# Friction and Other Applications of Newton's Laws

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HAPTER   5   Friction and Other
Applications of
Newton’s Laws

magine a skier descending a slope covered with fresh powder, gliding back and forth,
I   leaving gracefully curving tracks in the snow. The skier experiences motion and forces
that we will describe in this chapter. The skier’s “center of mass” follows a curving path
determined by the forces of gravity and friction. In this chapter we shall study friction,
motion along circular arcs, and motion of a body’s center of mass as applications of
Newton’s laws of motion.

5–1      Friction
Friction is a force acting between two surfaces, tending to prevent the surfaces from
sliding over each other. In some cases friction is desirable. For example, most forms
of transportation depend on friction. If a road is covered with ice, it is difﬁcult to walk
or drive a car on it because the frictional force between the ice and either your feet or
the tires of your car may not be great enough to prevent slipping.
Often friction is not desirable. For example, it can cause machine parts to wear. For
this reason, oil and other lubricants are used to reduce friction. Friction in human joints
is very low because our bodies contain a natural lubricating system. Consequently,
though our bones rub against each other at the joints as we move, bones do not
normally wear out, even after many years of use.
We can begin to understand friction by using Newton’s laws to analyze some
simple experiments. As our ﬁrst case, consider a book lying on a table. Since the book
is at rest, the resultant force on it must be zero. So we know that the book’s weight
must be canceled by an opposing force S supplied by the table’s surface* (Fig. 5–1a).

*This surface force arises from an elastic deformation of the surface. For a rigid surface, this deformation
is so slight as to be undetectable by the casual observer. In the case of a heavy object on a nonrigid
surface, however—a dumbbell on a rubber mat at the gym, for example—the effect is easy to see.

118
5–1 Friction                                       119

Now suppose you push very lightly on the book, applying a horizontal force P
directed to the right (Fig. 5–1b). If this force is small enough, the book will remain at
rest. To keep the book from moving, the surface must now exert a force that is directed
to the left. In other words, there will be a horizontal component of S to cancel P and
maintain a resultant force of zero. The vector component of S along the surface is
called friction and is denoted by f. The vector component of S normal (perpendicular)
to the surface is called the normal force and is denoted by N. In force diagrams, we
usually draw f and N, the vector components of S, rather than S itself.
If you now slowly increase the horizontal force P, at ﬁrst the book does not move.                            (a)
But when you make P large enough, the book begins to move. Once it is moving, the
force needed to maintain the motion is smaller than the force needed to begin move-
ment. The forces acting on the book for various values of P are shown in Fig. 5–2. In
Fig. 5–2a to Fig. 5–2c, as P increases, f does also, and so equilibrium is maintained.
In Fig. 5–2c, the force of friction has reached its maximum fmax. When P exceeds fmax,
the book accelerates. Once the book is moving, the frictional force usually decreases
to a value below fmax; thus a force P only slightly greater than fmax can produce a
signiﬁcant acceleration (Fig. 5–2d). In Fig. 5–2e, the magnitude of P has been reduced
so that it just balances the magnitude of the reduced frictional force; therefore the net
force is again zero. In accordance with Newton’s second law, the book now continues                              (b)
to move at constant velocity.
Friction between surfaces that slide over each other is called kinetic friction,          Fig. 5–1 (a) A surface supports a book,
whereas friction between surfaces at rest with respect to each other is called static fric-   exerting a force S equal in magnitude
tion. The kinetic frictional force between two surfaces is often less than the maximum        to the book’s weight w. (b) When you
static frictional force, as we have assumed in our description of the friction between        exert a force P on the book, the surface
book and table.                                                                               pushes back with a frictional force, f,
The maximum frictional force that the table can exert on the book depends on the          that is tangent to the surface.
two surfaces in contact: the kind of material each is made of, their smoothness,
whether they are dry or lubricated, and so on. The force fmax also depends on the
magnitude of the forces pressing the surfaces together. You can demonstrate this by
repeating the experiment described above while someone presses down on the book.
To move the book now, you must exert a greater horizontal force. The normal force
also increases, as it must to balance the additional downward force as well as the
weight of the book.

(a)                                               (b)                                      (c)

(d)                                       (e)
Fig. 5–2 The frictional force f varies as an applied force P varies.
120                                               CHAPTER 5         Friction and Other Applications of Newton’s Laws

Table 5–1    Coefﬁcients of Static                   We ﬁnd that the maximum frictional force increases in direct proportion to the
Friction                             magnitude of the normal force. The constant of proportionality is called the coefﬁ-
cient of static friction and denoted by s, ( is the Greek letter mu, pronounced
Surfaces                     s            /myou/), and so we can write

Bone joints in             0.002–0.01
mammals                                                                                fmax            s   N                       (static friction)   (5–1)
Ski wax on snow,           0.04
0° C, dry                                       The coefﬁcient s depends on the surfaces in contact and their condition. Some repre-
Ski wax on snow,           0.1                   sentative values of s are given in Table 5–1.
0° C, wet                                           The magnitude of the static frictional force f depends on the situation. Provided that
Ski wax on snow,           0.2                   its magnitude is less than fmax, f will always be such that the body remains at rest with
–20° C                                          a resultant force of zero. That is, if the external horizontal force trying to move the
Teﬂon on steel             0.04                  body has a magnitude less than fmax, then friction will be able to balance the external
Graphite on steel          0.1                   force, and the object will remain at rest.
Steel on steel, lubri-     0.2                       The force of kinetic friction is also proportional to N, with a coefﬁcient k that is
cated with motor oil                            usually less than or equal to s and depends on the sliding velocity:
Steel on steel, clean      0.7
Brake material on          0.4                                                              f       k   N                          (sliding friction)   (5–2)
cast iron
Wood on wood, wet          0.2
Eqs. 5–1 and 5–2 are the force laws that allow us to compute the force of friction
Wood on wood, dry          0.25–0.5
whenever we know the normal force. They are not fundamental laws and they are
Wood on leather            0.3–0.4
not exact, but they are useful approximate expressions that work very well in most
Wood on brick              0.6
applications.
Glass on glass, clean      0.9
Rubber on solids           1–4

EXAMPLE 1      Forces on a Sliding Book
The book in Fig. 5–1 has a weight of 8.00 N. The coefﬁcient of              Inserting this result into the equation for Pmin, we ﬁnd
static friction between the book’s surface and the table’s surface
Pmin         s   N       (0.400)(8.00 N)         3.20 N
is 0.400, and the coefﬁcient of kinetic friction between the two
surfaces is 0.300. (a) Find the magnitude of the minimum hori-
(b) Here we use Fig. 5–2e as our free-body diagram. Since the
zontal force P required to move the book, which is initially at
book’s velocity is constant, a 0. Thus
rest. (b) Find the magnitude of the force P required to move the
book at constant velocity.                                                                                       Fx   max       0
P        f   0
SOLUTION (a) Fig. 5–2c shows the appropriate free-body                                                           P    f
diagram. Motion begins when the force P barely exceeds the
Since this is kinetic or sliding friction, we apply Eq. 5–2:
maximum force of static friction, fmax. So the minimum force
required equals fmax , the magnitude of which is found using                                                     P    f     k   N
Eq. 5–1 ( fmax   s N ).

