
ground.
STOP TO THINK ANSWERS
Stop to Think 4.1: c. Stop to Think 4.4: d.

r
F1 r
F2 r r r
F1 F1 1 F 2
r
F3 r
y-component of F3 r r r
cancels y-component of F1. F2 Fnet ar
r r
r F3 F3
x-component of F3 is to
the left and larger than r r
the x-component of F2.
r
First
r
addr Then add F3. This is Fnet. ar is in the same
r
F1 and F2. direction as Fnet.
Stop to Think 4.2: a, b, and d. Friction and the normal force are Stop to Think 4.5: c. The acceleration vector points downward as
r r
the only contact forces. Nothing is touching the rock to provide a the elevator slows. F net points in the same direction as ar, so F net
“force of the kick.” We’ve agreed to ignore air resistance unless a also points down. This will be true if the tension is less than the
problem specifically calls for it. weight: T , w.
Stop to Think 4.3: b. Acceleration is proportional to force, so
doubling the number of rubber bands doubles the acceleration of
the original object from 2 m/s2 to 4 m/s2. But acceleration is also
inversely proportional to mass. Doubling the mass cuts the acceler-
ation in half, back to 2 m/s2.
5 Dynamics I:
Motion Along a Line
This skydiver may not know it, but he
is testing Newton’s second law as he
plunges toward the ground below.
Looking Ahead
The goal of Chapter 5 is to learn
how to solve problems about
motion in a straight line. In this
chapter you will learn to:
■ Solve static and dynamic
equilibrium problems by
applying a Newton’s-first-law
strategy.
■ Solve dynamics problems by
applying a Newton’s-second-
law strategy.
■ Understand how mass, weight,
and apparent weight differ.
■ Use simple models of friction
and drag.
Looking Back
This chapter pulls together
A skydiver accelerates until reaching a terminal speed of about 140 mph. To
many strands of thought from
understand the skydiver’s motion, we need to look closely at the forces exerted on
Chapters 1–4. Please review:
him. We also need to understand how those forces determine his motion.
■ Sections 2.5–2.7 Constant In Chapter 4 we learned what a force is and is not. We also discovered the fun-
acceleration kinematics, damental relationship between force and motion: Newton’s second law. Chapter 5
including free fall. begins to develop a strategy for solving force and motion problems. Our strategy
■ Sections 3.3–3.4 Working with is to learn a set of procedures, not to memorize a set of equations.
vectors and vector components. This chapter focuses on objects that move in a straight line, such as runners,
■ Sections 4.2, 4.3, and 4.7 bicycles, cars, planes, and rockets. Weight, tension, thrust, friction, and drag
Identifying forces and drawing forces will be essential to our understanding. Projectile motion and motion in a
free-body diagrams. circle are the topics of the next two chapters.
122
5.1 . Equilibrium 123
5.1 Equilibrium
An object on which the net force is zero is said to be in equilibrium. The object
might be at rest in static equilibrium, or it might be moving along a straight line
with constant velocity in dynamic equilibrium. Both are identical from a Newton-
r r r
ian perspective because Fnet 5 0 and ar 5 0.
Newton’s first law is the basis for a four-step strategy for solving equilibrium
problems.
PROBLEM-SOLVING STRATEGY 5.1 Equilibrium problems The concept of equilibrium is essential

for the engineering analysis of stationary
MODEL Make simplifying assumptions. objects such as bridges.
VISUALIZE
Physical representation. Identify all forces acting on the object and show
them on a free-body diagram.
Pictorial representation. The free-body diagram is usually sufficient as a
picture for equilibrium problems, but you still must translate words to sym-
bols and identify what the problem is trying to find.
It’s OK to go back and forth between these two steps as you visualize the
situation.
SOLVE The mathematical representation is based on Newton’s first law
r r r
Fnet 5 a Fi 5 0
i
The vector sum of the forces is found directly from the free-body diagram.
ASSESS Check that your result has the correct units, is reasonable, and
answers the question.
Newton’s laws are vector equations. Recall from Chapter 3 that the vector
equation in the step labeled Solve is a shorthand way of writing two simultaneous
equations:
(Fnet ) x 5 a (Fi ) x 5 0
i
(5.1)
(Fnet ) y 5 a (Fi ) y 5 0
i
r
In other words, each component of Fnet must simultaneously be zero. Although
real-world situations often have forces pointing in three dimensions, thus requir-
r
ing a third equation for the z-component of Fnet, we will restrict ourselves for now
to problems that can be analyzed in two dimensions.
Equilibrium problems occur frequently, especially in engineering applications.
Let’s look at a couple of examples.
Static Equilibrium
EXAMPLE 5.1 Three-way tug-of-war 200 N. How hard, and in which direction, should you pull to
You and two friends find three ropes tied together with a single keep the knot from moving?
knot and decide to have a three-way tug-of-war. Alice pulls to MODEL We’ll treat the knot in the rope as a particle in static
the west with 100 N of force while Bob pulls to the south with equilibrium.
124 CHAPTER 5 . Dynamics I: Motion Along a Line
Physical representation Pictorial representation

A free-body diagram usually suffices
N y Establish a coordinate in equilibrium problems. All you need
Identify the knot system with the +x-axis to do is add the known information
Note that there’s and identify what you are trying to find.
as the system. to the east.
no net force.
r r
Fnet 5 0 r Known
T3
T1 5 100 N
x
r T2 5 200 N
T1
Name and labelrthe Find
Three tension forces r
T2 angle between T3 T3 and
act on the knot.
and the x-axis.
Identify forces. Draw free-body diagram. List knowns and unknowns.
FIGURE 5.1 Physical and pictorial representations for a knot in static equilibrium.
VISUALIZE Figure 5.1 shows how to carry out the physical and These are two simultaneous equations for the two unknowns T3
pictorial representations. Notice that we’ve defined angle u to and u. We will encounter equations of this form on many occa-
indicate the direction of your pull. sions, so make a note of the method of solution. First, rewrite
r r
SOLVE The free-body diagram shows tension forces T1, T2, and the two equations as
r
T3 acting on the knot. Newton’s first law, written in component T1 5 T3 cos u
form, is
T2 5 T3 sin u
(Fnet ) x 5 a (Fi ) x 5 T1x 1 T2x 1 T3x 5 0
i Next, divide the second equation by the first to eliminate T3:
(Fnet ) y 5 a (Fi ) y 5 T1y 1 T2y 1 T3y 5 0 T2 T3 sin u
i 5 5 tan u
T1 T3 cos u
NOTE You might have been tempted to write 2T1x in the first
1 2 1 2
r Then solve for u:
equation because T1 points in the negative x-direction. But the
net force, by definition, is the sum of all the individual forces. T2 200 N
r u 5 tan21 5 tan21 5 63.4°
The fact that T1 points to the left will be taken into account T1 100 N
when we evaluate the components.
Finally, use u to find T3:
The components of the force vectors can be evaluated
T1 100 N
directly from the free-body diagram: T3 5 5 5 224 N
cos u cos 63.4°
T1x 5 2T1 T1y 5 0
The force that maintains equilibrium and prevents the knot
T2x 5 0 T2y 5 2T2
from moving is thus
T3x 5 1T3 cos u T3y 5 1T3 sin u r
T3 5 ( 224 N, 63.4° north of east )
This is where the signs enter, with T1x being assigned a negative
r
value because T1 points to the left. Similarly, T2y 5 2T2. With ASSESS Is this result reasonable? Because your friends pulled
these components, Newton’s first law becomes west and south, you expected to pull in a generally northeast
direction. You also expected to pull harder than either of them
2T1 1 T3 cos u 5 0 but, because they didn’t pull in the same direction, less than the
r
2T2 1 T3 sin u 5 0 sum of their pulls. The result for T3 meets these expectations.
Dynamic Equilibrium
EXAMPLE 5.2 Towing a car up a hill MODEL We’ll treat the car as a particle in dynamic equilibrium.
A car with a weight of 15,000 N is being towed up a 20° slope at We’ll ignore friction.
constant velocity. Friction is negligible. The tow rope is rated at VISUALIZE This problem asks for a yes or no answer, not a num-
6000 N maximum tension. Will it break? ber, but we still need a quantitative analysis. Part of our analysis
5.2 . Using Newton’s Second Law 125
Physical representation y Pictorial representation
r Known
Tension T nr
r x 5 20°
T
w 5 15,000 N
r r Find
r
Fnet 5 0
r r w T
Weight w Normal force n
FIGURE 5.2 The physical and pictorial representations of a car being towed up a hill.
of the problem statement is to determine which quantity or quan- With these components, the first law becomes
tities allow us to answer the question. In this case the answer is
T 2 w sin u 5 0
clear: We need to calculate the tension in the rope. Figure 5.2
shows the physical and pictorial representations. Note the simi- n 2 w cos u 5 0
larities to Examples 4.2 and 4.6 in Chapter 4, which you may The first of these can be rewritten as
want to review.
r T 5 w sin u
SOLVE The free-body diagram shows forces T , nr, and w
r
acting
on the car. Newton’s first law is 5 (15,000 N) sin 20° 5 5130 N
(Fnet ) x 5 a Fx 5 Tx 1 n x 1 wx 5 0 Because T , 6000 N, we conclude that the rope will not break.
It turned out that we did not need the y-component equation in
(Fnet ) y 5 a Fy 5 Ty 1 n y 1 wy 5 0 this problem.
Notice that we dropped the label i from the sum. From here on, ASSESS Because there’s no friction, it would not take any ten-
we’ll use gFx and gFy as a simple shorthand notation to sion force to keep the car rolling along a horizontal surface
indicate that we’re adding all the x-components and all the (u 5 0°). At the other extreme, u 5 90°, the tension force
y-components of the forces. would need to equal the car’s weight (T 5 w 5 15,000 N) to
We can deduce the components directly from the free-body lift the car straight up at constant velocity. The tension force for
diagram: a 20° slope should be somewhere in between, and 5130 N is a
little less than half the weight of the car. That our result is rea-
Tx 5 T Ty 5 0
sonable doesn’t prove it’s right, but we have at least ruled out
nx 5 0 ny 5 n careless errors that give unreasonable results.
wx 5 2w sin u wy 5 2w cos u
NOTE The weight has both x- and y-components in this coor-

dinate system, both of which are negative due to the direction of
r
the vector w . You’ll see this situation often, so be sure you
understand where wx and wy come from.
5.2 Using Newton’s Second Law

Equilibrium is important, but it is a special case of motion. Newton’s second law
is a more general link between force and motion. We now need a strategy for
using Newton’s second law to solve dynamics problems.
The essence of Newtonian mechanics can be expressed in two steps:
r ONLINE 2.1, 2.2, 2.3, 2.4
■ The forces on an object determine its acceleration ar 5 Fnet /m.
■ The object’s trajectory can be determined by using ar in the equations of
kinematics.
These two ideas are the basis of a strategy for solving dynamics problems.
PROBLEM-SOLVING STRATEGY 5.2 Dynamics problems
MODEL Make simplifying assumptions.

