Download Ch 13 Equilibrium

Chapter 13
Equilibrium
CHAPTER 13
EQUILIBRIUM
When does a structure fall over, when does a bridge
collapse, how do you lift a weight in a way that
prevents serious injury to your back? We begin to
answer such questions by applying Newton’s laws to
an object that has neither linear nor angular acceleration. The most interesting special case is when an
object is at rest and will stay that way, when it is not
about to tip over or collapse.
13-2
Equilibrium
EQUATIONS FOR EQUILIBRIUM
If the center of mass of an object is not accelerating,
then we know that the vector sum of the external forces
acting on it is zero. If the object has no angular
acceleration, then the sum of the torques about any axis
must be zero. These two conditions
sum of
external
forces
Σ Fi external = 0
(1)
sum of
torques
about
any axis
Σ τ i external = 0
(2)
i
i
are what we will consider to be the required conditions
for an object to be in equilibrium.
Example 1 Balancing Weights
As our first example, suppose we have a massless rod
of length L and suspend two masses m1and m2from the
ends of the rod as shown in Figure (1). The rod is then
suspended from a string located a distance x from he
left end of the rod. What is the distance x and how
strong a force F must be exerted by the string?
Solution: The first step is to sketch the situation and
draw the forces involved, as we did in Figure (1). Our
system will be the rod and the two masses. The external
forces acting on this system are the two gravitational
forces m 1g and m 2g , and the force of the string F .
Since all the forces are y directed, when we set the
vector sum of these external forces to zero we have
ΣFy = F – m1 g – m2 g = 0
(3)
Thus we get for F
Equations 1 and 2 are a complete statement of the basic
physics to be discussed in this chapter. Everything else
will be examples to show you how to effectively apply
these equations in order to understand and predict when
an object will be in equilibrium. In particular we wish
to show you some techniques that make it quite easy to
apply these equations.
F
x
L–x
F = m1 + m2 g
(4)
and we see that F must support the weight of the two
masses.
To figure out where to suspend the rod, we use the
condition that the net torque produced by the three
external forces must be zero. Since a torque is a force
times a lever arm about some axis, you have to choose
an axis before you can calculate any torques. The
important point in equilibrium problems is that you can
choose the axis you want. We will see that by
intelligently selecting an axis, we can simplify the
problem to a great extent.
Our definition of a torque τ caused by a force F is
O
m2
m1
m2 g
m1 g
Figure 1
Masses m1 and m2 suspended from a massless
rod. At what position x do we suspend the rod
in order for the rod to balance?
τ = r ×F
(5)
where r is a vector from the axis O to the point of
application of the force F as shown in Figure (2).
In this chapter we do not need the full vector formalism
for torque that we used in the discussion of the gyroscope. Here we will use the simpler picture that the
magnitude of a torque caused by a force F is equal to
the magnitude of F times the lever arm r⊥ , which is the
distance of closest approach between the axis and the
line of action of the force F as shown in Figure (2). If
the torque tends to cause a counter clockwise rotation,
as it is in Figure (2), we will call this a positive torque.
If it tends to cause a clockwise rotation, we will call that
a negative torque.
13-3
(In Figure (2), the vector r × F points up out of the
page. If the vector F were directed to cause a clockwise
rotation, then r × F would point down into the paper.
[It is good practice to check this for yourself.] Thus we
are using the convention that torques pointing up are
positive, and those pointing down are negative.)
Returning to our problem of the rod and weights shown
in Figure (1), let us take as our axis for calculating
torques, the point of suspension of the rod, labeled
point O in Figure (1). With this choice, the force F
which passes though the point of suspension, has no
lever arm about point O and therefore produces no
torque about that point.
The gravitational force m 1g has a lever arm x about
point O and is tending to rotate the rod counter clockwise. Thus m 1g produces a positive torque of magnitude m1 gx. The other gravitational force m 2g has a
lever arm (L – x) and is tending to rotate the rod
clockwise. Thus m 2g is producing a negative torque
magnitude – m2 g L – x . Setting the sum of the
torques about point O equal to zero gives
m1 gx – m2 g L – x = 0
x =
(6)
m2
L
m1 + m2
(7)
We obtained Equation 7 by setting the torques about
the balance point equal to zero. This choice had the
advantage that the suspending force F had no lever arm
and therefore did not appear in our equations. We
mentioned earlier that the condition for equilibrium
was that the sum of the torques be zero about any axis.