As in part (a), Fy 0 and so forces N and w balance; that is,
Pmin     fmax   s   N
N w 0, or N w. Inserting this into the preceding equation
Inspecting our free-body diagram (Fig 5–2c), we see that the                for P, we ﬁnd
only vertical forces are N and w. Applying Newton’s second law,
P       kw              (0.300)(8.00 N)        2.40 N
we ﬁnd
Fy      may 0
N      w       0
N       w 8.00 N
5–1 Friction              121

EXAMPLE 2       Walking on Ice
A person tries to walk up a 15° incline on a street covered with
ice (Fig. 5–3a). Is this possible if the coefﬁcient of static friction
between shoes and ice is 0.2?

SOLUTION The forces acting on the person are shown in
Fig. 5–3b. We choose the y-axis perpendicular to the incline
because with this choice of axes, the motion will be along the
x-axis and there will be no component of acceleration along the
y-axis. We need to investigate the x components of force in
order to determine whether there can be a net force and therefore
an acceleration along the positive x-axis. As each foot pushes
backwards in the attempt to move forward, the ice surface ex-
erts a forward frictional force on the foot. This force is opposed
by the x component of the person’s weight, w sin . Thus
Fx          f   w sin
This relation tells us that, in order for the person to move
forward, f must be greater than w sin . This may or may not be
possible, depending on the value of s, since
fmax       s   N
The normal force is related to w and through the condition
Fy   0 (since a y    0). The normal force just cancels the
component of the weight perpendicular to the surface:
N           w cos
(It is a common error to assume that N always equals w. Notice
that this is not true here.)
Combining the three preceding equations, we obtain an
expression for the maximum value of the resultant force ( Fx):
( Fx)max      s(w cos ) w sin
(w cos )( s tan )
Only if s tan is positive can ( Fx)max be positive. Other-
wise it will not be possible to walk up the incline. For this
example,      15°. Thus

s   tan          0.2       tan 15°   0.07                               Fig. 5–3

So ( Fx)max is negative, and it is impossible to go up the incline.
In fact, it would be impossible even to stand on the icy incline.
122                                                                     CHAPTER 5         Friction and Other Applications of Newton’s Laws

EXAMPLE 3          Time to Complete a Ski Run
A skier starts with an initial velocity of 4.0 m/s, directed down
a 22° slope. The slope is 130 m long, and k is approximately
independent of velocity and equal to 0.10. Find the skier’s ﬁnal
velocity and the time she takes to reach the bottom. Neglect air
resistance.

SOLUTION The ﬁrst step is to ﬁnd the skier’s acceleration
by considering the forces in the free-body diagram (Fig. 5–4).

Fx                  w sin           f
ax
m                         m
From Eq. 5–2, we know that
f               k   N
Since ay     0, we have
Fy               0
N          w cos                 0
N               w cos
Combining these expressions for ax, f, and N, we obtain
w sin                         k   (w cos )
ax
m
or, using the relation w                 mg,
ax    g [sin       k (cos )]
(9.8 m/s2)(sin 22° 0.10 cos 22°)                                  2.8 m/s2

Now we may use the equations for linear motion at constant
acceleration from Chapter 2. First we ﬁnd the velocity vx at the
bottom of the slope:
vx2   vx02  2axx
(4.0 m/s)2 2(2.8 m/s2)(130 m)                                   740 m2/s2
vx    27 m/s

Next we relate the time t to the distance x, using Eq. 2–10:
1
x            2   (vx 0           vx)t
Solving for t, we ﬁnd
2x                      2(130 m)
t
vx0        vx             4.0 m/s 27 m/s                                                       Fig. 5–4
8.4 s

(Actually, air resistance would be a signiﬁcant factor in this
situation, and so the time required would be somewhat longer.)
5–1 Friction                    123

EXAMPLE 4        Connected Blocks Accelerating Together
Determine how long the hanging block in Fig. 5–5a takes to
reach the ﬂoor. The coefﬁcient of kinetic friction between m1
and the table is 0.40; the pulley is frictionless and has negligible
mass; m1 1.0 kg and m2 0.50 kg.

SOLUTION First we must ﬁnd the acceleration of m2. We
draw a separate free-body diagram for each block (Figs. 5–5b
and 5–5c) and apply Newton’s second law to each. The two
blocks are coupled together by the string; therefore both have
the same magnitudes of acceleration, a, and tension, T.
For m1, both the acceleration a1 and the tension T1 are along
the positive x-axis. Applying Newton’s second law, we obtain
Fx   m1a1x
T     f   m1a                          (1)
(a)
But f     kN    0.40 N, and N w m1g (since                           Fy    0).
Thus substituting f 0.40m1g into Eq. 1, we ﬁnd
T       0.40m1g         m1a                    (2)

In applying Newton’s second law to m2, we choose the positive
x-axis in the downward direction, so that a2x  a:
Fx   m 2 a 2x
m 2g        T    m 2a                         (3)

Adding Eqs. 2 and 3, we ﬁnd
(b)                   (c)
m2 g       0.40 m1g           m1a    m 2a
Fig. 5–5
m2 0.40m1
a                 g
m1 m2
0.50 kg       0.40 kg
(9.8 m/s2)
1.0 kg      0.50 kg
0.65 m/s2

Finally, we use Eq. 2–11, which relates distance and time for
linear motion at constant acceleration, and solve for t:
1
x      2   axt 2
t           2x/ ax