VISUALIZE
Pictorial representation. Show important points in the motion with a

sketch, establish a coordinate system, define symbols, and identify what the
problem is trying to find. This is the process of translating words to symbols.
Physical representation. Use a motion diagram to determine the object’s
acceleration vector ar. Then identify all forces acting on the object and show
them on a free-body diagram.
It’s OK to go back and forth between these two steps as you visualize the
situation.
SOLVE The mathematical representation is based on Newton’s second law
r r
Fnet 5 a Fi 5 mar
i
The vector sum of the forces is found directly from the free-body diagram.
Depending on the problem, either
■ Solve for the acceleration, then use kinematics to find velocities and
positions, or
■ Use kinematics to determine the acceleration, then solve for unknown
forces.
ASSESS Check that your result has the correct units, is reasonable, and
answers the question.
Newton’s second law is a vector equation. To apply the step labeled Solve, you
must write the second law as two simultaneous equations:
(Fnet ) x 5 a Fx 5 max
(5.2)
(Fnet ) y 5 a Fy 5 may
The primary goal of this chapter is to illustrate the use of this strategy. Let’s start
with some examples.
EXAMPLE 5.3 Speed of a towed car (Fnet ) x 5 a Fx 5 Tx 1 fx 1 n x 1 wx 5 max

A 1500 kg car is pulled by a tow truck. The tension in the tow
(Fnet ) y 5 a Fy 5 Ty 1 fy 1 n y 1 wy 5 may
rope is 2500 N, and a 200 N friction force opposes the motion.
If the car starts from rest, what is its speed after 5.0 seconds? All four forces acting on the car have been included in the vec-
MODEL We’ll treat the car as an accelerating particle. We’ll tor sum. The equations are perfectly general, with 1 signs
r
assume, as part of our interpretation of the problem, that the everywhere, because the four vectors are added to give Fnet. We
road is horizontal and that the direction of motion is to the right. can now “read” the vector components directly from the free-
body diagram:
VISUALIZE Figure 5.3 shows the pictorial and physical represen-
tations. We’ve established a coordinate system and defined sym- Tx 5 1T Ty 5 0
bols to represent kinematic quantities. We’ve identified the
nx 5 0 ny 5 1n
speed v1, rather than the velocity v1x, as what we’re trying to find.
fx 5 2f fy 5 0
SOLVE We begin the mathematical representation with New-
ton’s second law: wx 5 0 wy 5 2w
5.2 . Using Newton’s Second Law 127
Pictorial representation Physical representation

ax
x vr y
0
x0, v0x, t0 x1, v1x, t1 r
a nr
Known r
Tension T r r
x0 5 v0x 5 t0 5 0 T 5 2500 N f T
x
t1 5 5 s f 5 200 N
r r
m 5 1500 kg Weight w Normal n
r r
Friction f r
w Fnet
Find
v1
FIGURE 5.3 Pictorial and physical representations of a car being towed.
The signs of the components depend on which way the vectors along the y-axis. The requirement ay 5 0 allows us to conclude
point. Substituting these into the second-law equations and that n 5 w. Although we do not need n for this problem, it will
dividing by m gives be important in many future problems.
1 We can finish by using constant-acceleration kinematics to

ax 5 (T 2 f ) find the velocity:
m
1 v1x 5 v0x 1 ax Dt
5 (2500 N 2 200 N) 5 1.53 m/s2
1500 kg
5 0 1 (1.53 m/s2 )(5.0 s)
1
ay 5 (n 2 w) 5 7.65 m/s
m
NOTE Newton’s second law has allowed us to determine ax The problem asked for the speed after 5.0 s, which is v1 5
exactly but has given only an algebraic expression for ay. How- 7.65 m/s.
ever, we know from the motion diagram that ay 5 0! That is, ASSESS 7.65 m/s < 15 mph, a reasonable speed after 5 s of
the motion is purely along the x-axis, so there is no acceleration acceleration.
r
EXAMPLE 5.4 Altitude of a rocket The fact that vector w points downward—and which might have
A 500 g model rocket with a weight of 4.9 N is launched straight tempted you to use a minus sign in the y-equation—will be
up. The small rocket motor burns for 5.0 s and has a steady taken into account when we evaluate the components. None
thrust of 20 N. What maximum altitude does the rocket reach? of the vectors in this problem has an x-component, so only the
Assume that the mass loss of the burned fuel is negligible. y-component of the second law is needed. We can use the free-
MODEL We’ll treat the rocket as an accelerating particle. Air body diagram to see that
resistance will be neglected. (Fthrust ) y 5 1Fthrust
VISUALIZE The pictorial representation of Figure 5.4 on the wy 5 2w
next page finds that this is a two-part problem. First, the rocket
accelerates straight up. Second, the rocket continues going up This is the point at which the directional information about the
as it slows down, a free-fall situation. The maximum altitude is force vectors enters. The y-component of the second law is then
at the end of the second part of the motion.
1
SOLVE We now know what the problem is asking, have estab- a0y 5 (F 2 w)
m thrust
lished relevant symbols and coordinates, and know what the
forces are. We begin the mathematical representation by writing 20.0 N 2 4.9 N
5 5 30.2 m/s2
Newton’s second law, in component form, as the rocket acceler- 0.500 kg
ates upward. The free-body diagram shows two forces, so
Notice that we converted the mass to SI units of kilograms
(Fnet ) x 5 a Fx 5 (Fthrust ) x 1 wx 5 ma0x before doing any calculations and that, because of the definition
of the newton, the division of newtons by kilograms automati-
(Fnet ) y 5 a Fy 5 (Fthrust ) y 1 wy 5 ma0y cally gives the correct SI units of acceleration.

y
Stop
Max altitude Known y
y2, v2y, t2
y0 5 0 m Fthrust 5 20 N
v0y 5 0 m/s
a1y w 5 4.9 N ar
t0 5 0 s
x
t1 5 5 s a1y 5 29.8 m/s2
v2y 5 0 (top) m 5 500 g 5 0.5 kg r
Fuel out w r
y1, v1y, t1 r Fnet
Weight w
Find
y2 After burnout
y
r
Fthrust
r
a0y Thrust Fthrust r
ar Fnet
r
Start w
0 y0, v0y, t0
vr
Before burnout
FIGURE 5.4 Pictorial and physical representations of a rocket launch.
The acceleration of the rocket is constant until it runs out of We can use free-fall kinematics to find the maximum
fuel, so we can use constant-acceleration kinematics to find the altitude:
altitude and velocity at burnout (Dt 5 t1 5 5.0 s):
v2y2 5 0 5 v1y2 2 2g Dy 5 v1y2 2 2g(y2 2 y1 )
y1 5 y0 1 v0y Dt 1 12 a0y (Dt) 2 which we can solve to find
5 12 a0y (Dt) 2 5 377 m v1y2 (151 m/s) 2
y2 5 y1 1 5 377 m 1
v1y 5 v0y 1 a0y Dt 5 a0y Dt 5 151 m/s 2g 2(9.80 m/s2 )
5 1540 m 5 1.54 km
The only force on the rocket after burnout is gravity, so the sec-
ond part of the motion is free-fall with a1y 5 2g. We do not ASSESS The maximum altitude reached by this rocket is
know how long it takes to reach the top, but we do know that 1.54 km, or just slightly under one mile. While this is fairly high,
the final velocity is v2y 5 0. it does not seem unreasonable for a high-acceleration rocket.
These first examples have shown all the details. Our purpose has been to show
how the problem-solving strategy is put into practice. Future examples will be
briefer, but the basic procedure will remain the same.
STOP TO THINK 5.1 A Martian lander is approaching the surface. It is slowing

its descent by firing its rocket motor. Which is the correct free-body diagram for
the lander?
Descending
and slowing
(a) (b) (c) (d) (e)
5.3 . Mass and Weight 129
5.3 Mass and Weight

When the doctor asks what you weigh, what does she really mean? We do not
make a large distinction in our ordinary use of language between the terms weight
and mass, but in physics their distinction is of critical importance.
Mass, you’ll recall from Chapter 4, is a scalar quantity that describes an object’s
inertia. Loosely speaking, it also describes the amount of matter in an object. y
Mass, measured in kilograms, is an intrinsic property of an object; it has the same
value wherever the object may be and whatever forces might be acting on it.
Weight, on the other hand, is a force. Specifically, it is the gravitational force x
exerted on an object by a planet. Weight is a vector, not a scalar, and the vector’s
direction is always straight down. Weight is measured in newtons. Weight is the only
r
Mass and weight are not the same thing, but they are related. We can use w force acting on this
r r
Galileo’s discovery about free fall to make the connection. Figure 5.5 shows the object, so Fnet 5 w.
free-body diagram of an object in free fall. The only force acting on this object is FIGURE 5.5 The free-body diagram of an
r r
its weight, the downward pull of gravity, so Fnet 5 w . Newton’s second law is thus object in free fall.
r r
Fnet 5 w 5 mar (5.3)
Recall Galileo’s discovery that any object in free fall, regardless of its mass,
has the same acceleration:
ar free fall 5 (9.80 m/s2, downward) 5 (g, downward) (5.4)
2
where g 5 9.80 m/s is the acceleration due to gravity at the Earth’s surface. If
we use the free-fall acceleration in Equation 5.3, the object’s weight is
r
w 5 (mg, downward) (5.5)
The magnitude of the weight force, which we call simply “the weight,” is directly
proportional to the mass, with g as the constant of proportionality:
w 5 mg (5.6) Insert 386851013 here
Because an object’s weight depends on g, and the value of g varies from planet
to planet, weight is not a fixed, constant property of an object. The value of g at
the surface of the moon is about one-sixth its earthly value, so an object on the
moon would have only one-sixth its weight on Earth. The object’s weight on A spring scale, such as the familiar
Jupiter would be larger than its weight on Earth. Its mass, however, would be the bathroom scale, measures weight,
same. The amount of matter has not changed, only the gravitational force exerted not mass.
on that matter.
So when the doctor asks what you weigh, she really wants to know your mass.
That’s the amount of matter in your body. You can’t really “lose weight” by going If the unknown mass differs
to the moon, even though you would weigh less there! from the known masses, the
beam will rotate about the pivot.
Known Unknown
Measuring Mass and Weight masses mass
A pan balance, shown in Figure 5.6, is a device for measuring mass. You may
have used a pan balance to “weigh” chemicals in a chemistry lab. An unknown
mass is placed in one pan, then known masses are added to the other until the
pans balance. Gravity pulls down on both sides, effectively comparing the
masses, and the unknown mass equals the sum of the known masses that balance
it. Although a pan balance requires gravity in order to function, it does not depend
r Pivot
on the value of g. Consequently, the pan balance would give the same result on w known
wr unknown
The pans balance when
another planet. the masses are equal.
Spring scales, such as the two shown in Figure 5.7 on the next page, measure
weight, not mass. Hanging an item on the scale in Figure 5.7a, which might be Both pans are pulled down
by the force of gravity.
used to weigh items in the grocery store, stretches the spring. The spring in the
“bathroom scale” in Figure 5.7b is compressed when you stand on it. FIGURE 5.6 A pan balance measures mass.
(a) A spring scale can be understood on the basis of Newton’s first law. The object
0
being weighed is at rest, in static equilibrium, so the net force on it must be zero.
The stretched spring in Figure 5.7a pulls up; the compressed spring in Figure 5.7b
r r
10 5 pushes up. But in both cases, in order to have Fnet 5 0, the upward spring force
must exactly balance the downward weight force:
r Cut-away detail
Fsp of spring Fsp 5 w 5 mg (5.7)
The reading of a spring scale is Fsp, the magnitude of the force that the spring
is exerting. If the object is in equilibrium, then Fsp is exactly equal to the object’s
weight w. The scale does not “know” the weight of the object. All it can do is to
r
w
measure how much the spring is stretched or compressed. On a different planet,
with a different value for g, the expansion or compression of the spring would be
different and the scale’s reading would be different.
(b)
The unit of force in the English system is the pound. We noted in Chapter 4
r that the pound is defined as 1 lb ; 4.45 N. An object whose weight w 5 mg is
Fsp
4.45 N has a mass
w 4.45 N
m5 5 5 0.454 kg 5 454 g
g 9.80 m/s2
r
w
You may have learned in previous science classes that “1 pound 5 454 grams” or,
equivalently, that “1 kg 5 2.2 lb.” Strictly speaking, these well-known “conversion
FIGURE 5.7 A spring scale measures factors” are not true. They are comparing a weight (pounds) to a mass (kilograms).
weight. The correct statement would be, “A mass of 1 kg has a weight on earth of 2.2
pounds.” On another planet, the weight of a 1 kg mass would be something other than
2.2 pounds.
EXAMPLE 5.5 Mass and weight on Jupiter

