A large suspension bridge is a remarkable structure, a precise blend of
engineering and imagination. In its construction, its designers must ensure its
stability and endurance. To do so, they must understand how to balance the
many forces acting on it so that it remains motionless. They must recognize
how the bridge responds to the enormous forces acting on it—how its parts
stretch or flex under the bridge’s load. And the designers must certainly be
able to calculate the maximum forces that can be applied to bridge members,
such as beams or cables, before such members fail.
In this chapter, we study how any structure—from a bridge to a tree to a knee
joint—must be designed in order to support itself and external loads. We will
see under what conditions an object under load will remain motionless, and
how it responds to changes in the forces acting on it. Is the structure stable if
outside forces disturb it, or does it come crashing down? How do its members
bend and stretch as forces act? And are the members strong enough to resist
such bending, or is their ultimate strength exceeded, with disastrous results?
8.1
Static Equilibrium
In the preceding chapters, we have made an in-depth study of motion and its
causes. In many disciplines, it is just important to understand under what
conditions objects do not move. In structural engineering, buildings and
dams must be designed such that they remain motionless, even when huge
forces act on them. In sports science, a correct stationary position is often the
starting point for a successful athletic event. And a study of, say, the knee
joint must understand the enormous forces that the joint must sustain, even
when the subject is standing at rest.
Recall from Section 5.2 that when any object is at rest, we say that it is in
static equilibrium. In that section we also found the condition necessary for a
particle to be in static equilibrium: the net force Fnet on the particle must be
zero, so that it does not accelerate. Such a situation is shown in Figure 8.1a,
where the two forces applied to the block balance, and the block can remain
at rest. But in Chapter 7 we moved beyond the particle model to study
extended objects that could rotate as well as move translationally. For an
extended object, having Fnet  0 is by itself not enough to ensure static
equilibrium. In Figure 8.1b, the same two forces are applied, but this time off
center. The net force is still zero, so the block will not undergo an overall
acceleration. But the block will begin to rotate, because now there is a net

torque applied to it. So there is a second condition for static equilibrium: the
net torque  net on the object must be zero.
Figure 8.1 A block with no net force acting on it may still be out of equilibrium.
For an object to be in static equilibrium, then, it must have both zero net force
acting and zero net torque acting on it. If we write the net force in component
form, these conditions are
 Fx  0
 Fy  0
  0
Conditions for static equilibrium
STOP TO THINK 8.1 Whichof the objects below could be in static
equilibrium?
EXAMPLE 8.1 Lifting a weight
Weightlifting can exert extraordinary forces on the body’s joints. In the strict
curl event, a standing athlete lifts the barbell using only the forearms. The
record weight lifted in the strict curl is over 200 pounds (about 900 N). Figure
8.2a shows the arrangement of the arm bones and the main lifting muscles
when the forearm is horizontal. With the arms held stationary in this
position, what is the tension in the lifting biceps muscle when 900 N is lifted?
Figure 8.2 Forces and torques in holding a barbell.
Prepare Figure 9.2b shows a simplified model of the arm, indicating the
forces acting on the forearm. Fm is the tension force due to the muscle, Fw is
the downward force of the weight, and Fe is the force of the elbow joint on
the forearm.
 to be in static equilibrium, the net forceand net torque
Solve For the forearm
on it must be zero. Setting thenet force to zero gives
F  F
y
m
 Fe  Fw  0
We know the magnitude of the weight force, Fw = 450 N (assuming that each
arm supports half the 900 N total weight). We don’t know either of the other
two forces Fm and Fe , nor does the force equation above give us enough
information to find them. Using the
fact that the torque must be zero gives us
that extra information. Choosing the elbow joint to be the axis of rotation, we
have

 net  Fw d  Fm dm  0
Note that Fe makes no contribution to the torque, since it acts directly at the
pivot. We can solve the above equation for Fm to find
Fm  Fw
d
35cm
 (450N)
 3150N  700pounds .
dm
5cm

Assess We see that we need to use the fact that both the net force and the net
torque are zero for an object in equilibrium. The short distance d m from the
muscle to the elbow joint means that the force of the muscle has to be very
large in order to counter the torque generated by a force applied at the end of
the forearm.
Choosing the Pivot Point
In Example 8.1 above, we calculated the net torque using the elbow joint as
the axis. But we have learned that the torque on an object depends on which
point is chosen as the axis of rotation. Was there something special about our
choice of the elbow joint as the axis?
Consider the hammer shown in Figure 8.3, supported on a pegboard by two
pegs A and B. Because it’s in static equilibrium, the net torque around the
pivot at A must be zero: the clockwise torque due to the weight w is exactly
balanced by the counterclockwise torque due to the force nB of peg B. But if
instead we take B as the pivot, the net torque is still zero. The
counterclockwise torque due to w (with a large force but small moment arm)
balances the clockwise torque due to n A (with a small force but large
moment arm). Indeed, for an object in static equilibrium, the net torque about
every axis must be zero. This means that when you calculate the torque, you
can pick any point you wish as an axis. Even if we use the arbitrary point C,
which is not a real pivot point for the hammer, we would find the torque to be
zero.
Figure 8.3 The net torque on an object in static equilibrium is zero when calculated
about any axis..
Although any choice of axis will work, some choices are better because they
simplify the calculations. Often, there is a “natural” axis of rotation in the
problem, an axis about which rotation would occur if the object were not in
static equilibrium. Example 8.1 is of this type, with the elbow joint as a
natural axis of rotation.
If no point naturally suggests itself as an axis, you should look for a point on
the object where a force acts. Good choices are where either several forces
act at once, or where a force whose magnitude you don’t know acts. Such
points are good choices because any force acting at the axis does not
contribute to the torque. By choosing the axis where a force acts, that force
will not appear in the torque equation.
Don’t agonize too much about this choice of axis. If you can’t decide, pick
any point on the object where a force acts. You will still be able to solve the
problem.
PROBLEM-SOLVING STRATEGY 25.1 Static equilibrium problems
Prepare Model the object as a simple shape. Draw a pictorial representation
that shows all forces and distances. List known information.

