### Video Transcript

In this video, weβre talking about
the moment of a couple. This means something a little bit
different from what we see on screen. But as weβll learn, it does involve
vectors that point in opposite directions. Letβs jump right in.

We can recall that the mathematical
definition of a moment involves a force, weβve called it π
, as well as a point. If our force is exerted a
perpendicular distance π away from this point weβve picked, then we say that about
this point it exerts a moment π. And itβs helpful to recall that in
this equation the force weβre talking about is explicitly perpendicular to the
distance between where the force is applied and the point of rotation. So, this variable π, we say, is
the moment of the force π
about this point we picked. But now letβs expand this idea to
talk about the moment of a couple.

A couple is a pair of forces that
are equal in magnitude but point in opposite directions. And importantly, for a pair of
forces like π
one and π
two to form a couple, they canβt act on the same line of
action. For example, if we had two forces
acting like this, π
π΄ and π
π΅, even though they had the same magnitude and act
in opposite directions, theyβre not a couple because they do lie along the same line
of action. So, coming back to our point here,
if we were to apply a force couple about this point, then out a distance π to the
right of our point, we would apply this force, weβve called it π
prime, which is
equal in magnitude but opposite in direction to π
.

π
prime and π
form a force
couple, and we can see what effect this would have on the moment π about our
point. Whenever we have a couple of forces
acting about a point, the moment created by one of the forces is equal to the moment
created by the other. And this equality is not just in
magnitude but also in direction or sign. For example, notice that the force
π
creates a counterclockwise moment about our point that would be positive, and the
force π
prime creates a moment in the same direction. This will always be true when we
work with couples because recall that force couples always act in opposite
directions. The total moment then created by
our couple is simply two times the moment created by either one of the forces in the
couple.

Knowing all this, letβs get some
practice with these ideas now through a few examples.

In the figure below, π΄π΅πΆπ· is a
rectangle in which π΄π΅ equals five centimeters and π΅πΆ equals four
centimeters. The forces that are shown in the
figure are in newtons, and the system is in equilibrium. Find the value of πΉ one plus πΉ
two.

Okay, looking at our figure, we see
this rectangle with corners, π΄, π΅, πΆ, π·. And that each of the four sides has
a force acting on it. On the left side, a force of
magnitude πΉ two acts upward, and on the right a force of that same magnitude acts
downward. And then on the top, a force of
magnitude πΉ one acts to the left, and on the bottom a force of magnitude 35 newtons
acts to the right.

A key piece of information weβre
given is that our system is in equilibrium. This means that even though these
four forces are acting on it, it doesnβt translate and it also doesnβt rotate. The forces acting on the four sides
of our rectangle effectively cancel one another out. This tells us that these four
forces comprise two force couples. The first couple is force πΉ one
and this 35-newton force. And the second is made of these two
forces of magnitude πΉ two.

The fact that these pairs of forces
do form couples tells us that the magnitude of force πΉ one must equal 35
newtons. This is because the forces in a
couple have the same magnitude but act in opposite directions. So, we can say right away that πΉ
one equals 35 newtons, which means that now to answer our question, we just need to
solve for πΉ two.

If we consider the point at the
very center of our rectangle to be the axis of rotation for these two force couples,
then we can see that the moment about this point created by our πΉ two force couple
would point in a clockwise direction. Weβll call that π two. And then notice that the moment
created by our other force couple acts in opposition to this. And that actually makes sense as we
recall that our system is an equilibrium. That means these two moments must
cancel one another out, and therefore they have to point in opposite directions.

If we think just in terms of
magnitudes, then we can write that π one is equal in magnitude to π two. And now letβs recall that, in
general, the moment created by a couple of forces, weβll call it π sub c, is equal
to two times the perpendicular component of one of those forces multiplied by the
distance away from the axis of rotation.

Going back to our sketch, we could
say, for example, that the distance between the force πΉ one at the top of our
rectangle and the axis of rotation is π one and the distance between πΉ two and
that center point is π two, meaning that we can write our moment balance equation
like this. We see that the factors of two
cancel from each side of our equation.

And now, since itβs πΉ two that we
want to solve for, letβs rearrange so that that is the subject of our equation. If we divide both sides by π two,
that cancels out on the right. And we can see that πΉ two equals
πΉ one multiplied by this distance ratio π one to π two. We know the force πΉ one. And letβs recall that weβre given
the side lengths of this rectangle. π΄π΅ is equal to five centimeters,
which, we can see is equal to two times π two and π΅πΆ is four centimeters or two
times π one. These relationships tell us that π
two equals five half centimeters and π one equals two centimeters.