What is the kilograms-to-pounds conversion factor on Jupiter, where the accelera-
tion due to gravity is 25.9 m/s2?
SOLVE Consider an object with a mass of 1 kg. Its weight on Jupiter is
wJupiter 5 mgJupiter 5 (1 kg)(25.9 m/s2 )
1 lb
5 25.9 N 3 5 5.82 lb
4.45 N
If you had gone to school on Jupiter, you would have learned that 1 kg 5 5.82 lb.
Apparent Weight
The weight of an object is the force of gravity on that object. You may never have
thought about it, but gravity is not a force that you can feel or sense directly. Your
sensation of weight—how heavy you feel—is due to contact forces pressing
against you. Surfaces touch you and activate nerve endings in your skin. As you
read this, your sensation of weight is due to the normal force exerted on you by
the chair in which you are sitting. When you stand, you feel the contact force of
the floor pushing against your feet. If you hang from a rope, your sensation of
weight is due to the tension force pulling up on you.
r
If you stand at rest, with ar 5 0, then the force you feel, the normal force press-
ing against your feet, is exactly equal in magnitude to your weight. This would be
r
a trivial conclusion if objects were always at rest, but what happens if ar 2 0?
Recall the sensations you feel while being accelerated. You feel “heavy” when
an elevator suddenly accelerates upward or when an airplane accelerates for take-
off. This sensation vanishes as soon as the elevator or airplane reaches a steady
cruising speed. Your stomach seems to rise a little and you feel lighter than nor-
mal as the upward-moving elevator brakes to a halt or a roller coaster goes over
5.3 . Mass and Weight 131
the top. Your true weight w 5 mg has not changed during these events, but your The man feels heavier
apparent weight has. than normal while
accelerating upward.
To investigate this, imagine a man weighing himself by standing on a spring
ar
scale in an elevator as it accelerates upward. What does the scale read? How does
the scale reading correspond to the man’s sensation of weight?
As Figure 5.8 shows, the only forces acting on the man are the upward spring
force of the scale and the downward weight force. This seems to be the same situ-
ation as Figure 5.7b, but there’s one big difference. The man is accelerating; he’s
not in equilibrium. Thus, according to Newton’s laws, there must be a net force
acting on the man in the direction of ar.
r
For the net force Fnet to point upward, the magnitude of the spring force must
be greater than the magnitude of the weight force. That is, Fsp . w. This conclu-
sion has major implications. Looking at the free-body diagram in Figure 5.8, we
see that the y-component of Newton’s second law is
Spring scale
(Fnet ) y 5 (Fsp ) y 1 wy 5 Fsp 2 w 5 Fsp 2 mg 5 may (5.8)
where m is the man’s mass. y

The scale reading is the value of Fsp, the magnitude of the force that the scale
1 2 1 2
exerts on the man. Solving Equation 5.8 for Fsp gives
r r
ay ay Fsp Fnet
Fsp 5 scale reading 5 mg 1 may 5 mg 1 1 5w 11 (5.9)
g g x
r
If the elevator is either at rest or moving with constant velocity, then ay 5 0 and w
the man is in equilibrium. In that case, Fsp 5 w and the scale correctly reads his
weight. But if ay 2 0, the scale’s reading is not the man’s true weight.
Let’s define an object’s apparent weight wapp as the magnitude of the contact FIGURE 5.8 A man weighing himself in an
1 2
force that supports the object. From Equation 5.9, this is accelerating elevator.
ay
wapp 5 w 1 1 (5.10)
g
We’ve found wapp for a situation where a contact force supports the object from
below. A homework problem will let you show that the expression is also valid
when the object is supported from above by the tension in a rope or cable.
An object appears to weigh whatever the scale reads, although, as the saying
goes, appearances can be deceiving. As the elevator accelerates upward, ay . 0
and wapp . w. You feel heavier than normal, and a scale would read more than
your true weight. The acceleration vector ar points downward and ay , 0 when
the elevator brakes, so wapp , w. You feel lighter and, if you were standing on a
scale, it would read less than your true weight.
An object doesn’t have to be on a scale for its apparent weight to differ from its
true weight. An object’s apparent weight is the magnitude of the contact force
supporting it. It makes no difference whether this is the spring force of the scale
or simply the normal force of the floor.
The idea of apparent weight has important applications. Astronauts are nearly
crushed by their apparent weight during a rocket launch when ay W g. Much of
the thrill of amusement park rides, such as roller coasters, comes from rapid
changes in your apparent weight. In Chapter 7, the concept of apparent weight
will help us understand how you can swing a bucket of water over your head
without the water falling out!
Weightlessness
One last issue before leaving this topic: Suppose the elevator cable breaks and the
elevator, along with the man and his scale, plunges straight down in free fall!
What will the scale read? The acceleration in free fall is ay 5 2g. When this
acceleration is used in Equation 5.10, we find that wapp 5 0! In other words, the
man has no sensation of weight.
Think about this carefully. Suppose, as the elevator falls, the man inside
releases a ball from his hand. In the absence of air resistance, as Galileo discov-
ered, both the man and the ball would fall at the same rate. From the man’s per-
spective, the ball would appear to “float” beside him. Similarly, the scale would
float beneath him and not press against his feet. He is what we call weightless.
Surprisingly, “weightless” does not mean “no weight.” An object that is weight-
less has no apparent weight. The distinction is significant. The man’s weight is still
mg, because gravity is still pulling down on him, but he has no sensation of weight
as he free falls. The term “weightless” is a very poor one, likely to cause confusion
because it implies that objects have no weight. As we see, that is not the case.
But isn’t this exactly what happens to astronauts orbiting the earth? You’ve
seen films of astronauts and various objects floating inside the Space Shuttle. If
an astronaut tries to stand on a scale, it does not exert any force against her feet
and reads zero. She is said to be weightless. But if the criterion to be weightless is
to be in free fall, and if astronauts orbiting the earth are weightless, does this
Astronauts are weightless as they orbit mean that they are in free fall? This is a very interesting question to which we
the earth. shall return in Chapter 7.
STOP TO THINK 5.2 An elevator that has descended from the 50th floor is com-
ing to a halt at the 1st floor. As it does, your apparent weight is
a. More than your true weight. b. Less than your true weight.
c. Equal to your true weight. d. Zero.
5.4 Friction
In everyday life, friction is everywhere. Friction is absolutely essential for many
things we do. Without friction you could not walk, drive, or even sit down (you
would slide right off the chair!). It is sometimes useful to think about idealized
frictionless situations, but it is equally necessary to understand a real world where
friction is present. Although friction is a complicated force, many aspects of fric-
tion can be described with a simple model.
Static Friction
r
Chapter 4 defined static friction f s as the force on an object that keeps it from
r
slipping. Figure 5.9 shows a person pushing on a box with horizontal force Fpush.
If the box remains at rest, “stuck” to the floor, it must be because of a static fric-
tion force pushing back to the left. The box is in static equilibrium, so the static
friction must exactly balance the pushing force:
fs 5 Fpush (5.11)
Pushing
force
r nr
vr 5 0 r r
Fpush
fs
Object
is at rest.
r
w r r
Fnet 5 0
Static friction is opposite Free-body diagram

the push to prevent motion.
FIGURE 5.9 Static friction keeps an object from slipping.

5.4 . Friction 133
r r r
To determine the direction of f s, decide which way the object would move if there fs Fpush
r
were no friction. The static friction force f s points in the opposite direction to pre- r r
vent the motion. fs balances Fpush and
the box does not move.
Unlike weight, which has the precise and unambiguous magnitude w 5 mg,
r
the size of the static friction force depends on how hard you push. The harder the r
fs Fpush
person in Figure 5.9 pushes, the harder the floor pushes back. Reduce the pushing
r r
force, and the static friction force will automatically be reduced to match. Static fs grows as Fpush increases . . .
friction acts in response to an applied force. Figure 5.10 illustrates this idea.
r r
But there’s clearly a limit to how big fs can get. If you push hard enough, the fs Fpush
object slips and starts to move. In other words, the static friction force has a maxi-
mum possible size fs max. fs 5 fs max
r
■ An object remains at rest as long as fs , fs max. . . . until fs reaches fs max. Now, if Fpush gets
any bigger, the object will start to move.
■ The object slips when fs 5 fs max.
■ A static friction force fs . fs max is not physically possible. FIGURE 5.10 Static friction acts in response
to an applied force.
Experiments with friction (first done by Leonardo da Vinci) show that fs max is
proportional to the magnitude of the normal force. That is,
fs max 5 msn (5.12)
where ms is called the coefficient of static friction. The coefficient is a dimen- TABLE 5.1 Coefficients of friction
sionless number that depends on the materials of which the object and the surface Static Kinetic Rolling
are made. Table 5.1 shows some typical values of coefficients of friction. It is to Materials ms mk mr
be emphasized that these are only approximate. The exact value of the coefficient
Rubber on
depends on the roughness, cleanliness, and dryness of the surfaces. concrete 1.00 0.80 0.02
NOTE Equation 5.12 does not say fs 5 msn. The value of fs depends on the Steel on steel
force or forces that static friction has to balance to keep the object from mov- (dry) 0.80 0.60 0.002
ing. It can have any value from 0 up to, but not exceeding, msn. Steel on steel
(lubricated) 0.10 0.05
Wood on wood 0.50 0.20
Kinetic Friction Wood on snow 0.12 0.06
Once the box starts to slide, in Figure 5.11, the static friction force is replaced by Ice on ice 0.10 0.03
r
a kinetic friction force f k. Experiments show that kinetic friction, unlike static
friction, has a nearly constant magnitude. Furthermore, the size of the kinetic fric-
tion force is less than the maximum static friction, f k , fs max, which explains
why it is easier to keep the box moving than it was to start it moving. The direc-
r
tion of f k is always opposite to the direction in which an object slides across the
surface.
The kinetic friction force is also proportional to the magnitude of the normal
force:
Pushing
fk 5 mkn (5.13) force ar
where mk is called the coefficient of kinetic friction. Table 5.1 includes typical Object is
values of mk. You can see that mk , ms, causing the kinetic friction to be less than accelerating.
the maximum static friction.
Kinetic friction is opposite the motion.
Rolling Friction
If you slam on the brakes hard enough, your car tires slide against the road sur- nr
face and leave skid marks. This is kinetic friction. A wheel rolling on a surface r r
fk Fpush
also experiences friction, but not kinetic friction. The portion of the wheel that
contacts the surface is stationary with respect to the surface, not sliding. To see
this, roll a wheel slowly and watch how it touches the ground. r
r
w Fnet
Textbooks draw wheels as circles, but no wheel is perfectly round. The weight
of the wheel, and of any object supported by the wheel, causes the bottom of the FIGURE 5.11 The kinetic friction force is
wheel to flatten where it touches the surface, as Figure 5.12 on the next page opposite the direction of motion.
Molecular bonds shows. The contact area between a car tire and the road is fairly large. The contact
break as the wheel area between a steel locomotive wheel and a steel rail is much less, but it’s not zero.
rolls forward.
Molecular bonds are quickly established where the wheel presses against the
surface. These bonds have to be broken as the wheel rolls forward, and the effort
needed to break them causes rolling friction. (Think how it is to walk with a wad
of chewing gum stuck to the sole of your shoe!) The force of rolling friction can
be calculated in terms of a coefficient of rolling friction mr:
Contact area
r
The wheel flattens where it touches f r 5 (mr n, direction opposite the motion) (5.14)
the surface, giving a contact area
rather than a point of contact. Rolling friction acts very much like kinetic friction, but values of mr (see Table 5.1)
are much less than values of mk. This is why it is easier to roll an object on wheels
FIGURE 5.12 Rolling friction is due to
than to slide it.
the contact area between a wheel and
the surface.
A Model of Friction
These ideas can be summarized in a model of friction:
r
Static: f s # (ms n, direction as necessary to prevent motion)
2.5, 2.6 ONLINE
r
Kinetic: f k 5 (mk n, direction opposite the motion) (5.15)
r
Rolling: f r 5 (mr n, direction opposite the motion)
Here “motion” means “motion relative to the surface.” The maximum value of
static friction fs max 5 ms n occurs at the point where the object slips and begins to
move.
NOTE Equations 5.15 are a “model” of friction, not a “law” of friction.
These equations provide a reasonably accurate, but not perfect, description of
how friction forces act. For example, we’ve ignored the surface area of the
object because surface area has little effect. Likewise, our model assumes that
the kinetic friction force is independent of the object’s speed. This is a fairly
good, but not perfect, approximation. Equations 5.15 are a simplification of
reality that works reasonably well, which is what we mean by a “model.” They
are not a “law of nature” on a level with Newton’s laws.
Figure 5.13 summarizes these ideas graphically by showing how the friction
r
force changes as the magnitude of an applied force Fpush increases.
Static friction Kinetic friction
f The object slips when

fs 5 fs max. The friction
force decreases as the object
begins to move.
fs max 5 sn The kinetic friction force
remains constant once the
object is moving.
fk 5 constant
sh
pu
kn
F
5
f s
Slope 5 1
At first the object doesn’t move,
so the static friction force increases
to match the pushing force. This Fpush
causes the graph to increase with
a slope of 1. At rest Accelerating
r r r r
fs Fpush fk Fpush
r
ar 5 0 ar
FIGURE 5.13 The friction force response to an increasing applied force.