Pick a point about which the torques will be calculated.

Determine the moment arms of all forces about this pivot point.

Determine the sign of each torque about this pivot point.
Solve The mathematical representation is based on the fact that an object in
static equilibrium has no net force and no net torque.
Fnet  0 and  net  0

Write equations for  Fx  0,

Solve the resulting equations.
 F  0,   0 .
y
Assess Check that your result is reasonable and answers the question.
EXAMPLE 8.2 A board on sawhorses
A board weighing 100 N sits on two sawhorses, as shown in Figure 8.4a.
What are the magnitudes of the two forces F1 and F2 of the sawhorses acting
on the board?


Figure 8.4 A board sitting on two sawhorses.

Prepare The board and the forces acting on it are shown in Figure 8.4b. F1
and F2 are the forces on the board due to the sawhorses, and w is the weight
of the board, acting at the center of gravity.
As discussed above, a good choice for the pivot is a point where
an unknown
force acts, since then the unknown force contributesnothing to the torque.
Two such points are Point A in Figure 8.4b, where force F1 acts, or point B,
where F2 acts. Either will work; let’s choose point A for this example.
With this choice of pivot, we see that the moment arm for w, which acts
halfway down the board, is L / 2. Since w tendsto rotate the board
 clockwise, its torque is negative. The moment arm for F is the distance d2 ,
2
and, since F2 tends to rotate the board counterclockwise, its torque is
positive.
Solve Since the board is in static equilibrium, the net force Fnet and the net
torque  net must both be zero. The forces acting have only y components, so
we needn’t worry about the x components. So we have
F
y
 F1  w  F 2  0
,
from which we can solve for F1 :
F1  w  F 2 .
Next we compute the torque around axis A, and set it to zero. We have
L
2
 net   w  d2 F2  0 ,
from which we find
F2 
Lw (3.0 m)(100 N)

 75.0 N
2d2
2(2.0 m)
From the force equation above, we also have
F1  w  F2  100 N  75.0 N  25.0 N
Assess It seems reasonable that F1  F2 since more of the board sits over the
right sawhorse.
EXAMPLE 8.3 Choosing a different axis
Let’s repeat Example 8.2, but with the axis at a different point. Let’s put the
axis at the center of gravity where the weight w acts. Will this change our
results?
Prepare In Figure 8.5 we define new distances d1 and d2 from the axis to
where the forces F1 and F2 act. Moment arm d1 and force F1 produce a
negative torque, while d2 and F2 produce a positive torque.
Figure 8.5 A board on two sawhorses, with axis at the center of gravity.
Solve The force equation still reads F1  w  F2  0 , from which we have, as
before,
F1  w  F2
Now we need to calculate the net torque about the center of gravity. The
weight w, acting at the pivot, now does not contribute to the torque. We have
 net  d1F1  d2 F2  0
If we insert the expression for F1 from the force equation into the torque
result, we get
d1 w  F2  d2 F2  0 .
We can solve this for F2 to get
F2 
d1w
(1.5 m)(100N)

 75N
d2  d1 1.5 m  0.5 m
as before. Again, we can use the force equation to find that F1 = 25 N.
Assess Even with a very different choice of axis from Example 8.2, we get
the same result for the forces.
STOP TO THINK 8.2 A uniform beam is suspended from a wire, and has a
pivot on its left end. In in which direction is the force of the pivot on the
beam?
A.
B.
C.
D.
E.
An interesting application of the principle of static equilibrium is the
determination of the position of the center of gravity of the human body. As
we’ll see later, the position of a person’s center of gravity plays an important
role in many aspects of sports and athletics. Because the human body is
highly flexible, the position of the center of gravity is quite variable, and
depends on just how the body is posed.
Figure 8.6 Board-and-scale method of finding a subject's center of gravity.
The position of the body’s center of gravity can be located accurately from
simple measurements with a reaction board and a scale. Figure 8.6 shows the
method. The subject lies or stands on the board in the desired posture. The
board has a pivot at its left end, while its right end is supported on a spring
scale. The spring scale pushes up on the board with a force F; the magnitude
F of this force is the scale reading. The following example shows how the
position of the subject’s center of gravity can be using this method.
EXAMPLE 8.4 Finding the center of gravity of the human body
A subject, weighing 730 N, lies on a reaction board that is 2.5 m long and
weighs 60 N. The scale on the right reads 300 N. Find the distance d from
the pivot to the subject’s center of gravity.
Figure 8.7 Finding the center of gravity of the human body.
Prepare The forces and distances in the problem are drawn in Figure 8.7.
We’ll calculate all torques around the pivot at the left end of the board. Then
the torque due to F is positive, and those due to w and wb are negative. The
torque due to n, which acts at the pivot, is zero.
Solve Since the board and subject are in static equilibrium, we require that
the net force and net torque on the board be zero. The force conditions reads
F
y
 n  wb  w  F  0
The net torque must also be zero in static equilibrium, so we have
L
   2 w
b
 dw  LF  0
Here we assumed that the board was uniform, so that its weight w b acts at a
point halfway along the board, that is, a distance L 2 from the pivot.
We can now solve the torque condition for d :
LF  12 Lwb (2.5 m)(300 N)  12 (2.5 m)(60 N)

