16.

17.

18.

19.

Motion in Two Dimensions

building does the ball strike the ground? (b) Find the
height from which the ball was thrown. (c) How long
does it take the ball to reach a point 10.0 m below the
level of launching?
A landscape architect is planning an artificial waterfall in
a city park. Water flowing at 1.70 m/s will leave the end of
a horizontal channel at the top of a vertical wall 2.35 m
high, and from there the water falls into a pool. (a) Will
the space behind the waterfall be wide enough for a
pedestrian walkway? (b) To sell her plan to the city council, the architect wants to build a model to standard scale,
one-twelfth actual size. How fast should the water flow in
the channel in the model?
ᮡ A placekicker must kick a football from a point 36.0 m
(about 40 yards) from the goal, and half the crowd hopes
the ball will clear the crossbar, which is 3.05 m high.
When kicked, the ball leaves the ground with a speed of
20.0 m/s at an angle of 53.0° to the horizontal. (a) By
how much does the ball clear or fall short of clearing the
crossbar? (b) Does the ball approach the crossbar while
still rising or while falling?
A dive-bomber has a velocity of 280 m/s at an angle u

Chapter 4

Bill Lee/Dembinsky Photo Associates

94

(a)

(b)

3 = challenging;

Ⅺ = SSM/SG;

x f ϭ 0 ϩ 111.2 m>s 2 1cos 18.5°2 t
0.360 m ϭ

0.840 m ϩ 111.2 m>s 2 1sin 18.5°2t Ϫ 12 19.80 m>s2 2t 2

where t is the time at which the athlete lands after taking
off at t ϭ 0. Identify (a) his vector position and (b) his
vector velocity at the takeoff point. (c) The world longjump record is 8.95 m. How far did the athlete jump in
this problem? (d) Describe the shape of the trajectory of
his center of mass.
23. A fireworks rocket explodes at height h, the peak of its
vertical trajectory. It throws out burning fragments in all
directions, but all at the same speed v. Pellets of solidified
metal fall to the ground without air resistance. Find the
smallest angle that the final velocity of an impacting fragment makes with the horizontal.


of mass, as we will study in Chapter 9. The components of
the position of an athlete’s center of mass from the beginning to the end of a certain jump are described by the
two equations

ᮡ

= ThomsonNOW;

Ⅵ = symbolic reasoning;

ⅷ = qualitative reasoning


Problems

95

speed in a horizontal circle. Determine the rotation rate,
in revolutions per second, required to give an astronaut a
centripetal acceleration of 3.00g while in circular motion
with radius 9.45 m.
27. Young David who slew Goliath experimented with slings
before tackling the giant. He found he could revolve a
sling of length 0.600 m at the rate of 8.00 rev/s. If he
increased the length to 0.900 m, he could revolve the
sling only 6.00 times per second. (a) Which rate of rotation gives the greater speed for the stone at the end of
the sling? (b) What is the centripetal acceleration of the
stone at 8.00 rev/s? (c) What is the centripetal acceleration at 6.00 rev/s?

Section 4.6 Relative Velocity and Relative Acceleration


35. A river has a steady speed of 0.500 m/s. A student swims
upstream a distance of 1.00 km and swims back to the starting point. If the student can swim at a speed of 1.20 m/s
in still water, how long does the trip take? Compare this
answer with the time interval required for the trip if the
water were still.

a ϭ 15.0 m/s2
v
2.50 m

30.0Њ

a

Figure P4.31

32. A race car starts from rest on a circular track. The car
increases its speed at a constant rate at as it goes once
around the track. Find the angle that the total acceleration of the car makes—with the radius connecting the
center of the track and the car—at the moment the car
completes the circle.

2 = intermediate;

3 = challenging;

Ⅺ = SSM/SG;

ᮡ

shape of the can’s trajectory as seen by the boy? (d) An
observer on the ground watches the boy throw the can
and catch it. In this observer’s ground frame of reference,
describe the shape of the can’s path and determine the
initial velocity of the can.
39. A science student is riding on a flatcar of a train traveling
along a straight horizontal track at a constant speed of
10.0 m/s. The student throws a ball into the air along a
path that he judges to make an initial angle of 60.0° with
the horizontal and to be in line with the track. The student’s professor, who is standing on the ground nearby,
observes the ball to rise vertically. How high does she see
the ball rise?
40. ⅷ A bolt drops from the ceiling of a moving train car that
is accelerating northward at a rate of 2.50 m/s2. (a) What
is the acceleration of the bolt relative to the train car?
(b) What is the acceleration of the bolt relative to the
Earth? (c) Describe the trajectory of the bolt as seen by