In Exercises 1 and 2 we have you select different axes
about which to set the torques equal to zero. With these
other choices, you will still get the same answer for x,
namely Equation 7, but you will have two unknowns,
F and x, and have to solve two simultaneous equations.
You will see that we simplified the work by choosing
the suspension point as the axis and thereby eliminating
F from our equation.
Exercise 1
If we choose the left end of the rod as our axis, as shown
in Figure (3), then only the forces F and m2g produce
a torque
(a) Is the torque produced by F positive or negative?
(b) Is the torque produced by m2g positive or negative?
(c) Write the equation setting the sum of the torques
about the left end equal to zero. Then combine that
equation with Equation 4 for F and solve for x. You
should get Equation 7 as a result.
Exercise 2
Let us check to see that Equation 7 is a reasonable
result. If m1 = m2 , then we get x = L/2 which says
that with equal weights, the rod balances in the center.
If, in the extreme, m1 = 0 , then we get x = L, which
tells us that we must suspend the rod directly over m 2 ,
also a reasonable result. And if m2 = 0 we get x = 0 as
expected.
Obtain two equations for x and F in Figure (1) by first
setting the torques about the left end to zero, then by
setting the torques about the right end equal to zero.
Then solve these two equations for x and see that you
get the same result as Equation 7
F = m1 g + m2 g
r
o
axis
τo = r x F
r
F
Figure 2
The torque τ = r × F has a magnitude τ = r⊥ × F .
x
axis
L
m2
Figure 3
Torques about the left end of the rod.
m2 g
13-4
Equilibrium
But the sum Σ m ix i is by definition equal to M times
the x coordinate of the center of mass of the object
GRAVITATIONAL FORCE ACTING
AT THE CENTER OF MASS
When we are analyzing the torques acting on an
extended object, we can picture the gravitational force
on the whole object as acting on the center of mass
point. To prove this very convenient result, let us
conceptually break up a large object of mass M into
many small masses m i as shown in Figure (4), and
calculate the total gravitational torque about some
arbitrary axis O. An individual particle m i located a
distance xi down the x axis from our origin O produces
a gravitational torque τi given by
τi = m i g x i
(8)
where mig is the gravitational force and xi the lever
arm. Adding up the individual torques τi to obtain the
total gravitational torque τO gives
τO =
Σ τi = Σ migxi
i
i
(9)
= gΣ mixi
i
MXcom ≡
Σi mixi
(10)
Using Equation 10 in 9 gives
τO = MgXcom
(11)
Equation 11 says that the gravitational torque about any
axis O is equal to the total gravitational force Mg times
the horizontal coordinate of the center of mass of the
object. Thus the gravitational torque is just the same as
if all of the mass of the object were concentrated at the
center of mass point.
Exercise 3
A wheel and a plank each have a mass M. The center
of the wheel is attached to one end of a uniform beam
of length L. A nail is driven through the center of mass
of the plank and nailed into the other end of the beam as
shown in Figure (5). Where do you attach a rope around
the beam so that the beam will balance? Explain how
you got your answer.
rope
mi
O
axis
plank mass m
xi
mi g
beam
length L
Figure 4
Conceptually break the large object of mass M into
many small pieces of mass m i , located a distance x i
down the x axis from our arbitrary origin O.
wheel
mass m
Figure 5
A wheel and a plank are attached
to the ends of a uniform beam.
13-5
TECHNIQUE OF SOLVING
EQUILIBRIUM PROBLEMS
In our discussion of the balance problem shown in
Figure (1), we saw that there were several ways to solve
the problem. We always have the condition that for
equilibrium the vector sum of the forces is zero
ΣiFi = 0 and sum of the torques τ i about any axis is
zero Σi τ i = 0 . By choosing various axes we can
easily get enough, or more than enough equations to
solve the problem. If we are not careful about the way
we do this however, we can end up with a lot of
simultaneous equations that are messy to solve.