 0.92 m
w
730 N
If the woman is 5’ 6” (1.68 m) tall, her center of gravity is about
(0.92 m)/(1.68 m)  55% of her height.
d
Assess Note that in this solution we didn’t actually need the information
from the  Fy  0 equation. This is because in this method we measure w,
wb , and F. The  Fy  0 equation serves only to determine the normal force
n of the pivot on the board, which we don’t need for our analysis.
9.2 Stability and Balance
Picture several common items—a dinner plate, a wine glass, a tall vase with
flowers—sitting on a table. All three are motionless, and so in static
equilibrium, but they differ in an important respect. The top-heavy vase
could be easily toppled by a small bump against the table; the wine glass
might take a larger blow from an errant elbow to knock it over; the plate,
however, would take a great disturbance to flip it. All these objects, then,
differ in their resistance to being toppled; we say that the plate is very stable
while the vase is only marginally so.
Why is the tall candlestick easy to knock over, but the squat bowl is very stable?
These examples lead us to think about the stability of an object in the
following way: how far can the object be tilted without it falling over? For a
vase this might be only a few degrees; for a plate, you’d have to lift its edge
more than 90 to get it to fall all the way over. The greater the angle to which
an object can be tilted before falling over, the greater its stability.
What determines how far we can tilt an object before it topples? Consider the
pop can in Figure 8.8a, sitting at rest on a table. Now slightly tilt the can, as
in Figure 8.8b, and release it. There is then a torque on the can produced by
its weight w that tends to rotate the can back towards its original equilibrium
position, where it would again end up sitting upright. If, however, we tilt the
can to a large angle and release it, as in Figure 8.8c, the torque acts so as to
rotate the can even farther to the right: the can will now fall over.
Figure 8.8 How far can a pop can be tilted before it falls?
From Figure 8.8, we can see the essential difference between angles at which
the can falls back, and angles at which it falls over. When the can was sitting
upright, it sat on a base of support that spanned the region between points A
and B. The base of support, then, is the region between the pivots about
which the object would tilt if pushed from various directions. In Figure 8.8b,
where the can will return to its original position, we note that a vertical line
dropped down from the center of gravity will fall within the base of support,
while in Figure 8.8c, where the can will fall over, the vertical line from the
center of gravity falls outside the base of support.
Finally, there is a special angle, shown in Figure 8.8d, at which the center of
gravity is directly over the edge of the base of support B. This is the largest
angle to which the can can be tilted without falling over. When this maximum
angle is large, an object is very stable; when this angle is small, an object is
only slightly so.
Example 8.5 How far can the block be tilted?
A block of wood is 7 cm high and 2 cm long. It is placed on a board, and the
board slowly tilted. How far can the board be tilted before the block falls?
Prepare From the discussion above, we know that an object becomes
unstable when its center of gravity is directly over the outside edge of its base
of support, that is, the point that it will rotate about when it starts to fall. In
Figure 8.9 we draw the block at this angle.
Figure 8.9 A block at the angle at which it is just about to fall.
Solve As shown in the inset to Figure 8.9, the small triangle on the block has
short sides of lengths 1.0 cm and 3.5 cm. We have then
1.0 cm
 0.286
3.5 cm
  arctan(0.286)  17.7
tan 
Assess This angle looks about right from the drawing.
TRY IT YOURSELF
Balancing pop can Try to balance a full—or empty—pop can on the
narrow bevel at the bottom. It can’t be done, because when full or empty, the
center of gravity is near the center of the can,XXX as in Figure 8.8b. But if
you put about 2 ounces (60 ml) of water in the can, the center of gravity will
be right over the bevel when the can is sitting on the bevel, and the can can
be balanced.
Relative Stability
We’ve seen that an object becomes unstable when its center of gravity
Just how stable and object is depends on the size of its base of support, and
how high its center of gravity is. In Figure 8.10 are shown two blocks with
equally high centers of gravity, but with different bases of support. The wide
block in (a) can be tilted substantially farther before its center of gravity (CG)
is over its pivot point than can the narrow block in (b). So we have out first
criterion for stability: The broader an object’s base of support, the more
stable it will be.
Figure 8.10 A broader base of support increases stability.
In Figure 8.11 we have two blocks with identical bases of support, but whose
centers of gravity differ in their height above the base. (These blocks are
obviously not of uniform composition.) We see that the block in (a), with a
high CG, can be tilted only slightly before it starts to fall. Conversely, the
block in (b), with a very low CG, can be tilted quite far before it will fall.
This illustrates our second stability criterion: The lower an object’s CG, the
more stable it is.
Figure 8.11 A lower center of gravity increases stability.
We can quantify these observations by examining how far an object such as
the one shown in Figure 8.12a can be tilted before falling. When the object is
sitting in its equilibirum position, the center of gravity is a height h above the
ground. We’ll call d the distance from the point on the ground directly below
the center of gravity to the pivot point about which the object will be tilted.
As we’ve seen, once the object is at the maximum angle  m to which it can
be tilted without falling, the center of gravity is directly below the pivot point.
Then, from the small triangle in Figure 8.12b, we see that tan  m  d / h, or
 m  arctan(d / h)
(8.1)
Figure 8.12 Finding the maximum angle to which an object can be tilted without
falling.
One area where stability plays a vital role is in the design of motor vehicles.
With the increasing popularity of sports-utility vehicles (SUVs), this topic has
become one of major concern. SUVs are particularly prone to rollover
accidents, in which the vehicle flips and rolls several times before coming to
rest. These type of accidents are especially deadly, being responsible for
some 25% of all traffics deaths each year.
SUVs are prone to such accidents because their centers of gravity are
generally quite high compared to their widths. Figure 8.13 shows front views
of a large SUV and a passenger car with their measured centers of gravity
indicated. The SUV’s center of gravity is at a height above the pavement that
is about 47% of its width, while the passenger car’s center of gravity height is
only about 33% of its width. The following example shows how this
difference affects the angle at which the two vehicles can be tilted (usually by
hitting a stationary object such as a curb) before rolling over.
Figure 8.13 Compared to passenger cars, SUVs have high centers of gravity relative
to their widths.
Example 8.6 Stability of vehicles
At what angles would the two vehicles in Figure 8.13 tip over?
Prepare We’ll use Equation (8.1) to find the maximum possible angle of tilt
 m . To do so, we’ll need to find the distances d and h for each car. We know
that the SUV in Figure 8.13 has its center of gravity at a height h that is 47%
of its width. But its width is twice the distance d defined in Figure 8.12. Thus
for the SUV we have h  0.47w  0.47(2d), so that
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Similarly we have that for the car, d / h  1 / (2  0.33)  1.51.
Solve We can use Equation (8.1) to find that, for the SUV,
 m  arctan(1.06)  46.7
while for the car,
 m  arctan(1.51)  56.5
Assess Rember that  m is the maximum angle an object can be tilted before
tipping, so a greater value of  m implies a more stable object. The car can be
tilted nearly 10 further than can the SUV, making it significantly more
stable against rolling over.
Stability and Balance of the Human Body
The human body is remarkable for its ability to constantly adjust its stance so
that its stability is optimized. This ability of humans (and other animals) to
In many cases we wish to maximize our stability, but there are also cases
where we trade off some stability for improved mobility. A shortstop’s
crouch, for instance, is a fairly stable position, but also allows for quick
movement to either side to field the ball. And in all athletic contests,
