­C H A P T E R 5

Force and Motion—I

5-1 NEWTON’S FIRST AND SECOND LAWS
Learning Objectives
After reading this module, you should be able to . . .
5.01 Identify that a force is a vector quantity and thus has acting on it as vectors with their tails anchored on the
both magnitude and direction and also components. particle.
5.02 Given two or more forces acting on the same par- 5.06 Apply the relationship (Newton’s second law)
ticle, add the forces as vectors to get the net force. between the net force on an object, the mass of the
5.03 Identify Newton’s first and second laws of motion. object, and the acceleration produced by the net force.
5.04 Identify inertial reference frames. 5.07 Identify that only external forces on an object can
5.05 Sketch a free-body diagram for an object, show- cause the object to accelerate.
ing the object as a particle and drawing the forces

Key Ideas
● The velocity of an object can change (the object can ● The mass of a body is the characteristic of that body
accelerate) when the object is acted on by one or more that relates the body’s acceleration to the net force caus-
forces (pushes or pulls) from other objects. Newtonian ing the acceleration. Masses are scalar quantities.
mechanics relates accelerations and forces. →
● The net force Fnet on a body with mass m is related to
● Forces are vector quantities. Their magnitudes are the body’s acceleration →a by
defined in terms of the acceleration they would give the → →
standard kilogram. A force that accelerates that standard Fnet = m a ,
body by exactly 1 m/s2 is defined to have a magnitude of
which may be written in the component versions
1 N. The direction of a force is the direction of the accel-
eration it causes. Forces are combined according to the Fnet,x = max Fnet,y = may and Fnet,z = maz.
rules of vector algebra. The net force on a body is the
vector sum of all the forces acting on the body. The second law indicates that in SI units
● If there is no net force on a body, the body remains
1 N = 1 kg · m/s2.
at rest if it is initially at rest or moves in a straight line at
constant speed if it is in motion. ● A free-body diagram is a stripped-down diagram in
● Reference frames in which Newtonian mechanics holds which only one body is considered. That body is repre-
are called inertial reference frames or inertial frames. Refer- sented by either a sketch or a dot. The external forces on
ence frames in which Newtonian mechanics does not hold the body are drawn, and a coordinate system is superim-
are called noninertial reference frames or noninertial frames. posed, oriented so as to simplify the solution.

What Is Physics?
We have seen that part of physics is a study of motion, including accelerations,
which are changes in velocities. Physics is also a study of what can cause an ­object
to accelerate. That cause is a force, which is, loosely speaking, a push or pull on the
object. The force is said to act on the object to change its velocity. For example,
when a dragster accelerates, a force from the track acts on the rear tires to cause
the dragster’s acceleration. When a defensive guard knocks down a quarterback, a
force from the guard acts on the quarterback to cause the quarterback’s backward
acceleration. When a car slams into a telephone pole, a force on the car from the

94
 5-1 NEWTON’S FIRST AND SECOND LAWS 95

pole causes the car to stop. Science, engineering, legal, and medical journals are
filled with articles about forces on objects, including people.
A Heads Up. Many students find this chapter to be more challenging than
the preceding ones. One reason is that we need to use vectors in setting up
­equations—we cannot just sum some scalars. So, we need the vector rules from
Chapter 3. Another reason is that we shall see a lot of different arrangements:
objects will move along floors, ceilings, walls, and ramps. They will move upward
on ropes looped around pulleys or by sitting in ascending or descending eleva-
tors. Sometimes, objects will even be tied together.
However, in spite of the variety of arrangements, we need only a single key
idea (Newton’s second law) to solve most of the homework problems. The pur-
pose of this chapter is for us to explore how we can apply that single key idea to
any given arrangement. The application will take experience—we need to solve
lots of problems, not just read words. So, let’s go through some of the words and
then get to the sample problems.

Newtonian Mechanics
The relation between a force and the acceleration it causes was first understood
by Isaac Newton (1642 –1727) and is the subject of this chapter. The study of that
relation, as Newton presented it, is called Newtonian mechanics. We shall focus
on its three primary laws of motion.
Newtonian mechanics does not apply to all situations. If the speeds of the
interacting bodies are very large —an appreciable fraction of the speed of light —
we must replace Newtonian mechanics with Einstein’s special theory of relativity,
which holds at any speed, including those near the speed of light. If the inter­acting
bodies are on the scale of atomic structure (for example, they might be electrons in
an atom), we must replace Newtonian mechanics with quantum mechanics. Physi-
cists now view Newtonian mechanics as a special case of these two more compre-
hensive theories. Still, it is a very important special case ­because it applies to the
motion of objects ranging in size from the very small (almost on the scale of atomic
structure) to astronomical (galaxies and clusters of galaxies).

Newton’s First Law

Before Newton formulated his mechanics, it was thought that some influence,
a “force,” was needed to keep a body moving at constant velocity. Similarly, a
body was thought to be in its “natural state” when it was at rest. For a body to
move with constant velocity, it seemingly had to be propelled in some way, by
a push or a pull. Otherwise, it would “naturally” stop moving.
These ideas were reasonable. If you send a puck sliding across a wooden
floor, it does indeed slow and then stop. If you want to make it move across the
floor with constant velocity, you have to continuously pull or push it.
Send a puck sliding over the ice of a skating rink, however, and it goes a lot
farther. You can imagine longer and more slippery surfaces, over which the puck
would slide farther and farther. In the limit you can think of a long, extremely
slippery surface (said to be a frictionless surface), over which the puck would
hardly slow. (We can in fact come close to this situation by sending a puck s­ liding
over a horizontal air table, across which it moves on a film of air.)
From these observations, we can conclude that a body will keep moving with
constant velocity if no force acts on it. That leads us to the first of Newton’s three
laws of motion:

 ewton’s First Law: If no force acts on a body, the body’s velocity cannot
N
change; that is, the body cannot accelerate.
96 CHAPTER 5 FORCE AND MOTION—I

In other words, if the body is at rest, it stays at rest. If it is moving, it continues to

move with the same velocity (same magnitude and same direction).
F

Force
a Before we begin working problems with forces, we need to discuss several fea-
→ tures of forces, such as the force unit, the vector nature of forces, the combining
Figure 5-1 A force F on the standard
→ of forces, and the circumstances in which we can measure forces (without being
­kilogram gives that body an acceleration a .
fooled by a fictitious force).
Unit. We can define the unit of force in terms of the acceleration a force would
give to the standard kilogram (Fig. 1-3), which has a mass defined to be ex­actly
1 kg. Suppose we put that body on a horizontal, frictionless surface and pull hori-
zontally (Fig. 5-1) such that the body has an acceleration of 1 m/s2. Then we can
define our applied force as having a magnitude of 1 newton (abbreviated N). If we
then pulled with a force magnitude of 2 N, we would find that the acceleration is
2 m/s2. Thus, the acceleration is proportional to the force. If the standard body of
1 kg has an acceleration of magnitude a (in meters per second per second), then the
force (in newtons) producing the acceleration has a magnitude equal to a. We now
have a workable definition of the force unit.
Vectors. Force is a vector quantity and thus has not only magnitude but
also direction. So, if two or more forces act on a body, we find the net force (or
resultant force) by adding them as vectors, following the rules of Chapter 3. A
single force that has the same magnitude and direction as the calculated net force
would then have the same effect as all the individual forces. This fact, called the
­principle of superposition for forces, makes everyday forces reasonable and pre-
dictable. The world would indeed be strange and unpredictable if, say, you and a
friend each pulled on the standard body with a force of 1 N and somehow the net
pull was 14 N and the resulting acceleration was 14 m/s2.
In this book, forces are most often represented with a vector symbol such
→ →
as F , and a net force is represented with the vector symbol Fnet. As with other
­vectors, a force or a net force can have components along coordinate axes.
When forces act only along a single axis, they are single-component forces.
Then we can drop the overhead arrows on the force symbols and just use signs
to indicate the directions of the forces along that axis.
The First Law. Instead of our previous wording, the more proper statement
of Newton’s First Law is in terms of a net force:

→
 ewton’s First Law: If no net force acts on a body ( Fnet = 0), the body’s velocity
N
­cannot change; that is, the body cannot accelerate.

There may be multiple forces acting on a body, but if their net force is zero, the
body cannot accelerate. So, if we happen to know that a body’s velocity is con-
stant, we can immediately say that the net force on it is zero.

Inertial Reference Frames

Newton’s first law is not true in all reference frames, but we can always find
­reference frames in which it (as well as the rest of Newtonian mechanics) is true.
Such special frames are referred to as inertial reference frames, or simply inertial
frames.

An inertial reference frame is one in which Newton’s laws hold.

For example, we can assume that the ground is an inertial frame provided we can
neglect Earth’s astronomical motions (such as its rotation).
 5-1 NEWTON’S FIRST AND SECOND LAWS 97

That assumption works well if, say, a puck is sent sliding along a short N
strip of frictionless ice — we would find that the puck’s motion obeys Newton’s W E
laws. However, suppose the puck is sent sliding along a long ice strip extending
from the north pole (Fig. 5-2a). If we view the puck from a stationary frame in S
space, the puck moves south along a simple straight line because Earth’s rota-
tion around the north pole merely slides the ice beneath the puck. However,
if we view the puck from a point on the ground so that we rotate with Earth,
the puck’s path is not a simple straight line. Because the eastward speed of the
ground beneath the puck is greater the farther south the puck slides, from our
ground-based view the puck appears to be deflected westward (Fig. 5-2b). How- (a)
ever, this apparent deflection is caused not by a force as required by ­Newton’s
laws but by the fact that we see the puck from a rotating frame. In this situa-
tion, the ground is a noninertial frame, and trying to explain the deflection in
terms of a force would lead us to a fictitious force. A more common example
of inventing such a nonexistent force can occur in a car that is rapidly increas-
ing in speed. You might claim that a force to the rear shoves you hard into the
seat back. Earth’s rotation
In this book we usually assume that the ground is an inertial frame and that causes an
measured forces and accelerations are from this frame. If measurements are made apparent deflection.
in, say, a vehicle that is accelerating relative to the ground, then the measurements
(b)
are being made in a noninertial frame and the results can be surprising.
Figure 5-2 (a) The path of a puck sliding
from the north pole as seen from a station-
Checkpoint 1 ary point in space. Earth rotates to the east.
→ (b) The path of the puck as seen from the
Which of the figure’s six arrangements correctly show the vector addition of forces F1
→ → ground.
and F2 to yield the third vector, which is meant to represent their net force Fnet?

F1 F1 F1

(a) F2 (b) F2 (c) F2

F2

F1 F1 F1

(d) (e) (f )
F2 F2

Mass
From everyday experience you already know that applying a given force to bod-
ies (say, a baseball and a bowling ball) results in different accelerations. The com-
mon explanation is correct: The object with the larger mass is accelerated less.
But we can be more precise. The acceleration is actually inversely related to the
mass (rather than, say, the square of the mass).
Let’s justify that inverse relationship. Suppose, as previously, we push on the
standard body (defined to have a mass of exactly 1 kg) with a force of magnitude
1 N. The body accelerates with a magnitude of 1 m/s2. Next we push on body
X with the same force and find that it accelerates at 0.25 m/s2. Let’s make the
­(correct) assumption that with the same force,
mX a0
= ,
m0 aX
98 CHAPTER 5 FORCE AND MOTION—I

and thus
a0 1.0 m/s2
mX = m0 = (1.0 kg) = 4.0 kg.
aX 0.25 m/s2
Defining the mass of X in this way is useful only if the procedure is consistent.
Suppose we apply an 8.0 N force first to the standard body (getting an accel-
eration of 8.0 m/s2) and then to body X (getting an acceleration of 2.0 m/s2). We
would then calculate the mass of X as
a0 8.0 m/s2
mX = m0 = (1.0 kg) = 4.0 kg,
aX 2.0 m/s2
which means that our procedure is consistent and thus usable.
The results also suggest that mass is an intrinsic characteristic of a body—it
automatically comes with the existence of the body. Also, it is a scalar quantity.
However, the nagging question remains: What, exactly, is mass?
Since the word mass is used in everyday English, we should have some intui-
tive understanding of it, maybe something that we can physically sense. Is it
a body’s size, weight, or density? The answer is no, although those characteristics
are sometimes confused with mass. We can say only that the mass of a body is
the characteristic that relates a force on the body to the resulting acceleration. Mass
has no more familiar definition; you can have a physical sensation of mass only
when you try to accelerate a body, as in the kicking of a baseball or a bowling ball.

Newton’s Second Law

All the definitions, experiments, and observations we have discussed so far can
be summarized in one neat statement:

Newton’s Second Law: The net force on a body is equal to the product of the
body’s mass and its acceleration.

In equation form,
→ →
Fnet = m a   (Newton’s second law). (5-1)

Identify the Body. This simple equation is the key idea for nearly all the
homework problems in this chapter, but we must use it cautiously. First, we must
→
be ­certain about which body we are applying it to. Then Fnet must be the vector
sum of all the forces that act on that body. Only forces that act on that body are
to be included in the vector sum, not forces acting on other bodies that might be
­involved in the given situation. For example, if you are in a rugby scrum, the net
force on you is the vector sum of all the pushes and pulls on your body. It does not
include any push or pull on another player from you or from anyone else. Every
time you work a force problem, your first step is to clearly state the body to which
you are applying Newton’s law.
Separate Axes. Like other vector equations, Eq. 5-1 is equivalent to three
component equations, one for each axis of an xyz coordinate system:

Fnet, x = max,   Fnet, y = may,   and   Fnet, z = maz. (5-2)

Each of these equations relates the net force component along an axis to the
­acceleration along that same axis. For example, the first equation tells us that
the sum of all the force components along the x axis causes the x component ax
of the body’s acceleration, but causes no acceleration in the y and z directions.
Turned around, the acceleration component ax is caused only by the sum of the
 5-1 NEWTON’S FIRST AND SECOND LAWS 99

force components along the x axis and is completely unrelated to force compo-
nents along another axis. In general,

 he acceleration component along a given axis is caused only by the sum of the force
T
components along that same axis, and not by force components along any other axis.