2(1 .0 m )/ (0 .6 5 m /s 2)   1.8 s
A          Closer Look

Microscopic Description of Friction
Frictional forces arise from electromag-                   N = N1 + N2 + N3 + …
netic interaction between the molecules
of the two surfaces in contact. We shall
now describe a simple model that accounts
for the main features of macroscopic fric-
tion. Fig. 5–A is a sketch of a highly magni-
fied region of interaction between two
surfaces for a block resting on a horizontal
surface when no frictional force acts. As
the blowup shows, the surfaces are in
contact over only a small fraction of their
total area. The total normal force on the
upper surface does not act at a single
point; instead, it is the sum of a large
number of very small upward forces, acting
at the various points of contact.
If we exert a horizontal force P on the
block, attempting to slide it to the right, a     Fig. 5–A The normal force is the resultant     Fig. 5–B At A, where the surfaces are
frictional force f arises, which is directed to   of a large number of forces arising from       being pulled apart, the lower surface resists
the left, opposing P. Like the normal force,      molecular interaction at points of micro-      motion of the upper surface to the right
this frictional force is also the resultant of    scopic contact between the two surfaces.       by exerting an attractive molecular force
fA, directed to the left. At R, where the
many small forces, occurring only where
Although the magnitude of the frictional   surfaces are being pushed into each other,
the surfaces are in true microscopic
force depends on the microscopic area of       the lower surface resists motion of the
contact. These microscopic frictional forces
contact, it does not depend on the appar-      upper surface to the right by exerting a
can be either attractive or repulsive, as
ent, or macroscopic, area of contact. Thus     repulsive force fR, also directed to the left.
indicated in Fig. 5–B.                                                                           Both fA and fR contribute to the total fric-
the same force would be required to move
The normal force changes when the                                                            tional force.
any of the three identical blocks shown in
microscopic area of contact between the
Fig. 5–C, even though they have different
two surfaces changes. If someone pushes
orientations and therefore different
down on the block, there will be contact
apparent areas of contact with the table.
between the two surfaces at more places,
The area of microscopic contact is the
and the total microscopic area of contact
same for all three, since they have the same
will increase. This increased area of contact
weight and must therefore have the same
allows more molecules of the lower sur-
number of molecules collectively exerting
face to exert upward forces, thereby con -
the normal force that balances the weight.
tributing to a larger total normal force,
Polishing a rough surface will at ﬁrst
which is necessary to balance the increased
tend to reduce the frictional force because    Fig. 5–C The same force is required to
downward force. The increased area of
of a decrease in the number of places like     move any of the three identical blocks, irre-
contact also allows for an increased fric-                                                       spective of orientation. In other words the
R in Fig. 5–B. When two identical blocks
tional force. Again, the reason is simply that                                                   force of friction is independent of the
have highly polished surfaces and are free
there are more molecules in contact. The                                                         apparent area of contact.
of adsorbed gases, oils, and other impuri-
magnitudes of both N and fmax are approx-
ties, however, the frictional force can be
imately proportional to the number of
many times larger than the normal force.
molecules in close contact, or to the
microscopic area of contact. Therefore fmax
is approximately proportional to N.
A    Closer Look

In this case, the surface irregularities   When you are moving, the increased pres-
are so reduced that a much larger number       sure on the cartilage squeezes out more
of the molecules on each surface are           synovial ﬂuid, thereby reducing friction and
within range of the electromagnetic forces     allowing the bones to slide.
exerted by molecules on the other surface.
Then the same forces that hold the parts
of each block together internally begin to
hold the two blocks to each other. The
same effect is present in a stack of micro-
scope slides (Fig. 5–D).
Lubricants greatly reduce friction
between solids. The lubricant separates the
surfaces of the solids beyond the range of
their molecular forces, leaving only the
weaker frictional force between lubricant                                                           Fig. 5–E Synovial ﬂuid lubricates human
joints.
and solid. The lubrication of human knee
joints (Fig. 5–E) varies with need. When       Fig. 5–D Slipping of these glass micro-
you are standing still, much of the lubri-     scope slides is prevented by the strong
cating synovial ﬂuid is absorbed into the      frictional force acting between their very
cartilage, so that friction between your       clean surfaces.
bones is great enough to prevent slipping.

5–2      Centripetal Force
We saw in Chapter 3 that a body undergoing circular motion is continuously subject to
an acceleration, called centripetal acceleration, directed toward the center of the
circle. Because there is an acceleration, Newton’s second law implies that there must
be acting on the body a resultant force that is also directed toward the center of the
circle. Using Eq. 3–12 (a v 2/r), we obtain an expression for the magnitude of this net
force acting on a body in circular motion:

 F     ma

 F
mv2
(5–3)
r

Any net force that produces this centripetal acceleration is termed centripetal force.
Eq. 5–3 should not be confused with a force law. Force laws always express a partic-
ular kind of force (weight, spring force, friction) in terms of some physical object in
the environment (the earth, a spring, a surface in contact). But centripetal force can be
a force of any kind. Or it can result from a combination of different kinds of forces. In
the examples and problems that follow, we shall see several kinds of centripetal
forces: friction, normal force, tension, and weight.
126                                         CHAPTER 5           Friction and Other Applications of Newton’s Laws

EXAMPLE 5      Rounding a Curve at Maximum Speed
Find an expression for the maximum speed with which a car
can travel without slipping around a level, circular curve of

SOLUTION The maximum speed of the car is determined
by the maximum centripetal force that can be exerted on it. As
Fig. 5–6 shows, the forces acting on the car are its weight and
the surface force exerted by the road on the tires. The only
force that acts in the horizontal plane is friction. Therefore the
frictional force must be the centripetal force:
mv2
f
r
Fig. 5–6
Since there is to be no motion along the radius of the curve, this
is a static frictional force, whose magnitude will vary as
required to maintain the motion. The maximum frictional force          Solving for vmax, we obtain
determines the maximum velocity:                                                                   vmax               gr
s
mvmax2
fmax                                        If s 1.0, a typical value for tires on a dry road, vmax     gr.
r
A curve of radius 10 m could be safely driven at a speed no
But fmax     s N, and it follows from cancellation of the vertical
greater than 9.9 m/s (22 mi/h), but a curve of radius 100 m
components of force that N mg, and therefore fmax            s mg.
could be traveled at speeds up to 31 m/s (70 mi/h) without any
Thus
skidding.
mvmax2
s mg
r

