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 𝜃 is equal to
four-fifths.
So, we have a vertical rod, and
we’re told that this is in equilibrium. It has two known forces acting on
it, but these are the only forces. Consider the point where the rod
meets the surface that it rests on; that’s point 𝐴. There’s a downward-acting weight
force exerted on this point. But by Newton’s third law of
motion, there’s also an oppositely directed reaction force; let’s call that 𝑅. And along with these forces, there
is indeed one more force that acts on the rod. We know this is true because
consider the rotation of the rod about either point where the two horizontal forces
act.
Considering only the forces we
drawn in so far, we can see the rod won’t be in rotational equilibrium; that is, it
will have a nonzero moment. This issue is resolved if we assume
some frictional force, let’s call that 𝐅 sub 𝑟, acting on our rod to the left at
point 𝐴. And now, we have a free-body
diagram showing all the forces acting on our rod.
Now we know that for the body to be
in equilibrium, the sum of the forces must be equal to zero and the sum of the
moments about some point — let’s call that 𝑃 — must also be equal to zero. Now, we’ve written this using
vector notation. But of course, we know that we can
simply resolve horizontally and vertically and achieve the same result. Because there are several unknown
forces here, we’re going to focus on the sum of the moments. And we’re going to form an equation
involving them, taking the sum of the moments about point 𝐴.
Since the moment is force times
perpendicular distance from the point 𝐴 to the line of action of the force, the
moment of our seven-newton force is seven cos 30 times 4.7 plus 2.1. That’s 47.6 cos 30. Similarly, the moment of the
unknown force 𝐅 will be 𝐅 cos 𝜃 times 2.1. Next, we’ll take the sum of the
moments, taking the counterclockwise direction to be positive. And since we know that sum is equal
to zero, we obtain 47.6 cos 30 minus 2.1𝐅 cos 𝜃 equals zero.
In fact, we haven’t yet used the
information that cos 𝜃 is equal to four-fifths. So, let’s substitute that into our
equation, and we’ll be able to solve for 𝐅. When we do, the second part of our
expression becomes negative 2.1𝐅 times four-fifths, which is 1.68𝐅. Rearranging for 𝐅 from we have,
it’s equal to 47.6 cos 30 divided by 1.68. That gives us 85 over three times
root three over two, which simplifies to 85 root three over six newtons. And so, we’ve obtained the
magnitude of the force 𝐅 that makes our rod in equilibrium. It’s 85 root three over six
newtons.