Forces in Equilibrium. Equation 5-1 tells us that if the net force on a body
→
is zero, the body’s a­ cceleration a = 0. If the body is at rest, it stays at rest; if it
is moving, it continues to move at constant velocity. In such cases, any forces on
the body ­balance one another, and both the forces and the body are said to be in
equilibrium. Commonly, the forces are also said to cancel one another, but the
term “cancel” is tricky. It does not mean that the forces cease to exist (canceling
forces is not like canceling dinner reservations). The forces still act on the body
but cannot change the velocity.
Units. For SI units, Eq. 5-1 tells us that
1 N = (1 kg)(1 m/s2) = 1 kg · m/s2.(5-3)
Some force units in other systems of units are given in Table 5-1 and Appendix D.
Diagrams. To solve problems with Newton’s second law, we often draw a
free-body ­diagram in which the only body shown is the one for which we are
summing forces. A sketch of the body itself is preferred by some teachers but,
to save space in these chapters, we shall usually represent the body with a dot.
Also, each force on the body is drawn as a vector arrow with its tail anchored on
the body. A coordinate system is usually included, and the acceleration of the
body is sometimes shown with a vector arrow (labeled as an acceleration). This
whole procedure is designed to focus our attention on the body of interest.

Table 5-1 Units in Newton’s Second Law (Eqs. 5-1 and 5-2)
System Force Mass Acceleration

SI newton (N) kilogram (kg) m/s2

CGSa dyne gram (g) cm/s2
Britishb pound (lb) slug ft/s2
a
1 dyne = 1 g · cm/s2.
b
1 lb = 1 slug · ft/s2.

External Forces Only. A system consists of one or more bodies, and any
force on the bodies inside the system from bodies outside the system is called
an external force. If the bodies making up a system are rigidly connected to one
→
another, we can treat the system as one composite body, and the net force Fnet on
it is the vector sum of all external forces. (We do not include internal forces — that
is, forces between two bodies inside the system. Internal forces cannot accelerate
the system.) For example, a connected railroad engine and car form a system.
If, say, a tow line pulls on the front of the engine, the force due to the tow line
acts on the whole engine – car system. Just as for a single body, we can relate
the net external force on a system to its acceleration with Newton’s second law,
→ →
Fnet = m a , where m is the total mass of the system.

Checkpoint 2 3N 5N
The figure here shows two horizontal forces acting
on a block on a frictionless floor. If a third horizon-
→ →
tal force F3 also acts on the block, what are the magnitude and direction of F3 when
the block is (a) stationary and (b) moving to the left with a constant speed of 5 m/s?
100 CHAPTER 5 FORCE AND MOTION—I

Sample Problem 5.01 One‑ and two‑dimensional forces, puck

Here are examples of how to use Newton’s second law for a A

puck when one or two forces act on it. Parts A, B, and C of The horizontal force
F1
Fig. 5‑3 show three situations in which one or two forces act x causes a horizontal
on a puck that moves over frictionless ice along an x axis, acceleration.
in one-dimensional m ­ otion. The puck’s mass is m = 0.20 kg. (a)
→ →
Forces F1 and F2 are directed along the axis and have
→ This is a free-body
­magnitudes F1 = 4.0 N and F2 = 2.0 N. Force ­F3 is directed Puck F1
x diagram.
at angle θ = 30° and has ­magnitude F3 = 1.0 N. In each situ-
ation, what is the ­acceleration of the puck?
(b)

KEY IDEA B
→ These forces compete.
In each situation we can r­ elate the acceleration a to the net force
→ → → F2 F1
Fnet acting on the puck with Newton’s s­ econd law, Fnet = m a . x Their net force causes
However, ­because the motion is along only the x axis, we can a horizontal acceleration.
simplify each situation by writing the second law for x compo- (c)
nents only:
F2 F1 This is a free-body
Fnet, x = max.(5-4) x
diagram.
The free-body diagrams for the three situations are also (d)
given in Fig. 5-3, with the puck represented by a dot.
C
Situation A: For Fig. 5-3b, where only one horizontal force
Only the horizontal
acts, Eq. 5‑4 gives us F2
x component of F3
θ
F1 = max, F3 competes with F2.
(e)
which, with given data, yields
F1 4.0 N F2 This is a free-body
ax = = = 20 m/s2. (Answer) θ
x
diagram.
m 0.20 kg  F3
The positive answer indicates that the acceleration is in the (f )
positive direction of the x axis. Figure 5-3 In three situations, forces act on a puck that moves
along an x axis. Free-body diagrams are also shown.
Situation B: In Fig. 5-3d, two horizontal forces act on the
→ →
puck, F1 in the positive direction of x and F2 in the negative
direction. Now Eq. 5-4 gives us
one-­dimensional.) Thus, we write Eq. 5-4 as
F1 − F2 = max, F3,x − F2 = max. (5-5)
which, with given data, yields From the figure, we see that F3,x = F3 cos θ. Solving for the
F1 − F2 4.0 N − 2.0 N a­ cceleration and substituting for F3,x yield
ax = = = 10 m/s2.
m 0.20 kg F3,x − F2 F3 cos θ − F2
ax = =
(Answer) m m
Thus, the net force accelerates the puck in the positive direc- (1.0 N)(cos 30°) − 2.0 N
tion of the x axis. = = −5.7 m/s2.
0.20 kg
→ (Answer)
Situation C: In Fig. 5-3f, force F3 is not directed along
the ­direction of the puck’s acceleration; only x component Thus, the net force accelerates the puck in the negative
→
F3, x is. (Force F3 is two-dimensional but the m
­ otion is only direction of the x axis.

Additional examples, video, and practice available at WileyPLUS

 5-1 NEWTON’S FIRST AND SECOND LAWS 101

Sample Problem 5.02 Two‑dimensional forces, cookie tin

Here we find a missing force by using the acceleration. In x components: Along the x axis we have
the overhead view of Fig. 5-4a, a 2.0 kg cookie tin is acceler- F3,x = max − F1,x − F2,x
ated at 3.0 m/s2 in the direction shown by →
a , over a friction-
= m(a cos 50°) − F1 cos(−150°) − F2 cos 90°.
less horizontal surface. The acceleration is caused by three
→
horizontal forces, only two of which are shown: F1 of magni- Then, substituting known data, we find
→
tude 10 N and F2 of magnitude 20 N. What is the third force F3,x = (2.0 kg)(3.0 m/s2) cos 50° − (10 N) cos(−150°)
→
F3 in unit-vector notation and in magnitude-angle n­ otation?
− (20 N) cos 90°
KEY IDEA = 12.5 N.
→ y components: Similarly, along the y axis we find
The net force Fnet on the tin is the sum of the three forces
and is related to the ­acceleration →
a via Newton’s second law F3,y = may − F1,y − F2,y
→
( Fnet = m→a ). Thus, = m(a sin 50°) − F1 sin(−150°) − F2 sin 90°
= (2.0 kg)(3.0 m/s2) sin 50° − (10 N) sin(−150°)
→ → → →
F1 + F2 + F3 = m a , (5-6) − (20 N) sin 90°
which gives us = −10.4 N.

→ → → Vector: In unit-vector notation, we can write

F3 = m→
a − F 1 − F 2. (5-7) →
F3 = F3,x î + F3,y ĵ = (12.5 N)î − (10.4 N)ĵ
Calculations: Because this is a two-dimensional p
→
­ ro­b­lem, ≈ (13 N)î − (10 N)ĵ. (Answer)
we cannot find F3 merely by substituting the ­magnitudes for
We can now use a vector-capable calculator to get the mag-
the vector quantities on the right side of Eq. 5-7. Instead, →
→ → → nitude and the angle of F3. We can also use Eq. 3-6 to obtain
we must vectorially add m a , − F1 (the reverse of F1), and
→ → the magnitude and the angle (from the positive direction of
− F2 (the reverse of F2), as shown in Fig. 5‑4b. This addition
the x axis) as
can be done directly on a vector-capable calculator because
we know both magnitude and angle for all three vectors.
F3 = √F 23,x + F 23,y = 16 N
However, here we shall evaluate the right side of Eq. 5-7 in
terms of components, first along the x axis and then along
F3,y
the y axis. Caution: Use only one axis at a time. and θ = tan−1 = −40°.(Answer)
F3, x

y
We draw the product
These are two F2 This is the resulting y
of mass and acceleration
of the three horizontal acceleration
as a vector.
horizontal force vector.
vectors. a
– F1

50° – F2
ma
x x
30°

F1 F3
(a) (b)
Then we can add the three
vectors to find the missing
third force vector.
Figure 5-4 (a) An overhead view of two of three horizontal forces that act on a cookie
→ → →
tin, resulting in acceleration →
a . F3 is not shown. (b) An arrangement of vectors m a , − F1,
→ →
and − F2 to find force F3.

Additional examples, video, and practice available at WileyPLUS

102 CHAPTER 5 FORCE AND MOTION—I

5-2 SOME PARTICULAR FORCES

Learning Objectives
After reading this module, you should be able to . . .
5.08 Determine the magnitude and direction of the gravi­ 5.11 Determine the magnitude and direction of the nor-
tational force acting on a body with a given mass, at a loca- mal force on an object when the object is pressed or
tion with a given free-fall acceleration. pulled onto a surface.
5.09 Identify that the weight of a body is the magnitude of 5.12 Identify that the force parallel to the surface is a
the net force required to prevent the body from ­falling frictional force that appears when the object slides or
freely, as measured from the reference frame of the ground. attempts to slide along the surface.
5.10 Identify that a scale gives an object’s weight when the 5.13 Identify that a tension force is said to pull at both
measurement is done in an inertial frame but not in an ends of a cord (or a cord-like object) when the cord
accelerating frame, where it gives an apparent weight. is taut.

Key Ideas
→ →
● A gravitational force F g on a body is a pull by another ● A normal force F N is the force on a body from a sur-
body. In most situations in this book, the other body face against which the body presses. The normal force is
is Earth or some other astronomical body. For Earth, always perpendicular to the surface.
the force is directed down toward the ground, which is →
● A frictional force f is the force on a body when the
assumed to be an inertial frame. With that assumption, body slides or attempts to slide along a surface. The force
→
the magnitude of F g is is always parallel to the surface and directed so as to
Fg = mg, oppose the sliding. On a frictionless surface, the frictional
force is negligible.
where m is the body’s mass and g is the magnitude of the ● When a cord is under tension, each end of the cord pulls
free-fall acceleration. on a body. The pull is directed along the cord, away from
the point of attachment to the body. For a massless cord (a
● The weight W of a body is the magnitude of the upward
cord with negligible mass), the pulls at both ends of the cord
force needed to balance the gravitational force on the have the same magnitude T, even if the cord runs around a
body. A body’s weight is related to the body’s mass by massless, frictionless pulley (a pulley with negligible mass
W = mg. and negligible friction on its axle to oppose its rotation).