Our first solution of the equilibrium condition for
Figure (1) suggests a technique for simplifying the
solution of equilibrium problems. In Equation 6 we set
to zero the sum of the torques about the balance point
O shown in Figure (1) reproduced here. We wanted to
calculate the position x of the balance point, and were
not particularly interested in the magnitude of the force
F . By taking the torques about the point O where F has
no lever arm, F does not appear in our equation. As a
result the only variable in Equation 6 is x, which can be
immediately solved to give the result in Equation 7. As
we saw in Exercises 1 and 2, if we chose the torques
about any other point, both variables x and F appear in
our equations, and we have to solve two simultaneous
equations.
We will now consider some examples and exercises
that look hard to solve, but turn out to be easy if you take
the torques about the correct point. The trick is to find
a point that eliminates the unknown forces you do not
want to know about.
Example 3 Wheel and Curb
A boy is trying to push a wheel up over a curb by
applying a horizontal force Fboy as shown in Figure
(6a). The wheel has a mass m, radius r, and the curb a
height h as shown. How strong a force does the boy
have to apply?
Solution: We will consider the wheel to be the object
in equilibrium, and as a first step sketch all the forces
acting on the wheel as shown in Figure (6b). We can
treat this as an equilibrium problem by noting that as the
wheel is just about to go up over the curb, there is no
force between the bottom of the wheel and the road.
There is, however, the force of the curb on the wheel,
labeled Fcurb in Figure (6b). We know the point at
which Fcurb acts but we do not know off hand either the
magnitude or direction of Fcurb, nor are we asked to find
Fcurb.
Fboy
h
mg
Figure 6a
A boy, exerting a horizontal force on the axle of a
wheel, is trying to push the wheel up over a curb.
How strong a force must the boy exert?
Fboy
F
x
r
Fcurb
O
L–x
O
mg
no force
m2
m1
m2 g
m1 g
Figure 1 (Repeated)
We can eliminate any force by a proper choice of axis.
Figure 6b
Forces on the wheel as the wheel is
just about to go up over the curb.
13-6
Equilibrium
We can eliminate the unknown force Fcurb by setting to
zero the torques about the point O where the curb
touches the wheel. Since Fcurb has no lever arm about
this point, it will not appear in the resulting equation.
In Figure (6c) we have sketched the geometry of the
problem. About the point O the force Fboy has a lever
arm (r – h) and is tending to cause a clockwise rotation
about point O. Thus Fboy is producing (by our convention) a negative torque, of magnitude Fboy (r – h). The
gravitational force mg has a lever arm shown in
Figure (6c), and is tending to produce a counter clockwise rotation. Thus it is producing a positive torque of
magnitude + mg about point O. Since there are no
other torques about point O, setting the sum of the
torques equal to zero gives
– Fboy(r – h) + mg = 0
Fboy = mg
The final slip in solving this problem is to relate the
distance to the wheel radius r and curb height h. As
shown in Figure (6d), the right triangle from the axle to
the curb has sides , (r – h) and hypotenuse r. By the
Pythagorean theorem we get
r2 =
2
+ r– h
2
or
= 2 rh– h2
which finishes the problem.
(13)
Exercise 4
The direction of Fcurb is slightly off in Figure (6b). Explain
what would happen to the wheel if Fcurb pointed as
shown.
(12)
r–h
r
We immediately see that if the curb is as high as the
axle, if r = h, there is no finite force that will get the
wheel over the curb.
2r
Figure 7a
A frictionless rod is placed in a
hemispherical frictionless bowl. What is
the equilibrium position of the rod?
Fboy
r
(r – h)
O
mg
Figure 6c
Geometry of the problem.
axel
r
(r – h)
curb
Figure 6d
Figure 7a
We simulated a frictionless rod in a hemispherical
frictionless bowl by placing ball bearing rollers at
one end of the rod and the edge of the “bowl”. The
rod always comes to rest at this angle.