= ThomsonNOW;

Ⅵ = symbolic reasoning;

ⅷ = qualitative reasoning


Chapter 4

Motion in Two Dimensions

an observer inside the train car. (d) Describe the trajectory of the bolt as seen by an observer fixed on the Earth.

the basketball hits the rim of the basket, 3.05 m above the
floor. It bounces straight up with one-half the speed with
which it hit the rim. What height above the floor does the
basketball reach on this bounce?
44. ⅷ (a) An athlete throws a basketball toward the east, with
initial speed 10.6 m/s at an angle of 55.0° above the horizontal. Just as the basketball reaches the highest point of
its trajectory, it hits an eagle (the mascot of the opposing
team) flying horizontally west. The ball bounces back horizontally west with 1.50 times the speed it had just before
their collision. How far behind the player who threw it
does the ball land? (b) This situation is not covered in the

45Њ nose high

31000

rule book, so the officials turn the clock back to repeat
this part of the game. The player throws the ball in the
same way. The eagle is thoroughly annoyed and this time
intercepts the ball so that, at the same point in its trajectory, the ball again bounces from the bird’s beak with 1.50
times its impact speed, moving west at some nonzero
angle with the horizontal. Now the ball hits the player’s
head, at the same location where her hands had released
it. Is the angle necessarily positive (that is, above the horizontal), necessarily negative (below the horizontal), or
could it be either? Give a convincing argument, either
mathematical or conceptual, for your answer.
45. Manny Ramírez hits a home run so that the baseball just
clears the top row of bleachers, 21.0 m high, located 130 m
from home plate. The ball is hit at an angle of 35.0° to
the horizontal, and air resistance is negligible. Find (a) the
initial speed of the ball, (b) the time interval required for


Zero g

1.8g

1.8g

0

Courtesy of NASA

96

65
Maneuver time, s
(a)

(b)
Figure P4.42

2 = intermediate;

3 = challenging;

Ⅺ = SSM/SG;

ᮡ

= ThomsonNOW;


Figure P4.51

Figure P4.47

48. An astronaut on the surface of the Moon fires a cannon
to launch an experiment package, which leaves the barrel
moving horizontally. (a) What must be the muzzle speed
of the package so that it travels completely around the
Moon and returns to its original location? (b) How long
does this trip around the Moon take? Assume the free-fall
acceleration on the Moon is one-sixth of that on the
Earth.
49. ⅷ A projectile is launched from the point (x ϭ 0, y ϭ 0)
with velocity 112.0ˆi ϩ 49.0ˆj 2 m/s, at t ϭ 0. (a) Make a
S
table listing the projectile’s distance 0 r 0 from the origin at
the end of each second thereafter, for 0 Յ t Յ 10 s. Tabulating the x and y coordinates and the components of
velocity vx and vy may also be useful. (b) Observe that the
projectile’s distance from its starting point increases with
time, goes through a maximum, and starts to decrease.
Prove that the distance is a maximum when the position
vector is perpendicular to the velocity. Suggestion: Argue
S
S
S
that if v is not perpendicular to r , then 0 r 0 must be
increasing or decreasing. (c) Determine the magnitude
of the maximum distance. Explain your method.
50. ⅷ A spring cannon is located at the edge of a table that
is 1.20 m above the floor. A steel ball is launched from


52. A truck loaded with cannonball watermelons stops suddenly to avoid running over the edge of a washed-out
bridge (Fig. P4.52). The quick stop causes a number of
melons to fly off the truck. One melon rolls over the edge
with an initial speed vi ϭ 10.0 m/s in the horizontal direction. A cross section of the bank has the shape of the bottom half of a parabola with its vertex at the edge of the
road and with the equation y2 ϭ 16x, where x and y are
measured in meters. What are the x and y coordinates of
the melon when it splatters on the bank?

