### Video Transcript

π΄π΅πΆπ· is a rectangle, in which
π΄π΅ equals 45 centimeters, π΅πΆ equals 55 centimeters, and π·πΈ equals 28
centimeters. Forces of magnitudes 225, 275, 265,
and 135 newtons act along π΄π΅, π΅πΆ, πΆπΈ, and πΈπ΄ respectively. If the system of forces is
equivalent to a couple, determine the magnitude of the moment of the forces.

Okay, so here we have this
rectangle π΄π΅πΆπ·. And weβre told that side length
π΄π΅ equals 45 centimeters, π΅πΆ is 55 centimeters, and that side length π·πΈ is 28
centimeters. Weβre Also told the magnitude of
four forces acting on this rectangle and that this system of forces is equivalent to
a couple. This means we can effectively
replace the four forces with two equal and opposite forces that donβt lie along the
same line of action. Knowing all this, we want to
determine the magnitude of the moment of these four forces.

To start doing that, letβs clear a
bit of space on screen. And letβs consider what it means
that these four forces are equivalent to a couple. It means if we were to draw the
lines of action of these four forces, then if we were to pick two intersection
points of these lines so that these points included all four lines of action, then
we could effectively model all four forces as though they originated from those two
points. So for example, if we pick the
points πΆ and π΄ in our rectangle, then from point π΄ we can consider our 225-newton
force to be acting down and our 135-newton force to be acting to the right, likewise
with point πΆ where we say our 275-newton force acts to the left and our 265-newton
force acts up and to the right. Because our system of forces is
equivalent to a couple, we can say that the total force acting at point π΄ is equal
and opposite that total force acting at point πΆ.

And actually we can go further than
this. Since forces can be separated into
independent vertical and horizontal components, we can say that the net horizontal
force at point π΄ is equal and opposite that at point πΆ, and similarly for the net
vertical force at these two points. As a side note, we didnβt need to
use the points π΄ and πΆ as a model for the origins of our four forces. Taken together, these two points
meet our condition of overlapping all four lines of action, but, looking back to our
rectangle, so did the points, say, πΈ and π΅. Either of these pairs of points
would work for analyzing this scenario.

But anyway, knowing that the total
forces at πΆ and π΄ form a couple, our goal is to solve for the moment created by
that couple. We can recall that the moment
created by a couple of forces is equal to two times the force component
perpendicular to the distance between where the force is applied and where the axis
of rotation is located. The idea here is that each of the
forces in the couple contributes equally to the moment overall. Thatβs why if we solve for this
perpendicular component of one of those two forces and then multiply it by the
distance π, we need only multiply that result by two to solve for the overall
moment.

Using π΄ and πΆ as the points of
origin of the two forces in our couple, we can think of those forces as acting on a
line like this that joins at the two points. So the moment weβre solving for
acts about this midpoint of that line. In our rectangle, that line and
midpoint would look like this. So hereβs the idea. If we can find the net force thatβs
perpendicular to this line acting either at point π΄ or point πΆ, then we will have
solved for πΉ perpendicular in this equation. We could choose either point πΆ or
point π΄ at which to solve for this force.

And since the forces at point π΄
act in what we could call the purely vertical and purely horizontal directions,
letβs choose that point. Our goal then, is to solve for the
components of the 225- and 135-newton forces that are perpendicular to this dashed
orange line. Adding those together will give us
πΉ perpendicular here. Looking at the right triangle
created by our 225-newton force, we can call this interior angle of that triangle π
sub one. Weβre interested in this angle
because the sin of π sub one times 225 equals the perpendicular component of this
force.

Now, if we go back to our original
diagram, this angle in that diagram is also equal to π sub one. And we see that indeed this is an
interior angle in this right triangle drawn in orange. The shorter two sides of the
triangle have lengths of 45 and 55 centimeters, respectively. And by the Pythagorean theorem, we
can say then that the hypotenuse has a length of the square root of 45 squared plus
55 squared. Thatβs equal to the square root of
5050.

At this point, we can recall that
given a right triangle where π is another interior angle of that triangle, then the
sin of π equals the ratio of the opposite side length to the hypotenuse. And that means when it comes to our
angle of interest, the sin of π sub one, this is equal to 55 divided by the square
root of 5050. For π sub one, thatβs the opposite
side length to the hypotenuse length ratio. So then we now have an expression
for the perpendicular component of this 225-newton force. And we can start to keep a tally of
this using a variable weβll call πΉ perpendicular. This is exactly that variable we
see in our equation for π sub c. And we now know that itβs equal to
this term plus a second term weβll figure out soon.

That second term is equal to this
component of our 135-newton force. To solve for it, we can use a
similar approach. Letβs say that this angle interior
in this right triangle is π sub two. On our rectangle, the angle π sub
two would look like this. And notice that this is the same as
this angle here in our triangle. Once more, to calculate the
component of this force perpendicular to the dashed line, weβll want to use the sin
of this angle. We want to calculate 135 times the
sin of π sub two. The sin of this angle equals the
opposite side length, 45 centimeters, divided by the hypotenuse.

We now know the component of the
135-newton force thatβs perpendicular to our dashed orange line. This then is the second and final
term in our equation for πΉ perpendicular. All right, so to find π sub c, the
only thing that remains for us now is to solve for this distance π and then
multiply it by πΉ perpendicular and two. Looking at our original diagram,
that distance is half the distance from point πΆ to point π΄. In other words, itβs this distance
here. We see right away that that equals
one-half, our hypotenuse length. In other words, π is equal to the
square root of 5050 over two.

Knowing all this, weβre now ready
to calculate π sub c. Plugging in the values we
calculated for πΉ perpendicular and π, notice that this leading factor of two will
cancel out with this denominator of two and that also this denominator of the square
root of 5050 that appears in both terms in πΉ perpendicular will be multiplied by
that same value and therefore become equal to one. We find then that π sub c equals
225 times 55 plus 135 times 45. And this equals 18450. We recall that the units of our
forces are newtons and the units of our distances in this example are
centimeters. The moment then created by this
system of forces, which is equivalent to a couple, equals 18450 newton
centimeters.