### Video Transcript

In the given figure, determine the magnitude of the force 𝐹 that makes the rod in
equilibrium, given that the magnitude of the given force is seven newtons and cos of
𝜃 equals four-fifths.

Looking at our figure, we see this vertical rod, which we’re told is in
equilibrium. Two forces are drawn in acting on the rod, one here of magnitude seven newtons and an
unknown force 𝐹. We know, though, that these aren’t the only forces that act on this rod. If we consider the point where the rod meets the surface that it rests on, this is
labeled as point 𝐴. We know that there’s a downward acting weight force being exerted on this point, and
there’s also an oppositely directed reaction force; we’ll call it 𝑅. Note that the magnitude of this reaction force is greater than that of 𝑤 because it
needs to counteract the downward acting components of the seven-newton force and the
unknown force 𝐹.

Along with these forces, there’s actually one more that acts on this rod. We know this because if we were to consider the rotation of the rod about either this
point here or this point here, where one of our two horizontal forces act, then
considering only the forces we’ve drawn in so far, we can see that actually this rod
would not be in rotational equilibrium. That is, it would have a nonzero moment. This issue is resolved if we assume a frictional force, we’ll call it 𝑧, acting on
our rod to the left at point 𝐴. And now our free body diagram showing all the forces acting on our rod is
complete.

Knowing this, we can recall the two conditions that must both be satisfied for a
rigid body to be in equilibrium. First, the net force acting on the body must be zero. This ensures translational equilibrium. And then the net moment acting on a body in equilibrium must also be zero. This guarantees rotational equilibrium. Applying these conditions to our given situation will yield force and moment
equations. For example, we can see that we have force components that are both vertical and
horizontal, which would therefore yield two force balanced equations.

Furthermore, seeing that several of our forces would create a moment about the
vertical axis of this rod, we can apply the second condition to yield another
equilibrium equation. Let’s recall that it’s the magnitude of this unknown force 𝐹 that we want to solve
for. We could start by applying either one of these equilibrium conditions. But notice that if we apply the second condition that the net moment of this rod must
be zero and if we choose point 𝐴 as the place about which we calculate moments,
then that choice of location will eliminate from consideration both the weight, the
friction, and the reaction forces. And that’s because all three of these forces act at this point and therefore have a
distance of zero between their line of action and this rotation axis.

That means if we’re considering the net moment of the forces on this rod about point
𝐴, we only need to take into account our seven-newton force, which acts on the rod
at this location, and our unknown force 𝐹, which acts here. Recalling that the moment due to a certain force is equal to the distance from where
that force is applied to the axis of rotation multiplied by the component of the
force perpendicular to that line, we can see that we’ll be interested in the
horizontal component of our seven-newton force and that of our unknown force 𝐹
along with the perpendicular distances from these forces to point 𝐴.

As we start writing out our moment balance equation that the net moment on this rod
is zero, let’s set up the convention that any moment that leads to a
counterclockwise rotation about point 𝐴 we’ll consider positive. This means that a moment causing rotation in the other direction is considered
negative. Set up this way, we can see that the moment caused by the seven-newton force will be
positive, while that caused by our unknown force 𝐹 will be negative. We can write that seven newtons times the cos of 30 degrees — by taking this cos, we
isolate the horizontal component of the seven-newton force — multiplied by the
distance from where this force is applied to point 𝐴 — which we can see is 4.7
meters plus 2.1 meters, that’s a total of 6.8 meters — minus, because it generates a
moment we’ve called negative, the force magnitude 𝐹 times the cos of the unknown
angle 𝜃 multiplied by 2.1 meters, that’s the distance from where 𝐹 is applied to
the point 𝐴, is all equal to zero.

Just to be clear about why we’ve multiplied our force 𝐹 by the cos of 𝜃, notice
that 𝜃 is this given angle here and that as an alternate interior angle, this is 𝜃
as well. Therefore, the horizontal component of 𝐹 is indeed 𝐹 times that cos of 𝜃. Now, in our problem statement, we’re told that the cos of 𝜃 is equal to
four-fifths. And we can also recognize that the cos of 30 degrees is the square root of three over
two. Making these substitutions, we’re now ready to rearrange and solve for the force
𝐹. If we add 𝐹 times four-fifths times 2.1 meters to both sides of this equation, then
we get this result. Shifting this up a bit, let’s now divide both sides of this equation by four divided
by five multiplied by 2.1 meters.

From here, we see that the units of meters cancel from numerator and denominator. And if we then multiply both numerator and denominator of our left-hand side by five
divided by four, we then have this expression for the force magnitude 𝐹. This is equal to 35 divided by eight multiplied by 68 divided by 21 times the square
root of three newtons. And if we multiply together and simplify as far as possible these two fractions, then
we find they simplify to 85 divided by six, which means that the magnitude of the
force 𝐹 is eighty-five sixths times the square root of three newtons.