vi ϭ 10 m/s

Figure P4.52

53. Your grandfather is copilot of a bomber, flying horizontally over level terrain, with a speed of 275 m/s relative to
the ground, at an altitude of 3 000 m. (a) The bombardier releases one bomb. How far will the bomb travel
horizontally between its release and its impact on the
ground? Ignore the effects of air resistance. (b) Firing
from the people on the ground suddenly incapacitates
the bombardier before he can call, “Bombs away!” Consequently, the pilot maintains the plane’s original course,
altitude, and speed through a storm of flak. Where will
the plane be when the bomb hits the ground? (c) The
plane has a telescopic bombsight set so that the bomb hits
the target seen in the sight at the moment of release. At
what angle from the vertical was the bombsight set?
54. A person standing at the top of a hemispherical rock of
radius R kicks a ball (initially at rest on the top of the
S
rock) to give it horizontal velocity vi as shown in Figure
P4.54. (a) What must be its minimum initial speed if the
ball is never to hit the rock after it is kicked? (b) With this

3.00 m above the ground. (a) Assuming no air resistance
acts on the mouse, find the diving speed of the hawk.
(b) What angle did the hawk make with the horizontal
during its descent? (c) For how long did the mouse
“enjoy” free fall?
56. The determined coyote is out once more in pursuit of the
elusive roadrunner. The coyote wears a pair of Acme jetpowered roller skates, which provide a constant horizontal
acceleration of 15.0 m/s2 (Fig. P4.56). The coyote starts at
rest 70.0 m from the brink of a cliff at the instant the roadrunner zips past in the direction of the cliff. (a) Assuming the roadrunner moves with constant speed, determine
the minimum speed it must have to reach the cliff before
the coyote. At the edge of the cliff, the roadrunner
escapes by making a sudden turn, while the coyote continues straight ahead. The coyote’s skates remain horizontal
and continue to operate while the coyote is in flight, so
its acceleration while in the air is 115.0ˆi Ϫ 9.80ˆj 2 m>s2.
(b) The cliff is 100 m above the flat floor of a canyon. Determine where the coyote lands in the canyon.
(c) Determine the components of the coyote’s impact
velocity.
Coyote Roadrunner
stupidus delightus EP
BE
BEE
P

Figure P4.56

57.

ᮡ

A car is parked on a steep incline overlooking the

jumpers lean forward in the shape of an airfoil, with their
hands at their sides, to increase their distance. Why does
this method work?)

10.0 m/s
15.0Њ

50.0Њ
Figure P4.59

60. An angler sets out upstream from Metaline Falls on the
Pend Oreille River in northwestern Washington State. His
small boat, powered by an outboard motor, travels at a
constant speed v in still water. The water flows at a lower
constant speed vw. He has traveled upstream for 2.00 km
when his ice chest falls out of the boat. He notices that
the chest is missing only after he has gone upstream for
another 15.0 min. At that point, he turns around and
heads back downstream, all the time traveling at the same
speed relative to the water. He catches up with the floating ice chest just as it is about to go over the falls at his
starting point. How fast is the river flowing? Solve this
problem in two ways. (a) First, use the Earth as a reference frame. With respect to the Earth, the boat travels
upstream at speed v Ϫ vw and downstream at v ϩ vw. (b) A
second much simpler and more elegant solution is
obtained by using the water as the reference frame. This
approach has important applications in many more complicated problems; examples are calculating the motion
of rockets and satellites and analyzing the scattering of
subatomic particles from massive targets.
61. An enemy ship is on the east side of a mountainous island
as shown in Figure P4.61. The enemy ship has maneuvered to within 2 500 m of the 1 800-m-high mountain


62. In the What If? section of Example 4.5, it was claimed
that the maximum range of a ski jumper occurs for a
launch angle u given by

u ϭ 45° Ϫ

where f is the angle that the hill makes with the horizontal in Figure 4.14. Prove this claim by deriving this
equation.

f
2

Answers to Quick Quizzes
4.1 (a). Because acceleration occurs whenever the velocity
changes in any way—with an increase or decrease in
speed, a change in direction, or both—all three controls
are accelerators. The gas pedal causes the car to speed up;
the brake pedal causes the car to slow down. The steering
wheel changes the direction of the velocity vector.
4.2 (i), (b). At only one point—the peak of the trajectory—
are the velocity and acceleration vectors perpendicular to
each other. The velocity vector is horizontal at that point,
and the acceleration vector is downward. (ii), (a). The
acceleration vector is always directed downward. The
velocity vector is never vertical and parallel to the acceleration vector if the object follows a path such as that in Figure 4.8.
4.3 15°, 30°, 45°, 60°, 75°. The greater the maximum height,
the longer it takes the projectile to reach that altitude
and then fall back down from it. So, as the launch angle
increases, the time of flight increases.