Some Particular Forces

The Gravitational Force
→
A gravitational force F g on a body is a certain type of pull that is directed t­ oward
a second body. In these early chapters, we do not discuss the nature of this force
and usually consider situations in which the second body is Earth. Thus, when
→
we speak of the gravitational force F g on a body, we usually mean a force that
pulls on it directly toward the center of Earth — that is, directly down ­toward the
ground. We shall assume that the ground is an inertial frame.
Free Fall. Suppose a body of mass m is in free fall with the free-fall accelera-
tion of magnitude g. Then, if we neglect the effects of the air, the only force acting
→
on the body is the gravitational force F g . We can relate this downward force and
downward acceleration with Newton’s second law ( F = m→
→
a ). We place a ver-
tical y axis along the body’s path, with the positive direction upward. For this
axis, Newton’s second law can be written in the form Fnet,y = may , which, in our
­situation, becomes
−Fg = m(−g)

or Fg = mg. (5-8)

In words, the magnitude of the gravitational force is equal to the product mg.
 5-2 SOME PARTICULAR FORCES 103

At Rest. This same gravitational force, with the same magnitude, still acts on
the body even when the body is not in free fall but is, say, at rest on a pool table
or moving across the table. (For the gravitational force to disappear, Earth would
have to disappear.)
We can write Newton’s second law for the gravitational force in these vector
forms: mL mR
→ →
F g = −Fg ĵ = −mgĵ = m g , (5-9)
where ĵ is the unit vector that points upward along a y axis, directly away from
→
the ground, and g is the free-fall acceleration (written as a vector), directed FgL = mL g FgR = mR g
downward.
Figure 5-5 An equal-arm balance. When
Weight the device is in balance, the gravitational
→
The weight W of a body is the magnitude of the net force required to prevent the force FgL on the body being weighed (on
body from falling freely, as measured by someone on the ground. For example, the left pan) and the total gravitational
→
to keep a ball at rest in your hand while you stand on the ground, you must pro- force FgR on the reference bodies (on the
vide an upward force to balance the gravitational force on the ball from Earth. right pan) are equal. Thus, the mass mL
Suppose the magnitude of the gravitational force is 2.0 N. Then the ­magnitude of of the body being weighed is equal to the
your upward force must be 2.0 N, and thus the weight W of the ball is 2.0 N. We total mass mR of the reference bodies.
also say that the ball weighs 2.0 N and speak about the ball weighing 2.0 N.
A ball with a weight of 3.0 N would require a greater force from you —
namely, a 3.0 N force — to keep it at rest. The reason is that the gravitational
force you must balance has a greater magnitude — namely, 3.0 N. We say that this
second ball is heavier than the first ball.
Now let us generalize the situation. Consider a body that has an acceleration
→
a of zero relative to the ground, which we again assume to be an iner­tial frame.
→
Two forces act on the body: a downward gravitational force Fg and a balancing
upward force of magnitude W. We can write Newton’s second law for a vertical y
axis, with the positive direction upward, as
Fnet,y = may.
In our situation, this becomes
W − Fg = m(0)(5-10)
or W = Fg (weight, with ground as inertial frame).(5-11)

This equation tells us (assuming the ground is an inertial frame) that

Scale marked
 he weight W of a body is equal to the magnitude Fg of the gravitational force
T in either
on the body. weight or
mass units

Substituting mg for Fg from Eq. 5-8, we find

W = mg   (weight), (5-12)

which relates a body’s weight to its mass. Fg = mg

Weighing. To weigh a body means to measure its weight. One way to do this
is to place the body on one of the pans of an equal-arm balance (Fig. 5-5) and Figure 5-6 A spring scale. The reading is
proportional to the weight of the object
then place reference bodies (whose masses are known) on the other pan until
on the pan, and the scale gives that weight
we strike a balance (so that the gravitational forces on the two sides match). The
if marked in weight units. If, instead, it is
masses on the pans then match, and we know the mass of the body. If we know marked in mass units, the reading is the
the value of g for the location of the balance, we can also find the weight of the object’s weight only if the value of g at the
body with Eq. 5-12. location where the scale is being used is
We can also weigh a body with a spring scale (Fig. 5-6). The body stretches the same as the value of g at the location
a spring, moving a pointer along a scale that has been calibrated and marked in where the scale was calibrated.
104 CHAPTER 5 FORCE AND MOTION—I

e­ ither mass or weight units. (Most bathroom scales in the United States work
this way and are marked in the force unit pounds.) If the scale is marked in
mass units, it is accurate only where the value of g is the same as where the scale
was calibrated.
The weight of a body must be measured when the body is not accelerating
vertically relative to the ground. For example, you can measure your weight on
a scale in your bathroom or on a fast train. However, if you repeat the measure-
ment with the scale in an accelerating elevator, the reading differs from your
weight because of the acceleration. Such a measurement is called an apparent
weight.
Caution: A body’s weight is not its mass. Weight is the magnitude of a force
and is related to mass by Eq. 5-12. If you move a body to a point where the value
of g is different, the body’s mass (an intrinsic property) is not different but the
weight is. For example, the weight of a bowling ball having a mass of 7.2 kg is
71 N on Earth but only 12 N on the Moon. The mass is the same on Earth and
Moon, but the free-fall acceleration on the Moon is only 1.6 m/s2.

The Normal Force

If you stand on a mattress, Earth pulls you downward, but you remain stationary.
The reason is that the mattress, because it deforms downward due to you, pushes
up on you. Similarly, if you stand on a floor, it deforms (it is compressed, bent, or
buckled ever so slightly) and pushes up on you. Even a seemingly rigid concrete
floor does this (if it is not sitting directly on the ground, enough people on the
floor could break it).
→
The push on you from the mattress or floor is a normal force FN. The name
comes from the mathematical term normal, meaning perpendicular: The force on
you from, say, the floor is perpendicular to the floor.

 hen a body presses against a surface, the surface (even a seemingly rigid
W
→
one) ­deforms and pushes on the body with a normal force FN that is perpen-
dicular to the surface.

Figure 5-7a shows an example. A block of mass m presses down on a table,

→
deforming it somewhat because of the gravitational force Fg on the block. The
→
table pushes up on the block with normal force FN. The free-body diagram for the
→ →
block is given in Fig. 5-7b. Forces Fg and FN are the only two forces on the block
and they are both vertical. Thus, for the block we can write Newton’s ­second law
for a positive-upward y axis (Fnet, y = may) as
FN − Fg = may.

The normal force Normal force FN

is the force on
the block from the
supporting table. FN
Block
Block
x

The gravitational Fg
The forces
force on the block Fg balance.
is due to Earth’s
downward pull. (a) (b)

→
Figure 5-7 (a) A block resting on a table experiences a normal force FN perpendicular to
the tabletop. (b) The free-body diagram for the block.
 5-2 SOME PARTICULAR FORCES 105

From Eq. 5-8, we substitute mg for Fg, finding

FN − mg = may.
Then the magnitude of the normal force is
FN = mg + may = m(g + ay) (5-13)
for any vertical acceleration ay of the table and block (they might be in an accel-
erating elevator). (Caution: We have already included the sign for g but ay can be
positive or negative here.) If the table and block are not accelerating relative to
the ground, then ay = 0 and Eq. 5-13 yields
FN = mg. (5-14)

Checkpoint 3
→
In Fig. 5-7, is the magnitude of the normal force FN greater than, less than, or equal
to mg if the block and table are in an elevator moving upward (a) at constant speed
and (b) at increasing speed?

Friction
If we either slide or attempt to slide a body over a surface, the motion is resisted Direction of
by a bonding between the body and the surface. (We discuss this bonding more in attempted
→ slide
the next chapter.) The resistance is considered to be a single force f , called either
the frictional force or simply friction. This force is directed along the s­urface, f
→
opposite the direction of the intended motion (Fig. 5-8). Sometimes, to simplify Figure 5-8 A frictional force f opposes the
a situation, friction is assumed to be negligible (the surface, or even the body, is attempted slide of a body over a surface.
said to be ­frictionless).

Tension
When a cord (or a rope, cable, or other such object) is attached to a body and
→
pulled taut, the cord pulls on the body with a force T directed away from the
body and along the cord (Fig. 5-9a). The force is often called a tension force
­because the cord is said to be in a state of tension (or to be under tension), which
means that it is being pulled taut. The tension in the cord is the magnitude T of
the force on the body. For example, if the force on the body from the cord has
magnitude T = 50 N, the tension in the cord is 50 N.
A cord is often said to be massless (meaning its mass is negligible compared
to the body’s mass) and unstretchable. The cord then exists only as a connection
between two bodies. It pulls on both bodies with the same force magnitude T,

T T T

T
T
The forces at the two ends of T
the cord are equal in magnitude.

(a) (b ) (c )
Figure 5-9 (a) The cord, pulled taut, is under tension. If its mass is negligible, the cord
→
pulls on the body and the hand with force T , even if the cord runs around a massless,
­frictionless pulley as in (b) and (c).
106 CHAPTER 5 FORCE AND MOTION—I

even if the bodies and the cord are accelerating and even if the cord runs around
a massless, frictionless pulley (Figs. 5-9b and c). Such a pulley has negligible mass
compared to the bodies and negligible friction on its axle opposing its rotation. If
the cord wraps halfway around a pulley, as in Fig. 5-9c, the net force on the pulley
from the cord has the magnitude 2T.

Checkpoint 4
The suspended body in Fig. 5-9c weighs 75 N. Is T equal to, greater than, or less than
75 N when the body is moving upward (a) at constant speed, (b) at increasing speed,
and (c) at decreasing speed?

5-3 APPLYING NEWTON’S LAWS

Learning Objectives
After reading this module, you should be able to . . .
5.14 Identify Newton’s third law of motion and third-law 5.16 For an arrangement where a system of several
force pairs. objects moves rigidly together, draw a free-body dia-
5.15 For an object that moves vertically or on a horizontal gram and apply Newton’s second law for the individual
or inclined plane, apply Newton’s second law to a free- objects and also for the system taken as a composite
body diagram of the object. object.

Key Ideas
→ →
● The net force Fnet on a body with mass m is related to ● If a force FBC acts on body B due to body C, then there
→
the body’s acceleration →
a by is a force FCB on body C due to body B:
→
F m→a, net = → →
FBC = FCB.
which may be written in the component versions
Fnet,x = max Fnet,y = may and Fnet,z = maz. The forces are equal in magnitude but opposite in directions.

Newton’s Third Law

Two bodies are said to interact when they push or pull on each other — that is,
Book B Crate C when a force acts on each body due to the other body. For example, suppose you
position a book B so it leans against a crate C (Fig. 5-10a). Then the book and
→
crate interact: There is a horizontal force FBC on the book from the crate (or due
(a) →
to the crate) and a horizontal force FCB on the crate from the book (or due to the
book). This pair of forces is shown in Fig. 5-10b. Newton’s third law states that
FBC FCB

B C Newton’s Third Law: When two bodies interact, the forces on the bodies from
(b) each other are always equal in magnitude and opposite in direction.
The force on B
due to C has the same
magnitude as the For the book and crate, we can write this law as the scalar relation
force on C due to B. FBC = FCB    (equal magnitudes)
Figure 5-10 (a) Book B leans against crate or as the vector relation
→ → →
C. (b) Forces FBC (the force on the book
→ FBC =  − FCB    (equal magnitudes and opposite directions), (5-15)
from the crate) and FCB (the force on the
crate from the book) have the same mag- where the minus sign means that these two forces are in opposite directions. We
nitude and are opposite in direction. can call the forces between two interacting bodies a third-law force pair. When
 5-3 APPLYING NEWTON’S LAWS 107

Cantaloupe

FCE
Cantaloupe C

FEC These are

Table T
third-law force
Earth pairs.

Earth E

(a) (c )
F CT
F CT (normal force from table)
These forces So are these.
F TC
just happen
F CE (gravitational force)
to be balanced.
(b) (d )

Figure 5-11 (a) A cantaloupe lies on a table that stands on Earth. (b) The forces on
→ →
the cantaloupe are FCT and FCE. (c) The third-law force pair for the cantaloupe – Earth
interaction. (d) The third-law force pair for the cantaloupe – table interaction.

any two bodies interact in any situation, a third-law force pair is present. The
book and crate in Fig. 5-10a are stationary, but the third law would still hold if
they were moving and even if they were accelerating.
As another example, let us find the third-law force pairs involving the can-
taloupe in Fig. 5-11a, which lies on a table that stands on Earth. The cantaloupe
interacts with the table and with Earth (this time, there are three bodies whose
interactions we must sort out).
Let’s first focus on the forces acting on the cantaloupe (Fig. 5-11b). Force
→ →
FCT is the normal force on the cantaloupe from the table, and force FCE is the
gravitational force on the cantaloupe due to Earth. Are they a third-law force
pair? No, because they are forces on a single body, the cantaloupe, and not on
two interacting bodies.
To find a third-law pair, we must focus not on the cantaloupe but on the
­interaction between the cantaloupe and one other body. In the cantaloupe – Earth
interaction (Fig. 5-11c), Earth pulls on the cantaloupe with a gravitational force
→ →
FCE and the cantaloupe pulls on Earth with a gravitational force FEC. Are these
forces a third-law force pair? Yes, because they are forces on two interacting bod-
ies, the force on each due to the other. Thus, by Newton’s third law,
→ →
FCE =  − FEC    (cantaloupe – Earth interaction).
Next, in the cantaloupe – table interaction, the force on the cantaloupe from
→ →
the table is FCT and, conversely, the force on the table from the cantaloupe is FTC
(Fig. 5-11d). These forces are also a third-law force pair, and so
→ →
FCT =  − FTC    (cantaloupe – table interaction).

Checkpoint 5
Suppose that the cantaloupe and table of Fig. 5-11 are in an elevator cab that begins
→ →
to accelerate upward. (a) Do the magnitudes of FTC and FCT increase, decrease,
or stay the same? (b) Are those two forces still equal in magnitude and opposite in
→ →
­direction? (c) Do the magnitudes of FCE and FEC increase, ­decrease, or stay the same?
(d) Are those two forces still equal in magnitude and ­opposite in direction?
108 CHAPTER 5 FORCE AND MOTION—I

Applying Newton’s Laws

The rest of this chapter consists of sample problems. You should pore over
them, learning their procedures for attacking a problem. Especially important is
knowing how to translate a sketch of a situation into a free-body diagram with
appropriate axes, so that Newton’s laws can be applied.