13-7
Example 4 Rod in a Frictionless Bowl
We include the problem here, first because it gives
some practice with what we mean by a frictionless
surface, but more importantly it is an example where
we can gain considerable insight without solving any
equations.
You place a frictionless rod of length 2r in a frictionless
hemispherical bowl of radius r. Where does the rod
come to rest? (Put in just enough friction to have it
come to rest.) The situation is diagramed in Figure (7a).
In Figure (7b), we have made a reasonably accurate
simulation of the problem by using a semi circular
piece of plastic for the bowl and placing small rollers on
one end of the rod and one rim of the bowl to mimic the
frictionless surfaces. In Figure (7c) we have sketched
the forces acting on the rod. There is the downward
force of gravity mg that acts at the center of mass of the
rod, the force Fb exerted by the bowl on the end of the
rod, and the force Fr exerted by the rim.
The idealization that we have a frictionless surface is
equivalent to the statement that the surface can only
exert normal forces, forces perpendicular to the surface. Thus the force Fb exerted by the frictionless
surface of the bowl is normal to the bowl and points
toward the center of the circle defining the bowl. The
force Fr between the rim of the bowl and the frictionless rod must be perpendicular to the rod as shown.
Off hand we know nothing about the magnitude of the
forces Fb and Fr , only their directions. If we extend the
lines of action of Fb and Fr they will intersect at some
point above the rod as shown in Figure (7c). If we set
to zero the sum of the torques about this intersection
point, where neither Fb or Fr has a lever arm, then
neither Fb or Fr will contribute. The only remaining
torque is that produced by the gravitational force mg .
If the rod is in equilibrium, then the torque produced by
mg about the intersection point must also be zero, with
the result that the line of action of mg must also pass
through the intersection point as shown in Figure (7d).
Thus the rod will come to rest when the center of mass
of the rod lies directly below the intersection point of
Fb and Fr . This result is nicely demonstrated by
comparing the prediction, Figure (7d), with the experiment, Figure (7b).
Fr
Fr
Fb
Fb
mg
mg
Figure 7c
Figure 7d
Forces acting on the rod. Because the bowl is
frictionless, Fb is perpendicular to the surface of
the bowl. Because the rod is frictionless Fr is
perpendicular to the rod.
For equilibrium, the center of mass must lie directly
below the intersection point of Fb and Fr .
13-8
Equilibrium
Exercise 5
A spherical ball of mass m, radius r, is suspended by a
string of length attached to a frictionless wall as shown
in Figure (8).
Exercise 6 Ladder Problem
A ladder is leaning against a frictionless wall at an angle
θ as shown in Figure (9). Assume that the ladder is
massless, and that a person of mass m is on the ladder.
(a) show that the line of action of the tension force T (the
line of the string) passes through the center of the ball.
The force between the ground and the bottom of the
ladder can be decomposed into a normal component
Fn , and a horizontal component Ff that can exist only
if there is friction between the ladder and the ground.
(b) find the tension T.
It is traditional in introductory texts to say that the ladder
will start to slip if the friction force Ff exceeds a value of
µFn where µ is called the “coefficient of static friction”.
This idea is reasonable in that as the normal force Fn
increases, so does the gripping or friction force Ff .
However the coefficient µ depends so much upon the
circumstances of the particular situation, that the theory
is not particularly useful. What, for example, should you
use for the value of µ if the ends of the ladder sink down
into the ground?
T
r
Figure 8
Ball suspended from a frictionless wall.
However for the sake of this problem, assume that the
ladder will just start to slip when Ff = µFn . Assume that
µ has the value µ = 1/ 3 = .557 .
(a) at what angle θ would you place the ladder so that it
will not start to slip until the person climbing it just
reaches the top?
(b) at what angle θ would you place the ladder so that
it will not start to slip until the person has gone half way
up?
frictionless surface
θ
person of
mass m
massless ladder
mg
Fn
Fn
Ff = µF
Figure 9
Ladder leaning against a frictionless wall.
