LECTURE 24 Bending Moments (Torques) and ... - McGill Physics

"Give me a lever long enough and a place
to stand and I will move the earth"
Archimedes (circa 340 BC)
Introduction
Previously I introduced the concept of
equilibrium of forces; that for an object to be
stationary any force acting on it must be balanced
by another force that makes the net force zero.
However, it was known from antiquity that
this was not enough to make sure that the object
did not move. For example, two objects sitting
on opposite ends of a balance bar will only result
in the balance bar not moving when the two have
the same weight.
A
TO BALANCE A AND B MUST BE THE SAME WEIGHT
This device was well known since it is the
basis of the balance from which the word
"balance" itself has been derived and which has
been used in trade from the very beginning of
history.
Clearly, in such a device the requirement is
not that the forces balance as in the previous
lecture. The forces are both downward, i.e. in the
same direction. They are actually balanced by the
force on the pivot point, which must support both
weights. For example, if each weight was 250 N
(i.e. about 25 kg but remember my promise not to
use kg as force), then the pivot point would push
up with a force of 500 N.
250 N 250 N
500 N
LECTURE 24
Bending Moments (Torques)
and Forces in the Human Body
B
In what sense then do the 250 N forces
themselves balance?
Another Demonstration
As an example of the importance of this
question, consider another demonstration. I
clamp a board to the end of the table so that it
overhangs by about 80 cm. Then, with the help
of a student who sits on the other end of the table
to keep it steady, I ease myself sideways out
along the board past the edge of the table.
You will clearly see that the board bends more
and more as I move out from the table edge.
When I am about 60 cm from the table edge, the
board actually breaks. That there is a very large
force involved is indicated by the loud crack when
the board broke.
From the point of view of simple equilibrium
of forces, this makes no sense. I did not get any
heavier as I moved out on the board. Therefore
the board was not asked to provide an increased
reaction force as I moved out. It had always to
support only my weight, which it was seen to
easily do when I was close to the edge of the
table. Neither was there a change in the direction of
the forces so as to give some amplification due to
force triangulation, as there was in the case of my
pulling on a cable in the demonstration of Lecture
19. My weight and the force of the board on my
feet were always in opposite direction to each
other.
Where then did the forces that broke the
board come from?

Physics 101A - Physics for the life sciences 2
Turning and Torque
A clue can be found in the bending of the
board as I moved out on it. Clearly, the board
was being turned downward by my weight and it
was this turning which the board had to resist.
But how do you resist a turn? Obviously by
trying to turn the other way.
But what is a turning effort and how is it
balanced?
This can be seen by considering the case of a
balance in which the weights are not equal. To
keep the case simple, have one of the weights
twice that of the other.
d 2d
A B
IF B IS HALF THE WEIGHT OF A THEN IT MUST BE TWICE
AS FAR FROM THE PIVOT POINT FORIT TO BALANCE B
This situation will be balanced when the
lighter weight is exactly twice as far from the
pivot as the heavier weight.
Again this fact was known from antiquity. It
also seems to be fairly obvious since even preschool
children learn it. In fact, it is used as a
Piaget-type test of the intellectual development of
young children.
The same is true for any combination of
weights. Thus the "balancing power" of a weight
is the product of the weight times its distance
from the pivot point. This “balancing power" is
often referred to as the "moment" of the weight
about the pivot, and the distance of a force from a
turning point is referred to as its “moment arm”.
The moment of a force is also often referred
to as a "torque" about the pivot. Both words are
synonymous. (You might wonder at this strange
use of the word moment. It is based on the old
English meaning of "moment" as importance,
rather than as in its other meaning as a short
period of time. Thus the moment of a force can
be thought of as the importance of a force, where
its importance is related to how far it is from a
pivot point. The word torque is easier in that it
derives from the Latin word for twist.)
Thus, for an object such as a balance beam to
be in true equilibrium, the torques (i.e. moments)
on it must balance. This means that a torque
tending to turn the beam clockwise, i.e. th etorque
caused by B in the above diagram, must be
balanced by a torque tending to turn the beam
counterclockwise, i.e. from A in the diagram..
But where were the forces that were creating
the counterclockwise torque that was resisting my
clockwise torque on the board?
The Torque of Couples
A further clue as to where these forces were is
that when the board is bent downward from my
weight the top surface becomes stretched while
the bottom surface becomes compressed. There
must then be a tension in the top part of the board
resisting it from being pulled apart and a
compression in the bottom part resisting it from
being crushed.
TOP BEING STRETCHED
BOTTOM BEING CRUSHED
Now consider the board as being of two parts
each applying these forces to each other.
Consider the free body diagram for the piece I
was standing on. The tension and compression
forces are acting on this piece of board as shown
in the diagram below
Such a set of forces form what is called a
"couple". In such a couple the forces are indeed
equal and opposite but their lines of force are
separated. The forces themselves then actually
balance but the torques do not. A "couple"
therefore provide a pure torque, equal to either
one of the forces (remember they are both of the
same magnitude) multiplied by the distance
between their lines of action.
As an exercise in the concept of a couple,
consider two parallel forces of 200 N in opposite
directions being applied to a bar at points which
are 0.25 m apart.

