### Video Transcript

In this video, weβll learn how to
calculate the moment of a couple of two forces and the resultant of two or more
couples. We define a couple as a system of
two forces of equal magnitudes and opposite directions which act in different lines
of action. And a couple produces a rotation
due to the moments of the forces in that couple.

Letβs consider a pair of equal
magnitude forces acting on a rod perpendicularly to the length of that rod. Since the magnitude of these forces
are equal and they act in opposite directions and in different lines of action, this
is a couple. Letβs explore the moment produced
by the couple at the midpoint between these two lines of action, which weβve labeled
as point π΄.

The forces π
one and π
two
produce a clockwise rotation. Itβs typical to take
counterclockwise to be positive, so theyβre producing a rotation in the negative
direction. The moment is the product of the
force and the perpendicular distance from the pivot to the line of action of the
force. The force π
sub one is acting at a
distance π over two units from π΄. It acts perpendicular to the line
that joins point π΄ to the line of action. And so the moment of force π
one
about point π΄ is negative π
one times π over two, where π
one here is simply the
magnitude of the force. We can simplify this a little and
write it as negative π
sub one π over two. And of course itβs negative because
itβs acting in that clockwise direction.

In a similar way, we define the
magnitude of force π
sub two as simply π
sub two. And the moment of this force about
point π΄ is negative π
sub two π over two. Now, remember, we said that the
magnitude of these two forces was equal. So this is equivalent to saying
that the moment of the second force is negative π
sub one π over two about π΄. The sum then of the moments about
π΄ is negative π
sub one π over two minus π
sub two π over two. But since negative π
sub π over
two is equal to negative π
sub one π over two, we can equivalently write this as
negative two π
sub one π over two or negative two π
sub two π over two. We can then simplify by dividing
through by two to give us either negative π
sub one π or negative π
sub two
π.

So what happens if we consider the
moment about point π΅, the endpoint of our rod? Well, if we define the length of
the rod to be π units, the moment about point π΅ by force π
sub one is negative
ππ
sub one. Similarly, the moment of force two
is π minus π times π
sub two. Now, this is positive because this
is trying to rotate the object in a counterclockwise direction. Now, of course, we said that the
magnitude of force π
sub one and π
sub two is equal. So we can write this as π minus π
times π
sub one. Then, the sum of the moments about
π΅ is negative ππ
sub one plus π minus π π
sub two, which weβve just seen can
be equivalently written as π minus π π
sub one.

Distributing our parentheses, and
we see that this simplifies to negative π
sub one π or equivalently negative π
sub two π. And here is where it gets a little
bit clever. We notice that this is the same as
the net moment about π΄. In fact, this will be the same
about any point on the rod as the moment due to a couple is in fact independent of
the point about which the moments due to the couple are taken.

Letβs generalize this. The moment of a couple is the sum
of the moments of the two forces of that couple about a point in space. The magnitude of the moment of a
couple is in fact the product of the magnitude of one of the forces and the distance
between them. We define this as π equals π
times π, where π
is the magnitude of the force and π is called the arm of the
couple.

Now here itβs worth reminding
ourselves what we mean by the arm of the couple. The arm is the length between a
joint axis and the line of force acting on that joint when that force acts
perpendicularly. Finally, the moment of a couple is
independent of the point about which we take the moments of the two forces. So with all this in mind, letβs
look at an example of how to calculate the length of the moment arm.

If the norm of the moment of a
couple is 750 newton meters and the magnitude of one of its two forces is 50
newtons, determine the length of the moment arm.

Itβs worth recalling what we mean
by the norm of the moment. The norm of the moment is simply
the magnitude of that moment. And so weβre given information
about the norm of the moment and the magnitude of one of its two forces. And so we can link these by using
the formula that determines the magnitude of the moment of a couple. Itβs the product of the magnitude
of either of the forces in the couple and π, the length of the moment arm. In this case, weβre told that the
norm of the moment or its magnitude is 750 newton meters. And the magnitude of one of its two
forces is 50 newtons. We can therefore substitute
everything we know about the moment of this couple into the formula, and we get 750
equals 50 times π.

To solve for π, we divide both
sides of this equation by 50. So π is 750 divided by 50, or
15. And since weβre dividing newton
meters by meters, the units for our length are simply meters. And so the length of the moment arm
in this case is 15 meters.

Now, the forces donβt necessarily
act perpendicularly to the line which connects the points from which they act. In these cases though, there is a
technique that will help us determine the moment due to the couple. We saw that π is the arm of the
couple. And itβs the length between a joint
axis and the line of force acting on that joint when the force acts
perpendicularly. Now thatβs the important part. This definition allows us to
consider the components of each force which act perpendicular to the line that joins
the points at which they act. If we define the distance between
the points of the lines of action of each force as π, we can say that the component
of the force that acts perpendicular to this length is the magnitude of π
times sin
π. And so weβre now able to say that
the magnitude of the moment due to the couple is equal to the magnitude of either
force π
times sin π times π.

In our next example, weβll
demonstrate an application of this.

In the figure below, π
sub one is
equal to three newtons and π
sub one and π
sub two form a couple. Find the algebraic measure of the
moment of that couple.

In order to find the algebraic
measure of the moment of the couple, we begin by recalling that if the forces form a
couple, then their magnitudes must be equal. If we define π
sub one and π
sub
two to be their respective magnitudes, then we can say that π
sub one equals π
sub
two, which equals three newtons. Then, to find the magnitude of the
moment of the couple, we find the product of the magnitude of one of the forces
times sin π and π, where π is the distance between the line of action of each
force. And π is the angle that each force
makes with the line connecting the points that force one and force two act from.