Substituting in these values and
leaving out the units of centimeters, we see that πΉ two equals 35 newtons times two
over five-halves, which simplifies to four-fifths times 35 newtons or 28
newtons. So, since πΉ two equals 28 newtons
and πΉ one equals 35 newtons, the sum of these must be 63 newtons. Thatβs the sum of the magnitudes of
the forces in our two couples.

Letβs now look at an example where
our forces are not at 90 degrees to the axis of our system.

In the figure below, πΉ one equals
three newtons, and πΉ one and πΉ two form a couple. Find the algebraic measure of the
moment of that couple.

Looking at our figure, we see the
forces πΉ two and πΉ one, which are acting at 45 degrees to the axis of this
system. Weβre told that these forces form a
couple, and we see that they act at a distance of seven root two centimeters from
one another. We want to calculate the moment
that this couple of forces produces, and that moment will be calculated about the
midpoint of this black line.

Based on the directions of πΉ one
and πΉ two, we can see that they will create a counterclockwise moment about this
point. By convention, moments in this
direction are considered positive. So now, to answer our question, we
need to calculate the magnitude of this moment. Letβs recall that the moment
created by a couple of forces, we can call it π sub c, is equal to two times the
perpendicular component of the magnitude of one of those forces in the couple
multiplied by the distance between where that force is applied and the axis of
rotation.

In our case, that distance would be
the length of this pink line here. Whether weβre talking about force
πΉ one or πΉ two, that distance is the same. And since πΉ one and πΉ two form a
couple, we can work with just one of them and use that in our calculation for π sub
c. Letβs choose to work with the force
πΉ sub two. Since this force forms a couple
with πΉ sub one, we know that its magnitude must be three newtons. And what we want to do is multiply
the component of this force perpendicular to our black line by the distance, weβll
call it π, between where the force is applied and our rotation point.

As for that perpendicular component
of our force πΉ two, because this angle here is marked as 45 degrees, this one here
must be the same. Thatβs because we know that they
add up to 90 degrees. So, if we call this component of πΉ
sub two πΉ sub two perpendicular, then we can say that itβs equal to three newtons
times the cos of 45 degrees. The cos of 45 degrees equals the
square root of two over two. So, the component of πΉ two
perpendicular to the axis of our system is three root two over two newtons. This is the value weβll use in our
equation to calculate the moment of our couple of forces.

π sub c equals two times the
perpendicular component of πΉ sub two multiplied by π, where π is this
distance. And we see that distance is seven
root two over two centimeters. With all these values plugged in to
calculate π sub c, we find that all of our factors of two and root two cancel one
another out. So, we end up with three times
seven or 21 newton centimeters. This is the moment of this couple
of forces.

Letβs now look at an example where
our forces are given as vectors.

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

Okay, to get started here, letβs
plot the two points π΄ and π΅ where these forces π
one and π
two are acting. Here, we have our π₯- and π¦-axes
marked out in arbitrary units. And we know that the point π΄ is
here; thatβs at two, two. And the point π΅ is located here at
negative two, negative two. Weβre told that at point π΄ a force
π
one is acting. And based on its components, we can
say that π
one acts along this line, passing through the point one, four. When it comes to plotting the force
π
two, weβre told that this force, along with π
one, forms a couple. And we can recall that forces in a
couple have the same magnitude but act in opposite directions.

So, starting at point π΅ and moving
in the opposite direction to the force π
one, π
two would look like this where it
passes through the point negative one, negative four. Our question then says, what is the
perpendicular distance between these two forces? What this is saying is that if we
draw the lines of action of these two forces, hereβs the line of action of π
two,
and hereβs the line of action of π
one, then we want to solve for the perpendicular
distance between these lines. We could measure that distance here
or here or here or anywhere along these lines so long as that distance is
perpendicular to these two parallel lines of action.

Now, since these two lines of
action are lines, letβs give them names according to that property. Letβs call the line of action from
force π
one π¦ one and that of force π
two π¦ two. And we can recall that in an
π₯π¦-plane, the general formula for a line is that the π¦-coordinate is equal to the
slope of the line or gradient multiplied by the π₯-coordinate added to the π¦-axis
intercept of the line.

So, for π¦ one and π¦ two then, the
lines of action of our two forces, letβs solve for the equations of those lines. First, letβs consider the slope or
gradient of the lines. And thankfully because theyβre
parallel, they have the same slope. As we consider again the components
of the force π
one, this tells us that for every increase of one in our π₯-value,
our π¦-value correspondingly decreases by two. Therefore, the slope of this line
is negative two. And as we mentioned, this is a
match for the slope of π¦ two since the lines are parallel.