5.1

The Concept of Force

5.6

Newton’s Third Law

5.2

Newton’s First Law and
Inertial Frames

5.7

Some Applications of
Newton’s Laws

5.3

Mass

5.8

Forces of Friction

5.4

Newton’s Second Law


What force (if any) causes the Moon to orbit the Earth? Newton answered this
and related questions by stating that forces are what cause any change in the velocity of an object. The Moon’s velocity is not constant because it moves in a nearly
circular orbit around the Earth. This change in velocity is caused by the gravitational force exerted by the Earth on the Moon.
100


Section 5.1

The Concept of Force

101

Contact forces

(a)

(b)

(c)

Field forces

m

M

(d)

–q


forces between objects, (2) electromagnetic forces between electric charges, (3) strong
forces between subatomic particles, and (4) weak forces that arise in certain radioactive
decay processes. In classical physics, we are concerned only with gravitational and
electromagnetic forces. We will discuss strong and weak forces in Chapter 46.

The Vector Nature of Force
It is possible to use the deformation of a spring to measure force. Suppose a vertical force is applied to a spring scale that has a fixed upper end as shown in Figure
5.2a (page 102). The spring elongates when the force is applied, and a pointer on
the scale reads the value
of the applied force. We can calibrate the spring by definS
ing a reference force F1 as the force that produces
a pointer reading of 1.00 cm. If
S
we now apply a different
downward
force
whose
magnitude is twice that of the
F
2
S
reference force F1 as seen in Figure 5.2b, the pointer moves to 2.00 cm. Figure
5.2c shows that the combined effect of the two collinear forces is the sum of the
effects of the individual forces.
S
Now suppose the two forces are applied simultaneously with F1 downward and
S
F2 horizontal as illustrated in Figure 5.2d. In this case, the pointer reads 2.24 cm.

Giraudon/Art Resource

1
2
3
4

4

0
1
2
3
4

0

The Laws of Motion

1

Chapter 5

2

102

F2
u
F1

F1

to obtain the net force on an object.

5.2

Newton’s First Law and Inertial Frames

We begin our study of forces by imagining some physical situations involving a
puck on a perfectly level air hockey table (Fig. 5.3). You expect that the puck will
remain where it is placed. Now imagine your air hockey table is located on a train
moving with constant velocity along a perfectly smooth track. If the puck is placed
on the table, the puck again remains where it is placed. If the train were to accelerate, however, the puck would start moving along the table opposite the direction
of the train’s acceleration, just as a set of papers on your dashboard falls onto the
front seat of your car when you step on the accelerator.
As we saw in Section 4.6, a moving object can be observed from any number of
reference frames. Newton’s first law of motion, sometimes called the law of inertia,
defines a special set of reference frames called inertial frames. This law can be
stated as follows:

Air flow

Electric blower
Figure 5.3 On an air hockey table,
air blown through holes in the surface allows the puck to move almost
without friction. If the table is not
accelerating, a puck placed on the
table will remain at rest.

Newton’s first law

ᮣ

model the Earth as an inertial frame, along with any other frame attached to it.
Let us assume we are observing an object from an inertial reference frame. (We
will return to observations made in noninertial reference frames in Section 6.3.)
Before about 1600, scientists believed that the natural state of matter was the state
of rest. Observations showed that moving objects eventually stopped moving.
Galileo was the first to take a different approach to motion and the natural state of
matter. He devised thought experiments and concluded that it is not the nature of
an object to stop once set in motion: rather, it is its nature to resist changes in its
motion. In his words, “Any velocity once imparted to a moving body will be rigidly
maintained as long as the external causes of retardation are removed.” For example, a spacecraft drifting through empty space with its engine turned off will keep
moving forever. It would not seek a “natural state” of rest.
Given our discussion of observations made from inertial reference frames, we
can pose a more practical statement of Newton’s first law of motion:
In the absence of external forces and when viewed from an inertial reference
frame, an object at rest remains at rest and an object in motion continues in
motion with a constant velocity (that is, with a constant speed in a straight
line).