EXAMPLE 6      Car Swinging on a Cable
One of the rides at an amusement park consists of small cars                                                   v2
that are supported by steel cables and rotated in a circle at                                   tan
g sin
gradually increasing speed (Fig. 5–7a). As the speed increases,        or
the cars rise. Find the speed of a car when its center is 5.00 m
v      g sin tan
from the cable’s point of support and the cable makes an angle
of 35.0° with the vertical.                                                          (9 .8 0 m /s 2) (5 .0 0 m )( si n 3 5.0 °) (t an 3 5.0 °)

4.44 m/s
SOLUTION          We choose the car as the free body. The only
forces acting on the car are its weight w and the tension T in the
cable (Fig. 5–7b). The component of T in the horizontal direc-
tion is the centripetal force:
mv2
T sin
r
Because there is no more vertical motion once the angle is
reached, the vertical forces now cancel:
T cos         mg
Dividing this equation into the preceding one, we obtain
v2
tan
gr
But it is evident from Fig. 5–7a that r is itself a function of :
r     sin

where is the length of the cable. Substituting this value for r                           (a)                                           (b)
into the equation above, we ﬁnd                                                                        Fig. 5–7
5–3 Center of Mass                                           127

5–3      Center of Mass
A System Moving as a Single Particle
Newton’s second law ( F ma) applies to single particles and also to systems of
particles in which the particles all have the same acceleration a. For example, when a
car moves in a straight line, all parts of the car’s body have the same velocity and
acceleration. Another example is a ski jumper who, after leaving the ramp, maintains
body and skis in a ﬁxed, rigid position (Fig. 5–8). In both these cases, we can apply
Newton’s second law to the whole system because each system acts as a single
particle.

Motion of the Center of Mass
Even when the parts of a system have different velocities and accelerations, however,
there is still one point in the system whose acceleration we may ﬁnd by applying                     Fig. 5–8 Each part of the system of
Newton’s second law. This one point is called the center of mass of the system, and                  skier and skis has the same velocity and
its acceleration is denoted by acm. Thus                                                             acceleration, during the interval shown.

F     Macm                                       (5–4)

where F is the vector sum of the external forces acting on the system and M is the
mass of the whole system. This equation implies that the motion of the center of
mass is the same as that of a particle of mass M subject to the same external forces
as those acting on the system. For example, suppose a wheel rolls along a horizontal
frictionless surface with no resultant force on the wheel. According to Eq. 5–4, the
wheel’s center of mass (its geometric center) will experience no acceleration. Although
all other parts of the wheel are accelerated because of rotation, the center of mass
moves at constant velocity—just like a particle on which no force acts. Other examples
of center-of-mass motion are shown in Fig. 5–9.

Fig. 5–9 (a) A rotating wrench slides along a frictionless horizontal surface. There is no
resultant force on the wrench, and therefore its center of mass follows a linear path at constant
speed. (b) A whirling baton tossed into the air experiences only the external force of gravity—
in other words, its weight W Mg. Its center-of-mass acceleration a cm W/M Mg/M g,
and the center of mass follows the characteristic parabolic path of a simple projectile.                                (b)
128                                            CHAPTER 5         Friction and Other Applications of Newton’s Laws

Deﬁnition of the Center of Mass
Next we shall deﬁne the center of mass and then, after computing the center of mass
for some simple systems, we shall apply Eq. 5–4 (derived at the end of this section).
For a system of n particles (Fig. 5–10), the coordinates of the center of mass relative
to an arbitrary origin O is deﬁned to be the following weighted average of the particles’
coordinates:

mx      m1x1 m 2 x 2 … m n x n
xcm                                                               (5–5)
m          m1 m2 … mn

Fig. 5–10 Finding the center of mass of        and
particles.
my       m1y1 m 2 y2 … m n yn
ycm                                                               (5–6)
m          m1 m2 … mn

EXAMPLE 7          Center of Mass of Two Particles
Find the center of mass of a system made up of two particles A             Thus we see that the center of mass is the same physical point
and B, 1.0 m apart, if each particle has a mass of 25 g.                   in the system, independent of the choice of origin. It is possible
to prove this result in general for any system.
SOLUTION        If we choose as origin the position of particle A
(Fig. 5–11a), we have

mx       mA xA   mB xB
xcm
m         mA     mB
(a)
(25 g)(0) (25 g)(1.0 m)
0.50 m
25 g 25 g
that is, the point midway between the two particles is the center
of mass.
To see that the choice of origin is completely arbitrary, con-
(b)
sider what happens when we choose the midpoint as our origin
(Fig. 5–11b). In this case, we would compute xcm as follows:                                           Fig. 5–11

mx       mA xA   mB xB
xcm
m         mA     mB
(25 g)( 0.50 m)    (25 g)( 0.50 m)
0
25 g   25 g
5–3 Center of Mass                                           129

Finding the center of mass of a homogeneous sphere or other symmetrical body is
particularly simple. Fig. 5–12 shows three symmetrical bodies in which the center of
symmetry is chosen as the origin. For each particle of mass m and coordinate x          d,
there is a second particle of mass m and coordinate x        d. Thus the contributions of
these two particles to mx cancel. Since the entire body can be divided into such pairs,
xcm 0. We can apply the same argument to the y coordinates to show that ycm 0.
Thus the center of mass of each body is at the origin—the center of symmetry.                (a) Flat circular plate (b) Sphere
Suppose you wish to ﬁnd the center of mass of a system consisting of two or
more extended bodies. If you know the center of mass of each body, you can ﬁnd the
center of mass of the system simply by treating each body of mass M as a particle of
mass M located at the body’s center of mass. For example, you can ﬁnd the center of
mass of the system of the earth and the moon by treating the earth as a particle
located at the center of the earth and the moon as a particle located at the center of the
moon. Problem 51 outlines a proof of this result.
In a rigid body, the center of mass is ﬁxed with respect to the body. If a system of
particles is not rigidly bound together, however, its center of mass will not necessarily
be ﬁxed with respect to any point in the system. For example, a human body is clearly                 (c) Rectangular block
not rigid. If a person stands upright, the center of mass is located within the midsec-      Fig. 5–12 Finding the center of mass of
tion of the body. But if a person bends sufﬁciently at the waist, the center of mass can     a symmetric body.
be outside the body, as we shall see in the following example.