5.4 . Friction 135
STOP TO THINK 5.3 Rank in order, from largest to smallest, the size of the
r r
friction forces f a to f e in these 5 different situations. The box and the floor are
made of the same materials in all situations.
At rest On the verge of slipping Speeding up Constant speed Slowing down
ar r r ar
a50
No Push Push
Push Push
push
r r r r r
fa fb fc fd fe
(a) (b) (c) (d) (e)
EXAMPLE 5.6 How far does a box slide? where we’ve used w 5 mg for the weight. The negative sign
r
Carol pushes a 10.0 kg wood box across a wood floor at a occurs in the first equation because f k points to the left and thus
steady speed of 2.0 m/s. How much force does Carol exert on the component is negative: ( f k ) x 5 2 f k . Similarly, wy 5 2w
the box? If she stops pushing, how far will the box slide before because the weight vector points down. In addition to Newton’s
coming to rest? laws, we also have our model of kinetic friction:
MODEL We model the box as a particle and we describe the fric- f k 5 mkn
tion forces with the model of static and kinetic friction. This is a
two-part problem: first while Carol is pushing the box, then as it Altogether we have three simultaneous equations in the three
slides after she releases it. unknowns Fpush, f k, and n. Fortunately, these equations are
easy to solve. The y-component of Newton’s law tells us that
VISUALIZE This is a fairly complex situation, one that calls for
n 5 mg. We can then find the friction force to be
careful visualization. Figure 5.14 shows the pictorial and physi-
r
cal representations both while Carol pushes, when ar 5 0, and f k 5 mkmg
after she stops. We’ve placed x 5 0 at the point where she stops
Substitute this into the x-component of the first law, giving
pushing because this is the point where the kinematics calcula-
tion for “How far?” will begin. Notice that each part of the Fpush 5 f k 5 mk mg
motion needs its own free-body diagram. The box is moving
until the very instant that the problem ends, so only kinetic fric- 5 (0.20)(10.0 kg)(9.80 m/s2 ) 5 19.6 N
tion is relevant. where m k for wood on wood was taken from Table 5.1. This is
SOLVE We’ll start by finding how hard Carol has to push to how hard Carol pushes to keep the box moving at a steady
keep the box moving at a steady speed. The box is in dynamic speed.
r
equilibrium ( ar 5 0), and Newton’s first law is The box is not in equilibrium after Carol stops pushing it.
Our strategy for the second half of the problem is to use New-
a Fx 5 Fpush 2 f k 5 0 ton’s second law to find the acceleration, then use kinematics to
a Fy 5 n 2 w 5 n 2 mg 5 0 find how far the box moves before stopping. We see from the

Box released Stops
Pushes Releases Stops While pushing After releasing
vr
r
ar 5 0 ar
ax y y
nr nr
r r r
fk Fpush fk
x
0 x x
x0, v0, t0 x1, v1, t1
Known Find r
wr w
x0 5 0 m v0x 5 2 m/s t0 5 0 s Fpush and x1
v1x 5 0 m/s m 5 10 kg k 5 0.20 While pushing After releasing
FIGURE 5.14 Pictorial and physical representations of a box sliding across a floor.
motion diagram that ay 5 0. Newton’s second law, applied to Now we are left with a problem of constant-acceleration
the second free-body diagram of Figure 5.14, is kinematics. We are interested in a distance, rather than a time
interval, so the easiest way to proceed is
a Fx 5 2 f k 5 max
v1x2 5 0 5 v0x2 1 2ax Dx 5 v0x2 1 2ax x1
a Fy 5 n 2 mg 5 may 5 0
from which the distance that the box slides is
We also have our model of friction,
2v0x2 2(2.0 m/s) 2
f k 5 mkn x1 5 5 5 1.02 m
2ax 2(21.96 m/s2 )
We see from the y-component equation that n 5 mg, and thus
We get a positive answer because the two negative signs cancel.
f k 5 mkmg. Using this in the x-component equation gives
ASSESS Carol was pushing at 2 m/s < 4 mph, which is fairly
max 5 2 f k 5 2mkmg fast. The box slides 1.02 m, which is slightly over 3 feet. That
This is easily solved to find the box’s acceleration: sounds reasonable.
NOTE We needed both the horizontal and the vertical com-
ax 5 2mkg 5 2(0.20) (9.80 m/s2 ) 5 21.96 m/s2
ponents of the second law even though the motion was entirely
The acceleration component ax is negative because the accel- horizontal. This need is typical when friction is involved
eration vector ar points to the left, as we see from the motion because we must find the normal force before we can evaluate
diagram. the friction force.
EXAMPLE 5.7 Dumping a file cabinet From the free-body diagram we see that fs has only a negative
A 50.0 kg steel file cabinet is in the back of a dump truck. The x-component and that n has only a positive y-component. The
r r
truck’s bed, also made of steel, is slowly tilted. What is the size weight vector can be written w 5 1w sin u i^ 2 w cos u j^, so w
of the static friction force on the cabinet when the bed is tilted has both x- and y-components in this coordinate system. Thus
20°? At what angle will the file cabinet begin to slide? the first law becomes
MODEL We’ll model the file cabinet as a particle. We’ll also use
a Fx 5 w sin u 2 fs 5 mg sin u 2 fs 5 0
the model of static friction. The file cabinet will slip when the
static friction force reaches its maximum value fs max. a Fy 5 n 2 w cos u 5 n 2 mg cos u 5 0
VISUALIZE Figure 5.15 shows the pictorial and physical repre- where we’ve used w 5 mg. The x-component equation allows
sentations when the truck bed is tilted at angle u. This is a static us to determine the size of the static friction force when
equilibrium problem, so the free-body diagram suffices for a pic- u 5 20°:
torial representation. We can make the analysis easier if we tilt
fs 5 mg sin u 5 (50.0 kg)(9.80 m/s2 ) sin 20°
the coordinate system to match the bed of the truck. To prevent
the file cabinet from slipping, the static friction force must point 5 168 N
up the slope.
This value does not require knowing ms. We simply have to find
SOLVE The file cabinet is in static equilibrium. Newton’s first the size of the friction force that will balance the component of
law is r
w that points down the slope. The coefficient of static friction
(Fnet ) x 5 a Fx 5 nx 1 wx 1 (fs ) x 5 0 only enters when we want to find the angle at which the file
(Fnet ) y 5 a Fy 5 ny 1 wy 1 (fs ) y 5 0
Known nr
s 5 0.80 m 5 50 kg
k 5 0.60
r
Find fs
Normal nr fs where 5 20°
r
Weight wr where cabinet slips
Friction fs x
r
w
FIGURE 5.15 The pictorial and physical representations of a file cabinet in a tilted dump truck.
5.5 . Drag 137
cabinet slips. Slipping occurs when the static friction reaches its The mg in both terms cancels, and we find
maximum value
sin u
fs 5 fs max 5 msn 5 tan u 5 ms
cos u
From the y-component of Newton’s law we see that u 5 tan21 ms 5 tan21 ( 0.80 ) 5 38.7°
n 5 mg cos u. Consequently,
ASSESS Steel doesn’t slide all that well on unlubricated steel, so
fs max 5 msmg cos u a fairly large angle is not surprising. The answer seems reason-
able. It is worth noting that n 5 mg cos u in this example. A
Substituting this into the x-component of the first law gives
common error is to use simply n 5 mg. Be sure to evaluate the
mg sin u 2 msmg cos u 5 0 normal force within the context of each specific problem.
The angle at which slipping begins is called the angle of repose. Figure 5.16
shows that knowing the angle of repose can be very important because it is the
angle at which loose materials (gravel, sand, snow, etc.) begin to slide on a moun-
tainside, leading to landslides and avalanches.
Causes of Friction
It is worth a brief pause to look at the causes of friction. All surfaces, even those
quite smooth to the touch, are very rough on a microscopic scale. When two
objects are placed in contact, they do not make a smooth fit. Instead, as Fig-
ure 5.17 shows, the high points on one surface become jammed against the high
points on the other surface while the low points are not in contact at all. Only a FIGURE 5.16 The angle of repose is the
very small fraction (typically 1024) of the surface area is in actual contact. The angle at which loose materials, such as
amount of contact depends on how hard the surfaces are pushed together, which gravel or snow, begin to slide.
is why friction forces are proportional to n.
At the points of actual contact, the atoms in the two materials are pressed
closely together and molecular bonds are established between them. These bonds
are the “cause” of the static friction force. For an object to slip, you must push it
hard enough to break these molecular bonds between the surfaces. Once they are
broken, and the two surfaces are sliding against each other, there are still attrac-
tive forces between the atoms on the opposing surfaces as the high points of the
materials push past each other. However, the atoms move past each other so
quickly that they do not have time to establish the tight bonds of static friction. Two surfaces
That is why the kinetic friction force is smaller. in contact
Occasionally, in the course of sliding, two high points will be forced together
so closely that they do form a tight bond. As the motion continues, it is not this
surface bond that breaks but weaker bonds at the base of one of the high points.
When this happens, a small piece of the object is left behind “embedded” in the
surface. This is what we call abrasion. Abrasion causes materials to wear out as a
result of friction, be they the piston rings in your car or the seat of your pants. In Very few points
machines, abrasion is minimized with lubrication, a very thin film of liquid are actually
in contact.
between the surfaces that allows them to “float” past each other with many fewer
points in actual contact.
Friction, as seen at the atomic level, is a very complex phenomenon. A detailed
understanding of friction is at the forefront of engineering research today, where
it is especially important for designing highly miniaturized machines and nano-
structures.
Molecular bonds form

5.5 Drag between the two materials.
These bonds have to be
The air exerts a drag force on objects as they move through the air. You experi- broken as the object slides.
ence drag forces every day as you jog, bicycle, ski, or drive your car. The drag FIGURE 5.17 An atomic-level view
force is especially important for the skydiver at the beginning of the chapter. of friction.
r
The drag force D
r
D
■ Is opposite in direction to rv.
■ Increases in magnitude as the object’s speed increases.
Figure 5.18 illustrates the drag force.
Drag is a more complex force than ordinary friction because drag depends on
the object’s speed. At relatively low speeds the drag force is small and can usually
be neglected, but drag plays an important role as speeds increase.
Experimental studies have found that the drag force depends on an object’s
FIGURE 5.18 The drag force on a high- speed in a complicated way. Fortunately, we can use a fairly simple model of drag
speed motorcyclist is significant. if the following three conditions are met:
■ The object’s size (diameter) is between a few millimeters and a few meters.
■ The object’s speed is less than a few hundred meters per second.
■ The object is moving through the air near the earth’s surface.
These conditions are usually satisfied for balls, people, cars, and many other
objects of the everyday world. Under these conditions, the drag force can be written
r
D < ( 14 Av2, direction opposite the motion ) (5.16)
where A is the cross-section area of the object. The size of the drag force is pro-
portional to the square of the object’s speed. This model of drag fails for objects
that are very small (such as dust particles), very fast (such as jet planes), or that
move in other media (such as water). We’ll leave those situations to more
advanced textbooks.
NOTE Let’s look at this model more closely. You may have noticed that an
area multiplied by a speed squared does not give units of force. Unlike the 21 in
r
Dx 5 12 a(Dt) 2, which is a “pure” number, the 41 in the expression for D has
units. This number depends on the air’s density and viscosity, and it’s actually
1 3
4 kg/m . We’ve suppressed the units in Equation 5.16, but doing so gives us an
A falling sphere
expression that works only if A is in m2 and v is in m/s. Equation 5.16 cannot
be converted to other units. And the number is not exactly 14 , which is why
Equation 5.16 has an < sign rather than an 5 sign, but it’s close enough to
r allow Equation 5.16 to be a reasonable yet simple model of drag.
The area in Equation 5.16 is the cross section of the object as it “faces into the
wind.” Figure 5.19 shows how to calculate the cross-section area for objects of
different shape. It’s interesting to note that the magnitude of the drag force, 14 Av 2,
The cross section is A 5 pr 2
an equatorial circle. depends on the object’s size and shape but not on its mass. We will see shortly
that the irrelevance of the mass has important consequences.
A cylinder falling end down A cylinder falling side down A box falling end down
L a
r b
2r
The cross section A 5 2rL

is a rectangle.
A 5 pr 2 A 5 ab
The cross section

is a circle. The cross section
is a rectangle.
FIGURE 5.19 Cross-section areas for objects of different shape.

