### Video Transcript

In the figure shown, π΄π΅πΆπ· is a
right-angled trapezoid at π΄, where π΄π΅ equals 12 centimeters, π΅πΆ equals 32
centimeters, and π΄π· equals 16 centimeters. The shown forces are measured in
newtons and are represented completely by the sides of the trapezoid, where the
magnitude of the forces is proportional to their corresponding side lengths. If the system of forces is
equivalent to a couple, find πΉ one, πΉ two, and πΉ three.

All right, so here we see this
trapezoid with corners π΄, π΅, πΆ, and π·. And we also see the three unknown
forces πΉ one, πΉ two, and πΉ three we want to solve for. Each of these forces acts along a
side of the trapezoid along with this known force of 30 newtons. Our problem statement tells us that
the side length π΄π΅ is 12 centimeters, π΅πΆ is 32 centimeters, and side length π΄π·
is 16. Weβre told further that the
magnitude of the forces acting on this trapezoid is proportional to their
corresponding side lengths.

This means, for example, that since
side length π΅πΆ is twice as long as side length π΄π·, then the magnitude of the
force πΉ three must be twice as great as the magnitude of πΉ one. This same proportionality
relationship applies to the other sides. And weβre also told that this
system of four forces is equivalent to a couple. That means we can equate these four
forces to two equal and opposite forces that act along different lines of
action. Knowing all this, itβs the unknown
forces πΉ one, πΉ two, and πΉ three we want to solve for.

To begin doing that, letβs clear
some space on screen and letβs think about what it means that these four forces are
equivalent to a couple. If we draw the lines of action of
these four forces, they effectively trace out the sides of this trapezoid. Because these four forces are
equivalent to a couple, that means we can pick two points of intersection of these
lines of action. And so long as those points lie
along all four lines of action, then we can say that a force couple effectively
originates from those points.

Hereβs what we mean by that. Say we picked the corners of the
trapezoid π΅ and π·. At point π΅, the lines of action of
forces πΉ two and πΉ three meet. And point π·, we see, is an
intersection of the lines of action of our 30-newton force and πΉ one. In that sense then, weβve accounted
for all four forces. And so if we sketch in these forces
as though they originate at these two corners of our trapezoid, they would look
something like this.

So weβre modeling all four forces
as though they start at these two corners of our trapezoid. Note that we could just as well
have picked the corners π΄ and πΆ since those two corners intersect all four lines
of action. But either way, we can analyze
these four forces as a couple. All this means that the net force
acting at point π· is equal in magnitude but opposite in direction to the net force
acting at point π΅. And actually since forces in what
we could call the vertical direction are independent of forces in the horizontal
direction, we can also say that the total vertical force at π· is equal and opposite
that at π΅, and likewise for the horizontal components.

This means there are two force
balance equations we can write out, one for the vertical direction and one for the
horizontal. If we decide that forces to the
right and forces pointing upward are positive, then when it comes to the vertical
forces acting at points π΅ and π·, we can write that the vertical component of our
30-newton force, weβll call it 30 sub v, minus the force πΉ two acting at point π΅
equals zero. Rearranging this equation slightly,
we see then that if we can solve for that vertical component of our 30-newton force,
weβll know πΉ two.

Coming back to our sketch, we see
that this 30-newton force is effectively the hypotenuse of a right triangle. If we think of this triangle though
not in terms of forces but in terms of distances, we know that this side is equal to
12 centimeters, this side is 32 minus 16 or 16 centimeters. And that means the hypotenuse is
equal to the square root of 12 squared plus 16 squared, or 20 centimeters. Knowing these side lengths is
helpful because if we call this interior angle of our right triangle π, then we can
say that the vertical component of our 30-newton force is equal to 30 times the sin
of π.

We recall that given a right
triangle where another interior angle is π, the sin of that angle is equal to the
ratio of the opposite side length to the hypotenuse length. This means that in our right
triangle of interest, the sin of π is equal to the ratio of 12 divided by 20. 30 times 12 over 20 equals 18. And so we now know the magnitude of
force πΉ two. And we know that this is in units
of newtons. Weβll record this result and now
move on to solving for the two remaining unknown forces πΉ one and πΉ three. Both of these forces, we see, are
directed horizontally.

For our force balance equation, we
can write that πΉ three minus πΉ one minus the horizontal component of the 30-newton
force, 30 times the cos of π, all equal zero. Coming back to our example right
triangle, we can say that the cos of this angle π is equal to the adjacent side
length divided by the hypotenuse. In the triangle formed from our
trapezoid, this ratio is equal to 16 divided by 20. Even knowing this though, we can
see that here we have one equation but two unknowns. Weβll need to bring in some other
information then to solve for πΉ one and πΉ three.

At this point, we can recall that
the forces involved in this scenario are proportional to one another according to
their corresponding side lengths. This means, for example, that the
ratio of side length π΄π·, 16 centimeters, to the side length π΄π΅ is equal to the
ratio of πΉ one to πΉ two, the forces on those respective sides. This is great because it means that
πΉ one equals 16 over 12 times πΉ two, which we know. So πΉ one then equals 16 over 12
times 18 newtons, or 24 newtons.

Recalling this proportionality
method then, there are now two ways we could go about solving for πΉ three. We could use the same method we
used to solve for πΉ one. Or, now that we know πΉ one, we
could use this force balance equation here. Since weβve already written this
equation out, letβs continue on with it. Adding πΉ one and 30 times 16 over
20 to both sides, we get that πΉ three equals πΉ one plus 30 times 16 over 20. And knowing that πΉ one without
units is 24 and that 16 over 20 equals four-fifths, we find that πΉ three is equal
to 48 with units of newtons.

We find then that the three forces
πΉ one, πΉ two, and πΉ three acting on the sides of this trapezoid have magnitudes
of 24, 18, and 48 newtons, respectively. These magnitudes mean that all four
forces involved with the trapezoid act as a couple.