participants are in constant motion. To keep her balance, the athlete must
continuously (and unconsciously) adjust her pose so that her CG is over her
(constantly-changing) base of support.
A simple example of how the body naturally realigns its center of gravity is
found in the act of standing up on tiptoes. Figure 9.13(a) shows the body in
its normal standing position. The position of the center of gravity, found
using the board-and-scale method, is also shown. Notice that the CG is wellcentered over the base of support (the feet), ensuring maximal stability. If the
subject were now to stand on tiptoes, without otherwise adjusting the body
position, the center of gravity would fall behind the base of support, which is
now the balls of the feet, and he would fall backwards. So the body naturally
leans forward, regaining stability by moving the CG over the balls of the feet
(Figure 9.13(b).
Figure 9.14 (Left) When standing normally, the center of gravity is centered over the
feet. (Right) On tiptoes, the body must lean forward so that the CG is over the small
portion of the feet still on the ground.
TRY IT YOURSELF
Noisy magnets The striped pole structure of refrigerator magnets can be
demonstrated by using two identical such magnets.
You can try this yourself: stand facing a wall with your toes touching the base
of the wall. Now rise onto your tiptoes. You will be unable to stand without
falling backwards. As we see from Figure 9.13(b), in order to stand on
tiptoes the body has to lean significantly far forward. With the wall in your
way, you cannot lean far enough forward to maintain your balance, and you
will begin to topple backwards.
Even the simple act of walking can be broken down into a sequence of stable
and unstable positions. If you walk slowly forward, you will notice that as
you extend your leg forward to initiate the next step, a point is reached where
the body is no longer stable—the CG has moved forward of the base of
support, the rear foot. At this point, the body begins to fall forward—onto the
leading foot. Then the body leans forward, setting its CG over the front foot,
so that the rear foot can be lifted without falling. As the rear foot swings
forward, the process begins again. Note that this description is
oversimplified, and ignores details such as the side-to-side loss of stability as
each foot is raised.
Fields such as kinesiology and sports mechanics are often concerned with the
stability and balance of human subjects whose CG can move with respect to
the body. In our everyday movements, we are constantly shifting our centers
of gravity from stable to unstable positions. In doing so, the body naturally
adjusts so that the CG is moved back to a stable position. This can occur by
either repositioning the body so that the CG moves back within the base of
support (this happens, for instance, when you bend over), or by changing the
base of support itself, as happens constantly as you walk.
STOP TO THINK 9.4 Rank order the three shapes shown according to how
stable they are, from least stable to most stable. The dots show their centers
of gravity.
This is because all matter is, at an atomic level, in some sense “springy.” To
understand what this means, we will first take a look at springs work, and
then make a model for solids that incorporates this information.
Many materials, and particularly many biological materials, are soft or
“stretchy,” changing their dimensions considerably when forces act on them.
Even very stiff objects like metal beams deflect when forces are applied to
them. The top of a very tall building may sway back and forth by several feet
in a strong wind. No matter how stiff an object might appear, it will bend, if
only microscopically, under an applied force.
9.4 Springs & Elastic Materials
In considering objects in equilibrium, we have assumed that they maintain
their shape as forces and torques are applied to them. In reality this is an
oversimplification. Every solid object stretches or compresses, if only
slightly, when a force acts upon it.
. If you stretch a rubber band, a force appears that tries to pull the rubber band
back to its equilibrium, or unstretched, length. A force that restores a system
to an equilibrium position is called a restoring force. Systems that exhibit
restoring forces are called elastic. The most basic examples of elasticity are
things like springs and rubber bands. If you stretch a spring, a tension-like
force pulls back. Similarly, a compressed spring tries to re-expand to its
equilibrium length. Other examples of elasticity and restoring forces abound.
The steel beams bend slightly as you drive your car over a bridge, but they
are restored to equilibrium after your car passes by. Your leg bones flex a bit
during each step you take. Nearly everything that stretches, compresses,
flexes, bends, or twists exhibits a restoring force and can be called elastic.
Elasticity has many important applications, but we have another motive for
introducing elasticity at this point. One of the goals of this textbook is to
understand the atomic structure of matter. We have already introduced a
simple model of a solid in which particle-like atoms are held together by
spring-like molecular bonds. Figure 10.14 reminds you of this model. We
devoted the first seven chapters to understanding the dynamics of particles.
Now it’s time to look more closely at the elastic behavior of the bonds, the
“glue” that holds matter together.
Springs
We’re going to use a simple spring as a prototype of elasticity. Suppose you
have a spring whose equilibrium length is L 0 . This is the length of the
spring when it is neither pushing nor pulling. If you now stretch the spring to
length L, how hard does it pull back? One way to find out is to attach the
spring to a bar, as shown in Figure 8.15, then to hang a mass m from the
spring. The mass stretches the spring to length L. Lengths L 0 and L are easily
measured with a meter stick.
Figure 8.15 A spring of equilibrium length L0 is stretched to length L by a hanging
mass.
The mass hangs in static equilibrium, so the upward spring force Fsp exactly
r
balances the downward weight force w to give Fnet  0 That is,
Fsp  w  mg
(8.2)
By using different masses to stretch the spring to different lengths, we can
determine how Fsp , the magnitude of the spring’s restoring force, depends on
the length L.
Figure 8.16 shows measured data for the restoring force of a real spring.
Figure 8.16 Measured data for the restoring force of a real spring.
Notice that the quantity graphed along the horizontal axis is y  L  L0 .
This is the distance that the end of the spring has moved, which we call the
displacement from equilibrium. The graph shows that the restoring force is
proportional to the displacement. Recall from Math Model XX that this
means that we can represent the relationship between the restoring force and
the displacement as
Fsp  ky
(8.3)
If the motion of the spring had been horizontal, moving in the x-direction, we
would write instead
Fsp  kx
(8.4)
The proportionality constant k, which is the slope of the force-versusdisplacement graph, is called the spring constant. The units of the spring
constant are N/m.
The force does not depend on the spring’s length L, but on
the displacement x or y of the end of the spring. b
NOTE c
The spring constant k is a property that characterizes a spring, just as mass m
characterizes a particle. If k is large, it takes a large pull to cause a significant
stretch, and we call the spring a “stiff ” spring. If k is small, the spring can be
stretched with very little force, and we call it a “soft” spring. Every spring has
its own, unique value of k that remains constant for that spring. The spring
constant for the spring in Figure 8.16 can be determined from the slope of the
straight line to be k  35 N/m
Just as we used massless strings, we will adopt the
idealization of a massless spring. While not a perfect description, it is
a good approximation if the mass attached to a spring is much larger
than the mass of the spring itself. b
NOTE c
Hooke’s Law
Experiments show that Equation 10.25 is valid whether the spring is stretched
or compressed. The only difference is the direction of the force—pulling in
one case, pushing in the other. But because the spring force really is a vector,
we do need to write Equation 10.25 in a form that gives the correct sign to the
vector component of Fsp 
Figure 10.17 shows a spring along a generic s-axis. The equilibrium position
of the end of the spring is denoted se  This is the position, or coordinate, of
the free end of the spring, not the spring’s equilibrium length L 0 
FIGURE 10.17 The direction of Fsp is always opposite the displacement
r
s
When the spring is stretched, the displacement from equilibrium s  s  se
 