5.5 . Drag 139
EXAMPLE 5.8 Air resistance compared to rolling Drag due to air

friction resistance Car’s cross-
The profile of a typical 1500 kg passenger car, as seen from section area A
the front, is 1.6 m wide and 1.4 m high. At what speed does
the magnitude of the drag equal the magnitude of the rolling
friction? vr
MODEL Treat the car as a particle. Use the models of rolling Rolling friction
friction and drag. due to road
VISUALIZE Figure 5.20 shows the car and a free-body diagram.
A full pictorial representation is not needed because we won’t y
be doing any kinematics calculations.
nr
SOLVE Drag is less than friction at low speeds, where air resis-
r
tance is negligible. But drag increases as v increases, so there D
will be a speed at which the two forces are equal in size. Above x
r
this speed, drag is more important than rolling friction. fr
The magnitudes of the forces are D < 14 Av2 and fr 5 mrn.
There’s no motion and no acceleration in the vertical direction, wr
so we can see from the free-body diagram that n 5 w 5 mg.
Thus fr 5 mr mg. Equating friction and drag, we have
FIGURE 5.20 A car experiences both
1 2 rolling friction and drag.
Av 5 mr mg
4
Solving for v, we find
4mrmg 4(0.02) (1500 kg) (9.8 m/s2 ) ASSESS 23 m/s is approximately 50 mph, a reasonable result.
Å A Å
v5 5 5 23 m/s This calculation shows that our assumption that we can ignore
(1.4 m) (1.6 m)
air resistance is really quite good for car speeds less than 30 or
where the value of mr for rubber on concrete was taken from 40 mph. Calculations that neglect drag will be increasingly
Table 5.1. inaccurate as speeds go above 50 mph.
Figure 5.21 shows a ball moving up and down vertically. If there were no air
resistance, the ball would be in free fall with afree fall 5 2g throughout its flight.
Let’s see how drag changes this.
Referring to Figure 5.21:
r r
1. The drag force D points down as the ball rises. This increases the net force 3. At the highest point, vr 5 0 so there’s no drag.
on the ball and causes the ball to slow down more quickly than it would in a
a5g
vacuum. The magnitude of the acceleration, which we’ll calculate below, is
a . g.
2. The drag force decreases as the ball slows.
r
3. rv 5 0 at the highest point in the ball’s motion, so there’s no drag and the 2. Drag decreases as 4. Drag increases as
wr
acceleration is simply afree fall 5 2g. the ball slows. the ball speeds up.
r
4. The drag force increases as the ball speeds up. D
r
5. The drag force D points up as the ball falls. This decreases the net force on
the ball and causes the ball to speed up less quickly than it would in a r r r
w D w
vacuum. The magnitude of the acceleration is a , g.
We can use Newton’s second law to find the ball’s acceleration a c as it rises.
1 2
You can see from the forces in Figure 5.21 that r
D
Start
(Fnet ) y 2mg 2 D D
ac 5 5 52 g1 (5.17)
m m m
r
1. Drag adds to wr D r 5. Drag subtracts
The magnitude of a c , which is the ball’s deceleration as it rises, is g 1 D/m. Air w
weight as the from weight as
resistance causes the ball to slow down more quickly than it would in a vacuum. ball rises: the ball falls:
But Equation 5.17 tells us more. Because D depends on the object’s size but not a.g a,g
on its mass, drag has a larger effect (larger acceleration) on a less massive ball FIGURE 5.21 Drag force on a ball moving
than on a more massive ball of the same size. vertically.
A Ping-Pong ball and a golf ball are about the same size, but it’s harder to
throw the Ping-Pong ball than the golf ball. We can now give an explanation:
■ The drag force has the same magnitude for two objects of equal size.
■ According to Newton’s second law, the acceleration (the effect of the force)
depends inversely on the mass.
■ Therefore, the effect of the drag force is larger on a less massive ball than on a
more massive ball of equal size.
1 2
As the ball in Figure 5.21 falls, its acceleration a T is
(Fnet ) y 2mg 1 D D
aT 5 5 52 g2 (5.18)
m m m
The magnitude of a T is g 2 D/m, so the ball speeds up less quickly than it would
in a vacuum. Once again, the effect is larger for a less massive ball than for a
more massive ball of equal size.
Terminal Speed
r
D The drag force increases as an object falls and gains speed. If the object falls far
Terminal speed is
enough, it will eventually reach a speed, shown in Figure 5.22, at which D 5 w.
reached when the That is, the drag force will be equal and opposite to the weight force. The net force
r r
drag exactly balances
r at this speed is Fnet 5 0, so there is no further acceleration and the object falls with
the weight: ar 5 0.
a constant speed. The speed at which the exact balance between the upward drag
force and the downward weight force causes an object to fall without acceleration
r is called the terminal speed vterm. Once an object has reached terminal speed, it
w
will continue falling at that speed until it hits the ground.
FIGURE 5.22 An object falling at terminal It’s not hard to compute the terminal speed. It is the speed, by definition,
speed. at which D 5 w or, equivalently, 14 Av2 < mg. This speed is
4mg
Å A
vterm < (5.19)
A more massive object has a larger terminal speed than a less massive object of
equal size. A 10-cm-diameter lead ball, with a mass of 6 kg, has a terminal speed
of 170 m/s while a 10-cm-diameter Styrofoam ball, with a mass of 50 g, has a ter-
minal speed of only 15 m/s.
A popular use of Equation 5.19 is to find the terminal speed of a skydiver. A
The velocity starts at zero, then skydiver is rather like the cylinder of Figure 5.19 falling “side down.” A typical
becomes increasingly negative skydiver is 1.8 m long and 0.40 m wide (A 5 0.72 m2 ) and has a mass of 75 kg.
(motion in 2y-direction). His terminal speed is
As drag increases with
increasing speed, the slope 4mg 4(75 kg)(9.8 m/s2 )
Å A Å
decreases in magnitude. vterm < 5 5 64 m/s
v 0.72 m2
0 t
This is roughly 140 mph. A higher speed can be reached by falling feet first or
head first, which reduces the area A.
Figure 5.23 shows the results of a more detailed calculation for a falling
2vterm object. Without drag, the velocity graph is a straight line with slope 5 ay 5 2g.
When drag is included, the slope steadily decreases in magnitude and approaches
zero (no further acceleration) as the object reaches terminal speed.
The slope approaches zero
(no further acceleration) as
Although we’ve focused our analysis on objects moving vertically, the same
the object approaches ideas apply to objects moving horizontally. If an object is thrown or shot horizon-
r
terminal speed vterm. tally, D causes the object to slow down. An airplane reaches its maximum speed,
Without drag, the graph is a which is analogous to the terminal speed, when the drag is equal and opposite to
straight line with slope ay 5 2g. the thrust: D 5 Fthrust. The net force is then zero and the plane cannot go any
FIGURE 5.23 The velocity-versus-time
faster. The maximum speed of a passenger jet is about 550 mph.
graph of a falling object with and We will continue to neglect drag unless a problem specifically calls for drag to
without drag. be considered.
5.6 . More Examples of Newton’s Second Law 141
STOP TO THINK 5.4 The terminal speed of a Styrofoam ball is 15 m/s. Suppose
a Styrofoam ball is shot straight down with an initial speed of 30 m/s. Which
velocity graph is correct?
v (m/s) v (m/s) v (m/s) v (m/s) v (m/s)

0 t 0 t 0 t 0 t 0 t
230 230 230 230 230
(a) (b) (c) (d) (e)
5.6 More Examples of Newton’s

Second Law
We will finish this chapter with several additional examples in which we use the
problem-solving strategy in more complex scenarios. ONLINE 2.7, 2.8, 2.9
EXAMPLE 5.9 Stopping distances VISUALIZE Figure 5.24 shows pictorial and physical representa-
A 1500 kg car is traveling at a speed of 30 m/s when the driver tions. We’ve shown the car sliding uphill, but these representa-
slams on the brakes and skids to a halt. Determine the stopping tions work equally well for a level or downhill slide if we let u
distance if the car is traveling up a 10° slope, down a 10° slope, be zero or negative, respectively. We’ve used a tilted coordinate
or on a level road. system so that the motion is along one of the axes. We’ve
MODEL We’ll represent the car as a particle and we’ll use the assumed that the car is traveling to the right, although the prob-
model of kinetic friction. We want to solve the problem only lem didn’t state this. You could equally well make the opposite
once, not three separate times, so we’ll leave the slope angle u assumption, but you would have to be careful with negative val-
unspecified until the end. ues of x. The car skids to a halt, so we’ve taken the coefficient
of kinetic friction for rubber on concrete from Table 5.1.
Pictorial representation
x Known
x0 5 t0 5 0 v0x 5 30 m/s
ax m 5 1500 kg v1x 5 0 m/s
x1, v1x, t1
k 5 0.80
5 210°, 0°, 10°
Find
0 x0, v0x, t0
Dx 5 x1 2 x0 5 x1
This representation works for a downhill
slide if we let be negative.
Physical representation y
Stops
nr
r
x
fk
r
v ar

Normal nr r
r
Fnet
r
Weight w Friction fk r
w
FIGURE 5.24 Pictorial and physical representations of a skidding car.