Lecture 24 - Bending Moments and Forces in the Human Body 3
200 N
0.25 m
200 N
Take the torques about the point of
application of the force on the left. Having no
distance from that point, the left force provides no
torque (i.e. has no moment). The torque then is
all due to the force on the right. It is clockwise
which, as for angles, is considered negative. Its
value is −200N × 0.25 m = −50 N-m.
Now consider the same situation but calculate
the torques about the point of application of the
force on the right. Now the force on the right
provides no torque and all the torque is being
provided by the force on the left. This is also
seen to be clockwise and −50-N-m.
Now consider the torques about the middle of
the bar. Here each force provides a clockwise
torque but each torque is only −200N × 0.125 m,
or −25 N-m. However, again the total torque is
−50 N-m.
Finally consider the torques about a point at x
meters to the right of the right end of the bar.
200 N
0.25 m
200 N
Now the torques are 200N × x m= 200x N-m
for the right force (it is positive because it is now
counterclockwise) and −200N × (x +0.25)m for
the left force. The sum of these is again
−50 N–m.
Thus it does not matter where we take the
point about which to calculate the torque. For a
x
couple it is always the same. Since there is no net
force in a couple, a couple is a pure torque.
The Forces that Broke The Board
What held me up before the board broke was
the moment of the couple created by the tension
in the top of the board and the compression in the
bottom of the board.
Consider now the torque that I applied to the
board just when it broke. I am about 75 kg which
means that my weight is about 750 N. When the
board broke I was about 60 cm (0.6 m) from the
edge of the table. My torque on the board at the
edge of the table was therefore about _−750 N
times 0.6 meters, or about −450 N-m. If you
have had the misfortune of dealing with torque
before this lecture you may well have heard it
expressed in kg-cm, or if your were in the
backward part of the world called the US, in
pound-feet (lb-ft). If you have kg-cm then divide
by 10, roughly, to get N-m. If you have lb-ft then
divide by 7, very roughly.
The torque that must resist this torque, i.e.
must balance it, comes from a couple that has, at
most, a separation which is the thickness of the
board, or only 20 mm or 0.02 m. To provide a
torque of 450 N-m the forces in this couple must
be at least 22500 N each! (If you are still not
used to Newtons, this is just over 2 Tonnes!) It is
clear that such forces could easily break the
board. In fact it is remarkable that the board
could even support me out to the distance that it
did.
Actually, because the forces of tension and
compression are distributed throughout the
thickness of the board the average separation of
the forces is less than 2 cm. 22500 N would be
the force only if all the tension and compression
was restricted to the top and bottom of the board.
The ability of the board to withstand my weight is
therefore even more remarkable.
The Creation of Large Forces in Your Body
The building up of tremendous forces in the
board used in this demonstration might seem of
academic interest, and then only of interest to
physicists. However, it does illustrate an
extremely important phenomenon relating to the
human backbone. If you have ever worked in a
job that required lifting objects you will have
come across signs telling you to lift with your
legs, not you back. The reason for such signs is
that by bending over from your hips with your
back straight, even before you pick up the object,