EXAMPLE 8        Center of Mass of a Sitting Person
The mass distribution of a sitting person can be roughly approx-
imated by the rectangles shown in Fig. 5–13. Find the person’s
center of mass.

SOLUTION We ﬁnd the center of mass of the entire body
by treating each rectangle as though its mass were concen-
trated at its geometric center. The center of mass of the 65 kg
rectangle is at point P, with coordinates x 10 cm, y 45 cm,
and the center of mass of the 25 kg rectangle is at point P with
coordinates x 60 cm, y 6.0 cm. Applying Eqs. 5–7 and 5–8
we ﬁnd

(65 kg)(10 cm)   (25 kg)(60 cm)
xcm
65 kg   25 kg

24 cm

(65 kg)(45 cm)   (25 kg)(6.0 cm)
ycm
65 kg   25 kg
Fig. 5–13
34 cm

These coordinates indicate that the center of mass, shown by the
red dot in the ﬁgure, is outside the body.
130                                          CHAPTER 5         Friction and Other Applications of Newton’s Laws

The center of mass of a projectile follows a parabolic trajectory and is described by
the equations developed in Chapter 3 for a single particle (if air resistance is negli-
gible). A ﬁreworks display is a beautiful demonstration of this effect. Immediately after
exploding, the ﬁreworks form a symmetric pattern about the center of mass. Since the
forces on the fragments during the explosion are all internal forces, they cancel each
other and have no effect on the motion of the center of mass, which will continue to
follow an approximately parabolic trajectory, just as though there had been no explo-
sion. The center of mass of a long jumper or a high jumper also follows a parabolic
trajectory, though the shape of the body may change considerably during the jump, and
consequently the position of the center of mass relative to parts of the body may also
change (Fig. 5–14).

Fig. 5–14 The high jumper’s initial
velocity determines the maximum height
reached by the body’s center of mass.
To clear the greatest possible height
for a given center-of-mass height, the
jumper rolls over the bar, keeping as low
as possible the parts of his body that are
not directly over the bar at any instant.

EXAMPLE 9       Two Skaters Meet
Two ice skaters, John (mass 75 kg) and Michelle (mass                  This is the distance John moves to meet Michelle.
55 kg), initially standing motionless on frictionless ice, 8.0 m
apart, begin to pull on opposite ends of a rope (Fig. 5–15).
How far does John move before he meets Michelle?

SOLUTION Since no net external force acts on the system
of John and Michelle, their center of mass, which is initially at
rest, must remain at rest. When they meet, it must be at this
point. We can use Eq. 5–7 to compute the initial distance of the
center of mass from John:

mx
xcm
m
(75 kg)(0) (55 kg)(8.0 m)
75 kg 55 kg
3.4 m                                                                         Fig. 5–15

Derivation of the Dynamics of the Center of Mass
Now we shall derive Eq. 5–4 ( F Macm). We begin with the equation deﬁning the
x coordinate of the center of mass for an n particle system (Eq. 5–5):
m1x1 m2x2      …    mn xn
xcm                   …                                 (5–5)
m1 m2             mn
Denoting the sum of the particle masses in the denominator by M, and multiplying this
equation by M, we obtain
5–3 Center of Mass   131

Mxcm        m1x1        m2x2            …     mn xn                           (5–7)

When the particles move, their center of mass may move. The change in the center-of-
mass x coordinate is related to the particles’ changes in coordinates, x1, x2, … , by
an equation of the same form as the equation above.
M xcm        m1 x1           m2 x2            …       mn xn
Dividing this expression by t gives
xcm               x1                  x2       …               xn
M               m1               m2                           mn
t                t                   t                        t
In the limit as t approaches zero, this equation becomes a relationship between
instantaneous velocities:
Mvcm,x        m1v1x       m2v2x           …       mnvnx                         (5–8)

When the particles are accelerated, their center of mass may be accelerated. We can
obtain an expression for the x component of center-of-mass acceleration acm,x in terms
of the particles’ accelerations, a1x, a2x, … , by repeating the operations that led from Eq.
5–7 to Eq. 5–8:

M vcm,x         m1 v1x          m2 v2x            …       mn vnx
vcm,x              v1x                 v2x       …               vnx
M                 m1                m2                            mn
t                  t                   t                         t
Macm,x      m1a1x         m2 a2x        …        mn anx                            (5–9)

We can go through the same series of steps that led from Eq. 5–5 to Eq. 5–9, starting
with the expression for the y coordinate of the center of mass (Eq. 5–6), leading to the
following expression relating the y component of center-of-mass acceleration to the y
components of the particles’ accelerations:
Macm,y        m1 a1y      m2 a2y          …       mn any                       (5–10)

This equation has the same form as Eq. 5–9. Eqs. 5–9 and 5–10 give the x and y
components of the center-of-mass acceleration vector acm. Together these equations are
equivalent to the vector equation
Macm         m1a1        m2 a2           …     mn a n
We can apply Newton’s second law to each particle. Thus in the preceding equation we
may insert m1a1       F1, m2a2      F2, … , mnan   Fn, where F1, F2, … , Fn are
the resultant forces on the particles:

Macm           F1             F2        …        Fn         (5–11)

The right-hand side of this equation is the sum of all the forces acting on all of the
particles in the system. It includes forces acting between the particles (internal forces)
as well as forces exerted on the system from outside (external forces). From Newton’s
third law, we know that forces always occur in action-reaction pairs. Therefore, when
we sum over all forces, the internal forces all cancel, leaving only the sum of the
external forces, which we denote simply by F. And so Eq. 5–11 is equivalent to Eq.
5–4 ( F Macm ).
In the special case where all parts of the system have the same acceleration a,
this acceleration will also be the acceleration of the center of mass, and the equation
F Macm reduces to the statement of the second law for a single particle: F Ma.
C
HAPTER           5      SUMMARY

The magnitude of the force of static friction varies from           circular motion. This resultant force is directed toward the
zero to a maximum value proportional to the normal force            center of the circle and has magnitude
pushing the two surfaces together:                                                                     mv2
F
fmax       s   N                                                             r
The location of the center of mass is deﬁned by the
where s is the coefﬁcient of static friction. The force of
equations
kinetic friction is given by
mx                my
f         N                                             xcm                ycm
k                                                           m                 m
where k is the coefﬁcient of sliding friction.                      The motion of the center of mass of a system is determined
Centripetal force is the force or combination of forces          by the equation
that produces the centripetal acceleration associated with
F    Macm