Mass

PITFALL PREVENTION 5.1
Newton’s First Law
Newton’s first law does not say what
happens for an object with zero net
force, that is, multiple forces that
cancel; it says what happens in the
absence of external forces. This subtle
but important difference allows us
to define force as that which causes
a change in the motion. The

physics, we say that the bowling ball is more resistant to changes in its velocity than
the basketball. How can we quantify this concept?
Mass is that property of an object that specifies how much resistance an object
exhibits to changes in its velocity, and as we learned in Section 1.1 the SI unit of
mass is the kilogram. Experiments show that the greater the mass of an object, the
less that object accelerates under the action of a given applied force.
To describe mass quantitatively, we conduct experiments in which we compare
the accelerations a given force produces on different objects. Suppose a force actS
ing on an object of mass m1 produces an acceleration a1, and the same force acting

103


104

Chapter 5

The Laws of Motion
S

on an object of mass m2 produces an acceleration a2. The ratio of the two masses
is defined as the inverse ratio of the magnitudes of the accelerations produced by
the force:
m1
a2
ϵ
m2
a1

Mass and weight are

(5.1)

Newton’s Second Law

Newton’s first law explains what happens to an object when no forces act on it: it
either remains at rest or moves in a straight line with constant speed. Newton’s second law answers the question of what happens to an object that has one or more
forces acting on it.
Imagine performing an experiment in which you push a block of fixed mass
S
across a frictionless horizontal surface. When you exert some horizontal force F on
S
the block, it moves with some acceleration a. If you apply a force twice as great on
the same block,
the acceleration of the block doubles. If you increase the applied
S
force to 3F, the acceleration triples, and so on. From such observations, we conclude that
the acceleration of an object is directly proportional to the force acting
S
S
on it: F ϰ a. This idea was first introduced in Section 2.4 when we discussed the
direction of the acceleration of an object. The magnitude of the acceleration of an
object is inversely proportional to its mass, as stated in the preceding section:
0 Sa 0 ϰ 1>m.
These experimental observations are summarized in Newton’s second law:
When viewed from an inertial reference frame, the acceleration of an object
is directly proportional to the net force acting on it and inversely proportional to its mass:
S

aF
aϰ

second law, we have
S
indicated that the acceleration is due to the net force ͚ F acting on an object. The
net force on an object is the vector sum of all forces acting on the object. (We
sometimes refer to the net force as the total force, the resultant force, or the unbalanced force.) In solving a problem using Newton’s second law, it is imperative to
determine the correct net force on an object. Many forces may be acting on an
object, but there is only one acceleration.
Equation 5.2 is a vector expression and hence is equivalent to three component
equations:
a Fx ϭ max ¬¬a Fy ϭ may¬¬a Fz ϭ maz

(5.3)

Quick Quiz 5.2 An object experiences no acceleration. Which of the following
cannot be true for the object? (a) A single force acts on the object. (b) No forces
act on the object. (c) Forces act on the object, but the forces cancel.
Quick Quiz 5.3 You push an object, initially at rest, across a frictionless floor

with a constant force for a time interval ⌬t, resulting in a final speed of v for the
object. You then repeat the experiment, but with a force that is twice as large. What
time interval is now required to reach the same final speed v? (a) 4⌬t (b) 2⌬t
(c) ⌬t (d) ⌬t/2 (e) ⌬t/4
The SI unit of force is the newton (N). A force of 1 N is the force that, when
acting on an object of mass 1 kg, produces an acceleration of 1 m/s2. From this
definition and Newton’s second law, we see that the newton can be expressed in
terms of the following fundamental units of mass, length, and time:
1 N ϵ 1 kg # m>s2

(5.4)



(5.5)

A convenient approximation is 1 N Ϸ 14 lb.

E XA M P L E 5 . 1

An Accelerating Hockey Puck

A hockey puck having a mass of 0.30 kg slides on the horizontal, frictionless
surface of an ice rink. Two hockey sticks strike the puck simultaneously,
exertS
ing the forces on the puck
shown
in
Figure
5.4.
The
force
has
a
magnitude
F
1
S
of 5.0 N, and the force F2 has a magnitude of 8.0 N. Determine both the magnitude and the direction of the puck’s acceleration.
SOLUTION
Conceptualize Study Figure 5.4. Using your expertise in vector addition from
Chapter 3, predict the approximate direction of the net force vector on the
puck. The acceleration of the puck will be in the same direction.