Sample Problem 5.03 Block on table, block hanging

Figure 5-12 shows a block S (the sliding block) with mass FN Block S
M = 3.3 kg. The block is free to move along a horizontal
T
­frictionless surface and connected, by a cord that wraps over M
a frictionless pulley, to a second block H (the hanging block),
with mass m = 2.1 kg. The cord and pulley have negligible
masses compared to the blocks (they are “massless”). The FgS
T
hanging block H falls as the sliding block S accelerates to
the right. Find (a) the acceleration of block S, (b) the accel- m Block H
eration of block H, and (c) the tension in the cord.
FgH
Q   What is this problem all about?
You are given two bodies — sliding block and hanging
block — but must also consider Earth, which pulls on both Figure 5-13 The forces acting on the two blocks of Fig. 5-12.
bodies. (Without Earth, nothing would happen here.) A
total of five forces act on the blocks, as shown in Fig. 5-13:
­certain time, block S moves 1 mm to the right in that same
1. The cord pulls to the right on sliding block S with a force time. This means that the blocks move ­together and their
of magnitude T. ­accelerations have the same magnitude a.
2. The cord pulls upward on hanging block H with a force Q How do I classify this problem? Should it suggest a par-
of the same magnitude T. This upward force keeps block ticular law of physics to me?
H from falling freely. Yes. Forces, masses, and accelerations are involved,
3. Earth pulls down on block S with the gravitational force and they should suggest Newton’s second law of ­motion,
→ →
FgS, which has a magnitude equal to Mg. Fnet = m→ a . That is our starting key idea.
4. Earth pulls down on block H with the gravitational force Q If I apply Newton’s second law to this problem, to which
→
FgH, which has a magnitude equal to mg. body should I apply it?
→
5. The table pushes up on block S with a normal force FN. We focus on two bodies, the sliding block and the hang-
There is another thing you should note. We assume that ing block. Although they are extended objects (they are not
the cord does not stretch, so that if block H falls 1 mm in a points), we can still treat each block as a particle because
every part of it moves in exactly the same way. A second
key idea is to apply Newton’s second law s­ eparately to each
Sliding
block S
block.
Q What about the pulley?
M
We cannot represent the pulley as a particle ­because
­different parts of it move in different ways. When we discuss
rotation, we shall deal with pulleys in detail. Meanwhile,
Frictionless
surface we eliminate the pulley from consideration by assuming its
Hanging mass to be negligible compared with the masses of the two
m
block H blocks. Its only function is to change the cord’s orientation.
→
Q OK. Now how do I apply F = m→ net a to the sliding block?
Represent block S as a particle of mass M and draw all
Figure 5-12 A block S of mass M is connected to a block H of the forces that act on it, as in Fig. 5-14a. This is the block’s
mass m by a cord that wraps over a pulley. free-body diagram. Next, draw a set of axes. It makes sense
 5-3 APPLYING NEWTON’S LAWS 109

y We can now substitute mg for FgH and −a for ay (negative

­because block H accelerates in the negative direction of the
y y axis). We find
T − mg = −ma.(5-20)
a
FN Now note that Eqs. 5-18 and 5-20 are simultaneous equa-
T
M T m tions with the same two unknowns, T and a. Subtracting
x x
these equations eliminates T. Then solving for a yields
Hanging
Sliding a FgH block H m
FgS block S a= g.(5-21)
M+m
Substituting this result into Eq. 5-18 yields
(a) (b) Mm
T= g.(5-22)
Figure 5-14 (a) A free-body diagram for block S of Fig. 5-12. M+m
(b) A free-body diagram for block H of Fig. 5-12.
Putting in the numbers gives, for these two quantities,
m 2.1 kg
a= g= (9.8 m/s2)
M+m 3.3 kg + 2.1 kg
to draw the x axis parallel to the table, in the direction in
which the block moves. = 3.8 m/s2 (Answer)

Q Thanks, but you still haven’t told me how to apply Mm (3.3 kg)(2.1 kg)
→ and T= g= (9.8 m/s2)
Fnet = m→a to the sliding block. All you’ve done is M+m 3.3 kg + 2.1 kg
­explain how to draw a free-body diagram. = 13 N. (Answer)
You are right, and here’s the third key idea: The
→
­expression Fnet = M→
Q The problem is now solved, right?
a is a vector equation, so we can write
it as three component equations: That’s a fair question, but the problem is not really
finished until we have examined the results to see whether
Fnet, x = Max   Fnet, y = May   Fnet, z = Maz (5-16) they make sense. (If you made these calculations on the job,
in which Fnet, x, Fnet,y, and Fnet, z are the components of the wouldn’t you want to see whether they made sense before
net force along the three axes. Now we apply each compo- you turned them in?)
nent equation to its corresponding direction. Because block Look first at Eq. 5-21. Note that it is dimensionally
S does not accelerate vertically, Fnet, y = May becomes ­correct and that the acceleration a will always be less than
g (because of the cord, the hanging block is not in free fall).
FN − FgS = 0   or   FN = FgS. (5‑17) Look now at Eq. 5-22, which we can rewrite in the form
Thus in the y direction, the magnitude of the normal force is M
equal to the magnitude of the gravitational force. T= mg.(5-23)
M+m
No force acts in the z direction, which is perpendicular
to the page. In this form, it is easier to see that this equation is also
In the x direction, there is only one force component, ­ imensionally correct, because both T and mg have dimen-
d
which is T. Thus, Fnet, x = Max becomes sions of forces. Equation 5-23 also lets us see that the ten-
sion in the cord is always less than mg, and thus is always
T = Ma.(5-18) less than the gravitational force on the hanging block. That
This equation contains two unknowns, T and a; so we cannot is a comforting thought because, if T were greater than mg,
yet solve it. Recall, however, that we have not said anything the hanging block would accelerate ­upward.
about the hanging block. We can also check the results by studying special cases,
→
Q I agree. How do I apply F = m→ a to the hanging block?
net
in which we can guess what the answers must be. A simple
­example is to put g = 0, as if the experiment were carried out
We apply it just as we did for block S: Draw a free-body
→ in interstellar space. We know that in that case, the blocks
diagram for block H, as in Fig. 5-14b. Then apply Fnet = m→ a
would not move from rest, there would be no forces on the
in component form. This time, because the acceleration is
ends of the cord, and so there would be no tension in the
along the y axis, we use the y part of Eq. 5-16 (Fnet, y = may)
cord. Do the formulas predict this? Yes, they do. If you put
to write
g = 0 in Eqs. 5-21 and 5-22, you find a = 0 and T = 0. Two
T − FgH = may. (5-19) more special cases you might try are M = 0 and m → ∞.

Additional examples, video, and practice available at WileyPLUS

 110 CHAPTER 5 FORCE AND MOTION—I

Sample Problem 5.04 Cord accelerates box up a ramp

Many students consider problems involving ramps (inclined perpendicular to the plane), as expressed by Newton’s sec-
planes) to be especially hard. The difficulty is probably visual ond law (Eq. 5-1).
because we work with (a) a tilted coordinate system and (b) the
components of the gravitational force, not the full force. Here Calculations: We need to write Newton’s second law for
is a typical example with all the tilting and angles explained. motion along an axis. Because the box moves along the
(In WileyPLUS, the figure is available as an animation with inclined plane, placing an x axis along the plane seems rea-
voiceover.) In spite of the tilt, the key idea is to apply Newton’s sonable (Fig. 5-15b). (There is nothing wrong with using our
second law to the axis along which the motion occurs. usual coordinate system, but the expressions for compo-
In Fig. 5-15a, a cord pulls a box of sea biscuits up along nents would be a lot messier because of the misalignment of
a frictionless plane inclined at angle θ = 30.0°. The box has the x axis with the motion.)
mass m = 5.00 kg, and the force from the cord has magni- After choosing a coordinate system, we draw a free-
tude T = 25.0 N. What is the box’s acceleration a along the body diagram with a dot representing the box (Fig. 5-15b).
inclined plane? Then we draw all the vectors for the forces acting on the box,
with the tails of the vectors anchored on the dot. (Drawing
KEY IDEA the vectors willy-nilly on the diagram can easily lead to errors,
especially on exams, so always anchor the tails.)
→
The acceleration along the plane is set by the force Force T from the cord is up the plane and has magni-
→
components along the plane (not by force components tude T = 25.0 N. The gravitational force Fg is downward (of

A Figure 5-15 (a) A box is pulled up a plane

by a cord. (b) The three forces acting on the
y

→ Normal force
box: the cord’s force T , the gravitational
→ → FN x
force Fg, and the normal force FN. (c)–(i) Cord
­Finding the force components along the T
plane and perpendicular to it. In WileyPLUS, The box accelerates. Cord’s pull
this figure is available as an animation with
voiceover. Fg
θ θ Gravitational
force
(a) (b)

This is a right Perpendicular

triangle. 90° – θ This is also. 90° – θ component of Adjacent leg
Fg (use cos θ )
θ θ θ
Fg
Hypotenuse
θ

Parallel Opposite leg

(c) (d) (e) component of (f ) (use sin θ )
Fg

The net of these y

These forces
forces determines x merely balance.
the acceleration. T FN x
θ mg cos θ mg sin θ
mg

mg sin θ mg cos θ

(g) (h) (i)

 5-3 APPLYING NEWTON’S LAWS 111

course) and has magnitude mg = (5.00 kg)(9.80 m/s2) = 49.0 N. plane and thus cannot affect the motion along the plane. (It
That direction means that only a component of the force is has no component along the plane to accelerate the box.)
along the plane, and only that component (not the full force) We are now ready to write Newton’s second law for
affects the box’s acceleration along the plane. Thus, before we motion along the tilted x axis:
can write Newton’s second law for motion along the x axis, we
Fnet,x = max.
need to find an expression for that important component.
Figures 5-15c to h indicate the steps that lead to the The component ax is the only component of the acceleration
expression. We start with the given angle of the plane and (the box is not leaping up from the plane, which would be
work our way to a triangle of the force components (they are strange, or descending into the plane, which would be even
the legs of the triangle and the full force is the hypotenuse). stranger). So, let’s simply write a for the acceleration along the
→ →
Figure 5-15c shows that the angle between the ramp and Fg is plane. Because T is in the positive x direction and the compo-
90° − θ. (Do you see a right triangle there?) Next, Figs. 5-15d nent mg sin θ is in the negative x direction, we next write
→
to f show Fg and its components: One component is parallel
T − mg sin θ = ma. (5-24)
to the plane (that is the one we want) and the other is perpen-
dicular to the plane. Substituting data and solving for a, we find
Because the perpendicular component is perpendicu-
→ a = 0.100 m/s2. (Answer)
lar, the angle between it and Fg must be θ (Fig. 5-15d). The
component we want is the far leg of the component right The result is positive, indicating that the box accelerates up
triangle. The magnitude of the hypotenuse is mg (the mag- the inclined plane, in the positive direction of the tilted x
→
nitude of the gravitational force). Thus, the component we axis. If we decreased the magnitude of T enough to make
want has magnitude mg sin θ (Fig. 5-15g). a = 0, the box would move up the plane at constant speed.
→
We have one more force to consider, the normal force And if we decrease the magnitude of T even more, the
→
FN shown in Fig. 5-15b. However, it is perpendicular to the acceleration would be negative in spite of the cord’s pull.

When F1 is horizontal,
Sample Problem 5.05 Reading a force graph
the acceleration is
Here is an example of where you must dig information out 3.0 m/s2.
of a graph, not just read off a number. In Fig. 5-16a, two 3
forces are applied to a 4.00 kg block on a frictionless floor,
→ F1
2
but only force F1 is indicated. That force has a fixed mag- θ ax (m/s2)
nitude but can be applied at an adjustable angle θ to the x
→ 1
positive direction of the x axis. Force F2 is horizontal and (a)
fixed in both magnitude and angle. Figure 5-16b gives the
0
horizontal acceleration ax of the block for any given value of 0° 90°
θ from 0° to 90°. What is the value of ax for θ = 180°? θ
(b)
When F1 is vertical,
the acceleration is
0.50 m/s2.
KEY IDEAS
Figure 5-16 (a) One of the two forces applied to a block is shown.
(1) The horizontal acceleration ax depends on the net hori- Its angle θ can be varied. (b) The block’s acceleration component
zontal force Fnet, x, as given by Newton’s second law. (2) The ax versus θ.
net horizontal force is the sum of the horizontal components
→ → c­ orresponding acceleration is 0.50 m/s2. Thus, F2 = 2.00 N
of forces F1 and F2. →
and F2 must be in the positive direction of the x axis.
→ From Eq. 5-25, we find that when θ = 0°,
Calculations: The x component of F2 is F2 because the ­vector
→
is horizontal. The x component of F1 is F1 cos θ. Using these F1 cos 0° + 2.00 = 4.00ax. (5-26)
expressions and a mass m of 4.00 kg, we can write Newton’s
→ → From the graph we see that the corresponding accel­eration
second law ( Fnet = m a ) for motion along the x axis as
is 3.0 m/s2. From Eq. 5-26, we then find that F1 = 10 N.
F1 cos θ + F2 = 4.00ax. (5-25) Substituting F1 = 10 N, F2 = 2.00 N, and θ = 180° into
From this equation we see that when angle θ = 90°, F1 cos θ Eq. 5-25 leads to
is zero and F2 = 4.00ax. From the graph we see that the ax = −2.00 m/s2. (Answer)

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112 CHAPTER 5 FORCE AND MOTION—I

Sample Problem 5.06 Forces within an elevator cab

Although people would surely avoid getting into the ele- This tells us that the scale reading, which is equal to normal
vator with you, suppose that you weigh yourself while on force magnitude FN, d ­ epends on the vertical acceleration. Sub-
an elevator that is moving. Would you weigh more than, stituting mg for Fg gives us
less than, or the same as when the scale is on a stationary
FN = m(g + a) (Answer) (5-28)
floor?
In Fig. 5-17a, a passenger of mass m = 72.2 kg stands on for any choice of acceleration a. If the acceleration is upward,
a platform scale in an elevator cab. We are concerned with a is positive; if it is downward, a is negative.
the scale readings when the cab is stationary and when it is
(b) What does the scale read if the cab is stationary or
moving up or down.
­ oving upward at a constant 0.50 m/s?
m
(a) Find a general solution for the scale reading, whatever
the vertical motion of the cab. KEY IDEA