is positive while Fsp s , the s-component of the restoring force pointing to the
left, is negative. If the spring is compressed, the displacement from
equilibrium s is negative while the s-component of Fsp , which now points
to the right, is positive. No matter which way the end of the spring is
displaced from equilibrium, the sign of the force component Fsp s is always
 
opposite to the sign of the displacement s We can write this
mathematically as
F   ks
sp s
Hooke's law 
(10.26)
where s  s  se is the displacement of the end of the spring from
equilibrium. The minus sign is the mathematical indication of a restoring
force.
Equation 10.26 for the restoring force of a spring is called Hooke’s law. This
“law” was first suggested by Robert Hooke, a contemporary (and sometimes
bitter rival) of Newton. Hooke’s law is not a true “law of nature,” in the sense
that Newton’s laws are, but is actually just a model of a restoring force. It
works extremely well for some springs, as in Figure 10.16, but less well for
others. Hooke’s law will fail for any spring if it is compressed or stretched
too far.
Figure 8.17 The direction of Fsp is always opposite from the displacement ∆s..
Figure 8.17
Example 8.7 Pull until it slips
Figure 8.18a shows a spring attached to a 2.0 kg block. The other end of the
spring is pulled by a motorized toy train that moves forward at 5.0 cm/s. The
spring constant is 50 N/m and the coefficient of static friction between the
block and the surface is 0.60. The spring is at its equilibrium length at t  0 s
when the train starts to move. When does the block slip?
Figure 8.18 A toy train stretches the spring until the block slips.
Prepare Model the block as a particle and the spring as an ideal spring
obeying Hooke’s law. Figure 8.18b is a free-body diagram for the block.
Solve Recall that the tension in a massless string pulls equally at both ends of
the string. The same is true for the spring force: It pulls (or pushes) equally at
both ends. Imagine holding a rubber band with your left hand and stretching it
with your right hand. Your left hand feels the pulling force, even though it was
the right end of the rubber band that moved.
This is the key to solving the problem. As the right end of the spring moves,
stretching the spring, the spring pulls backward on the train and forward on
the block with equal strength. As the spring stretches, the static friction force
on the block increases in magnitude to keep the block at rest. The block is in
static equilibrium, so
   f   F
 Fnet x  Fsp
x
s x
sp
 fs  0
where Fsp is the magnitude of the spring force. The magnitude is Fsp  kx,
where x  vx t is the distance the train has moved. Thus
fs  Fsp  kx
The block slips when the static friction force reaches its maximum value
fs max  s n  s mg This occurs when the train has moved