Analyze Find the component of the net force acting
on the puck in the x direction:

a Fx ϭ F1x ϩ F2x ϭ F1 cos 1Ϫ20°2 ϩ F2 cos 60°
ϭ 15.0 N2 10.940 2 ϩ 18.0 N2 10.5002 ϭ 8.7 N

Find the component of the net force acting on the puck
in the y direction:

a Fy ϭ F1y ϩ F2y ϭ F1 sin 1Ϫ20°2 ϩ F2 sin 60°
ϭ 15.0 N2 1Ϫ0.3422 ϩ 18.0 N2 10.866 2 ϭ 5.2 N

Use Newton’s second law in component form (Eq. 5.3) to
find the x and y components of the puck’s acceleration:

Find the magnitude of the acceleration:
Find the direction of the acceleration relative to the
positive x axis:

ax ϭ

8.7 N
a Fx
ϭ
ϭ 29 m>s2
m
0.30 kg

ay ϭ

property of an object; rather, it is a
measure of the gravitational force
between the object and the Earth
(or other planet). Therefore,
weight is a property of a system of
items: the object and the Earth.

5.5

The Gravitational Force and Weight

All objects are attracted to the Earth. TheSattractive force exerted by the Earth on
an object is called the gravitational force Fg . This force is directed toward the center of the Earth,3 and its magnitude is called the weight of the object.
S
We saw in Section 2.6 that a freely falling object experiences an acceleration
g
S
S
acting toward the center of the Earth. Applying Newton’s
second
law
͚
F
ϭ
m
a
to
S
S
S

of 70.0 kg. The student’s weight in a location where g ϭ 9.80 m/s2 is 686 N (about
150 lb). At the top of a mountain, however, where g ϭ 9.77 m/s2, the student’s
3

This statement ignores that the mass distribution of the Earth is not perfectly spherical.


weight is only 684 N. Therefore, if you want to lose weight without going on a diet,
climb a mountain or weigh yourself at 30 000 ft during an airplane flight!
Equation 5.6 quantifies the gravitational force on the object, but notice that this
equation does not require the object to be moving. Even for a stationary object or
for an object on which several forces act, Equation 5.6 can be used to calculate the
magnitude of the gravitational force. The result is a subtle shift in the interpretation of m in the equation. The mass m in Equation 5.6 determines the strength of
the gravitational attraction between the object and the Earth. This role is completely different from that previously described for mass, that of measuring the
resistance to changes in motion in response to an external force. Therefore, we
call m in Equation 5.6 the gravitational mass. Even though this quantity is different
in behavior from inertial mass, it is one of the experimental conclusions in Newtonian dynamics that gravitational mass and inertial mass have the same value.
Although this discussion has focused on the gravitational force on an object
due to the Earth, the concept is generally valid on any planet. The value of g will
vary from one planet to the next, but the magnitude of the gravitational force will
always be given by the value of mg.

Quick Quiz 5.4 Suppose you are talking by interplanetary telephone to a
friend, who lives on the Moon. He tells you that he has just won a newton of gold
in a contest. Excitedly, you tell him that you entered the Earth version of the same
contest and also won a newton of gold! Who is richer? (a) You are. (b) Your friend
is. (c) You are equally rich.
CO N C E P T UA L E XA M P L E 5 . 2

107

To provide the acceleration upward, the floor or scale
must exert on your feet an upward force that is greater
in magnitude than your weight. It is this greater force
you feel, which you interpret as feeling heavier. The
scale reads this upward force, not your weight, and so its
reading increases.

Newton’s Third Law

If you press against a corner of this textbook with your fingertip, the book pushes
back and makes a small dent in your skin. If you push harder, the book does the
same and the dent in your skin is a little larger. This simple activity illustrates that
forces are interactions between two objects: when your finger pushes on the book,
the book pushes back on your finger. This important principle is known as Newton’s third law:
S

If two objects interact, the force F12 exerted by objectS1 on object 2 is equal
in magnitude and opposite in direction to the force F21 exerted by object 2
on object 1:
S

S

F12 ϭ ϪF21

(5.7)

When it is important to designate forces
as interactions between two objects, we
S

(b)
S

Figure 5.5 Newton’s third law. (a) The
force F12 exerted by object 1 on object 2 is equalS in magnitude
S
and opposite in direction to the force F21 exerted by object 2 on object 1.S(b) The force Fhn exerted by
the hammer on the nail is equal in magnitude and opposite to the force Fnh exerted by the nail on the
hammer.

PITFALL PREVENTION 5.6
n Does Not Always Equal mg
In the situation shown in Figure 5.6
and in many others, we find that
n ϭ mg (the normal force has the
same magnitude as the gravitational force). This result, however,
is not generally true. If an object is
on an incline, if there are applied
forces with vertical components, or
if there is a vertical acceleration of
the system, then n mg. Always
apply Newton’s second law to find
the relationship between n and mg.

