Video Transcript
In the given figure, if the forces
that act on a light rod 𝐴𝐵 are equivalent to a couple and the moment of this
couple is equal to 17 newton centimeters, find 𝐹 sub one and 𝐹 sub two.
Let’s begin by reminding ourselves
what we mean when we say that the forces are equivalent to a couple. Now, a couple consists of a pair of
parallel forces that are equal in magnitude, opposite in sense and do not share a
line of action. This means that the resultant of
the forces are equal to zero. And as such, a couple does not
produce a translation; it only produces a rotation. So this means that the resultant of
a couple is just a pure moment.
In this case, we’re told that the
moment of the couple is equal to 17 newton centimeters. And the most important fact that’s
going to help us answer this question, in fact, is that the moment of the couple is
independent of the reference location. In practice, this means that we can
take the sum of the moments about point 𝐴 or the sum of the moments about point
𝐶. Or, indeed, we can even calculate
the sum of the moments about point 𝐷 or point 𝐵. And we should always get a value of
17 newton centimeters.
Now, this is really useful because,
in fact, if we take moments about either point 𝐷 or point 𝐵, that will allow us to
calculate the value of either force 𝐹 sub one or 𝐹 sub two. Let’s take moments about 𝐷. This will allow us to find the
value of 𝐹 sub one. And the reason this happens is
because to calculate the moment of a force, we multiply the magnitude of that force
by the perpendicular distance from the line of action of the force to the pivot. Now, of course, force 𝐹 sub two
acts at a distance of zero units from 𝐷. So even if we attempted to
calculate the moment, we know it would be zero for force 𝐹 sub two.
Let’s begin then by looking at the
moment of the force acting at point 𝐴. This has a magnitude of five
newtons, and it’s looking to move the object 𝐴𝐵 in a counterclockwise
direction. So the moment is going to be
positive. The distance away from point 𝐷 is
one plus two. So the moment is five times one
plus two or five times three newton centimeters. We’ll evaluate this when we’ve
calculated the sum of all of our moments. Let’s next consider the moment of
the force that acts at point 𝐶 . This force is trying to rotate the object in a
clockwise direction. And so, its moment is going to be
negative. It has a magnitude of two newtons
and it acts at a point two centimeters away from 𝐷. So its moment is negative two times
two.
Finally, let’s consider the moment
of the force acting at point 𝐵. This is trying to rotate the object
in a counterclockwise direction, and so the moment is going to be positive. And it has a magnitude of 𝐹 sub
one and acts at a distance of one centimeter away from 𝐷. So the moment is 𝐹 sub one times
one. Now, of course, this is the sum of
all of the moments about point 𝐷. And we know that the sum of the
moments will be independent of the reference location. So the sum of these is equal to
17. Simplifying the left-hand side and
we get 15 minus four plus 𝐹 sub one, which simplifies further to 11 plus 𝐹 sub one
equals 17.
We’ll solve for 𝐹 sub one by
subtracting 11 from both sides. And that gives us 𝐹 sub one is 17
minus 11 or 𝐹 sub one is equal to six. So we’ve calculated the value of 𝐹
sub one. We could repeat this process but
take moments about point 𝐷 to find the value of 𝐹 sub two. Alternatively, we said earlier that
the sum of the forces acting on our body must be equal to zero. So let’s calculate the sum of the
forces and we’ll take upwards to be positive. In the upward direction, we have
two newtons and 𝐹 sub one, and in the downwards direction, we have five and 𝐹 sub
two. So the sum of our forces is two
plus 𝐹 sub one minus five minus 𝐹 sub two. And we know that that is equal to
zero.
In fact, we’ve already calculated
the value of 𝐹 sub one as well; it’s six. So our equation becomes two plus
six minus five minus 𝐹 sub two equals zero. That simplifies to three minus 𝐹
sub two equals zero. So adding 𝐹 sub two to both sides
of this equation and we get 𝐹 sub two equals three. We’re working in newtons, so we’re
done. The value of 𝐹 sub one and 𝐹 sub
two are six newtons and three newtons.