13-9
Example 5 A Bridge Problem
A bridge is constructed from massless rigid beams of length
. The ends of the beams are connected by a single large bolt
that acts more or less like a big hinge. As a result the only
forces you can have in each beam is either tension or
compression (i.e. each beam either pulls or pushes along the
length of the beam.) The idea is to be able to calculate the
tension or compression force in any of the beams when a
load is placed on the bridge.
In this example, we will place a mass m in the center of the
right most span as shown in Figure (10a). To illustrate the
process of calculating tension or compression in the beams,
we will calculate the force in the upper left hand beam. For
now we will assume that the beam is under tension and
exerts a force Td on joint d as shown. If it turns out that the
beam is under compression, then the magnitude of Td will
turn out to be negative. Thus we do not have to know ahead
of time whether the beam is under tension or compression.
Solution: When you have a statics problem involving an
object with a lot of pieces, and you want to calculate the
force in one of the pieces, the first step is to isolate part of the
object and consider that as a separate system with external
forces acting on it. In Figure (10b) we have chosen as our
isolated system the part of the bridge made from the girders
that have been drawn in heavy lines. The external forces
acting on this isolated system are the gravitational force mg
acting on the mass m, the supporting force F2 that holds up
the right end of the bridge, (we will assume that the ends of
the bridge are free to slide back and forth, so that the
supporting forces F1 and F2 point straight up). In addition
the tension (or compression) forces we have labeled Td ,
Tc1 and Tc2 are also acting on our isolated section of the
bridge.
Looking at Figure (10b), it is immediately clear that the
forces we do not want to know anything about are the
tension forces Tc1 and Tc2 . We can eliminate these forces
by setting to zero the torques about the joint labeled c. Using
our convention that counter clockwise torques are positive
and clockwise ones are negative, we get
ΣTabout c = F2 2 – mg
3
2
+ Td
3
2
In Equation 13 we have two unknowns, F2 and Td Thus
we need another equation. If we take the bridge as a whole,
and calculate the torques on about the point a (to eliminate
the force F1 ), we get
ΣT about a = F2 3 – mg
(14)
= 0
Solving Equation 14 for F2 gives
F2 =
5
mg
6
(15)
Using this value in Equation 13 gives
Td =
– mg
3 3
(18)
The minus (–) sign indicates that the beam is under compression.
Exercise 7
Find the tension (or compression in the beam that goes
from point (d) to point (e) in the bridge problem of Figure
(10a).
Td
b
d
f
a
g
m
c
e
mg
Figure 10a
Bridge with a truck on the last span.
Td d
b
f
h=
= 0
Tc2
a
Tc1 c
(13)
where we used the fact that Td's lever arm h is the altitude
of an equilateral triangle.
5
2
g
m
e
2
F1
mg
2
F2
Figure 10b
Finding the tension in the span from b to d.
3
2
13-10
Equilibrium
Exercise 8 Working with Rope
As most sailors know, if you use rope correctly, you can
create very large tension forces without exerting that
strong a force yourself. Suppose, for example, you wish
to make a raft out of two long logs with two short spacer
planks between them as shown in Figure (11a). You
wish to hold the raft together with a rope around the
center as shown.
The first step is to tie, as tightly as you can, the logs
together as shown in the end view of Figure (11b). Then
take another piece of rope and tie it as tightly as you can
as shown in Figure (11c). If you do a reasonably good
job, you can create a large tension in the rope holding
the logs together.
To analyze the problem, let T1 be the tension in the rope
holding the logs together, and T2 the tension in the line
between the ropes as shown in Figure (11d). For this
problem, assume that angle θ in Figure (11d) is 5
degrees, and the tension T2 that you could supply in
winding the line around the ropes was 200 newtons
(enough force to lift a 20 kilogram mass). What is the
tension T1 in the ropes holding the logs together?
Figure 11a
Constructing a raft by tying two logs together, with
wood spacers.
Figure 11b
End view of raft.
Figure 11c
Tightening the rope.
θ
T1
T1
T2
Figure 11d
Tensions in the ropes.
13-11
LIFTING WEIGHTS
AND MUSCLE INJURIES
The previous exercise on tying a raft together illustrates
the fact that with some leverage, you can create large
tensions in a rope. Similar large forces can exist in your
muscles when you lift weights, particularly if you do
not lift the weights properly.