KEY IDEAS For any constant velocity (zero or otherwise), the accelera-
tion a of the passenger is zero.
(1) The reading is equal to the magnitude of the normal
→
force FN on the passenger from the scale. The only other Calculation: Substituting this and other known values into
→
force ­acting on the passenger is the gravitational force Fg, Eq. 5-28, we find
as shown in the free-body diagram of Fig. 5-17b. (2) We
FN = (72.2 kg)(9.8 m/s2 + 0) = 708 N.
can relate the forces on the ­passenger to his acceleration → a
→
by using ­Newton’s second law ( Fnet = m→
(Answer)
a ). However, recall
that we can use this law only in an inertial frame. If the cab This is the weight of the passenger and is equal to the mag-
accelerates, then it is not an ­inertial frame. So we choose the nitude Fg of the gravitational force on him.
ground to be our inertial frame and make any measure of the (c) What does the scale read if the cab accelerates ­upward at
passenger’s acceleration relative to it. 3.20 m/s2 and downward at 3.20 m/s2?
Calculations: Because the two forces on the passenger and
Calculations: For a = 3.20 m/s2, Eq. 5-28 gives
his acceleration are all directed vertically, along the y axis
in Fig. 5-17b, we can use Newton’s second law written for y FN = (72.2 kg)(9.8 m/s2 + 3.20 m/s2)
components (Fnet, y = may) to get = 939 N, (Answer)
FN − Fg = ma and for a = −3.20 m/s2, it gives
or FN = Fg + ma.(5-27)
FN = (72.2 kg)(9.8 m/s2 − 3.20 m/s2)
= 477 N. (Answer)
y
For an upward acceleration (either the cab’s upward
speed is increasing or its downward speed is decreasing),
FN
the scale reading is greater than the passenger’s weight.
That reading is a measurement of an apparent weight,
because it is made in a noninertial frame. For a downward
acceleration (either d
­ ecreasing upward speed or increas-
ing downward speed), the scale reading is less than the
Passenger
passenger’s weight.
These forces
(d) During the upward acceleration in part (c), what is the
compete.
Fg magnitude Fnet of the net force on the passenger, and what
Their net force
is the magnitude ap,cab of his acceleration as measured in the
causes a vertical →
frame of the cab? Does Fnet = m→ ap,cab?
(a ) (b ) acceleration.
Figure 5-17 (a) A passenger stands on a platform scale that indi- Calculation: The magnitude Fg of the gravitational force on
cates either his weight or his apparent weight. (b) The free-body the passenger does not depend on the motion of the passen-
→
diagram for the passenger, showing the normal force FN on him ger or the cab; so, from part (b), Fg is 708 N. From part (c), the
→
from the scale and the gravitational force Fg. magnitude FN of the normal force on the passenger during the
 5-3 APPLYING NEWTON’S LAWS 113

upward acceleration is the 939 N reading on the scale. Thus, during the upward acceleration. However, his acceleration
the net force on the passenger is ap,cab relative to the frame of the cab is zero. Thus, in the
noninertial frame of the accelerating cab, Fnet is not equal to
Fnet = FN − Fg = 939 N − 708 N = 231 N, (Answer) map,cab, and Newton’s second law does not hold.

Sample Problem 5.07 Acceleration of block pushing on block

Some homework problems involve objects that move →

Dead-End Solution: Let us now include force FAB by writ-
together, because they are either shoved together or tied ing, again for the x axis,
together. Here is an example in which you apply Newton’s
second law to the composite of two blocks and then to the Fapp − FAB = mAa.
individual blocks. →
→ (We use the minus sign to include the direction of FAB.) Because
In Fig. 5-18a, a constant horizontal force Fapp of magni- FAB is a second unknown, we cannot solve this equation for a.
tude 20 N is applied to block A of mass mA = 4.0 kg, which
pushes against block B of mass mB = 6.0 kg. The blocks slide Successful Solution: Because of the direction in which
over a frictionless surface, along an x axis. →
force Fapp is applied, the two blocks form a rigidly connected
(a) What is the acceleration of the blocks? system. We can relate the net force on the system to the accel-
eration of the system with Newton’s second law. Here, once
→ again for the x axis, we can write that law as
Serious Error: Because force Fapp is applied directly to block
A, we use Newton’s second law to relate that force to the Fapp = (mA + mB)a,
­acceleration →
a of block A. Because the motion is along the
→
x axis, we use that law for x components (Fnet, x = max), writing where now we properly apply Fapp to the system with
it as total mass mA + mB. Solving for a and substituting known
values, we find
Fapp = mAa.
→ F app 20 N
However, this is seriously wrong because Fapp is not the only a= = = 2.0 m/s2.
mA + mB 4.0 kg + 6.0 kg
horizontal force acting on block A. There is also the force
→
FAB from block B (Fig. 5-18b). (Answer)

Thus, the acceleration of the system and of each block is in the

B positive direction of the x axis and has the magnitude 2.0 m/s2.
This force causes the
Fapp A →
acceleration of the full (b) What is the (horizontal) force FBA on block B from
x
two-block system. block A (Fig. 5-18c)?
(a)

KEY IDEA
These are the two forces
Fapp FAB
A x acting on just block A. We can relate the net force on block B to the block’s accel-
Their net force causes eration with Newton’s second law.
(b) its acceleration.
Calculation: Here we can write that law, still for compo-
nents along the x axis, as
B
This is the only force FBA = mBa,
FBA
x causing the acceleration
of block B. which, with known values, gives
(c)
→ FBA = (6.0 kg)(2.0 m/s2) = 12 N. (Answer)
Figure 5-18 (a) A constant horizontal force Fapp is applied to →
block A, which pushes against block B. (b) Two horizontal forces Thus, force FBA is in the positive direction of the x axis and
act on block A. (c) Only one horizontal force acts on block B. has a magnitude of 12 N.

Additional examples, video, and practice available at WileyPLUS

114 CHAPTER 5 FORCE AND MOTION—I

Review & Summary

Newtonian Mechanics   The velocity of an object can change A free-body diagram is a stripped-down diagram in which only
(the object can accelerate) when the object is acted on by one one body is considered. That body is represented by e­ ither a sketch
or more forces (pushes or pulls) from other o
­ bjects. Newtonian or a dot. The external forces on the body are drawn, and a coordi-
mechanics relates accelerations and forces. nate system is superimposed, oriented so as to simplify the solution.
→
Force  Forces are vector quantities. Their magnitudes are Some Particular Forces  A gravitational force Fg on a body
defined in terms of the acceleration they would give the standard is a pull by another body. In most situations in this book, the other
kilogram. A force that accelerates that standard body by exactly body is Earth or some other astronomical body. For Earth, the
1 m/s2 is defined to have a magnitude of 1 N. The ­direction of a force is directed down toward the ground, which is assumed → to be
force is the direction of the acceleration it causes. Forces are com- an inertial frame. With that assumption, the magnitude of Fg is
bined according to the rules of vector algebra. The net force on a Fg = mg,(5-8)
body is the vector sum of all the forces acting on the body.
where m is the body’s mass and g is the magnitude of the free-fall
Newton’s First Law   If there is no net force on a body, the acceleration.
body remains at rest if it is initially at rest or moves in a straight line The weight W of a body is the magnitude of the upward force
at constant speed if it is in motion. needed to balance the gravitational force on the body. A body’s
weight is related to the body’s mass by
Inertial Reference Frames  Reference frames in which W = mg.(5-12)
Newtonian mechanics holds are called inertial reference frames or →
inertial frames. Reference frames in which Newtonian mechanics A normal force FN is the force on a body from a surface
does not hold are called noninertial reference frames or noniner- against which the body presses. The normal force is always perpen-
tial frames. dicular to the surface.
→
A frictional force f is the force on a body when the body
Mass  The mass of a body is the characteristic of that body that slides or attempts to slide along a surface. The force is always par-
relates the body’s acceleration to the net force causing the accel- allel to the surface and directed so as to oppose the sliding. On a
eration. Masses are scalar quantities. frictionless surface, the frictional force is negligible.
When a cord is under tension, each end of the cord pulls on a
→ body. The pull is directed along the cord, away from the point of
Newton’s Second Law   The net force Fnet on a body with
→ attachment to the body. For a massless cord (a cord with negligible
mass m is related to the body’s acceleration a by
mass), the pulls at both ends of the cord have the same magnitude
→ → T, even if the cord runs around a massless, frictionless pulley (a pul-
Fnet = m a ,(5-1)
ley with negligible mass and ­negligible friction on its axle to oppose
which may be written in the component versions its rotation).

Fnet, x = max   Fnet, y = may   and   Fnet, z = maz.(5-2) →

Newton’s Third Law   If→a force FBC acts on body B due to
body C, then there is a force FCB on body C due to body B:
The second law indicates that in SI units
→ →
1 N = 1 kg · m/s2.(5-3) FBC = − FCB.

Questions
→
1  Figure 5-19 gives the free-body diagram for four situations (b) a y component? (c) In each situation, give the direction of a
in which an object is pulled by several forces across a friction- by naming either a quadrant or a direction along an axis. (Don’t
less floor, as seen from overhead. In which situations does reach for the calculator because this can be answered with a few
→
the acceleration a of the object have (a) an x component and mental calculations.)
y y y y
7N
6N 6N

3N
3N 3N 2N
2N 5N 3N
x x x x
2N 5N 2N 4N 5N
3N
4N 4N 4N 4N
5N

(1) (2) (3) (4)

Figure 5-19 Question 1.
 QUESTIONS 115

2  Two horizontal forces, 7 July 17, 1981, Kansas City: The newly opened Hyatt
→ → Regency is packed with people listening and dancing to a band
F1 = (3 N)î − (4 N)ĵ and F2 = −(1 N)î − (2 N)ĵ playing favorites from the 1940s. Many of the people are crowded
pull a banana split across a friction- onto the walkways that hang like bridges across the wide atrium.
y Suddenly two of the walkways collapse, falling onto the merry-
less lunch counter. Without u ­ sing a
1 4 makers on the main floor.
calculator, determine which of the
vectors in the ­free­‑­body ­diagram 2 3 The walkways were suspended one above another on verti-
→ cal rods and held in place by nuts threaded onto the rods. In the
of Fig. 5-20 best represent (a) F1
→ original design, only two long rods were to be used, each extend-
and (b) F2. What is the net-force
component along (c) the x axis and x ing through all three walkways (Fig. 5-24a). If each walkway and
(d) the y axis? Into which quad- the merrymakers on it have a combined mass of M, what is the
rants do (e) the net-force vector total mass supported by the threads and two nuts on (a) the low-
7 6 est walkway and (b) the highest walkway?
and (f) the split’s acceleration vec-
tor point? 8 5 Apparently someone responsible for the actual construc-
→ → tion realized that threading nuts on a rod is impossible except
3  In Fig. 5-21, forces F1 and F2 are at the ends, so the design was changed: Instead, six rods were
Figure 5-20 Question 2.
applied to a lunchbox as it slides at used, each connecting two walkways (Fig. 5-24b). What now is
constant velocity over a frictionless the total mass supported by the threads and two nuts on (c) the
floor. We are to decrease angle θ F1
lowest walkway, (d) the upper side of the highest walkway, and
without changing the magnitude of
→ (e) the lower side of the highest walkway? It was this design that
F1. For constant velocity, should we F2 θ failed on that tragic night—a simple engineering error.
­in­crease, decrease, or maintain the
→
magnitude of F2?
→ Figure 5-21 Question 3.
4  At time t = 0, constant F Rods Walkways
begins to act on a rock moving
through deep space in the +x
direction. (a) For time t > 0, which are possible f­unctions x(t)
for the rock’s position: (1) x = 4t − 3, (2) x = −4t 2 + 6t − 3,
→ Nuts
(3) x = 4t 2 + 6t − 3? (b) For which function is F ­directed oppo-
site the rock’s initial direction of motion?
5  Figure 5-22 shows overhead views of four situations in which
forces act on a block that lies on a frictionless floor. If the force
magnitudes are chosen properly, in which situations is it possible (a) (b)
that the block is (a) stationary and (b) moving with a constant Figure 5-24 Question 7.
velocity?
F1 8  Figure 5-25 gives three graphs of velocity component vx(t) and
three graphs of velocity component vy(t). The graphs are not to
(1) F2 (2) F1 F2 scale. Which vx(t) graph and which vy(t) graph best correspond to
each of the four situations in Question 1 and Fig. 5-19?

F1
F3 vx vx vx
F1
F2
(3) F2 (4)
F3
t t t
Figure 5-22 Question 5.