Questions

1 (a) What is the minimum possible value of          s?             9 You drive your car, ﬁrst along level pavement and then
(b) Is there a ﬁnite maximum possible value of s?                   up a hill.
2   (a) What is the magnitude of the normal force acting on             (a) Will the maximum possible force of static friction
each of the glass slides shown in Fig. 5–D?                          increase, decrease, or remain the same as you leave
(b) What is the value of s for these surfaces?                           the level part and start up the hill? Assume that the
3   Which would require a greater force in order to be                       surface of the road is the same.
moved: a 10 N brick on a horizontal wooden surface or               (b) Is it easier to spin your wheels on the hill or on the
a 100 N steel block on a horizontal Teﬂon surface?                       level part of the road?
4   A book lies on a table. One end of the table is gradually      10   What is the ratio of the maximum deceleration when
raised, so that the surface is inclined. When the incline is        braking to the maximum acceleration of a car with two-
great enough, the book begins to slide. Will sliding begin          wheel-drive on level ground? Assume that the car’s
sooner if the book lies ﬂat or stands on end, or does it            weight is equally distributed over the wheels and that
happen at the same point in either case? Assume that the            when the car has sufﬁcient power and good enough
book does not tip over when it is standing on end.                  brakes, the only limitation on the rates of acceleration or
5   You accelerate your car forward. What is the direction of           deceleration is the friction that can be provided by the
the frictional force on a package resting on the ﬂoor of            road.
the car?                                                       11   As a car rounds the top of a hill on a straight section of
6   A sheet of paper is initially at rest beneath a book on a           highway, it moves along an approximately circular path
table. You jerk the paper quickly to the right, exerting a          in a vertical plane. What force or combination of forces
large force on it. Explain what happens to the book and             provides the centripetal force necessary to keep the car
why.                                                                on this path: (a) friction; (b) normal force; (c) weight;
7   Your rear-wheel-drive car is stuck in snow. Would it be             (d) air resistance; (e) weight minus normal force;
better to (a) remove a heavy object from the trunk to               (f ) friction and weight?
reduce the weight on the rear wheels or (b) place a            12   Suppose you are riding a Ferris wheel at the amuse-
heavy object in the trunk to increase the weight on the             ment park. If the wheel rotates at a constant rate, is the
rear wheels?                                                        force exerted on you by your seat greatest at the top, the
8   Suppose you were on a perfectly frictionless ice pond.              bottom, or some point in between?
Would it be possible for you to walk or crawl very care-       13   A tall old redwood tree, almost devoid of branches,
fully to shore? Explain your answer.                                tapers gradually as it rises from the ground. Is the center
of mass of the tree closer to the top or the base?

132
Problems                                   133

14 A 3 m long javelin is thrown vertically upward, and the          Answers to Odd-Numbered Questions
tip reaches a maximum height of 10 m. What is the                1 (a) 0; (b) no; 3 the 10 N brick; 5 forward; 7 b;
maximum height of the javelin’s center of mass, assum-           9 (a) decrease; (b) easier on the hill; 11 e; 13 base; 15 at
ing the mass is uniformly distributed?                           the center of the hole
15 Where is the center of mass of a doughnut?
16 (a) In Example 9, will the place where John and Michelle
meet be affected by how they pull on the rope, that
is, by who pulls ﬁrst and by how hard or how long
each person pulls?
(b) Will the time it takes them to meet be affected by
how they pull on the rope?

Problems (listed by section)

7 Suppose you are standing on skis, headed straight down
5–1      Friction                                                      a slight incline covered with dry snow at 0 C. What is
1 Suppose the book in Fig. 5–2 weighs 9.0 N. To move the              the minimum angle of the incline that will cause you to
book from rest requires a minimum horizontal force P              slide?
of magnitude 3.0 N. The book will keep moving at                8 When you try to walk too fast on a slippery ﬂoor, you
constant velocity if the magnitude of the force P is              slip because walking fast requires you to exert a greater
reduced to 2.0 N. Find the coefﬁcients of static and              frictional force against the ﬂoor than the surface allows.
kinetic friction between the book and the table.                  What is the maximum frictional force you can exert
2   Suppose that while moving into an apartment you move              against the ﬂoor with one foot that bears your full
a refrigerator into place by sliding it across the ﬂoor. The      weight of 560 N, if s 0.20?
refrigerator weighs 1500 N and the coefﬁcient of static         9 What force must be applied to push a carton weighing
friction between the ﬂoor and the refrigerator is 0.45.           250 N up a 15 incline, if the coefﬁcient of kinetic fric-
What is the least force you could exert on the refriger-          tion is 0.40? Assume the force is applied parallel to the
ator to move it?                                                  incline and the velocity is constant.
3   You are holding a bulletin board weighing 6.0 N in             10 Two blocks, each of mass 1.0 kg, are pushed along the
place against a wall while your friend secures it to the          horizontal surface of a table by a horizontal force P of
wall. To keep it from slipping, you apply a force perpen-         magnitude 9.8 N, directed to the right, as shown in Fig.
dicular to the bulletin board, pressing it directly into          5–16. The blocks move together to the right at constant
the wall. How large must this force be if the coefﬁcient          velocity. Find (a) the frictional force exerted on the
of static friction with the wall is 0.40?                         lower block by the table; (b) the coefﬁcient of kinetic
4   To push a certain box across a level ﬂoor at constant             friction between the surface of the block and the table;
velocity requires that a horizontal force be applied to the       (c) the frictional force acting on the upper block.
box. The magnitude of this force is half as great as the
vertical force required to lift the box. Find the coefﬁ-
cient of kinetic friction between box and ﬂoor.
5   A friend on skis stands still on level ground, covered
with dry snow at 0 C. How hard would you have to
push your friend in the forward, horizontal direction to
move him if your friend weighs 800 N?
6   A wooden block weighing 5 N is at rest on top of a
wooden desk. The coefﬁcient of static friction between
the block and the desk is 0.5. Find the minimum
horizontal force required to move the block if (a) the                                  Fig. 5–16
block is being pressed into the surface with a 3 N force;
(b) the block is partially supported by a vertical string
under a tension of 3 N.
134                                 CHAPTER 5            Friction and Other Applications of Newton’s Laws