SOLVE Newton’s second law and the model of kinetic friction which we can solve for the stopping distance x1:
are
v0x2 v0x2
x1 5 2 5
a Fx 5 nx 1 wx 1 ( f k ) x 2ax 2g(sin u 1 mk cos u)
5 2mg sin u 2 f k 5 max Notice how the minus sign in the expression for a canceled the
minus sign in the expression for x1. Evaluating our result at the
a Fy 5 ny 1 wy 1 ( f k ) y
three different angles gives the stopping distances:
5 n 2 mg cos u 5 may 5 0
f k 5 m kn 48 m u 5 10° uphill
x1 5 c57 m u 5 0° level
We’ve written these equations by “reading” the motion diagram 75 m u 5 210° downhill
and the free-body diagram. Notice that both components of the
r
weight vector w are negative. ay 5 0, because the motion is The implications are clear about the danger of driving downhill
entirely along the x-axis. too fast!
The second equation gives n 5 mg cos u. Using this in the
ASSESS 30 m/s < 60 mph and 57 m < 180 feet on a level sur-
friction model, we find f k 5 mkmg cos u. Inserting this result
face. This is similar to the stopping distances you learned when
back into the first equation then gives
you got your driver’s license, so the results seem reasonable.
max 5 2mg sin u 2 mkmg cos u Additional confirmation comes from noting that the expression
for a x becomes 2g sin u if mk 5 0. This is what you learned in
5 2mg(sin u 1 mk cos u) Chapter 3 for the acceleration on a frictionless inclined plane.
ax 5 2g(sin u 1 mk cos u)
This is a constant acceleration. Constant-acceleration kinemat-

ics gives
v1x2 5 0 5 v0x2 1 2ax ( x1 2 x0 ) 5 v0x2 1 2ax x1
This is a good example for pointing out the advantages of working problems
algebraically. If you had started plugging in numbers early, you would not have
found that the mass eventually cancels out and you would have done several
needless calculations. In addition, it is now easy to calculate the stopping dis-
tance for different angles. Had you been computing numbers, rather than alge-
braic expressions, you would have had to go all the way back to the beginning for
each angle.
EXAMPLE 5.10 A dog sled race SOLVE We have enough information to calculate the accelera-
It’s dog sled race day in Alaska! A wooden sled, with rider and tion. We can then use ar to find the tension. We’ll assume that
supplies, has a mass of 200 kg. When the starting gun sounds, it the acceleration is constant during the first 15 m. Then
takes the dogs 15 meters to reach their “cruising speed” of
v1x2 5 v0x2 1 2ax (x1 2 x0 ) 5 2a x x1
5.0 m/s across the snow. Two ropes are attached to the sled, one
on each side of the dogs. The ropes pull upward at 10°. What v1x2 (5.0 m/s) 2
are the tensions in the ropes at the start of the race? ax 5 5 5 0.833 m/s2
2x1 2(15 m)
MODEL We’ll represent the sled as a particle and we’ll use the
model of kinetic friction. We interpret the question as asking for Newton’s second law can be written by “reading” the free-body
the magnitude T of the tension forces. We’ll assume that the diagram:
tensions in the two ropes are equal.
a Fx 5 nx 1 T1x 1 T2x 1 wx 1 ( f k ) x
VISUALIZE Figure 5.25 shows the pictorial and physical repre-
r r
sentations. Notice that the tensions T1 and T2 are tilted up, but 5 2T cos u 2 f k 5 max
the net force is directly to the right in order to match the accel-
eration ar of the motion diagram. a Fy 5 ny 1 T1y 1 T2y 1 wy 1 ( fk ) y
5 n 1 2T sin u 2 mg 5 may 5 0
5.6 . More Examples of Newton’s Second Law 143
Pictorial representation
Known
x0 5 v0x 5 t0 5 0
x1 5 15 m v1x 5 5 m/s
ax 5 10° k 5 0.06
m 5 200 kg
x
0 Find
x0, v0x, t0 x1, v1x, t1 T
Physical representation
y
r r
Tensions T1 and T2
nr
r r
T1 and T2
x
r
fk
r
Weight w Normal nr r
r Fnet
Friction fk r
Starts w
vr
ar
FIGURE 5.25 Pictorial and physical representations of an accelerating dog sled.
Make sure you understand where all the terms come from, Substituting the friction back into the x-equation gives
including their signs. We’ve used w 5 mg and our knowledge
r r 2T cos u 2 (mkmg 2 2mkT sin u)
that ar has only an x-component. The tensions T1 and T2 have
both x- and y-components. The assumption of equal tensions 5 2T(cos u 1 mk sin u) 2 mkmg 5 max
allows us to write T1 5 T2 5 T , and this introduces the factors
of 2. 1 m(ax 1 mkg)
T5
In addition, we have the model of kinetic friction 2 cos u 1 mk sin u
f k 5 m kn Using ax 5 0.833 from above with m 5 200 kg and u 5 10°,

we find that the tension is
From the y-equation and the friction equation,
T 5 143 N
n 5 mg 2 2T sin u
ASSESS It’s a bit hard to assess this result. We do know that the
f k 5 mk n 5 mkmg 2 2mkT sin u
weight of the sled is mg < 2000 N. We also know that the dogs
Notice that n is not equal to mg. The y-components of the can drag a sled over snow (small mk) but probably can’t lift the
tension forces support part of the weight, so the ground does sled straight up, so we anticipate that T V 2000 N. Our calcu-
not press against the bottom of the sled as hard as it would lation agrees.
otherwise.
EXAMPLE 5.11 Make sure the cargo doesn’t slide box. If the box does not slip, then there is no motion of the box
A 100 kg box of dimensions 50 cm 3 50 cm 3 50 cm is in the relative to the truck and the box must accelerate with the truck:
back of a flatbed truck. The coefficients of friction between the abox 5 atruck. As the box accelerates, it must, according to New-
box and the bed of the truck are ms 5 0.4 and mk 5 0.2. What is ton’s second law, have a net force acting on it. But from what?
the maximum acceleration the truck can have without the box Imagine, for a moment, that the truck bed is frictionless. The
slipping? box would slide backwards (as seen in the truck’s reference
MODEL This is a somewhat different problem from any we have
frame) as the truck accelerates. The force that prevents sliding
looked at thus far. Let the box, which we’ll model as a particle, is static friction, so the truck must exert a static friction force on
be the system. It contacts its environment only where it touches the box to “pull” the box along with it and prevent the box from
the truck bed, so only the truck can exert contact forces on the sliding relative to the truck.

y
Known
m 5 100 kg nr r
Box dimensions 50 cm 3 50 cm 3 50 cm fs
s 5 0.4 k 5 0.2 x
r
Find w
r
Normal nr r
Acceleration at which box slips Weight w r Fnet
Static friction fs
FIGURE 5.26 Pictorial and physical representations for the box in a flatbed truck.
VISUALIZE This situation is shown in Figure 5.26. There is only The y-equation of the second law and the friction model com-
r
one horizontal force on the box, f s, and it points in the forward bine to give fs 5 msmg. Substituting this into the x-equation,
direction to accelerate the box. Notice that we’re solving the and noting that ax is now amax, we find
problem with the ground as our reference frame. Newton’s laws
are not valid in the accelerating truck because it is not an iner- fs
amax 5 5 msg 5 3.9 m/s2
tial reference. There are no kinematics in this problem, so a list m
of known information next to the free-body diagram suffices as
The truck must keep its acceleration less than 3.9 m/s2 if slip-
a pictorial representation.
ping is to be avoided.
SOLVE Newton’s second law, which we can “read” from the
ASSESS 3.9 m/s2 is about one-third of g. You may have noticed
free-body diagram, is
that items in a car or truck are likely to tip over when you start
a Fx 5 fs 5 max or stop, but they slide only if you really floor it and accelerate
very quickly. So this answer seems reasonable. Notice that the
a Fy 5 n 2 w 5 n 2 mg 5 may 5 0 dimensions of the crate were not needed. Real-world situations
rarely have exactly the information you need, no more and no
Now, static friction, you will recall, can be any value between 0
less. Many problems in this textbook will require you to assess
and fs max. If the truck accelerates slowly, so that the box doesn’t
the information in the problem statement in order to learn
slip, then fs , fs max. However, we’re interested in the accelera-
which is relevant to the solution.
tion amax at which the box begins to slip. This is the acceleration
at which fs reaches its maximum possible value
fs 5 fs max 5 ms n
The mathematical representation of this last example was quite straightfor-

ward. The challenge was in the analysis that preceded the mathematics—that is,
in the physics of the problem rather than the mathematics. It is here that our
analysis tools—motion diagrams, force identification, and free-body diagrams—
prove their value.
Summary 145
SUMMARY
The goal of Chapter 5 has been to learn how to solve problems about motion in a straight line.
GENERAL STRATEGY
All examples in this chapter follow a four-part strategy. You’ll become a better problem solver if you adhere to it as you do the homework
problems. The Dynamics Worksheets will help you structure your work in this way.
Equilibrium Problems Dynamics Problems

Object at rest or moving with constant velocity. Object accelerating.
MODEL Make simplifying assumptions. MODEL Make simplifying assumptions.
VISUALIZE VISUALIZE
Physical representation: Pictorial representation:
Forces and free-body diagram Go back and forth Sketch to define situation.
Pictorial representation: between represen- Translate words to symbols.
Translate words to symbols. tations as needed. Physical representation:
Forces and free-body diagram
SOLVE Use Newton’s first law
r r r
SOLVE Use Newton’s second law
F net 5 a F i 5 0 r r
i Fnet 5 a Fi 5 mar
i
“Read” the vectors from the free-body diagram.
“Read” the vectors from the free-body diagram.
ASSESS Is the result reasonable?
Use kinematics to find velocities and positions.
ASSESS Is the result reasonable?
IMPORTANT CONCEPTS
Specific information about three important forces: Newton’s laws are vector expressions. You must
r write them out by components:
Weight w 5 (mg, downwards)
r (Fnet ) x 5 a Fx 5 max or 0
Friction f s 5 (0 to ms n, direction as necessary to prevent motion)
r
f k 5 (mk n, direction opposite the motion) (Fnet ) y 5 a Fy 5 may or 0
r
f r 5 (mr n, direction opposite the motion)
r
Drag D < ( 14 Av2, direction opposite the motion)
APPLICATIONS
Apparent weight is the magnitude of the contact force supporting an object. 4mg
Å A
It is what a scale would read, and it is your sensation of weight. It equals Terminal speed is vterm <
1 2
your true weight w 5 mg only when a 5 0.
ay
wapp 5 w 1 1
g
TERMS AND NOTATION

apparent weight, wapp coefficient of kinetic friction, mk coefficient of rolling friction, mr
weightless rolling friction terminal speed, vterm
coefficient of static friction, ms
EXERCISES AND PROBLEMS

The icon indicates that the problem can be done on a Dynamics 6. In each of the two free-body diagrams, the forces are acting on
Worksheet. a 2.0 kg object. For each diagram, find the values of ax and ay,
the x- and y-components of the acceleration.
(a) (b)
Exercises y y
2.82 N x
Section 5.1 Equilibrium 1.414 N
2N
1. The three ropes in the figure are tied to a small, very light 1N 5N 3N
ring. Two of the ropes are anchored to walls at right angles, 20° 15° x
and the third rope pulls as shown. What are T1 and T2, the 2N
3N
magnitudes of the tension forces in the first two ropes?
FIGURE EX5.6
7. Figure Ex5.7 shows the velocity graph of a 2.0 kg object as it

Rope 2
moves along the x-axis. What is the net force acting on this
object at t 5 1 s? At 4 s? At 7 s?
Rope 1 30° vx (m/s) Fx (N)
12 4
FIGURE EX5.1 100 N
2
0 t (s) 0 t (s)
2. The three ropes in the figure are tied to a small, very light 0 2 4 6 8 2 4 6
22
ring. Two of these ropes are anchored to walls at right angles
with the tensions shown in the figure. What are the magnitude
r FIGURE EX5.7 FIGURE EX. 5.8
and direction of the tension T3 in the third rope?
8. Figure Ex5.8 shows the force acting on a 2.0 kg object as it
moves along the x-axis. The object is at rest at the origin at
0.6 m t 5 0 s. What are its acceleration and velocity at t 5 6 s?
T2 5 80 N 9. A horizontal rope is tied to a 50 kg box on frictionless ice.
0.8 m What is the tension in the rope if:
T1 5 50 N a. The box is at rest?
b. The box moves at a steady 5.0 m/s?
r
T3 c. The box has vx 5 5.0 m/s and ax 5 5.0 m/s2?
FIGURE EX5.2
10. A 50 kg box hangs from a rope. What is the tension in the
3. A 20 kg loudspeaker is suspended 2.0 m below the ceiling by rope if:
two 3.0-m-long cables that angle outward at equal angles. a. The box is at rest?
What is the tension in the cables? b. The box moves up a steady 5.0 m/s?
4. A football coach sits on a sled while two of his players build c. The box has vy 5 5.0 m/s and is speeding up at 5.0 m/s2?
their strength by dragging the sled across the field with ropes. d. The box has vy 5 5.0 m/s and is slowing down at
The friction force on the sled is 1000 N and the angle between 5.0 m/s2?
the two ropes is 20°. How hard must each player pull to drag
the coach at a steady 2.0 m/s?
Section 5.3 Mass and Weight
11. An astronaut’s weight on earth is 800 N. What is his weight on
Section 5.2 Using Newton’s Second Law Mars, where g 5 3.76 m/s2?
5. In each of the two free-body diagrams, the forces are acting 12. A woman has a mass of 55 kg.
on a 2.0 kg object. For each diagram, find the values of ax and a. What is her weight on earth?
ay, the x- and y-components of the acceleration. b. What are her mass and her weight on the moon, where
g 5 1.62 m/s2?
(a) (b)
y y 13. It takes the elevator in a skyscraper 4.0 s to reach its cruising
speed of 10 m/s. A 60 kg passenger gets aboard on the ground
floor. What is the passenger’s apparent weight
3N 3N a. Before the elevator starts moving?
2N
4N
x x b. While the elevator is speeding up?
2N 4N
1N c. After the elevator reaches its cruising speed?
3N 2N
14. Figure Ex5.14 shows the velocity graph of a 75 kg passenger
in an elevator. What is the passenger’s apparent weight at
t 5 1 s? At 5 s? At 9 s?
FIGURE EX5.5
vy (m/s)
27. A 1000 kg steel beam is supported by two ropes. Each rope
has a maximum sustained tension of 6000 N. Does either rope
8
break? If so, which one(s)?
4
0 t (s)
FIGURE EX5.14 0 2 4 6 8 10
Rope 1 20° 30° Rope 2
Section 5.4 Friction