6  Figure 5-23 shows the same breadbox in four situations where

horizontal forces are applied. Rank the situations a­ ccording to (a) (b) (c)
the magnitude of the box’s acceleration, greatest first.
vy vy vy
3N 6N 58 N 60 N

(a) (b) t t t

13 N 15 N 43 N 25 N

20 N
(c ) (d) (d ) (e) (f )
Figure 5-23 Question 6. Figure 5-25 Question 8.
116 CHAPTER 5 FORCE AND MOTION—I

→
9  Figure 5-26 shows a train of four blocks being pulled across force F32 on block 3 from block 2? (d) Rank the blocks according
→
a frictionless floor by force F . What total mass is accelerated to to their acceleration magnitudes, greatest first. (e) Rank forces
→ → → →
the right by (a) force F , (b) cord 3, and (c) cord 1? (d) Rank the F , F21, and F32 according to magnitude, greatest first.
blocks according to their accelerations, greatest first. (e) Rank →
11  A vertical force F is applied to a block of mass m that lies on
the cords according to their tension, greatest first. →
a floor. What happens to the magnitude of the normal force F N
on the block from the floor as magnitude F is ­increased from
Cord Cord Cord →
zero if force F is (a) downward and (b) upward?
1 2 3 F
10 kg 3 kg 5 kg 2 kg 12  Figure 5-28 shows four choices for the direction of a force
of magnitude F to be applied to
Figure 5-26 Question 9. a block on an inclined plane. The b
directions are ­either horizontal or
10  Figure 5-27 shows three blocks 10 kg vertical. (For choice b, the force is
5 kg a
being pushed across a friction- 2 kg not enough to lift the block off the c
→ F
less floor by horizontal force F . plane.) Rank the choices accord-
What total mass is ­accelerated to 1 2 3 ing to the magnitude of the normal
→ 30° d
the right by (a) force F , (b) force force acting on the block from the
→
F21 on block 2 from block 1, and (c) Figure 5-27 Question 10. plane, greatest first. Figure 5-28 Question 12.

Problems
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual WWW Worked-out solution is at
http://www.wiley.com/college/halliday
• – ••• Number of dots indicates level of problem difficulty ILW Interactive solution is at
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com

Module 5-1   Newton’s First and Second Laws ••6  In a two-dimensional tug-of-
•1   Only two horizontal forces act on a 3.0 kg body that can move war, Alex, Betty, and Charles pull Alex
over a frictionless floor. One force is 9.0 N, acting due east, and the horizontally on an automobile tire Charles
other is 8.0 N, acting 62° north of west. What is the magnitude of at the angles shown in the overhead
the body’s acceleration? view of Fig. 5-30. The tire remains
­stationary in spite of the three pulls.
•2   Two horizontal forces act on a 2.0 kg chopping block that can →
Alex pulls with force FA of magni-
slide over a frictionless kitchen counter, which lies in an xy plane.
→ tude 220 N, and Charles pulls with
→
One force is F1 = (3.0 N)î + (4.0 N)ĵ. Find the acceleration of force FC of magnitude 170 N. Note Betty
→ 137°
the chopping block in unit-vector notation when the other force is
→ → that the direction of FC is not given.
(a) F2 = (−3.0 N)î + (−4.0 N)ĵ, (b) F 2 = (−3.0 N)î + (4.0 N)ĵ,
→ What is the magnitude of Betty’s
→
and (c) F2 = (3.0 N)î + (−4.0 N)ĵ. force FB?
•3  If the 1 kg standard body has an acceleration of 2.00 m/s2 ••7 SSM There are two forces on the Figure 5-30 Problem 6.
at 20.0° to the positive direction of an x axis, what are (a) the x 2.00 kg box in the overhead view of
component and (b) the y component of the net force acting on the Fig. 5-31, but only one is shown. For
body, and (c) what is the net force in unit-vector notation? F1 = 20.0 N, a = 12.0 m/s2, and θ = 30.0°, y
y find the second force (a) in unit-­vector
••4  While two forces act on it, a
particle is to move at the constant
­ notation and as (b) a magnitude and
F1
velocity →v = (3 m/s)î − (4 m/s)ĵ. One (c) an angle r­elative to the positive x
→ direction of the x axis.
of the forces is F1 = (2 N)î + F1
(−6 N)ĵ. What is the other force? θ1 ••8   A 2.00 kg object is subjected to θ
x a
••5 Three astronauts, propelled θ3 F2
three forces that give it an acceleration
→
by jet backpacks, push and guide a a = −(8.00 m/s2)î + (6.00 m/s2)ĵ. If
two of the three forces are Figure 5-31 Problem 7.
120 kg asteroid toward a processing F3 → →
dock, exerting the forces shown in F1 = (30.0 N)î + (16.0 N)ĵ and F2 =
Fig. 5-29, with F1 = 32 N, F2 = 55 N, Figure 5-29 Problem 5. −(12.0 N)î + (8.00 N)ĵ, find the third force.
F3 = 41 N, θ1 = 30°, and θ3 = 60°. ••9  A 0.340 kg particle moves in an xy plane according
What is the asteroid’s acceleration (a) in unit-vector notation and to x(t) = −15.00 + 2.00t − 4.00t3 and y(t) = 25.00 + 7.00t − 9.00t2,
as (b) a magnitude and (c) a direction relative to the positive direc- with x and y in meters and t in seconds. At t = 0.700 s, what
tion of the x axis? are (a) the magnitude and (b) the angle (relative to the positive
 PROBLEMS 117

direction of the x axis) of the net force on the particle, and (c) what reading on the scale? (This is the way by a deli owner who was once
is the angle of the particle’s direction of travel? a physics major.)
••10 A 0.150 kg particle moves along an x axis according
to x(t) = −13.00 + 2.00t + 4.00t2 − 3.00t3, with x in meters and t in Spring scale
seconds. In unit-vector notation, what is the net force acting on the
particle at t = 3.40 s?
••11   A 2.0 kg particle moves along an x axis, being propelled by
a variable force directed along that axis. Its position is given by Spring
x = 3.0 m + (4.0 m/s)t + ct 2 − (2.0 m/s3)t 3, with x in meters and t in scale N

O
GE

A
seconds. The factor c is a constant. At t = 3.0 s, the force on the

I
SA
LA

M
particle has a magnitude of 36 N and is in the negative direction of
the axis. What is c? (b)
→ →
•••12 Two horizontal forces F1 and F2 act on a 4.0 kg disk
that slides over frictionless ice, on which an xy coordinate system is Spring scale
→
laid out. Force F1 is in the positive direction of the x axis and has a N
→

O
GE
magnitude of 7.0 N. Force F2 has a magnitude of 9.0 N. Figure 5-32

A
I
SA
gives the x component vx of the velocity of the disk as a function of LA

M
time t during the sliding. What is the angle between the constant
→ →
directions of forces F1 and F2? N N

O
GE

GE
A

A
I

I
SA

SA
(a)
LA LA

M
(c)
vx (m/s)
Figure 5-34 Problem 15.
4
2
••16   Some insects can walk below Rod
t (s) a thin rod (such as a twig) by hang- Leg
0 1 2 3 ing from it. Suppose that such an joint Tibia
–2 θ
insect has mass m and hangs from a
–4 horizontal rod as shown in Fig. 5‑35,
with angle θ = 40°. Its six legs are all
Figure 5-32 Problem 12.
under the same tension, and the leg Figure 5-35 Problem 16.
sections nearest the body are hori-
Module 5-2   Some Particular Forces zontal. (a) What is the ratio of the
•13  Figure 5-33 shows an arrangement in tension in each tibia (forepart of a leg) to the insect’s weight? (b) If
which four disks are suspended by cords. The the insect straightens out its legs somewhat, does the tension in each
longer, top cord loops over a frictionless pul- tibia increase, ­decrease, or stay the same?
ley and pulls with a force of magnitude 98 N on Module 5-3   Applying Newton’s Laws
the wall to which it is attached. The tensions A •17 SSM WWW In Fig. 5-36,
in the three shorter cords are T1 = 58.8 N, T1 let the mass of the block be
T2 = 49.0 N, and T3 = 9.8 N. What are the B 8.5 kg and the angle θ be 30°.
masses of (a) disk A, (b) disk B, (c) disk C, T2 Find (a) the tension in the cord
and (d) disk D? C and (b) the normal force acting ess
tionl
•14  A block with a weight of 3.0 N is at T3 on the block. (c) If the cord is m Fric
rest on a horizontal surface. A 1.0 N upward D cut, find the magnitude of the
force is applied to the block by means of resulting acceleration of the θ
an attached vertical string. What are the (a) Figure 5-33   block.
mag­nitude and (b) direction of the force of Problem 13. •18 In April 1974, John Figure 5-36 Problem 17.
the block on the ­horizontal surface? Massis of Belgium managed
•15 SSM (a) An 11.0 kg salami is supported by a cord that runs to to move two passenger rail-
a spring scale, which is supported by a cord hung from the ceiling road cars. He did so by clamping his teeth down on a bit that was
(Fig. 5-34a). What is the reading on the scale, which is marked in SI attached to the cars with a rope and then leaning backward while
weight units? (This is a way to measure weight by a deli owner.) (b) pressing his feet against the railway ties. The cars together weighed
In Fig. 5-34b the salami is supported by a cord that runs around a 700 kN (about 80 tons). Assume that he pulled with a constant
pulley and to a scale. The opposite end of the scale is attached by a force that was 2.5 times his body weight, at an upward angle θ of
cord to a wall. What is the reading on the scale? (This is the way by 30° from the horizontal. His mass was 80 kg, and he moved the cars
a physics major.) (c) In Fig. 5-34c the wall has been replaced with a by 1.0 m. Neglecting any retarding force from the wheel rotation,
second 11.0 kg salami, and the assembly is stationary. What is the find the speed of the cars at the end of the pull.
118 CHAPTER 5 FORCE AND MOTION—I

•19 SSM A 500 kg rocket sled can be accelerated at a constant •24   There are two horizontal F1
rate from rest to 1600 km/h in 1.8 s. What is the magnitude of the forces on the 2.0 kg box in the over- x
required net force? head view of Fig. 5-38 but only one
•20   A car traveling at 53 km/h hits a bridge abutment. A passen- (of magnitude F1 = 20 N) is shown. Figure 5-38 Problem 24.
ger in the car moves forward a distance of 65 cm (with respect to The box moves along the x axis.
the road) while being brought to rest by an inflated air bag. What For each of the following values for the acceleration ax of the box,
magnitude of force (assumed constant) acts on the passenger’s find the second force in unit-vector notation: (a) 10 m/s2, (b) 20 m/s2,
upper torso, which has a mass of 41 kg? (c) 0, (d) −10 m/s2, and (e) −20 m/s2.
→ •25  Sunjamming. A “sun yacht” is a spacecraft with a large sail
•21   A constant horizontal force Fa pushes a 2.00 kg FedEx pack-
age across a frictionless floor on which an xy coordinate system that is pushed by sunlight. Although such a push is tiny in every-
has been drawn. Figure 5-37 gives the package’s x and y veloc- day circumstances, it can be large enough to send the spacecraft
ity components versus time t. What are the (a) magnitude and ­outward from the Sun on a cost-free but slow trip. Suppose that
→ the spacecraft has a mass of 900 kg and receives a push of 20 N.
(b) ­direction of Fa?
(a) What is the magnitude of the resulting ­acceleration? If the craft
starts from rest, (b) how far will it travel in 1 day and (c) how fast
vx (m/s) will it then be moving?

10
•26   The tension at which a fishing line snaps is commonly called the
line’s “strength.” What minimum strength is needed for a line that is
to stop a salmon of weight 85 N in 11 cm if the fish is initially drifting
at 2.8 m/s? Assume a constant deceleration.
5
•27 SSM An electron with a speed of 1.2 × 10 7 m/s moves horizon-
tally into a region where a constant vertical force of 4.5 × 10−16 N
acts on it. The mass of the electron is 9.11 × 10−31 kg. Determine
t (s) the vertical distance the electron is deflected ­during the time it has
0 1 2 3 moved 30 mm horizontally.
vy (m/s) •28   A car that weighs 1.30 × 10 4 N is initially moving at 40 km/h
when the brakes are applied and the car is brought to a stop in
0 t (s) 15 m. Assuming the force that stops the car is constant, find
0 1 2 3 (a) the magnitude of that force and (b) the time ­required for the
change in speed. If the initial speed is ­doubled, and the car expe-
–5 riences the same force during the braking, by what factors are
(c) the stopping distance and (d) the stopping time multiplied?
(There could be a lesson here about the danger of driving at high
–10 speeds.)
•29   A firefighter who weighs 712 N slides down a vertical pole
Figure 5-37 Problem 21. with an acceleration of 3.00 m/s2, directed downward. What are
the (a) magnitude and (b) direction (up or down) of the vertical
force on the firefighter from the pole and the (c) magnitude and
•22 A customer sits in an amusement park ride in which the (d) direction of the vertical force on the pole from the firefighter?
compartment is to be pulled downward in the negative direction of
•30 The high-speed winds around a tornado can drive pro-
a y axis with an acceleration magnitude of 1.24g, with g = 9.80 m/s2.
jectiles into trees, building walls, and even metal traffic signs. In
A 0.567 g coin rests on the customer’s knee. Once the motion
a laboratory simulation, a standard wood toothpick was shot by
begins and in unit-vector notation, what is the coin’s acceleration
pneumatic gun into an oak branch. The toothpick’s mass was 0.13 g,
relative to (a) the ground and (b) the customer? (c) How long does
its speed before entering the branch was 220 m/s, and its penetra-
the coin take to reach the c­ ompartment ceiling, 2.20 m above the
tion depth was 15 mm. If its speed was decreased at a uniform
knee? In unit-vector notation, what are (d) the actual force on the
rate, what was the magnitude of the force of the branch on the
coin and (e) the apparent force according to the customer’s mea-
toothpick?
sure of the coin’s acceleration?
•23   Tarzan, who weighs 820 N, swings from a cliff at the end of ••31 SSM WWW A block is projected up a frictionless inclined
a 20.0 m vine that hangs from a high tree limb and initially makes plane with initial speed v0 =
y
an angle of 22.0° with the vertical. Assume that an x axis extends 3.50 m/s. The angle of incline is
horizontally away from the cliff edge and a y axis extends upward. θ = 32.0°. (a) How far up the plane F1