2
fs max s mg 060 20 kg  98 m/s
x 


k
k
50 N/m

 0235 m  235 cm
The time at which the block slips is
t
x 235 cm

 47 s
v 50 cm/s
Assess
Elasticity: The Springiness of Matter
Imagine taking a very close up view of a thin metal rod, as shown in Figure
8.19. In Chapter 4 we introduced a simple model of matter. We picture a
solid material as being made up of particles—atoms and molecules—held
together by bonds that we can model as little springs. In fact, until the very
last chapters of this book, this model will serve us quite well.
Figure 8.19 An elastic solid can be modeled as particle-like atoms held together by
spring-like molecular bonds.
If you pull on the end of the rod, as shown in Figure 8.20a, you will slightly
stretch the spring-like bonds between the particles that make it up, and so the
rod itself will stretch. The stretched bonds pull back on your hands with a
restoring force that causes the rod to return to its original length when
released. In this sense, the entire rod acts as a (very stiff) spring: we say that
the rod is elastic.
If you instead bend the rod gently, as shown in Figure 8.20b, the top surface
of the rod will need to stretch a bit, and the lower edge compress. This is
possible because the molecular springs stretch on the top and compress on the
bottom. Because each spring has a restoring force that is trying to return the
spring to its equilibrium length, the entire rod will pull back on your hand
with a restoring force; if the rod is released it will spring back to its original
shape. Again, this
Figure 8.20 When an object is stretched or bent, atomic springs stretch and
compress.
All elastic systems follow the same general rule: if you compress, bend, twist,
or otherwise deform them, they develop a restoring force that acts to restore
them to their original shape. Let’s look at one quick example, and then we
will do a more thorough investigation in the next section of the chapter.
An Elastic System: The Diving Board
Figure 9.21 When a person stands on the end of a diving board, the board deflects
downward. The amount of the deflection is proportional to the weight of the person.
Suppose you are standing on the end of a diving board: the board will be bent
downward under your weight. In the previous section, we saw data for the
stretch of a spring as a function of the weight hung from it. You could take
similar data for a diving board, measuring the deflection of the board as a
function of the force applied to the end.
FIGURE 9.4.5: Force is applied to a diving board at the end. The deflection is
measured from the equilibrium position, as with a spring.
If you take real data for a real diving board, you would get a result like the
following:
Figure 9.22 Restoring force vs. displacement for a diving board.
As with a spring, the restoring force is linearly proportional to the
displacement. So the diving board is elastic: a deformation produces a
restoring force that has the same form as the restoring force of a spring:
F = kboard d
Many other systems exhibit similar rules, as we will see.
When a diver lands on the end of the board, it deflects. This produces an
upward force, but it also stores elastic potential energy in the deflection of the
board. A portion of this elastic potential energy will be returned to the diver
as the board rebounds.
9.5 Stress and Strain
The restoring force in springs has the form that it does because it is due to the
“springiness” of bonds between atoms. Since all matter is made of atoms, it
is no surprise to find that all matter has a certain amount of elasticity to it.
We think of materials such as rubber as being elastic, but concrete, steel,
copper—and bone—all are materials that have some elasticity. A bone can
bend a certain amount, as can a steel girder.
Let’s look at the restoring force that occurs when you stretch a metal rod as
an example.
Tensile Stress and Young’s Modulus
Suppose you clamp one end of a solid rod while using a strong machine to
pull on the other with force F. Figure 9.5.1 shows the experimental
arrangement.
Figure 9.23 Stretching a solid rod. In fact, machines that do just this are used to
measure the elastic properties of building materials—and biological materials as well!
When the force is applied, the solid rod will stretch. If the rod is made of
rubber, it will stretch a good deal; if it is made of steel, it will stretch just a
little. But the form of the stretch is the same, because it is due to the
stretching of spring-like molecular bonds. If we make a graph of the force
needed to produce a certain extension, we get a graph like this:
Figure 9.24 Force vs. extension for a solid rod. The shape of the graph and the slope
can vary significantly from material to material, but for most materials there is an
elastic region, an elastic limit and a breaking point more or less as shown.
Figure 9.5.2 shows graphically the amount of force needed to stretch the rod
by the amount ∆L. This graph contains several regions of interest. First is the
elastic region, ending at the elastic limit. As long as is less than the elastic
limit, the rod will return to its initial length L when the force is removed: it
works like a spring. A stretch beyond the elastic limit will permanently
deform the object; it will not return to its initial length when the force is
removed. And, not surprisingly, there comes a point when the rod breaks.
For most materials, the graph begins with a linear region, which is where we
will focus our attention. If is within the linear region, the force needed to
stretch the rod is:
F = k∆L
(9.5.1)
where k is the slope of the graph. Equation 9.5.1 has exactly the same form as
the equation for the stretch of a spring that we saw previously, as we would
expect.
The difficulty with Equation 9.5.1 is that the proportionality constant k
depends both on the composition of the rod—whether it is, say, plastic or
aluminum—and on the rod’s length and cross-sectional area. It would be
better to put the equation in a different form so that we can see whether the
stiffness of the rod results from its geometry (a thick steel rod will be stiffer
than a thin steel rod, of course) or the material (a steel rod will be stiffer than
a rubber rod of the same dimensions). Experiments show that the following
equation is a pretty good approximation for most materials in their elastic
region:
F
L
Y
A
L
(9.5.2)
The constant Y is called Young’s modulus. Young’s modulus is a constant

that is characteristic of a material:
all steel rods of any length or area will
have the same Young’s modulus. The area of the rod and the length of the
rod now appear as separate variables.
If we combine Equations 9.5.1 and 9.5.2, we can come up with an expression
for the k of the rod:
k
YA
L
(9.5.3)
When a force is applied to the rod, it will stretch. And the k value for the rod
will depend on the geometry (the length and the area) and the material (the
Young’s modulus) according to the above equation.