To illustrate the importance of lifting heavy objects
correctly, consider the sketch of Figure (12) showing a
shopper holding a funny looking 10 kg shopping bag
out at arms length. We wish to determine the forces that
must be exerted on the backbone and by the back
muscles in order to support this extra weight.
To analyze the forces, think of the upper body and arm
as essentially a rigid object supported by the backbone
and back muscles as shown. Since we are interested in
the extra forces required to lift the weight, we will
ignore the weight of the upper body itself.
The external forces acting on the upper body are the
upward compressional force Fb acting on the
backbone, and the downward force Fm acting at the
point where the back muscle is attached to the thighbone. (This is in reaction to the upward force exerted
on the thigh bone by the contacting back muscle.)
There is also the weight mg of the shopping bag. We
are letting L be the horizontal distance from the backbone to the shopping bag, and the separation between
the point where the thigh bone pushes up on the
backbone and pulls down on the back muscle.
If we want to solve for the force Fm exerted by the back
muscle we can eliminate Fb from our equation by
setting to zero the sum of the torques about point b, the
point where the thigh bone contacts the backbone. We
get
Σ τ about b = F m – MgL = 0
F m = mg L
(9)
We see that the back muscle has to pull down with a
force that is a factor of L / times greater than the
weight mg of the shopping bag. With = 2 cm , you
see that if you hold the shopping bag out at arms length,
say L = 80 cm, then Fm is L / = 40 times as great as
the weight of the shopping bag. For you to lift a 10 kg
bag at arms length, your back is essentially lifting a 10
kg × 40 = 400 kg mass, which has a weight of almost
half a ton! If instead you pulled your arm in so that L
was only 20 cm, then Fm drops 1/4 of its original value.
Do your back a favor and do not lift heavy objects out
at arms length.
Exercise 9
Write a single equation that allows you to solve for the
compressional force Fb exerted on the backbone, as
shown in Figure (12).
= 2 cm
L = 80 cm
funny
looking
shopping
bag
mg
back
muscle
b
Fb
Fm
Figure 12
Forces on the backbone .
13-12
Equilibrium
calf muscle
Exercise 10
Figure (13) shows the structure of the lower leg and foot.
Assume that this person is raising her heel a bit off the
ground, so that her foot is touching the floor at only one
place, namely the point labeled P in the diagram. Also
assume that the calf muscle is attached to the foot
bones at the point labeled (a), and that the leg bone acts
at the pivot point (b). If her mass is 60 kilograms, what
must be the forces exerted by the calf muscle and the
leg bone? Compare the strength of these forces with her
weight. (This and the next problem adapted from Halliday
& Resnick.)
leg bone
a
Exercise 11
The arm in Figure (14) is holding a 20 kilogram mass.
The arm pivots around the points marked with a small
black circle. What is the compressional force on the
bones in this joint? (Neglect the mass of the arm.)
P
5
cm
Figure 13
Foot
muscle
m
pivot
point
30 cm
3.5 cm
Figure 14
Arm
b
15 cm
13-13
Index
B
Balancing weights, equilibrium 13-2
C
Center of mass
Gravitational force acting on 13-4
E
Equilibrium
Balancing weights 13-2
Chapter on 13-1
Equations for 13-2
Example - bridge problem 13-9
Example - wheel and curb 13-5
How to solve equilibrium problems 13-5
Working with rope 13-10
G
Gravity
Gravitational force acting at center of mass 13-4
L
Lifting weights and muscle injuries 13-11
M
Muscle injuries lifting weights 13-11
R
Rope, working with 13-10
S
String forces
Working with rope 13-10
W
Weight
Lifting 13-11
X
X-Ch13
Exercise 1 13-3
Exercise 2 13-3
Exercise 3 13-4
Exercise 4 13-6
Exercise 5 13-8
Exercise 6 Ladder problem 13-8
Exercise 7 13-9
Exercise 8 Working with rope 13-10
Exercise 9 13-11
Exercise 10 13-12
Exercise 11 13-12