11 In the previous problem, suppose that the magnitude of           19 A cardboard box slides along a wood ﬂoor with an ini-
P is increased to 11.8 N, resulting in an acceleration of        tial velocity of 5.00 m/s. If the box comes to rest after
1.0 m/s2 of the two blocks to the right. Find (a) the fric-      traveling 3.00 m along the ﬂoor, what is the coefﬁcient
tional force exerted on the upper block by the lower             of kinetic friction between ﬂoor and box?
block; (b) the frictional force exerted on the lower block    20 A small package rests on the horizontal dashboard of a
by the upper block; (c) the frictional force exerted on          car. If s 0.30, what is the minimum acceleration of
the lower block by the table.                                    the car that will cause the package to slip off, assuming
12   A skier of mass 64 kg skis straight down a 12 slope at           that the car is on a level road?
constant velocity. Draw a free-body diagram of the skier
with the various external forces acting on her. Include
the force of air resistance, which is directed opposite the    5–2    Centripetal Force
velocity.
21 A 0.100 kg rock is attached to a 2.00 m long string and
(a) Find the value of the normal force.
(b) The force of air resistance has a magnitude of 75 N.          swung in a horizontal circle at a speed of 30.0 m/s. Find
Find the frictional force on the skis.                        the tension in the string. Neglect the effect of gravity.
22 A 1.00 kg mass is attached to a spring of force constant
(c) What is the coefﬁcient of kinetic friction?
13   A block weighing 10.0 N rests on a 30° inclined plane.            10.0 N/cm and placed on a frictionless surface. By how
Find (a) the normal force exerted by the plane on the             much will the spring stretch if the mass moves along a
block; (b) the frictional force exerted by the plane on           circular path of radius 0.500 m at a rate of 2.00 revolu-
the block; (c) the magnitude of the total force exerted           tions per second?
23 A passenger of mass 50.0 kg is in a car rounding a level
by the plane on the block; (d) the normal force and the
frictional force exerted by the block on the plane.               curve of radius 100.0 m at a speed of 20.0 m/s.
14   The coefﬁcient of static friction can be measured exper-          (a) Assuming that friction is the only horizontal force
imentally in the following way. Place an object on a                  acting on the passenger, ﬁnd the frictional force.
surface that is initially horizontal. Then incline the            (b) What would happen if s 0.3?
*24 Find the angle at which a curve of radius r should be
surface at an angle , which is gradually increased.
When reaches some maximum value max, the object                   banked in order that a car moving at a speed v will not
will begin to slide. This angle is easily measured. The           need any frictional force to round it (Fig. 5–17).
coefﬁcient of static friction is then found from the equa-
tion s tan max.
(a) Prove this result by analyzing the forces acting on
the block.
(b) What is max for a piece of dry, clean glass resting on
a glass surface?
15   A hockey puck is given an initial velocity of 40.0 m/s
along the ice. Find the speed of the puck 1.00 s later if
the coefﬁcient of kinetic friction between puck and ice is
0.600. (HINT: The result is independent of the mass of                                  Fig. 5–17
the puck.)
16   A car of mass 975 kg coasts on a level road, starting         25 A sprinter of mass 70.0 kg runs at a constant speed of
with an initial speed of 15.0 m/s and coming to rest             10.0 m/s in a straight line. Find the extra force the
15.0 s later. Find the frictional force acting on the car.       sprinter must exert on the ground to round a curve of
17   (a) Find the maximum rate of deceleration of a car on a          radius 10.0 m.
dry, level road, assuming the coefﬁcient of static        26 A sling carrying a rock of mass 0.200 kg is swung in a
friction between tires and pavement equals 1.0.              vertical circle of radius 1.50 m at a speed of 62.6 m/s
(b) Find the minimum stopping distance if the car has            (sufﬁcient to give a horizontal range of 400 m, as indi-
an initial speed of 25 m/s.                                  cated in Problem 17, Chapter 3). Find the force the hand
(c) Repeat parts a and b assuming a wet, oily surface for        must exert on the sling when the rock is (a) at the top if
which s 0.10.                                                its circular arc and (b) at the bottom of its circular arc.
18   Find the maximum acceleration of a car with two-wheel         27 A girl weighing 285 N is on a swing, supported by two
drive on a dry, level road with s 1.0, assuming that             chains of length 2.00 m. Find the tension in each chain
the car’s weight is equally distributed between the front        when the chains are vertical and her speed is 3.00 m/s.
and rear wheels.
Problems                                    135

*28 As a car rounds the top of a hill at a speed of 20.0 m/s,     32 A simple model of the mass distribution of a pole
it very brieﬂy loses contact with the pavement. This             vaulter is given in Fig. 5–21. The pole vaulter is shown
section of the road has an approximately circular shape          at the high point in the jump, at which point the midsec-
(Fig. 5–18). Find the radius r.                                  tion of his body is just over the bar (at the origin).
(a) Find the center-of-mass coordinates of the pole
vaulter’s body. Treat the rectangles as uniform distri-
butions of mass. The center of mass of each
rectangle is indicated by a dot, next to which is the
mass of the rectangle.
(b) If the center of mass is at an elevation of 5.10 m,
what is the height of the bar?

Fig. 5–18

5–3    Center of Mass
29 The mass of the earth is 5.98     1024 kg, the mass of the
moon is 7.36 10 kg, and the distance between the
22

centers of the earth and the moon is 3.82 108 m. How
far is the center of mass of the earth-moon system from
the center of the earth?
30 Find the center of mass of the particles in Fig. 5–19,
which are positioned at the corners of a square of edge
length 30.0 cm.

Fig. 5–19

31 Find the center of mass of the bricks shown in Fig. 5–20.                               Fig. 5–21

**33 Each side of an equilateral triangular plate has a length
of 1.00 m. Find the distance of the plate’s center of
mass from the center of one side of the plate. (HINT:
Use symmetry.)