Stop to Think 9.6.1 A 10 kilogram mass is hung from a cable of length 1
meter. This makes the cable stretch by 2 millimeters. Suppose a 10 kg mass
is hung from a similar cable of length 2 meters; how much does the cable
stretch?
a) 1 mm
b) 2 mm
c) 4 mm
Equation 9.5.2 is a bit different than other equations you have seen. The
force F is not alone, but shows up in the fraction F/A. This quantity is called
tensile stress. Notice that it will have the same units as pressure.
The change in length ∆L shows up in the ratio ∆L/L. This quantity is the
fractional increase in the length, and is called strain. Note that strain is
dimensionless. In practical calculations, the numerical values of strain are
generally very small because solids cannot generally be stretched very much
before reaching the breaking point.
With these definitions, Equation 9.5.2 can be written
stress = Y x strain
(9.5.4)
Because strain is dimensionless, Young’s modulus Y has the same dimensions
as stress: N/m2, or Pa. Table 9.5.1 gives values of average or typical values of
Young’s modulus for several common materials. Large values of Y
characterize materials that are stiff and rigid. “Softer” materials, at least
relatively speaking, have smaller values of Y. You can see that steel has a
larger Young’s modulus than aluminum: steel is in some sense stronger than
aluminum. Biological materials are an interesting case. Bones in your body
are made of two different kinds of bony material: cortical (or compact) bone
(which is dense and rigid) and cancellous (or spongy) bone (which is porous
and more flexible.) Most bones of your body are a mix of the two different
kinds of bony material:
Figure 9.25 A cross section of a long bone. The exterior is cortical bone; the interior
is cancellous bone.
The two kinds of bone in the body, cortical and cancellous, have very
different values of Young’s modulus.
TABLE 9.5.1: Young’s Modulus for Common Materials
Material
Young’s Modulus (1010 N/m2)
Aluminum
7
Bone (cancellous)
.02 – 0.3
Bone (cortical)
1.6
Cast Iron
20
Concrete
3
Copper
11
Glass
7
Plastic (polystyrene)
0.3
Steel
20
Wood (Douglas Fir)
1
The values of Young’s modulus in the table are large; it takes a large stress to
produce even a small strain. Let’s look at a practical example.
EXAMPLE 8.8 Stretching a wire
In the Physics Department of Colorado State University, a Foucault
pendulum was recently installed. (A Foucault pendulum is one that is
designed to illustrate the rotation of the earth. As time goes on, the plane of
swing of the pendulum changes from our point of view. This change is due
to the rotation of the earth: it’s not the pendulum that is changing orientation,
it is the earth!) This pendulum has a long cable, so that the pendulum can
have a long swing. A heavy steel ball is hung at the end of the cable. One of
the key factors to consider in the design was how much the cable would
stretch when the ball was hung from the cable. A pointer on the bottom of
the ball was designed to come within a few centimeters of the floor, so even a
small stretch could be a problem.
The cable has a diameter of 2.5 mm (giving it a cross-sectional area of
approximately 5 x 10-6 m2) and is made of steel, giving it a Young’s modulus
of 20 x 1010 N/m2. This is a stout cable, but the ball that is hung from it has a
mass of 122 kg, giving it a weight of approximately 1200 N.
When the ball was hung from the cable, how much did the cable
stretch?
PREPARE The ball is in static equilibrium, so the force on it due to the cable
is equal to its weight. This means that an equal force will be on the cable; it
is this force that will make the cable stretch.
SOLVE To compute how much the cable will stretch when the ball is hung
from it, we use Equation 9.5.2:
F
L
Y
A
L
Rearranging terms, we get:

L 
LF
AY
All the quantities on the right are known, so we can compute:
L =


(6.0 m)(1200 N)
= 0.0072 m = 7.2 mm
(5.0 x 10 -6 m2 )(20 x 10 10 N / m2 )
ASSESS
This stretch was quite noticeable when the ball was attached
to the cable, but not enough to cause a problem. When the ball’s weight was
released, it briefly bounced up and down on the cable due to the elastic
stretch of the cable.
Other Stress-Strain Relationships
Young’s modulus is one example of an elastic modulus: a measure of how an
object responds to an applied stress. As we saw, Young’s modulus is the
appropriate measure of the response of a solid to a force along its length, such
as when you stretch a wire. There are other kinds of stresses that can be
applied which produce other kinds of deformations. Table 9.5.2 shows some
of these.
TABLE 9.5.2 Different types of elastic modulus
Stretch along one axis:
Elastic
modulus:
Formula:
Young’s
modulus (Y)
Compress on all sides:
change volume:
{insert diagram: volume
strain, with ∆V noted}
Apply force parallel to side:
change shape:

Elastic
modulus:
Bulk
modulus (B)
Elastic
modulus:

Shear
modulus (S)
F
L
Y
A
L
Formula:
F
V
 B
A
V
Formula:
F
x
S
A
h

In each case, the basic form of the equation is exactly the same: a stress (the
force that produces the change) is proportional to a strain (a relative
deformation: change in length divided by length, change in volume divided
by volume):
stress = Y x strain
(9.5.4)
This similarity among the different relationships is because the variations are
due to the same root cause. In each case, the stress is defined as a force
divided by an area, and so will have units of N/m2. And in each case the
strain is dimensionless; it is defined as a length divided by a length or a
volume divided by a volume.
Usually, the elastic modulus that is most important is Young’s modulus.
When we speak of the elastic modulus we are generally speaking of Young’s
modulus.
9.6 Applications to Bodies and Buildings
In the previous section, we looked at the restoring forces in elastic materials
when they are deformed. In practical situations, the deformations that occur
are a bit more complicated. We are going to consider the case of the bending
of a beam. A beam is a structural member that has one long dimension.
Some examples are a diving board, a floor joist under the floor of your
house—and the long bones in your legs.
Bending Beams
In section 9.5, we looked at the example of a diving board: applying a force
to the end bent it downward. This is a case of deforming an elastic system,
but there isn’t a simple stretch as there is in a stretched spring or a stretched
rod: there is a change in shape.
This change in shape is what produces the restoring force. Suppose we have
a beam that is fixed at one end and free to move at the other; this particular
kind of beam is known as a cantilever.
Figure 9.26 A beam fixed at one end is known as a cantilever. A bending force
applied to the free end will cause a deflection. This deflection is a result of a change
in shape of the beam.
Let’s look in detail at the change in shape of the beam in Figure 9.6.1. In
Figure 9.6.2, you can see that the top of the beam stretches: it becomes
longer. The bottom of the beam compresses; it becomes shorter.
Figure 9.27 Look at a small piece of the beam in detail, we see that the bending of the
beam stretches the length of the top of the beam and compresses the length of the
bottom of the beam.
Bone: A Compound Material
When you bend a beam, as above, the top stretches and the bottom compresses. Your
bones, to be strong, must resist both kinds of deformations. But most pure materials
are good at resisting one or the other, but not both. Concrete is very good at resisting
compression, but it can be cracked if it is stretched. Plastics tend to be very resistant
to stretching, but they are easily compressed. To make a strong beam, it is often a
good idea to use a mix of two materials. Bone is a mix of calcium (a mineral, which
is good at resisting compression but poor at resisting stretching) and collagen (a
protein, a long-chain molecule that is very resistant to being stretched but easily
compressed.) The mix of the calcium and the collagen makes for quite a strong
material. Osteoporosis, a disease in which calcium in the bones is reduced, changes
the nature of the bone and can precipitate certain types of fractures, as the loss of
calcium makes the bones much weaker against compressive forces.
Since the deformation of a beam is more complicated than a simple stretch,
the equation relating the deformation and the applied force now takes on a bit
more character. Suppose our beam has length l, thickness t and width w:
The equation relating force and deflection is:
Y  w  t 3 
F  
d
3
 4  l 
(9.6.1)
The force is still proportional to the deflection, as for a spring. The quantity
in brackets plays the role that the spring constant k plays in Hooke’s Law: a
higher value means a stiffer beam.