Normal force

ᮣ

PITFALL PREVENTION 5.7
Newton’s Third Law

Because the Earth has such a large mass, however, its acceleration due to this reaction force is negligibly small.
S
Another example of Newton’s third law is shown in Figure 5.5b. The force Fhn
exerted
by the hammer on the nail is equal in magnitude and opposite the force
S
Fnh exerted by the nail on the hammer. This latter force stops the forward motion
of the hammer when it strikes the nail.
Consider a computer monitorS at rest
on a table as in Figure 5.6a. The
reaction
S
S
S
force to the gravitational force Fg ϭ FEm on the monitor is the force FmE ϭ ϪFEm
exerted by the monitor on the Earth. The monitor does not accelerate because
it
S
S
is held up by the table. The table exerts on the monitor an upward force n ϭ Ftm,
called the normal force.4 This force, which prevents the monitor from falling
through the table, can have any value needed, up to the point of breaking the
table. Because the monitor
has zero acceleration, Newton’s second law applied to
S
S
S
the monitor gives us ͚ F ϭ n ϩ mg ϭ 0, so nˆj Ϫ mgˆj ϭ 0, or n ϭ mg. The normal
force balances the gravitational force on the monitor, so the net force on the monS
itor is zero. The

109

n ϭ Ftm

n ϭ Ftm

Fg ϭ FEm
Fg ϭ FEm

Fmt
FmE

(a)

(b)

Figure 5.6 (a) When a computer monitorSis at rest on a table, the forces acting
on the monitor are the
S
S
S
normal force n and the gravitational
force Fg S. The reaction to n is the force Fmt exerted by the monitor
S
on the table. The reaction to Fg is the force FmE exerted by the monitor on the Earth. (b) The freebody diagram for the monitor.

analysis. This diagram can be simplified further by representing the object (such
as the monitor) as a particle simply by drawing a dot.

Quick Quiz 5.5 (i) If a fly collides with the windshield of a fast-moving bus,

which way it faced.) Therefore, the boy, having the

5.7

PITFALL PREVENTION 5.8
Free-Body Diagrams

smaller mass, experiences the greater acceleration. Both
individuals accelerate for the same amount of time, but
the greater acceleration of the boy over this time interval results in his moving away from the interaction with
the higher speed.
(B) Who moves farther while their hands are in contact?
SOLUTION
Because the boy has the greater acceleration and therefore the greater average velocity, he moves farther than
the man during the time interval during which their
hands are in contact.

Some Applications of Newton’s Laws

In this section, we discuss two analysis models for solving problems in which
S
objects are either in equilibrium 1a ϭ 0 2 or accelerating along a straight line
under the action of constant external forces. Remember that when Newton’s laws
are applied to an object, we are interested only in external forces that act on the
object. If the objects are modeled as particles, we need not worry about rotational
motion. For now, we also neglect the effects of friction in those problems involving


110


T

TЈ

Fg
(b)

(a)

Figure 5.7 (a) A lamp suspended
from a ceiling by a chain of negligible
mass. (b) The forces acting on the
S
lamp are the gravitational
force Fg
S
and the force T exerted by the chain.
(c) The forces
acting on the chain are
S
the force T
¿ exerted by the lamp and
S
the force T – exerted by the ceiling.

(a)
n

y


nitude to the magnitude of T ¿ and points in the opposite direction.

The Particle Under a Net Force
If an object experiences an acceleration, its motion can be analyzed with the particle under a net force model. The appropriate equation for this model is Newton’s
second law, Equation 5.2. Consider a crate being pulled to the right on a frictionless, horizontal surface as in Figure 5.8a. Suppose you wish to find the acceleration
of the crate and the force the floor exerts on it. The forces acting on the crate are
illustrated
in the free-body diagram in Figure 5.8b. Notice that the horizontal
S
S
force T being applied to the crate acts through the rope.S The magnitude of T is
equal to the tension in the rope. In addition Sto the force T, the free-body diagram
S
for the crate includes the gravitational force Fg and the normal force n exerted by
the floor on the crate.
We can now apply Newton’s second Slaw in component form to the crate. The
only force acting in the x direction is T. Applying ͚ Fx ϭ max to the horizontal
motion gives
T
a Fx ϭ T ϭ max¬or¬ax ϭ m

Fg
(b)
Figure 5.8 (a) A crate being pulled
to the right on a frictionless surface.
(b) The free-body diagram representing the external forces acting on the
crate.