Physics 101A - Physics for the life sciences 4
you will put about 4000N (400 kg) of
compression on your backbone. The diagram
below should illustrate why this happens.
HIP
0.30 m
400 N
(40 kg)
When you bend over the upper part of your
body applies a torque about your hips that tends
to turn the body downward (clockwise in the
diagram). This torque is about that which would
result from your upper body weight, typically
400N (40 kg), applied at a distance of about
300 mm (0.3 m) from your hips. The resulting
torque about your hips is therefore about
120 N–m.
This torque must be resisted by something, or
your upper body just falls down. That
something is, of course, the pull of your major
back muscle between your hip and your upper
backbone. This force, in turn, puts an equal force
of compression on your backbone. These forces,
the pull of your back muscle against the push of
your backbone, together form a couple of the
torque required to balance the torque from the
weight of your upper body.
The typical separation between your lower
back muscle, which provides the tension, and
your backbone, which provides the compression,
is only about 3cm. The counterclockwise torque
generated by this must balance the clockwise
torque from your upper body weight. It must
therefore be 120 N-m. At a couple separation of
0.03 m, the forces must be 4000N (400 kg). Such
forces are easily capable of either causing severe
strain on the back muscle, or fracture of the
backbone, or both, particularly if you are out of
shape or not used to lifting heavy weights.
SPINE
HIP
LOWER BACK MUSCLE
SPINE
LOWER BACK
MUSCLE
400 N
(40 kg)
3 cm
Another example of torques generating severe
forces within the human body is that of
supporting an object in your hand when you
stretch out your arm. Here there are two major
joints about which torques have to be resisted,
that of the elbow and that of the shoulder.
SHOULDER
ELBOW
Because of the greater distance of the force
from the shoulder, the torque at the shoulder will
be greater than the torque at the elbow. Nature
seems to know this by making the shoulder much
sturdier than the elbow.
As an example, suppose the distance from a
1 kg ball to your elbow joint was 35 cm and the
distance from your elbow joint to your shoulder
joint was another 35 cm. The torque that would
have to be generated by your forearm muscle
would then be about 3.5 N-m but the torque that
would have to be provided by your bicep would
be 7 N-m.
This also shows why the board broke exactly
at the edge of the table. (I knew this was going to
happen so that the piece of the board that broke
off would stay intact under my feet, providing a

Lecture 24 - Bending Moments and Forces in the Human Body 5
landing pad as I approached the floor.) The
torque required to prevent the board from turning
increases for increased distance from my feet.
The torque needed to overcome the torque I apply
to the board is therefore a maximum at the table
edge. Since this torque must be provided by
internal forces within the board, these forces are
also greatest at the table edge, and therefore that is
where it broke.
The Concept of Center of Mass
So far I have treated all masses as points at
specific places relative to a center of rotation.
How valid is this? For example, how valid was it
for me to assume that the mass of my upper body
when I bent over acted as if it was concentrated at
a point 30 cm from my hips?
As a start to this subject consider again the
two unequal masses on a balance. To make the
arithmetic simple, suppose they have masses of
1 kg and 2 kg and are balanced at distances of
1.0 m and 0.5 m respectively.
2 kg
0.5 m 1.0 m
1 kg
Now consider the torque of the two masses if
the “balance” point is under the 2.0 kg mass.
2 kg
0.5 m 1.0 m
1 kg
Now only the 1 kg mass provides any torque,
and that is −1.5 kg-m. Note that this is the same
as the total weight of the two masses together, i.e.
3 kg, at a moment arm of 0.5 m. This moment
arm is simply the distance of the actual balance
point from the new turning point.
Next consider the torque of the two masses if
the “balance” point is under the 1.0 kg mass.
2 kg
0.5 m 1.0 m
1 kg
Now only the 2 kg mass provides any torque,
and that is 3.0 kg-m. Note that this is the same as
the total weight of the two masses together, i.e.
3 kg, at a moment arm of 1.0 m. This moment
arm is again simply the distance of the actual
balance point from the new turning point.
Finally, consider the torque of the two masses
if the “balance” point was moved to an arbitrary
distance x to the left of the actual balance point.
2 kg
x
0.5 m 1.0 m
1 kg
Now both masses provide a negative torque, a
total of −1 kg × (1+x)m −2 kg × (x-0.5)m.
Doing the algebra this becomes simply 3x N-m
Yet again, this is just the total mass times the
distance of the original balance point from the
new turning point, i.e. x.
Thus, for the torques they produce, the two
masses always act as if their total mass was
concentrated at the balance point between them.
This point is therefore called the “center of
mass”.
Sometimes the center of mass is referred to
as the “center of gravity”. But there does not
have to be any gravity for the center of mass to
exist. It comes just from the distribution of the
masses making up a system. (This will be
important in the next lecture which will be about
rotational dynamics.)
How then, for a general case such as the
human body, do we find out where the center of
mass actually is? In principle, if you knew the
mass of all the body parts, and where they were
concentrated, you could calculate it, and in
modern ergonomics these sorts of calculations
are actually carried out, helped a great deal by
modern computers. However, it is usually much
simpler to determine the center of mass by
determining the balance point experimentally. In
the case of my body, this can be determined by
me stretching out on a bar and adjusting my
position on the bar until I am balanced.

