 ## Physics

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15. Rotational Equilibrium

# Torque & Equilibrium

1
concept

## Torque & Equilibrium 10m
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2
example

## Balancing a bar with a force 6m
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Hey, guys. So here we have an example of rotation of equilibrium. Let's check it out. So the bar below has a length of 4 m in a mass of kg. So I'm gonna draw here. L equals 4 m. M equals 10 kg. Um, it's masses distributed uniformly. What that means is that the center of mass, uh, of the bar is in the middle. And what that means is that that's where mg acts. Um, it says the bar is free to rotate about a fulcrum. This is the focus right here to support point positioned 1 m away from its left end. So this here is a distance of 1 m, Um, to the end. Okay. And you wanna push straight down on the left edge. So this is you with the force of F to try to balance the bar because if you didn't push on the left, the bar would tip over to the right. Okay. So what magnitude of force should you apply on the bar? In other words, what is f the magnitude of f and for part B. How much force is the fulcrum apply on the bar. Well, if the if the bar eyes rested on top of the focus and the focus is gonna push back with a force that's our normal force. And we want to know what is the magnitude of normal. So let's start with question A here. How do we find force? Well, we want to know how much force we need to balance. Uh, the bar, which means there will be a rotational equilibrium, were holding the bar by pushing down this way. So we wanna have rotational equilibrium, which means the sum of all torques will be zero. Okay, there are two torques. They're gonna act here one first you have m g going this way. So there's a torque do OMG. It is clockwise. So it's negative. And your force here is causing a torque this way because it's to the left of the center. So it's doing this to the bar, so torque of f is going to be counterclockwise positive. Um, the normal force acts at the axis of rotation. Therefore it has no. It produces no torque. Torque of normal would be normal. Our sign of data, but are zero because it's the force acts on the axis of rotation. So the whole thing is zero. So, really, what you have is thes two guys. Eso I can do this. I can say torque F plus negative torque. MGI equals zero. And if I send this to the other side, I get the torque f equals torque of M G. And they should make a ton of sense. Um, the torques. This basically just says that the talks are going opposite directions air canceling each other out. So next thing you do is you expand these two sides, so torque of f is going to be f r of f sign of data of F and on the right side, I have m g are of M G and Sign Fada of M. G. We're looking for F. So let's plug in everything here. The distance three are vector is the distance from the axis of rotation to the point where the force happens, so it's going to be this distance right here. This is our F, which is one, and the angle between F and R is 90 degrees. Okay, so this is our the our vector for F, and you can draw f like this or you could have kept it this way. It doesn't matter. Um, it's easy to see that it's 90 degrees sign of 90 is one um m g. I have the mass is 10 g is 9.8. What is our vector for G now? I didn't really draw this to scale here. But if the center of mass is in the middle, this means that this thing is 2 m and the entire right side is m. But the folk room is 1 m to the left. Therefore, this has to be another meter here. Okay, so 1 m from its 2 m from the left to the center mask is in the middle. But it's 1 m from the left to the folk room. So you've got another meter here and this is the distance for our m G. Okay, So RMG is this which is 1 m, um and ar f is this which is 1 m as well. So I'm gonna put one here and the sign will be a sign of data will be one as well because you can see how mg makes an angle of 90 degrees with its our vector. Okay, Everything cancels and we get that f equals Newtons f equals 98. Newts quote. That's it. So if you push with the force of 98 these things will exactly cancel each other. You might have seen from the fact that the distances were the same. If the distances are the same, the forces have to be the same. So that should make sense. Maybe you saw that on Ben for Part B very quickly. To find normal force, we have to use the fact that the sum of all forces is zero on the Y axis, right? And if you look at all the forces, all the forces, there's two forces going down, which is F plus F plus M g. They're both going down. And then there's one force going up, which is normal, and they all equal to zero. So I can say that normal equals F plus M g. This should also make sense right away because it's basically just says that all the forces going up equals the force is going down F nine and mgr, both 98. So when you add this thing up, you get 196 Newton's. Okay, So this is how much force you would need to keep this thing balanced. And this is how much you get a za result of doing that. That's how much normal folks you have as a result. Okay, so that's it for this one. Let me Do you have any questions?
3
Problem

A composite disc is made out of two concentric cylinders, as shown. The inner cylinder has radius 30 cm. The outer cylinder has radius 50 cm. If you pull on a light rope attached to the edge of the outer cylinder (shown left) with 100 N, how hard must you pull on a light rope attached to the edge of the inner cylinder (shown right) so the disc does not spin?   