Fig. 5–20
136                               CHAPTER 5            Friction and Other Applications of Newton’s Laws

34 The location of two particles at t     0 is shown in Fig.     37 An airplane of mass 1.0       104 kg falls vertically down-
5–22. The particles are initially at rest. Each particle is      ward, crashing to the ground. The plane explodes just
subject to a constant force, as indicated in the ﬁgure.          before impact. Investigators looking into the cause of the
(a) Find the position of the center of mass at t 0.              crash search for the wreckage. Large fragments of mass
(b) Find the resultant force acting on the system.               4.0 103 kg, 3.0 103 kg, and 1.0 103 kg are found
(c) Find the location of the center of mass at t 5.00 s.         respectively at the point of impact, 30 m north, and 50 m
east. Where should one look for the remainder of the
plane; in other words, where is the center of mass of the
remaining wreckage?
38 Olympic gold medalist Carl Lewis performs a long jump,
starting at rest on a 12.0 m long ﬂatbed railroad car of
mass 450.0 kg. Lewis, whose mass is 90.0 kg, runs along
the length of the car and jumps off the end. From the
Fig. 5–22
beginning of the run to the end of the jump, he travels a
total distance of 15.0 m relative to the earth. How far has
35 Spheres A and B are attached to the ends of a rod of
the railroad car moved by the time he hits the ground, if
negligible mass (Fig. 5–23). The spheres are initially at
no frictional force acts between car and earth?
rest on a frictionless, horizontal surface. A horizontal
string is attached to B at point P and is under a constant
tension of 6.00 N.                                            Additional Problems
(a) Find the initial location of the center of mass.          39 The two blocks shown in Fig. 5–25 are attached to
(b) How far will the center of mass move in 2.00 s?              opposite ends of a string, which passes over a friction-
less pulley of negligible mass. The pulley is attached to
a wall on the right. The top block is pulled to the left
with a force P of magnitude 10 N. As the top block
moves to the left, the bottom block moves to the right.
(a) Draw the appropriate free-body diagrams necessary
to solve for the acceleration of each block, but DO
NOT COMPLETE the problem.
(b) What additional information must be provided in
order to solve the problem?
Fig. 5–23

36 A boy of mass 60 kg is about to disembark from a canoe
of mass 40 kg. The canoe is initially at rest, with the
bow just touching the dock (Fig. 5–24). The center of
the canoe is 3.0 m behind the bow. As the boy moves
forward 6.0 m to the bow, the canoe moves away from
the dock.                                                                               Fig. 5–25
(a) How far is the boy from the dock when he reaches
the bow?                                                  40 Draw complete free-body diagrams for the block and the
(b) Is the canoe moving at this point? Assume that there         wedge shown in Fig. 5–26.
is no friction between the canoe and the water.

Fig. 5–24

Fig. 5–26
Problems                                  137

*41 Find the magnitude of the force P necessary to drag the    *46 Find the acceleration of the 5.0 kg block in Fig. 5–30.
crate shown in Fig. 5–27 at constant velocity. The crate
weighs 500 N, and k 1.0.

Fig. 5–27
Fig. 5–30

*42 Find the acceleration of the two blocks sliding down the
*47 A truck is traveling around a level curve of radius 100.0
incline in Fig. 5–28.
m at an instantaneous speed of 20.0 m/s. The truck’s
speed is increasing at the rate of 2.00 m/s2. Find the
magnitude of the frictional force on a 100.0 kg crate
resting on the ﬂoor of the truck.
48 A certain roller coaster design uses a vertical loop of
radius 8.00 m. Assuming that the roller coaster remains
on the track, what is the minimum speed of a car at the
top of the loop? (HINT: In general, the centripetal force
is the vector sum of the car’s weight w and the normal
force N exerted by the tracks. The value of N varies
with the speed. The minimum value of N is zero, which
Fig. 5–28
occurs when the roller coaster is just about to leave the
track.)
*43 Find the acceleration of the 2.0 kg block in Fig. 5–29.
49 Two books, each having a mass of 1.0 kg, are shown in
*44 The 1.0 kg block in Fig. 5–29 is replaced by a 4.0 kg
Fig. 5–31. If P     30 N and s          k   0.50 for all
block. Find the acceleration of the 2.0 kg block.
surfaces, ﬁnd the acceleration of each book.

Fig. 5–31

Fig. 5–29

**45 Find the minimum mass of a block to replace the 1.0 kg
block in Fig. 5–29, such that the 2.0 kg block moves
upward.
138                                  CHAPTER 5           Friction and Other Applications of Newton’s Laws

*50 A truck and its contents have a total mass of 4.00     103     **55 A board of uniform thickness is shown in Fig. 5–32.
kg. The truck is initially moving at a constant speed of           The board’s center of mass is initially at point O. If a
20.0 m/s. Some of the truck’s cargo, which is initially            hole centered at P is cut out of the board, how far to the
near the front end, shifts and moves toward the back. A            right of O is the new center of mass, if cutting the hole
mass of 4.00 102 kg moves 5.00 m in the backward                   removes 10.0% of the board’s mass? (HINT: Express
direction during a 1.00 s interval. How far does the truck         the center of mass of the entire board in terms of the
travel during this interval?                                       center of mass of the circular section and of the board
*51 The x coordinates of the centers of mass for two arbi-             with the hole.)
trary bodies A and B may be written
A mx
xA
A m

and
B mx
xB
B m

where the subscripts A and B indicate that the sums are
over the particles of the respective bodies. The system of
the two bodies has a center of mass with an x coordinate                                 Fig. 5–32
that you can ﬁnd by applying the deﬁnition
mx
xcm
m
taking the sums over all the particles in both bodies.
By separating the sums into two sums, each involving
only particles in one of the bodies, show that xcm may be
expressed as
MAxA MB xB
xcm
MA MB
where MA and MB are the respective masses of bodies A
and B. Show that the y coordinate of the center of
mass can be expressed in a similar way.
*52 Initially a wedge of mass 5.00 kg is at rest on a fric-
tionless horizontal surface, and a 1.00 kg block is at
rest near the top of the wedge (Fig. 5–26). The block
begins to slide down the incline. After moving 1.00 m
along the incline, the block again comes to rest. How far
does the wedge move to the right?
**53 An amusement park ride called the Rotor consists of a
room in the shape of a vertical cylinder (2.00 m in radius)
which, once the riders are inside, begins to rotate, forcing
them to the wall. When the room reaches a speed of one
rotation every 1.50 s, the ﬂoor suddenly drops out. What
is the minimum coefﬁcient of static friction between
riders and wall necessary to prevent them from sliding
down the wall?
*54 Suppose you suddenly discover that you are heading
your car toward the edge of a cliff. Would it be better to
try to stop while traveling straight ahead or to turn the
car to one side? Assume that, if you turn the car, your
path will be circular and that the coefﬁcient of static