Immediately after Tarzan steps off the cliff, the tension in the vine does the block go? (b) How long θ1
does it take to get there? (c) What x
is 760 N. Just then, what are (a) the force on him from the vine in
unit-vector notation and the net force on him (b) in unit-vector is its speed when it gets back to the
θ2
notation and as (c) a magnitude and (d) an angle relative to the bottom? F 2
­positive direction of the x axis? What are the (e) magnitude and ••32   Figure 5-39 shows an overhead
(f) ­angle of Tarzan’s acceleration just then? view of a 0.0250 kg lemon half and two Figure 5-39 Problem 32.
 PROBLEMS 119

of the three horizontal forces that act on it as it is on a frictionless as a function of time t, the component vx of the box’s velocity along an
→
table. Force F1 has a magnitude of 6.00 N and is at θ1 = 30.0°. Force x axis that extends directly up the ramp. What is the magnitude of the
→
F2 has a magnitude of 7.00 N and is at θ2 = 30.0°. In unit-vector normal force on the box from the ramp?
notation, what is the third force if the lemon half (a) is stationary, ••41   Using a rope that will snap if the tension in it exceeds 387 N,
(b) has the constant ve­loc­ity →
v = (13.0î − 14.0ĵ) m/s, and (c) has you need to lower a bundle of old roofing material weighing 449 N
the varying velocity →
v = (13.0t î − 14.0t ĵ) m/s2, where t is time? from a point 6.1 m above the ground. Obviously if you hang the
••33   An elevator cab and its load have a combined mass of 1600 kg. bundle on the rope, it will snap. So, you allow the bundle to acceler-
Find the tension in the supporting cable when the cab, originally ate downward. (a) What magnitude of the bundle’s acceleration will
moving downward at 12 m/s, is brought to rest with constant accel- put the rope on the verge of snapping? (b) At that acceleration, with
eration in a distance of 42 m. what speed would the bundle hit the ground?
••34 In Fig. 5-40, a crate of m ••42 In earlier days, horses pulled barges down canals in the
mass m = 100 kg is pushed at con- manner shown in Fig. 5-42. Suppose the horse pulls on the rope
stant speed up a frictionless ramp with a force of 7900 N at an angle of θ = 18° to the d ­ irection of
(θ = 30.0°) by a horizontal force F motion of the barge, which is headed straight along the positive
→
F . What are the magnitudes of direction of an x axis. The mass of the barge is 9500 kg, and the
→
(a) F and (b) the force on the crate magnitude of its acceleration is 0.12 m/s2. What are the (a) mag-
θ
from the ramp? nitude and (b) direction (relative to positive x) of the force on the
Figure 5-40 Problem 34. barge from the water?
••35   The velocity of a 3.00 kg par-
→
ticle is given by v = (8.00t î + 3.00t ĵ)
2

m/s, with time t in seconds. At the instant the net force on the par-
θ
ticle has a magnitude of 35.0 N, what are the direction (relative to
the positive direction of the x axis) of (a) the net force and (b) the
particle’s direction of travel?
••36   Holding on to a towrope moving parallel to a frictionless Figure 5-42 Problem 42.
ski slope, a 50 kg skier is pulled up the slope, which is at an angle
of 8.0° with the horizontal. What is the magnitude Frope of the force ••43 SSM In Fig. 5-43, a chain consisting of five
F
on the skier from the rope when (a) the magnitude v of the skier’s links, each of mass 0.100 kg, is lifted vertically with
velocity is constant at 2.0 m/s and (b) v = 2.0 m/s as v increases at constant acceleration of magnitude a = 2.50 m/s2.
a rate of 0.10 m/s2? Find the magnitudes of (a) the force on link 1 from 5
••37   A 40 kg girl and an 8.4 kg sled are on the frictionless ice link 2, (b) the force on link 2 from link 3, (c) the a
4
of a frozen lake, 15 m apart but connected by a rope of negligible force on link 3 from link 4, and (d) the force on
mass. The girl exerts a horizontal 5.2 N force on the rope. What are link 4 from link 5. Then find the magnitudes of 3
→
the acceleration magnitudes of (a) the sled and (b) the girl? (c) How (e) the force F on the top link from the person
2
far from the girl’s initial position do they meet? lifting the chain and (f) the net force accelerating
each link. 1
••38   A 40 kg skier skis directly down a frictionless slope angled
at 10° to the horizontal. Assume the skier moves in the negative ••44  A lamp hangs vertically from a cord in a
direction of an x axis along the slope. A wind force with compo- descending elevator that decelerates at 2.4 m/s2. Figure 5-43  
nent Fx acts on the skier. What is Fx if the magnitude of the skier’s (a) If the tension in the cord is 89 N, what is the Problem 43.
velocity is (a) constant, (b) increasing at a rate of 1.0 m/s2, and (c) lamp’s mass? (b) What is the cord’s tension when
increasing at a rate of 2.0 m/s2? the elevator ascends with an upward acceleration of 2.4 m/s2?
••39 ILW A sphere of mass 3.0 × 10−4 kg is suspended from a cord. ••45   An elevator cab that weighs 27.8 kN moves upward. What
A steady horizontal breeze pushes the sphere so that the cord is the tension in the cable if the cab’s speed is (a) increasing at a
makes a constant angle of 37° with the vertical. Find (a) the push rate of 1.22 m/s2 and (b) decreasing at a rate of 1.22 m/s2?
magnitude and (b) the tension in the cord. ••46  An elevator cab is pulled upward by a cable. The cab
••40 A dated box of dates, of mass 5.00 kg, is sent sliding up a and its single occupant have a combined mass of 2000 kg. When
frictionless ramp at an angle of θ to the horizontal. Figure 5-41 gives, that occupant drops a coin, its acceleration relative to the cab is
8.00 m/s2 downward. What is the tension in the ­cable?
vx (m/s)
4 ••47 The Zacchini family was renowned for their
human-cannonball act in which a family member was shot from a
2 ­cannon using either elastic bands or compressed air. In one version
of the act, Emanuel Zacchini was shot over three Ferris wheels to
t (s) land in a net at the same height as the open end of the cannon and
0 1 2 3 at a range of 69 m. He was propelled inside the barrel for 5.2 m and
–2 launched at an angle of 53°. If his mass was 85 kg and he underwent
constant acceleration inside the barrel, what was the magnitude of
–4
the force propelling him? (Hint: Treat the launch as though it were
Figure 5-41 Problem 40. along a ramp at 53°. Neglect air drag.)
120 CHAPTER 5 FORCE AND MOTION—I

••48 In Fig. 5-44, elevator cabs A and B are con- ••54 Figure 5-49 shows four penguins that are being playfully
nected by a short cable and can be pulled upward or pulled along very slippery (frictionless) ice by a curator. The masses
lowered by the ­cable above cab A. Cab A has mass of three penguins and the tension in two of the cords are m1 = 12 kg,
1700 kg; cab B has mass 1300 kg. A 12.0 kg box of cat- m3 = 15 kg, m4 = 20 kg, T2 = 111 N, and T4 = 222 N. Find the pen-
nip lies on the floor of cab A. The tension in the cable A guin mass m2 that is not given.
connecting the cabs is 1.91 × 10 4 N. What is the mag-
nitude of the normal force on the box from the floor? m4
••49   In Fig. 5-45, a block of mass m = 5.00 kg is m1 m3
pulled along a horizontal frictionless floor by a cord T2 T4
that exerts a force of magnitude F = 12.0 N at an B
angle θ = 25.0°. (a) What is the magnitude of the
block’s acceleration? (b) The force magnitude F is Figure 5-49 Problem 54.
slowly increased. What is its value just before the Figure 5-44  
block is lifted (completely) off the floor? (c) What is Problem 48.
the magnitude of the block’s acceleration just b­ efore ••55 SSM ILW WWW Two blocks are m1
it is lifted (completely) off the floor? in contact on a frictionless table. A hori-
m2
zontal force is applied to the larger block, F
as shown in Fig. 5‑50. (a) If m1 = 2.3 kg,
m2 = 1.2 kg, and F = 3.2 N, find the magni-
tude of the force between the two blocks. Figure 5-50  
m θ F (b) Show that if a force of the same mag- Problem 55.
nitude F is applied to the smaller block but
in the opposite direction, the mag­nitude of the force between the
blocks is 2.1 N, which is not the same value calculated in (a). (c)
Figure 5-45  
Explain the difference.
Problems 49 and 60. →
••56 In Fig. 5-51a, a constant horizontal force Fa is applied to
block A, which pushes against block B with a 20.0 N force directed
••50 In Fig. 5-46, three ballot A →
horizontally to the right. In Fig. 5-51b, the same force Fa is applied
boxes are connected by cords, one
to block B; now block A pushes on block B with a 10.0 N force
of which wraps over a pulley hav- B
­directed horizontally to the left. The blocks have a combined mass
ing negligible friction on its axle and
C of 12.0 kg. What are the→magnitudes of (a) their acceleration in
negligible mass. The three masses
Fig. 5-51a and (b) force Fa?
are mA = 30.0 kg, mB = 40.0 kg, and Figure 5-46 Problem 50.
mC = 10.0 kg. When the assembly is
A B B A
­released from rest, (a) what is the
tension in the cord connecting B and C, and (b) how far does A Fa Fa
move in the first 0.250 s ­(assuming it does not reach the pulley)?
••51 Figure 5-47 shows two blocks connected
(a) (b)
by a cord (of negligible mass) that passes over a
frictionless pulley (also of negligible mass). The Figure 5-51 Problem 56.
arrangement is known as Atwood’s ­machine. One
block has mass m1 = 1.30 kg; the other has mass ••57 ILW A block of mass m1 = 3.70 kg on a frictionless plane
m1 inclined at angle θ = 30.0° is connected by a cord over a mass-
m2 = 2.80 kg. What are (a) the magnitude of the
blocks’ acceleration and (b) the tension in the cord? m2 less, frictionless pulley to a second block of mass m2 = 2.30 kg
(Fig. 5-52). What are (a) the magnitude of the acceleration of
••52   An 85 kg man lowers himself to the ground each block, (b) the direction of the acceleration of the hanging
from a height of 10.0 m by holding onto a rope that block, and (c) the tension in the cord?
runs over a frictionless pulley to a 65 kg sandbag.
With what speed does the man hit the ground if he
Figure 5-47
started from rest?
Problems 51
••53  In Fig. 5-48, three connected blocks are and 65.
pulled to the right on a horizontal frictionless table m1
m2
by a force of ­magnitude T3 = 65.0 N. If m1 = 12.0 kg, m2 = 24.0 kg,
and m3 = 31.0 kg, calculate (a) the magnitude of the system’s θ
­acceleration, (b) the tension T1, and (c) the tension T2. Figure 5-52 Problem 57.

T1 T2 T3 ••58   Figure 5-53 shows a man sitting in a bosun’s chair that dan-
m1 m2 m3 gles from a massless rope, which runs over a massless, frictionless
pulley and back down to the man’s hand. The combined mass of
man and chair is 95.0 kg. With what force magnitude must the man
Figure 5-48 Problem 53. pull on the rope if he is to rise (a) with a constant velocity and
 PROBLEMS 121

(b) with an upward acceleration of the axis, with a speed of 3.0 m/s. What are its (a) speed and (b) direc-
1.30 m/s2? (Hint: A free-body dia- tion of travel at t = 11 s?
gram can really help.) If the rope
on the right extends to the ground Fx (N)
and is pulled by a co-worker, with
6
what force magnitude must the
co-worker pull for the man to rise
(c) with a constant velocity and
(d) with an upward acceleration of
1.30 m/s2? What is the magnitude 0 t (s)
of the force on the ceiling from the 2 4 6 8 10 12
pulley system in (e) part a, (f ) part
b, (g) part c, and (h) part d? –4

••59 SSM A 10 kg monkey climbs Figure 5-55 Problem 63.

up a massless rope that runs over Figure 5-53 Problem 58.
a frictionless tree limb and back
down to a 15 kg package on the •••64 Figure 5-56 shows a box of mass m2 = 1.0 kg on a fric-
ground (Fig. 5-54). (a) What is the tionless plane inclined at angle θ = 30°. It is connected by a cord of
magnitude of the least accelera- negligible mass to a box of mass m1 = 3.0 kg on a horizontal fric-
tion the monkey must have if it is tionless surface. The pulley is frictionless and massless. (a) If the
→
to lift the package off the ground? magnitude of horizontal force F is 2.3 N, what is the tension in the
→
If, after the package has been lifted, connecting cord? (b) What is the largest value the magnitude of F
the monkey stops its climb and may have without the cord becoming slack?
holds onto the rope, what are the
(b) magnitude and (c) direction m1
F
of the monkey’s acceleration and
(d) the tension in the rope?
m2
••60   Figure 5-45 shows a 5.00 kg
block being pulled along a friction- θ
less floor by a cord that applies
Figure 5-56 Problem 64.
a force of constant magnitude Bananas

