Question Video: Finding the Unknown Force Acting on a Fixed Vertical Rod to Achieve Equilibrium Mathematics

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 7 N and cos 𝜃 = 4/5.

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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.

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