The form of this equation has certain consequences:
The stiffness of a beam is inversely proportional to the cube of the length. If
you double the length of a beam, you will increase the deflection for a given
force by a factor of 8! If you have ever tried to carry or maneuver a long
board, you have noticed this.
The stiffness of a beam is proportional to the cube of the thickness. If you are
supporting the floor in a house, an 8-inch joist is 8 times as stiff as a 4-inch
joist.
Building codes for houses and other dwellings specify the necessary size of
beams to span different distances—based on this sort of equation. In a house,
there are certain expected loads on the floors: people, furniture, and so on.
You don’t want the floor to deflect too much when people walk on it, or when
furniture is placed on it. To keep this from happening, building codes specify
floor joists (beams that support the floor) of a minimum width and thickness
for a given span. Wider rooms with longer joists require thicker joists.
TABLE 9.6.1: Required joists for various spans of floors, from Uniform Building Code
Size of Joist
Maximum Span (if joists are
spaced 16” apart)
2” x 4”
4’11”
2” x 6”
7’9”
2” x 8”
10’3”
2” x 10”
13’1”
2” x 12”
15’11”
Stop to Think 9.6.1 2” x 6” joists span a 12’ living room area in an older home, but
the floor that they support is too “soft”: walking on the floor produces too much
deflection. Which of the following changes would produce the most stiffening of the
floor:
a) Replacing the joists with 2” x 8” joists
b) Doubling up the joists: putting another 2 x 6” joist right next to each existing joist
c) Adding a beam in the middle of the room to support the middle of each joist,
making the effective length of the joists 5’ instead of 10’
The long bones in your body work as beams as well. But, as you may know,
they aren’t uniform. Your bones are much less dense in the center. In other
animals, where weight is more of a factor, this effect is much more
pronounced. You may have heard it said that birds have “hollow bones.”
This isn’t quite true, but the cross-section of their bones shows much less
dense material in the center.
Figure 9.28 The long bones of birds are very dense on the outside, and much less
dense in the center. This results in great strength at reasonable weight.
You may have noticed that the cross-section of beams in bridges and other
such structures is not uniform either. Beams on bridges are often “I beams”
which have a cross section as in Figure 9.6.5.
Figure 9.29 A beam with rectangular cross section, and an I beam with the same
width and thickness.
When a beam bends, the restoring force comes largely from the top and the
bottom of the beam; as we have seen, there is very little stretching or
compression in the center of the beam. It makes sense to do away with the
parts of the beam that are contributing relatively little to the structural
strength—most of the mass in the middle—while keeping the top and bottom.
A similar reduction in mass while keeping most of the structural strength can
be obtained in a beam of circular cross section by eliminating the mass in the
center: turning a rod into a pipe—or making bones “hollow.”
Beyond the Elastic Limit
In section 9.5, we considered the elastic modulus of different materials. We
said that materials with a higher elastic modulus are stiffer. You might be
tempted to think of them as stronger, and in some sense this is true. But there
is more to the story. You can see in Table 9.5.1 that glass has the same
elastic modulus as aluminum; glass is actually pretty rigid. But glass will fail
at very small strains: it won’t bend very far before breaking, and so will not
support very much stress. In addition to the elastic modulus of building
materials, it is important to consider the ultimate stress of materials: how
much stress can they take before failing. The table below lists some ultimate
stresses for common materials; these numbers are averages and
approximations, as defects and other minor variations can greatly effect
ultimate stresses:
TABLE 9.2: Ultimate Stresses of Common Materials
Material
Ultimate Stress Possible Before
Failure
Cancellous Bone
5 x 106 N/m2
Oak
20 x 106 N/m2
Glass
60 x 106 N/m2
Cortical Bone
100 x 106 N/m2
Cast Iron
150 x 106 N/m2
Steel
1000 x 106 N/m2
Cast iron—which has about the same elastic modulus as steel—fails at a
much lower stress than steel. It is more brittle. The same is true of glass; it
will fail at a reasonably low ultimate stress. But note that it has an ultimate
stress that is higher than that of oak! In fact, if you made a glass floor that
was as thick as a typical oak floor, it would be just as sturdy.
Bone is an interesting case. The values listed in the above table are for static
forces. But bone can withstand significantly higher forces than the values
listed in the table if the forces are applied for a very short period of time. A
simple example: if you break a block with a blow from your hand, as we saw
in the energy chapter, the blow is very brief. Your hand can withstand a force
of more than 25,000 N during the blow!