Physics 101A - Physics for the life sciences 6
MY CENTER OF MASS MUST HAVE
BEEN ALONG THIS LINE OR I
WOULD HAVE NOT BALANCED
I then know that my "center of mass" is over
the pivot point. If it were not then the force vector
of my weight would not pass through the pivot
point and I would be unbalanced. I will ignore
snide remarks concerning that being my normal
state and just point out that my center of mass
seems to be somewhere around my pelvis.
Of course, it takes the crossing of two lines to
determine the position of any point such as my
center of mass. In the case of me being stretched
out, this second line is easy to determine.
Standing erect requires that my center of mass be
over the point I am standing on. Otherwise I
would topple over. In fact the great ability of
human beings compared to other animals is the
ease with which they can adjust their posture so
that their centers of mass are directly over their
feet. I can then draw a vertical line through my
body and the intersection of this line with the line
corresponding to the height of my pelvis will give
me my center of mass.
MY CENTER OF MASS MUST HAVE
BEEN ALONG THIS LINE OR I WOULD
NOT BE BALANCED ON MY FEET
CENTER OF MASS
The remarkable thing about the center of
mass concept is that it can be used to determine
the torque of an object about any pivot point just
by knowing the objects weight. All that you have
to do is take the total weight of the object and
multiply it by the distance of the line of action of
this weight from the pivot point, a distance which
I have already referred to as the "moment arm" of
the force.
MOMENT ARM
PIVOT POINT
FORCE
This was the basis of the statement earlier that
the torque of my upper body against my hip
when I bend over is equivalent to the weight of
my upper body being applied at about 30 cm
from my hips. This is because the center of mass
of my upper body is at this point, giving a
moment arm of 0.3 m when I bend over.
An intriguing thing about the center of mass
is that it can be outside your body! Consider, for
example, when I bend over as in the forbidden
pick up exercise. You can see from where I have
to keep by feet that my center of mass has now
moved forward from my pelvis. This is because it
must be somewhere along the vertical line
running to the soles of my feet.
MY CENTER OF MASS IS
ALONG THIS LINE
To determine where it actually is along this
line I must balance myself in this same
configuration but about a different point. I can

Lecture 24 - Bending Moments and Forces in the Human Body 7
choose to do so by again balancing myself on the
bar with the bar in my crotch.
MY CENTER OF MASS IS
ALONG THIS LINE
Combing these two diagrams into one gives
the intersection of the two lines and hence the
point of the center of mass.
MY CENTER OF MASS IS
ALONG THIS LINE
MY CENTER OF MASS IS
ALSO ALONG THIS LINE
CENTER OF MASS
My center of mass is obviously outside my
body. A remarkable consequence of this fact is
that by the proper movements a jumper can pass
over a bar while his or her center of mass actually
passes underneath it. This is shown in a
photograph of the so-called "Fosbury flop" in
Hecht.
A final point of importance in understanding
torque concerns the distance of the force from a
pivot point. This is not necessarily the distance
of the point of application of the force from the
pivot point. A familiar example of this should be
the torque you apply to the crankshaft of a
bicycle as the crankshaft turns. Suppose that you
are pedaling up a hill and to get the maximum
torque you stand up on the pedal that is forward.
You have probably experienced the fact that the
effect of your pedaling is much greater in the
middle of the stroke when the crankshaft is
horizontal then at the beginning of the stroke, or
near its end.
Maximum torque
W
W
W
Much reduced torque
The torque of a force applied to a point that is
not on a perpendicular line from the pivot point is
represented in the diagram below (and in figure
8.13 of Hecht).
MOMENT ARM
= r
LINE OF ACTION
OF FORCE
PIVOT POINT
r
θ
FORCE
The torque that this force applies at the pivot
point is not the product of the force and the
distance of its point of application from the pivot.
Rather it is the distance of the line of force from
the pivot point, or r in the diagram (previously
referred to as the "moment arm"). That is
τ = Frsinθ

Physics 101A - Physics for the life sciences 8
Another way to look at the torque of a force
applied to a point that is not on a perpendicular
line from the pivot point is to consider the force
as being resolved into two components, one along
the line to the pivot point and the other
perpendicular to that line.
PIVOT POINT
r
The force component along the line (F )
points directly away from the pivot point. It
therefore provides no torque. The torque is then
all provided by the perpendicular component,
which is Fsinθ . The torque is therefore again
τ = Frsinθ
The torque is therefore the same as that
obtained by taking the full force and multiplying
it by the moment arm.
F
θ
F
F
Summary
In summary, we have a second condition for
equilibrium; that the sum of all the torques on an
object must balance. This is the second
governing principle, along with the equilibrium of
forces, in the very important subject of "Statics"
in Engineering and Physics. It is very important
to realize that it is not just that one or the other of
these principles will apply to specific cases.
Rather, both principles apply all the time and
between them they form the basis of how forces
move objects.
Relevant material in Hecht
Section 8.4 subsection “Rotational
Equilibrium”. Appropriate problems - worked
examples in the text and 69 to 127 (59-97 2 nd
edition) at the end of the chapter.