20.0 N but with an angle θ(t) that

varies with time. When angle •••65 Figure 5-47 shows Atwood’s machine, in which two con-
θ = 25.0°, at what rate is the accel- Figure 5-54 Problem 59. tainers are connected by a cord (of negligible mass) passing over a
eration of the block changing if (a) frictionless pulley (also of negligible mass). At time t = 0, container
θ(t) = (2.00 × 10−2 deg/s)t and (b) θ(t) = −(2.00 × 10−2 deg/s)t? 1 has mass 1.30 kg and container 2 has mass 2.80 kg, but container 1
(Hint: The angle should be in radians.) is losing mass (through a leak) at the constant rate of 0.200 kg/s. At
what rate is the acceleration magnitude of the containers changing
••61 SSM ILW A hot-air balloon of mass M is descending
at (a) t = 0 and (b) t = 3.00 s? (c) When does the acceleration reach
vertically with downward acceleration of magnitude a. How
­
its maximum value?
much mass (ballast) must be thrown out to give the balloon an
­upward acceleration of magnitude a? Assume that the ­upward •••66 Figure 5-57 shows a section of a cable-car system. The
force from the air (the lift) does not change because of the maximum permissible mass of each car with occupants is 2800 kg.
decrease in mass. The cars, riding on a support cable, are pulled by a second cable
attached to the support tower on each car. Assume that the cables
•••62 In shot putting, many athletes elect to launch the shot
at an angle that is smaller than the theoretical one (about 42°) at
which the distance of a projected ball at the same speed and height Support cable
is greatest. One reason has to do with the speed the athlete can Pull cable
give the shot during the acceleration phase of the throw. Assume
that a 7.260 kg shot is accelerated along a straight path of length
1.650 m by a constant applied force of magnitude 380.0 N, starting
with an initial speed of 2.500 m/s (due to the athlete’s preliminary
motion). What is the shot’s speed at the end of the acceleration
phase if the angle between the path and the horizontal is (a) 30.00°
θ
and (b) 42.00°? (Hint: Treat the motion as though it were along a
ramp at the given angle.) (c) By what percent is the launch speed
decreased if the athlete increases the angle from 30.00° to 42.00°?
•••63 Figure 5-55 gives, as a function of time t, the force com-
ponent Fx that acts on a 3.00 kg ice block that can move only along
the x axis. At t = 0, the block is moving in the positive direction of Figure 5-57 Problem 66.
122 CHAPTER 5 FORCE AND MOTION—I

are taut and inclined at angle θ = 35°. What is the difference in 73 SSM In Fig. 5-61, a tin of
tension between adjacent sections of pull cable if the cars are at antioxidants (m1 = 1.0 kg) on a m1
the maximum permissible mass and are being accelerated up the frictionless inclined surface is con-
incline at 0.81 m/s2? nected to a tin of corned beef
(m2 = 2.0 kg). The pulley is massless β
•••67  Figure 5-58 shows three B and frictionless. An ­ upward force
blocks attached by cords that loop
of magnitude F = 6.0 N acts on the
over frictionless pulleys. Block B
corned beef tin, which has a down-
lies on a frictionless table; the masses
C ward acceleration of 5.5 m/s2. What m2
are mA = 6.00 kg, mB = 8.00 kg, and A
are (a) the tension in the connecting
mC = 10.0 kg. When the blocks are
cord and (b) angle β?
released, what is the t­ension in the Figure 5-58 Problem 67.
74   The only two forces acting on a F
cord at the right?
body have magnitudes of 20 N and
•••68 A shot putter launches a 7.260 kg shot by pushing it
35 N and directions that differ by Figure 5-61 Problem 73.
along a straight line of length 1.650 m and at an angle of 34.10°
80°. The resulting acceleration has
from the horizontal, accelerating the shot to the launch speed
a magnitude of 20 m/s2. What is the
from its initial speed of 2.500 m/s (which is due to the athlete’s
mass of the body?
preliminary motion). The shot leaves the hand at a height of 2.110 m
and at an angle of 34.10°, and it lands at a horizontal distance of 75   Figure 5-62 is an overhead view
15.90 m. What is the magnitude of the athlete’s average force on of a 12 kg tire that is to be pulled by x
the shot during the acceleration phase? (Hint: Treat the motion three horizontal ropes. One rope’s F1
during the acceleration phase as though it were along a ramp at force (F1 = 50 N) is indicated. The
the given angle.) forces from the other ropes are
to be oriented such that the tire’s Figure 5-62 Problem 75.
Additional Problems acceleration magnitude a is least.
69   In Fig. 5-59, 4.0 kg block A and 6.0 kg block B are connected What is that least a if (a) F2 = 30 N,
→
by a string of negligible mass. Force FA = (12 N)î acts on block A; F3 = 20 N; (b) F2 = 30 N, F3 = 10 N;
→
force FB = (24 N)î acts on block B. What is the tension in the string? and (c) F2 = F3 = 30 N? m
M F
76  A block of mass M is pulled
along a horizontal frictionless
A FA B FB
surface by a rope of mass m, as
­
Figure 5-63 Problem 76.
shown in Fig. 5-63. A horizontal
→
x force F acts on one end of the rope.
Figure 5-59 Problem 69. (a) Show that the rope must sag, even if only by an impercepti-
ble amount. Then, assuming that the sag is negligible, find (b) the
acceleration of rope and block, (c) the force on the block from the
70 An 80 kg man drops to a concrete patio from a window rope, and (d) the tension in the rope at its midpoint.
0.50 m above the patio. He neglects to bend his knees on landing, tak-
ing 2.0 cm to stop. (a) What is his average acceleration from when his 77 SSM A worker drags a crate across a factory floor by pulling
feet first touch the patio to when he stops? (b) What is the magnitude on a rope tied to the crate. The worker exerts a force of magni-
of the average stopping force exerted on him by the patio? tude F = 450 N on the rope, which is inclined at an upward angle
θ = 38° to the horizontal, and the floor exerts a horizontal force
71 SSM Figure 5-60 shows a box of dirty money (mass m1 = 3.0 kg) of magnitude f = 125 N that opposes the motion. Calculate the
on a frictionless plane inclined at angle θ1 = 30°. The box is con- magnitude of the acceleration of the crate if (a) its mass is 310 kg
nected via a cord of negligible mass to a box of laundered money and (b) its weight is 310 N.
(mass m2 = 2.0 kg) on a frictionless plane inclined at angle θ2 = 60°. →
The pulley is frictionless and has negligible mass. What is the ten- 78   In Fig. 5-64, a force F of mag-
m2 F
sion in the cord? nitude 12 N is applied to a FedEx
box of mass m2 = 1.0 kg. The force is m1
directed up a plane tilted by θ = 37°.
The box is connected by a cord to θ
m1 m2 a UPS box of mass m1 = 3.0 kg on
the floor. The floor, plane, and pul- Figure 5-64 Problem 78.
ley are frictionless, and the masses
of the pulley and cord are negligible. What is the tension in the
θ1 θ2 cord?
79  A certain particle has a weight of 22 N at a point where
Figure 5-60 Problem 71.
g = 9.8 m/s2. What are its (a) weight and (b) mass at a point where
72   Three forces act on a particle that moves with unchanging veloc- g = 4.9 m/s2? What are its (c) weight and (d) mass if it is moved to
→
ity →
v = (2 m/s)î − (7 m/s)ĵ. Two of the forces are F1 = (2 N)î + a point in space where g = 0?
→
(3 N)ĵ + (−2 N)k̂ and F2 = (−5 N)î + (8 N)ĵ + (−2 N)k̂. What is 80  An 80 kg person is parachuting and experiencing a down-
the third force? ward acceleration of 2.5 m/s2. The mass of the parachute is 5.0 kg.
 PROBLEMS 123

(a) What is the upward force on the open parachute from the air? acceleration in g units? (c) What force is required for the acceleration?
(b) What is the downward force on the parachute from the person? (d) If the engines are shut down when 0.10c is reached (the speed then
remains constant), how long does the ship take (start to finish) to jour-
81  A spaceship lifts off vertically from the Moon, where
ney 5.0 light-months, the distance that light travels in 5.0 months?
g = 1.6 m/s2. If the ship has an upward acceleration of 1.0 m/s2 as it
lifts off, what is the magnitude of the force e­ xerted by the ship on 91 SSM A motorcycle and 60.0 kg rider accelerate at 3.0 m/s2 up a
its pilot, who weighs 735 N on Earth? ramp inclined 10° above the horizontal. What are the magnitudes
of (a) the net force on the rider and (b) the force on the rider from
82   In the overhead view of Fig. y the motorcycle?
5-65, five forces pull on a box of
mass m = 4.0 kg. The force mag- 92   Compute the initial upward acceleration of a rocket of mass
nitudes are F1 = 11 N, F2 = 17 N, F5 1.3 × 10 4 kg if the initial upward force produced by its engine (the
F4
F3 = 3.0 N, F4 = 14 N, and F5 = 5.0 N, thrust) is 2.6 × 10 5 N. Do not neglect the gravitational force on the
θ4
and angle θ4 is 30°. Find the box’s x rocket.
F1 F3
acceleration (a) in unit-vector nota- 93 SSM Figure 5-66a shows a mobile hanging from a ceiling; it
tion and as (b) a magnitude and consists of two metal pieces (m1 = 3.5 kg and m2 = 4.5 kg) that are
(c) an angle relative to the positive F2 strung together by cords of negligible mass. What is the ­tension
direction of the x axis. in (a) the bottom cord and (b) the top cord? Figure 5‑66b shows
Figure 5-65 Problem 82.
83 SSM A certain force gives an a mobile consisting of three metal pieces. Two of the masses are
object of mass m1 an accelera- m3 = 4.8 kg and m5 = 5.5 kg. The tension in the top cord is 199 N.
tion of 12.0 m/s2 and an object of mass m2 an acceleration of What is the tension in (c) the lowest cord and (d) the middle cord?
3.30 m/s2. What acceleration would the force give to an object of
mass (a) m2 − m1 and (b) m2 + m1?
→
84   You pull a short refrigerator with a constant force F across
→
a greased (frictionless) floor, either with F horizontal (case 1) or
→ m3
with F tilted upward at an angle θ (case 2). (a) What is the ratio m1
of the refrigerator’s speed in case 2 to its speed in case 1 if you pull
for a certain time t? (b) What is this ratio if you pull for a certain
distance d? m2
85  A 52 kg circus performer is to slide down a rope that will m5
break if the tension exceeds 425 N. (a) What happens if the per-
former hangs stationary on the rope? (b) At what magnitude of (a) (b)
acceleration does the performer just avoid breaking the rope? Figure 5-66 Problem 93.
86  Compute the weight of a 75 kg space ranger (a) on Earth,
(b) on Mars, where g = 3.7 m/s2, and (c) in interplane­tary space,
94   For sport, a 12 kg armadillo runs onto a large pond of level,
where g = 0. (d) What is the ranger’s mass at each location?
frictionless ice. The armadillo’s initial velocity is 5.0 m/s along the
87   An object is hung from a spring balance attached to the ceiling positive direction of an x axis. Take its initial position on the ice as
of an elevator cab. The balance reads 65 N when the cab is standing being the origin. It slips over the ice while being pushed by a wind
still. What is the reading when the cab is moving upward (a) with with a force of 17 N in the positive direction of the y axis. In unit-
a constant speed of 7.6 m/s and (b) with a speed of 7.6 m/s while vector notation, what are the animal’s (a) velocity and (b) position
decelerating at a rate of 2.4 m/s2? ­vector when it has slid for 3.0 s?
88   Imagine a landing craft approaching the surface of Callisto, one 95   Suppose that in Fig. 5-12, the masses of the blocks are 2.0 kg
of Jupiter’s moons. If the engine provides an u ­ pward force (thrust) and 4.0 kg. (a) Which mass should the hanging block have if the
of 3260 N, the craft descends at constant speed; if the engine pro- magnitude of the acceleration is to be as large as possible? What
vides only 2200 N, the craft a­ ccelerates downward at 0.39 m/s2. then are (b) the magnitude of the acceleration and (c) the tension
(a) What is the weight of the landing craft in the vicinity of Callisto’s in the cord?
surface? (b) What is the mass of the craft? (c) What is the magni- 96   A nucleus that captures a stray neutron must bring the neu-
tude of the free-fall acceleration near the surface of Callisto? tron to a stop within the diameter of the nucleus by means of the
89   A 1400 kg jet engine is fastened to the fuselage of a ­passenger strong force. That force, which “glues” the ­nucleus together, is
jet by just three bolts (this is the usual practice). Assume that each approximately zero outside the nucleus. Suppose that a stray neu-
bolt supports one-third of the load. (a) Calculate the force on each tron with an initial speed of 1.4 × 10 7 m/s is just barely captured
bolt as the plane waits in line for clearance to take off. (b) During by a nucleus with diameter d = 1.0 × 10−14 m. Assuming the strong
flight, the plane encounters turbulence, which suddenly imparts an force on the neutron is constant, find the magnitude of that force.
upward vertical a­ cceleration of 2.6 m/s2 to the plane. Calculate the The neutron’s mass is 1.67 × 10−27 kg.
force on each bolt now. 97  If the 1 kg standard body is accelerated by only
→ →
90   An interstellar ship has a mass of 1.20 × 106 kg and is initially at F1 = (3.0 N)î + (4.0 N)ĵ and F2 = (−2.0 N)î + (−6.0 N)ĵ, then what
→
rest relative to a star system. (a) What constant ­acceleration is needed is Fnet (a) in unit-vector notation and as (b) a magnitude and
to bring the ship up to a speed of 0.10c (where c is the speed of light, (c) an angle relative to the positive x direction? What are the
→
3.0 × 108 m/s) relative to the star system in 3.0 days? (b) What is that (d) magnitude and (e) angle of a ?