15. Bonnie and Clyde are sliding a 300 kg bank safe across the FIGURE P5.27
floor to their getaway car. The safe slides with a constant
speed if Clyde pushes from behind with 385 N of force while 28. A 500 kg piano is being lowered into position by a crane while
Bonnie pulls forward on a rope with 350 N of force. What is two people steady it with ropes pulling to the sides. Bob’s
the safe’s coefficient of kinetic friction on the bank floor? rope pulls to the left, 15° below horizontal, with 500 N of ten-
16. A 4000 kg truck is parked on a 15° slope. How big is the fric- sion. Ellen’s rope pulls toward the right, 25° below horizontal.
tion force on the truck? a. What tension must Ellen maintain in her rope to keep the
17. A 1000 kg car traveling at a speed of 40 m/s skids to a halt on piano descending at a steady speed?
wet concrete where mk 5 0.6. How long are the skid marks? b. What is the tension in the main cable supporting the piano?
18. An Airbus A320 jetliner has a takeoff mass of 75,000 kg. It 29. In an electricity experiment, a 1.0 g plastic ball is suspended
reaches its takeoff speed of 82 m/s (180 mph) in 35 s. What is on a 60-cm-long string and given an electric charge. A
the thrust of the engines? You can neglect air resistance but charged rod brought near the ball exerts a horizontal electrical
r
not rolling friction. force Felec on it, causing the ball to swing out to a 20° angle
19. A 50,000 kg locomotive is traveling at 10 m/s when its engine and remain there.
r
and brakes both fail. How far will the locomotive roll before it a. What is the magnitude of Felec?
comes to a stop? b. What is the tension in the string?
20. A stubborn, 120 kg mule sits down and refuses to move. To drag 30. Henry gets into an elevator on the 50th floor of a building and
the mule to the barn, the exasperated farmer ties a rope around it begins moving at t 5 0 s. The figure shows his apparent
the mule and pulls with his maximum force of 800 N. The coef- weight over the next 12 s. wapp (N)
ficients of friction between the mule and the ground are a. Is the elevator’s initial
ms 5 0.8 and mk 5 0.5. Is the farmer able to move the mule? direction up or down? 900
750
Explain how you can tell. 600
b. What is Henry’s mass? 0 t (s)
0 2 4 6 8 10 12
Section 5.5 Drag c. How far has Henry trav-
eled at t 5 12 s? FIGURE P5.30
21. A 75 kg skydiver can be modeled as a rectangular “box” with
31. Zach, whose mass is 80 kg, is in an elevator descending at
dimensions 20 cm 3 40 cm 3 180 cm. What is his terminal
10 m/s. The elevator takes 3.0 s to brake to a stop at the first
speed if he falls feet first?
floor.
22. A 22-cm-diameter bowling ball has a terminal speed of
a. What is Zach’s apparent weight before the elevator starts
85 m/s. What is the ball’s mass?
braking?
b. What is Zach’s apparent weight while the elevator is
braking?
Problems
32. You’ve always wondered about the acceleration of the eleva-
tors in the 101-story-tall Empire State Building. One day,
23. Use information that you can find in Chapters 4 and 5 to esti-
while visiting New York, you take your bathroom scale into the
mate the acceleration of a car.
elevator and stand on them. The scales read 150 lb as the door
24. Use information that you can find in Chapters 4 and 5 to esti-
closes. The reading varies between 120 lb and 170 lb as the ele-
mate the size of the friction force on a baseball player sliding
vator travels 101 floors. What conclusions can you draw?
into second base.
33. An accident victim with a broken leg is being placed in trac-
25. A 5.0 kg object initially at rest at the origin is subjected to the
tion. The patient wears a spe-
time-varying force shown in Figure P5.25. What is the
cial boot with a pulley attached
object’s velocity at t 5 6 s?
to the sole. The foot and boot
Fx (N) Fx (N) together have a mass of 4.0 kg,
10 6 and the doctor has decided to
5 3 hang a 6.0 kg mass from the
0 t (s) 0 t (s) rope. The boot is held sus-

0 2 4 6 0 1 2 3 4 pended by the ropes and does 4 kg 15°
FIGURE P5.25 FIGURE P5.26 not touch the bed.

a. Determine the amount of
26. A 2.0 kg object initially at rest at the origin is subjected to the tension in the rope by using 6 kg
time-varying force shown in Figure P5.26. What is the Newton’s laws to analyze
object’s velocity at t 5 4 s? the hanging mass. FIGURE P5.33
b. The net traction force needs to pull straight out on the leg. 42. A 10 kg crate is placed on a horizontal conveyor belt. The
What is the proper angle u for the upper rope? materials are such that ms 5 0.5 and mk 5 0.3.
c. What is the net traction force pulling on the leg? a. Draw a free-body diagram showing all the forces on the
Hint: If the pulleys are frictionless, which we will assume, crate if the conveyer belt runs at constant speed.
the tension in the rope is constant from one end to the other. b. Draw a free-body diagram showing all the forces on the
34. Seat belts and air bags save lives by reducing the forces crate if the conveyer belt is speeding up.
exerted on the driver and passengers in an automobile colli- c. What is the maximum acceleration the belt can have with-
sion. Cars are designed with a “crumple zone” in the front of out the crate slipping?
the car. In the event of an impact, the passenger compartment 43. A baggage handler drops your 10 kg suitcase onto a conveyor
decelerates over a distance of about 1 m as the front of the car belt running at 2.0 m/s. The materials are such that ms 5 0.5
crumples. An occupant restrained by seat belts and air bags and mk 5 0.3. How far is your suitcase dragged before it is
decelerates with the car. By contrast, an unrestrained occupant riding smoothly on the belt?
keeps moving forward with no loss of speed (Newton’s first 44. Johnny jumps off a swing, lands sitting down on a grassy 20°
law!) until hitting the dashboard or windshield. These are slope, and slides 3.5 m down the slope before stopping. The
unyielding surfaces, and the unfortunate occupant then decel- coefficient of kinetic friction between grass and the seat of
erates over a distance of only about 5 mm. Johnny’s pants is 0.5. What was his initial speed on the grass?
a. A 60 kg person is in a head-on collision. The car’s speed at 45. A 2.0 kg wood block is launched up a wooden ramp that is
impact is 15 m/s. Estimate the net force on the person if he inclined at a 30° angle. The block’s initial speed is 10 m/s.
or she is wearing a seat belt and if the air bag deploys. a. What vertical height does the block reach above its start-
b. Estimate the net force that ultimately stops the person if he ing point?
or she is not restrained by a seat belt or air bag. b. What speed does it have when it slides back down to its
c. How do these two forces compare to the person’s weight? starting point?
35. A rifle with a barrel length of 60 cm fires a 10 g bullet with a 46. It’s moving day, and you need to push a 100 kg box up a 20°
horizontal speed of 400 m/s. The bullet strikes a block of ramp into the truck. The coefficients of friction for the box on
wood and penetrates to a depth of 12 cm. the ramp are ms 5 0.9 and mk 5 0.6. Your largest pushing
a. What frictional force (assumed to be constant) does the force is 1000 N. Can you get the box into the truck without
wood exert on the bullet? assistance if you get a running start at the ramp? If you stop on
b. How long does it take the bullet to come to rest? the ramp, will you be able to get the box moving again?
c. Draw a velocity-versus-time graph for the bullet in the wood. 47. It’s a snowy day and you’re pulling a friend along a level road
36. Compressed air is used to fire a 50 g ball vertically upward on a sled. You’ve both been taking physics, so she asks what
from a 1.0-m-tall tube. The air exerts an upward force of 2.0 N you think the coefficient of friction between the sled and the
on the ball as long as it is in the tube. How high does the ball snow is. You’ve been walking at a steady 1.5 m/s, and the
go above the top of the tube? rope pulls up on the sled at a 30° angle. You estimate that the
37. What thrust does a 200 g model rocket need in order to have a mass of the sled, with your friend on it, is 60 kg and that you’re
vertical acceleration of 10 m/s2 pulling with a force of 75 N. What answer will you give?
a. On Earth? 48. A horizontal rope pulls a 10 kg wood sled across frictionless
b. On the moon, where g 5 1.62 m/s2? snow. A 5.0 kg wood box rides on the sled. What is the largest
38. A 20,000 kg rocket has a rocket motor that generates tension force for which the box doesn’t slip?
3.0 3 105 N of thrust. 49. A pickup truck with a steel bed is carrying a steel file cabinet.
a. What is the rocket’s initial upward acceleration? If the truck’s speed is 15 m/s, what is shortest distance in
b. At an altitude of 5000 m the rocket’s acceleration has which it can stop without the file cabinet sliding?
increased to 6.0 m/s2. What mass of fuel has it burned? 50. You’re driving along at 25 m/s with your aunt’s valuable
39. A 2.0 kg steel block is at rest on a steel table. A horizontal antiques in the back of your pickup truck when suddenly you
string pulls on the block. see a giant hole in the road 55 m ahead of you. Fortunately,
a. What is the minimum string tension needed to move the your foot is right beside the brake and your reaction time is
block? zero! Will the antiques be as fortunate?
b. If the string tension is 20 N, what is the block’s speed after a. Can you stop the truck before it falls into the hole?
moving 1.0 m? b. If your answer to part a is yes, can you stop without the
c. If the string tension is 20 N and the table is coated with oil, antiques sliding and being damaged? Their coefficients of
what is the block’s speed after moving 1.0 m? friction are ms 5 0.6 and mk 5 0.3.
40. Sam, whose mass is 75 kg, takes off across level snow on his Hint: You’re not trying to stop in the shortest possible distance.
jet-powered skis. The skis have a thrust of 200 N and a coeffi- What’s your best strategy for avoiding damage to the antiques?
cient of kinetic friction on snow of 0.1. Unfortunately, the skis 51. A 2.0 kg wood box slides down a vertical wood wall while you
run out of fuel after only 10 s. push on it at a 45° angle. What magnitude of force should you
a. What is Sam’s top speed? apply to cause the box to slide down at a constant speed?
b. How far has Sam traveled when he finally coasts to a stop?
41. Sam, whose mass is 75 kg, takes off down a 50-m-high, 10°
slope on his jet-powered skis. The skis have a thrust of 200 N. 2 kg
Sam’s speed at the bottom is 40 m/s. What is the coefficient of r
kinetic friction of his skis on snow? Fpush
45°
FIGURE P5.51
52. A 1.0 kg wood block is pressed against a vertical wood wall 59. A machine has an 800 g steel
by the 12 N force shown. If the block is initially at rest, will it shuttle that is pulled along a
move upward, move downward, or stay at rest? square steel rail by an elastic
Elastic cord
cord. The shuttle is released
when the elastic cord has 20 N 45°
tension at a 45° angle. What is
1 kg Fixed steel rail
the initial acceleration of the
12 N shuttle? FIGURE P5.59
30°
FIGURE P5.52
60. A 1.0 kg ball hangs from the ceiling of a truck by a 1.0-m-long
string. The back of the truck, where you are riding with the
53. What is the terminal speed for an 80 kg skier going down a ball, has no windows and has been completely soundproofed.
40° snow-covered slope on wooden skis? Assume that the The truck travels along an exceedingly smooth test track, and
skier is 1.8 m tall and 0.40 m wide. you feel no bumps or bounces as it moves. Your only instru-
54. A 10 g Ping-Pong ball has a diameter of 3.5 cm. ments are a meter stick, a protractor, and a stopwatch.
a. The ball is shot straight up at twice its terminal speed. a. The driver tells you, over a loudspeaker, that the truck is
What is its initial acceleration? either at rest, or it is moving forward at a steady speed of
b. The ball is shot straight down at twice its terminal speed. 5 m/s. Can you determine which it is? If so, how? If not,
What is its initial acceleration? why not?
c. The ball is shot straight down at twice its terminal speed. b. Next, the driver tells you that the truck is either moving
Draw a plausible velocity-versus-time graph. forward with a steady speed of 5 m/s, or it is accelerating
55. Try this! Hold your right hand out with your palm perpendicu- at 5 m/s2. Can you determine which it is? If so, how? If
lar to the ground, as if you were getting ready to shake hands. not, why not?
You can’t hold anything in your palm this way because it c. Suppose the truck has been accelerating forward at 5 m/s2
would fall straight down. Use your left hand to hold a small long enough for the ball to achieve a steady position. Does
object, such as a ball or a coin, against your outstretched the ball have an acceleration? If so, what are the magni-
palm, then let go as you quickly swing your hand to the left tude and direction of the ball’s acceleration?
across your body, parallel to the ground. You’ll find that the d. Draw a free-body diagram that shows all forces acting on
object stays against your palm; it doesn’t slip or fall. the ball as the truck accelerates.
a. Is the condition for keeping the object against your palm e. Suppose the ball makes a 10° angle with the vertical. If
one of maintaining a certain minimum velocity vmin? Or possible, determine the truck’s velocity. If possible, deter-
one of maintaining a certain minimum acceleration amin? mine the truck’s acceleration.
Explain. 61. Imagine hanging from a big spring scale as it moves vertically
b. Suppose the object’s mass is 50 g, with ms 5 0.8 and with acceleration a. Show that Equation 5.10 is the correct
mk 5 0.4. Determine either vmin or amin, whichever you expression for your apparent weight.
answered in part a.
56. Suppose you use your hand to push a ball straight down Problems 62 through 65 show a free-body diagram. For each:
toward the floor. The ball’s weight is 1.0 N. a. Write a realistic dynamics problem for which this is the cor-
a. Draw the ball’s free-body diagram while you’re pushing it. rect free-body diagram. Your problem should ask a question
b. Is Fnet larger than, smaller than, or equal to the ball’s that can be answered with a value of position or velocity (such
weight w? Explain. as “How far?” or “How fast?”), and should give sufficient
c. Find an expression for the ball’s apparent weight when its information to allow a solution.
acceleration is ay. Evaluate wapp for ay equal to 2g, 21.5g, b. Solve your problem!
and 22g. 62. 63. y
y
57. You’ve been called in to investigate a construction accident in 14,500 N
which the cable broke while a crane was lifting a 4500 kg con-
9.8 N
tainer. The steel cable is 2.0 cm in diameter and has a safety x
4.9 N 12,000 N
rating of 50,000 N. The crane is designed not to exceed speeds x
20 N
of 3.0 m/s or accelerations of 1.0 m/s2, and your tests find 15°
9.8 N
that the crane is not defective. What is your conclusion? Did
the crane operator recklessly lift too heavy a load? Or was the 15,000 N
cable defective?
FIGURE P5.62 FIGURE P5.63
58. An artist friend of yours needs help hanging a 500 lb sculp-
ture from the ceiling. For artistic reasons, she wants to use 64. y 65. y
just two ropes. One will be 30° from vertical, the other 60°. 19.8 N
564 N
She needs you to determine the smallest diameter rope that
x
can safely support this expensive piece of art. On a visit to 34 N
4.9 N
the hardware store you find that rope is sold in increments x
100 N
30° 20°
of 18-inch diameter and that the safety rating is 4000 pounds
9.8 N 20 N
per square inch of cross section. What size rope should
you buy?
600 N
FIGURE P5.64 FIGURE P5.65
In Problems 66 through 69 you are given the dynamics equations a wax that’s too sticky will cause you to stop and be disquali-
that are used to solve a problem. For each of these, you are to fied. You know that a strong headwind will apply a 50 N hori-
a. Write a realistic problem for which these are the correct zontal force against you as you ski, and you know that your
equations. mass is 75 kg. Which wax do you choose?
b. Draw the free-body diagram and the pictorial representation 72. A testing laboratory wants
for your problem. to determine if a new wid- ay (m/s2)
c. Finish the solution of the problem. get can withstand large 19.6
66. 20.8n 5 (1500 kg)ax accelerations and decelera- 9.8