No acceleration occurs in the y direction because the crate moves only horizontally. Therefore, we use the particle in equilibrium model in the y direction. Applying the y component of Equation 5.8 yields
a Fy ϭ n ϩ 1ϪFg 2 ϭ 0¬or¬n ϭ Fg

Fg

S

P R O B L E M - S O LV I N G S T R AT E G Y

Applying Newton’s Laws

The following procedure is recommended when dealing with problems involving
Newton’s laws:

Figure 5.9 When a force F pushes
vertically downward on another
S
object, the normal force n on the
object is greater than the gravitational force: n ϭ Fg ϩ F.

1. Conceptualize. Draw a simple, neat diagram of the system. The diagram helps
establish the mental representation. Establish convenient coordinate axes for
each object in the system.
2. Categorize. If an acceleration component for an object is zero, the object is modeled as a particle in equilibrium in this direction and ͚ F ϭ 0. If not, the object
is modeled as a particle under a net force in this direction and ͚ F ϭ ma.
3. Analyze. Isolate the object whose motion is being analyzed. Draw a free-body diagram for this object. For systems containing more than one object, draw separate
free-body diagrams for each object. Do not include in the free-body diagram
forces exerted by the object on its surroundings.
Find the components of the forces along the coordinate axes. Apply the
appropriate model from the Categorize step for each direction. Check your
dimensions to make sure that all terms have units of force.
Solve the component equations for the unknowns. Remember that you must
have as many independent equations as you have unknowns to obtain a complete solution.


53.0Њ

37.0Њ

T3

x

SOLUTION
Conceptualize Inspect the drawing in Figure 5.10a.
Let us assume the cables do not break and that nothing
is moving.

Fg
(a)

(b)

T3
(c)

Figure 5.10 (Example 5.4) (a) A traffic light suspended by cables.
(b) The free-body diagram for the traffic light. (c) The free-body diagram for the knot where the three cables are joined.


112

Chapter 5


T1 sin 37.0°

T2 cos 53.0°

T2 sin 53.0°

S

T2
S

Apply the particle in equilibrium model to the knot:

S

Ϫ122 N

0

T3

(1)

a Fx ϭ ϪT1 cos 37.0° ϩ T2 cos 53.0° ϭ 0

(2)

a Fy ϭ T1 sin 37.0° ϩ T2 sin 53.0° ϩ 1Ϫ122 N2 ϭ 0
S


Answer We can argue from the symmetry of the problem that the two tensions T1 and T2 would be equal to each
other. Mathematically, if the equal angles are called u, Equation (3) becomes
T2 ϭ T1 a

cos u
b ϭ T1
cos u

which also tells us that the tensions are equal. Without knowing the specific value of u, we cannot find the values of
T1 and T2. The tensions will be equal to each other, however, regardless of the value of u.

CO N C E P T UA L E XA M P L E 5 . 5

Forces Between Cars in a Train

Train cars are connected by couplers, which are under
tension as the locomotive pulls the train. Imagine you
are on a train speeding up with a constant acceleration.
As you move through the train from the locomotive to
the last car, measuring the tension in each set of couplers, does the tension increase, decrease, or stay the

same? When the engineer applies the brakes, the couplers are under compression. How does this compression force vary from the locomotive to the last car?
(Assume only the brakes on the wheels of the engine
are applied.)


Section 5.7

SOLUTION
As the train speeds up, tension decreases from the front

Conceptualize Use Figure 5.11a to conceptualize the situation. From everyday experience, we
know that a car on an icy incline will accelerate
down the incline. (The same thing happens to a
car on a hill with its brakes not set.)

mg cos u

Categorize We categorize the car as a particle
under a net force. Furthermore, this problem
belongs to a very common category of problems
in which an object moves under the influence of
gravity on an inclined plane.

x

u

u

Fg = m g

(a)

(b)

Figure 5.11 (Example 5.6) (a) A car of mass m on a frictionless incline.
(b) The free-body diagram for the car.

Analyze Figure 5.11b shows the free-body diagram for the car. The only forces acting on the car are theS normal
S

the bottom of the incline is d. How long does it take the front bumper to reach the bottom of the hill, and what is
the car’s speed as it arrives there?
SOLUTION
Concepualize Imagine that the car is sliding down the hill and you use a stopwatch to measure the entire time
interval until it reaches the bottom.