We see from our diagram that π
here is equal to 45 degrees. And π, the distance between points
π΄ and π΅ here, which are the points where π
one and π
two act, is equal to seven
root two centimeters. And so the magnitude of the moment
of this couple is three times sin 45 degrees times seven root two. Now, of course, sin of 45 degrees
is an exact value that we should know by heart. Itβs root two over two. But of course, the square root of
two divided by two times the square root of two is two over two or simply one. And so the magnitude of the moment
of the couple is three times seven, which is 21 or 21 newton centimeters. Since the forces are trying to move
the body in a counterclockwise direction, we know that the algebraic measure of the
moment is going to be positive. So the answer is 21 newton
centimeters.

Now, so far, weβve considered a
couple of formulae. But an alternative way of
calculating the moment of a couple is to use the cross product. The moment of a force π
about a
point can be found using the cross product. If we take the vector π« to be the
position vector from the point about which the momentβs being taken to any point on
the line of action of the force, the moment of the force is the cross product of π«
and π
. If we consider a couple formed of
two forces π
sub one and π
sub two that act at π΄ and π΅, respectively, the moment
of the couple is then given by the cross product of the vector ππ and the vector
π
sub one, which is equivalent to finding the cross product of the vector ππ and
the vector π
sub two.

Essentially, in the equation for
the moment of the couple, the position vector π« is substituted by the vector
between the points of applications of the forces. We can think of this as taking
moments about π΅ in the first case and moments about π΄ in the second. So with this in mind, letβs look at
how to find the perpendicular distance between two force vectors that form a
couple.

Given that two forces π
sub one
equals negative π’ plus two π£ and π
sub two are acting at two points two, two and
negative two, negative two, respectively, to form a couple, find the perpendicular
distance between the two forces.

Weβve been given a vector
expression for one of the forces in our couple. Now, of course, for the forces to
form a couple, their magnitudes must be equal. They will also be acting in
opposite directions. So we could, if weβre required,
find the vector π
sub two. Theyβll sum to zero, so π
sub two
will be the negative of vector π
sub one. Thatβs the negative of negative π’
plus two π£, which is π’ minus two π£.

Now, in fact, itβs worth noting
that we donβt actually need to work this out to be able to answer the question. But it would allow us to check our
answer at the end. Then we know that for a couple
formed of two forces π
sub one and π
sub two that act at π΄ and π΅, respectively,
the moment of the couple is either given by the cross product of the vector ππ and
the vector π
sub one or equivalently the cross product of the vector ππ and the
vector π
sub two.

Now, weβre told that the force π
sub one acts at the point two, two and the vector π
sub two acts at the point
negative two, negative two. So we define π΄ and π΅ as
shown. Then we recall that the vector ππ
can be found by subtracting the vector ππ from the vector ππ. Since the vector ππ is the
position vector of π΄, itβs simply two π’ plus two π£. And similarly, the vector ππ,
which is the position vector for π΅, is negative two π’ minus two π£. Distributing the parentheses, and
we find that the component for π’ is two minus negative two, which is four. And similarly, the π£-component is
four as well. So the vector ππ is four π’ plus
four π£.

And so we can say that the cross
product of the vector ππ and the vector π
sub one is the cross product of the
vector four π’ plus four π£ plus zero π€ and the vector negative π’ plus two π£ plus
zero π€. And of course, we sometimes think
about this in terms of the determinant of a two-by-two matrix. We get four times two minus four
times negative one π€. Thatβs eight plus four π€, which is
12π€.

Now we want to find the
perpendicular distance between the two forces, in other words the length of the
moment arm. And so we use the formula that the
magnitude of the moment of a couple is equal to the magnitude of one of the forces
times the length of the moment arm. Now, in fact, since the moment is
acting in simply one direction, its magnitude is equal to 12. We take the magnitude of either
force. And to do so, we find the square
root of the sum of the squares of each component. Specifically, the magnitude of
force π
sub one is the square root of negative one squared plus two squared, which
is equal to root five. And so substituting what we know
about the moment of our couple into the formula, and we get 12 is equal to the
square root of five times π, which we can solve for π by dividing through by root
five.

Finally, weβll rationalize the
denominator to simplify this. And to do so, we multiply both the
numerator and denominator of the fraction by the square root of five, which gives us
12 root five over five. We can therefore say that the
perpendicular distance between the two forces is 12 root five over five length
units.

Weβll now recap some of the key
concepts from this lesson. In this video, we learned that a
couple is a system of two forces of equal magnitudes and opposite directions which
act in different lines of action. The moment of a couple is the sum
of the moments of the two forces of that couple about a point in space. And the magnitude of the moment is
equal to the magnitude of one of the two forces and the distance between them, or
the arm of the couple.

We learned that the moment of a
couple is a constant independent of the point about which we take the moments of the
two forces. And if the forces are not acting in
a direction perpendicular to the line which connects the points that they act from,
we can calculate the magnitude of the moment of the couple by finding the product of
the magnitude of one of the forces times sin π times π, where π is the
perpendicular distance between the points that the forces act from. And π is the angle between the
line of action of either force and this line.

Finally, for a couple formed of two
forces π
sub one acting at point π΄ and π
sub two acting at point π΅, the moment
of the couple could be calculated using the cross product. We say that the moment is equal to
the cross product of the vector ππ and the vector π
sub one, or equivalently the
cross product of the vector ππ and the vector π
sub two.