n 2 (1500 kg) (9.80 m/s2 ) 5 0 tions. To find out, they glue
0 t
2 a 5.0 kg widget to a test
67. T 2 0.2n 2 (20 kg) (9.80 m/s ) sin 20° 1s
2
stand that will drive it ver- 29.8
5 (20 kg) (2 m/s ) tically up and down. The
n 2 (20 kg) (9.80 m/s2 ) cos 20° 5 0 219.6
graph shows its accelera-
68. (100 N) cos 30° 2 fk 5 (20 kg)ax tion during the first second, FIGURE CP5.72
n 1 (100 N) sin 30° 2 (20 kg) (9.80 m/s2 ) 5 0 starting from rest.
a. Identify the forces acting on the widget and draw a free-
f k 5 0.2n
2 body diagram.
69. 2 f k 1 (20 kg) (9.80 m/s ) sin u 5 0 b. Determine the value of ny, the y-component of the normal
2
n 2 (20 kg) (9.80 m/s ) cos u 5 0 force acting on the widget, during the first second of motion.
f k 5 0.2n Give your answer as a graph of ny-versus-t.
c. Your answer to part b should show an interval of time dur-
ing which ny is negative. How can this be? Explain what it
Challenge Problems means physically for ny to be negative.
d. At what time is the apparent weight of the widget a maxi-
70. The figure shows an accelerometer, a device for measuring mum? What is the acceleration at this time?
the horizontal acceleration of cars and airplanes. A ball is e. Is the apparent weight of the widget ever zero? If so, at
free to roll on a parabolic what instant of time does this happen? What is the acceler-
y 5 x2
track described by the equation ation at that time?
y 5 x2, where both x and y f. Suppose the technician forgets to glue the widget to the
are in meters. A scale along the test stand. Will the widget remain on the test stand
bottom is used to measure the throughout the first second, or will it fly off the stand at
ball’s horizontal position x. some instant of time? If so, at what time will this occur?
a. Find an expression that 73. An object with cross section A is shot horizontally across fric-
allows you to use a mea- 2.5 m 0m .5 m tionless ice. Its initial velocity is v0x at t0 5 0 s. Air resistance
sured position x (in m) to FIGURE CP5.70 is not negligible.
compute the acceleration a x a. Show that the velocity at time t is given by the expression
(in m/s2). (For example, a x 5 3x is a possible expression.) v0x
b. What is the acceleration if x 5 20 cm? vx 5
1 1 Av0x t/4m
71. You’ve entered a “slow ski race” where the winner is the skier
who takes the longest time to go down a 15° slope without b. A 1.6 m wide, 1.4 m high, 1500 kg car hits a very slick
ever stopping. You need to choose the best wax to apply to patch of ice while going 20 m/s. If friction is neglected,
your skis. Red wax has a coefficient of kinetic friction 0.25, how long will it take until the car’s speed drops to 10 m/s?
yellow is 0.20, green is 0.15, and blue is 0.10. Having just fin- To 5 m/s?
ished taking physics, you realize that a wax too slippery will c. Assess whether or not it is reasonable to neglect kinetic
cause you to accelerate down the slope and lose the race. But friction.
STOP TO THINK ANSWERS
Stop to Think 5.1: a. The lander is descending and slowing. The acceleration. Kinetic friction is smaller than the maximum static
r
acceleration vector points upward, and so Fnet points upward. This friction that is exerted in b. fa 5 0 because no friction is needed to
can be true only if the thrust has a larger magnitude than the weight. keep the object at rest.
Stop to Think 5.2: a. You are descending and slowing, so your Stop to Think 5.4: d. The ball is shot down at 30 m/s, so
acceleration vector points upward and there is a net upward force v0y 5 230 m/s. This exceeds the terminal speed, so the upward
on you. The floor pushes up against your feet harder than gravity drag force is larger than the downward weight force. Thus the ball
pulls down. slows down even though it is “falling.” It will slow until
vy 5 215 m/s, the terminal velocity, then maintain that velocity.
Stop to Think 5.3: fb + fc 5 fd 5 fe + fa . Situations c, d, and e
are all kinetic friction, which does not depend on either velocity or

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4 Force and Motion

A force is a push or a pull.

A force acts on an object.

A force requires an agent.

A force can be either a contact force . . .

Let’s summarize these ideas as our definition of force:

TACTICS BOX 4.1 Drawing force vectors

1 Represent the object as a particle.

2 Place the tail of the force vector

3 Draw the force vector as an arrow pointing

The rope is the agent. The spring is the agent.

Pulling force of rope Pushing force of spring

(a) Combining Forces

FIGURE 4.2 Two forces applied to a box.

4.2 A Short Catalog of Forces

A compressed spring exerts A stretched spring exerts

FIGURE 4.4 The spring force.

Tension Force FIGURE 4.5 Tension.

Static friction acts

Force Notation Electric and Magnetic Forces

4.3 Identifying Forces

TACTICS BOX 4.2 Identifying forces

EXAMPLE 4.1 Forces on a bungee jumper

4 Locate the points where the environment touches

3 Draw a closed curve around the system.

6 Name and label long-range forces. Weight

EXAMPLE 4.2 Forces on a skier

1 Identify the system. Here

5 Name and label each contact force. The rope

6 Name and label long-range forces. Weight

FIGURE 4.14 Forces on a skier.

EXAMPLE 4.3 Forces on a rocket Air r

a. Gravity, acting downward.

4.4 What Do Forces Do? A Virtual Experiment

Maintain constant stretch.

a. 1 m/s2. b. 2 m/s2. c. 4 m/s2. d. 8 m/s2. e. 16 m/s2.

4.5 Newton’s Second Law

Thus we’ve found that a force of magnitude F causes an object of mass m to

4.6 Newton’s First Law

What Good Is Newton’s First Law?

Inertial Reference Frames

Common Misconceptions About Force

4.7 Free-Body Diagrams

TACTICS BOX 4.3 Drawing a free-body diagram

EXAMPLE 4.4 An elevator accelerates upward

Force identification Free-body diagram 2 Draw a coordinate

FIGURE 4.23 Free-body diagram of an elevator accelerating upward.

Motion diagram Force identification Free-body diagram

STOP TO THINK 4.5 An elevator suspended by a cable is moving upward and

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

TERMS AND NOTATION

EXERCISES AND PROBLEMS

Section 4.3 Identifying Forces 0 F (N)

1 b. The propulsion force of a bicycle.

7. Write a one-paragraph essay on the topic “Force and Motion.”

15. y 16. y 24. A force with x-component Fx acts on a 2.0 kg object as it

nr 25. A force with x-component Fx acts on a 500 g object as it

22. Redraw the two motion diagrams shown vr 3

27. A force with x-component Fx acts on a 500 g object as it