And now, letβs consider where these
two lines intercept the π¦-axis. Starting with line π¦ two, if we
follow this line of action, we see that it intercepts the π¦-axis at negative
six. So, we now have the complete
equation for the line π¦ two. For line π¦ one, this intersects
the π¦-axis off of the graph that weβve drawn. But knowing the points that this
line passes through, both point π΄ as well as the point one, four, we can see that
when π₯ is equal to zero, π¦ must be equal to positive six.

So, now, we have the equations of
our two parallel lines and we want to solve for the perpendicular distance between
them. To start doing that, we can write
the equation of a third line, weβll call it π¦ perpendicular that is perpendicular
to π¦ one and π¦ two. The key property of a line that is
perpendicular to another line, say π¦ two, is that the slope or gradient of either
one of the lines is the negative inverse of the slope of the other.

In our case then, since the slope
of our two lines of action is negative two, that means the slope of our
perpendicular line will be negative the inverse of this. In other words, it will be positive
one-half. And as we mentioned earlier, it
doesnβt matter where the π¦-intercept of our perpendicular line is because whether
itβs here or here or here or anywhere else, that distance we want to calculate will
be the same. So, we can leave π¦ perpendicular
as it is.

And now what we want to do is find
where π¦ perpendicular intercepts π¦ one and where it intercepts π¦ two. Letβs begin with where π¦
perpendicular and π¦ one meet. To find where this is, weβll set
them equal to one another. So, negative two π₯ plus six equals
one-half π₯. And if we add two π₯ to both sides
of this equation, we get the result that five-halves π₯ is equal to six, which
implies that π₯ is equal to twelve-fifths. This value weβve solved for is the
π₯-coordinate of the point where π¦ one is equal to π¦ perpendicular.

To find the corresponding
π¦-coordinate of that point, we just need to substitute this value of π₯ in for π₯
in either our π¦ perpendicular or π¦ one equations. If we do that here, then we find
that π¦ is equal to one-half times twelve-fifths or six-fifths. If we plot this point on our
sketch, it would be somewhere around here. As weβve seen, its coordinates are
twelve-fifths and six-fifths. The line that passes through this
point and is perpendicular to both of our lines of action will also intersect the
line π¦ two.

Now, we want to solve for where
that point of intersection is. Similar to earlier, weβll set π¦
two equal to π¦ perpendicular, which means that negative two π₯ minus six equals
one-half π₯. And now adding two π₯ to both
sides, we find that five-halves π₯ equals negative six, telling us that this
π₯-value of intersection is negative twelve-fifths. Substituting this value in for π₯
here, we find the corresponding π¦-value is negative six-fifths.

So, now, we have our two points of
intersection between our lines of action and a line perpendicular to them both. To calculate the distance π
separating our two lines of action, we just need to calculate the distance between
these two points. In general, the distance between
two points in the π₯π¦-plane is equal to the square root of the sum of the squares
of the difference between the two π₯-coordinates Ξπ₯ and the difference between the
two π¦-coordinates Ξπ¦.

In our case, Ξπ₯ is equal to the
π₯-coordinate of one point twelve-fifths minus the π₯-coordinate of the other point
negative twelve-fifths, giving us twenty-four fifths. And likewise, the change in the
π¦-coordinate, Ξπ¦, equals six-fifths minus negative six-fifths or
twelve-fifths. Writing these values off to the
side, we can say that π equals the square root of twenty-four fifths squared plus
twelve-fifths squared. We can factor a one-fifth out from
under the square root. And if we square 24 and 12 and add
those results together, we get 720.

And it turns out that 720 can be
written as 144 multiplied by five. And 144 we know is equal to 12
squared, which means we can factor 12 out from the square root and get a simplified
result of twelve-fifths times the square root of five. Recalling that this is a distance
that weβve calculated, we can give our final answer as twelve-fifths times the
square root of five length units. This is the perpendicular distance
between the lines of action of our two forces.

Letβs now finish up this lesson by
summarizing a few key points. In this lesson, weβve seen that a
couple is a pair of equal and opposite forces acting along different lines of
action. Along with this, we saw that a
couple creates a moment. That moment is equal to two times
the perpendicular component of one of the forces in the couple multiplied by the
distance between where that force is applied and the axis of rotation. This summarizes the moment of a
couple.