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# Torque & Equilibrium

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Sections
More 2D Equilibrium Problems
Equilibrium in 2D - Ladder Problems
Equilibrium with Multiple Supports
Torque & Equilibrium
Review: Center of Mass
Beam / Shelf Against a Wall
Center of Mass & Simple Balance
Equilibrium with Multiple Objects

Concept #1: Torque & Equilibrium

Transcript

Example #1: Balancing a bar with a force

Transcript

Hey guys! Here we have an example of rotational equilibrium. Let's check it out. The bar below has a length of 4 meters and a mass of 10 kilograms. I'm going to draw here l = 4 meters, m = 10 kilograms. Its mass is distributed uniformly. What that means is that the center of mass of the bar is in the middle. What that means is that that's where mg acts. It says the bar is free to rotate about a fulcrum. This is the fulcrum right here, the support point, positioned 1 meter away from its left end. This here is a distance of 1 meter on to the end. You want to 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. What magnitude of force should you apply on the bar? In other words, what is F, the magnitude of F? For Part B, how much force does the fulcrum apply on the bar? If the bar is resting on top of the fulcrum, the fulcrum is going to push back with a force that's our normal force. We want to know what is the magnitude of normal.

Let's start with question A here. How do we find force? We want to know how much force we need to balance the bar which means there will be rotational equilibrium. We're holding the bar by pushing down this way. We want to have rotational equilibrium which means the sum of all torques will be zero. There are two torques that are going to act here. One, first, you have mg going this way so there's a torque due to mg. 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. Torque of F is going to be counterclockwise positive. The normal force acts at the axis of rotation therefore it produces no torque. Torque of normal would be normal r sin_, but r is zero because the force acts on the axis of rotation so the whole thing is zero. Really what you have is these two guys. I can do this. I can say torque F + negative torque mg = 0. If I send this to the other side, I get that torque F = torque of mg. This should make a ton of sense, the torques. This basically just says that the torques are going opposite directions, are cancelling each other out. Next thing you do is you expand these two sides. Torque of F is going to be F r of F sin_ of F and on the right side I have mg r of mg and sin_ of mg. We're looking for F.

Let's plug in everything here. The r vector is the distance from the axis of rotation to the point where the force happens. It's going to be this distance right here. This is our F which is 1 and the angle between F and r is 90 degrees. This is the r vector for F. You can draw F like this or you could have kept it this way. It doesn't matter. It's easy to see that it's 90 degrees. Sin90 is 1. mg I have the mass is 10, g is 9.8. What is r vector for g? 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 meters and the entire right side is 2 meters but the fulcrum is 1 meter to the left therefore this has to be another meter here. It's 2 meters from the left to the center of mass because itÕs in the middle. But it's 1 meter from the left to the fulcrum, so you've got another meter here. This is the distance for r mg. r mg is this, which is 1 meter and r F is this which is 1 meter as well. I'm going to put 1 here. The sin_ will be 1 as well because you can see how mg makes an angle of 90 degrees with its r vector. Everything cancels and we get that F = 98 N. That's it. If you push with a 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. Then 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. If you look at all the forces, there's two forces going down which is F + mg. They're both going down and then there's one force going up which is normal and they all equal to zero. I can say that normal = F + mg. This should also make sense right away because this basically just says that all the forces going up equal the forces going down. F and mg are both 98, so when you add this thing up you get 196 N. This is how much force you would need to keep this thing balanced and this is how much youÕd get as a result of doing that. That's how much normal force you have as a result. That's it for this one. Let me know if you have any questions.

Practice: 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?

Example #2: Pin holding a horizontal bar

Transcript

Hey guys! Here in this example, we're trying to balance sort of a shelf on a wall by using a pin. Let's check it out. You got the red pin here and we're using it to try to balance the bar that has a mass of 20, so m = 20 and a length of 3 meters. It's held horizontally against the wall by the pin right here. The idea is that there's going to be an mg right in the middle that's pulling the bar down. This is the axis of rotation somewhere over here. This mg is producing a torque that would cause this to spin. But the pin which could be like a nail or something, is holding it up and the way it holds it up is by providing a counter torque. Here if we're holding it up, we're going to say that the sum of all torques equals zero because the bar is not going to accelerate. ItÕs not going to rotate. ItÕs not gonna have angular acceleration. The torque of mg is this way, which is negative so you'd want the torque of the pin to be positive so that they cancel. We're going to write torque of mg negative + torque of pin positive = 0. This gives us that IÕm going to send the negative to the other side, Torque of pin = torque of mg. This should make sense. All we're doing is getting these two guys to cancel each other out. What we're looking for here is what is the torque on the pin that's needed. The way we're going to calculate the torque on the pin is just by calculating the torque of mg. Torque of pin will be mg. IÕm gonna expand the right side, rsin_. The mass is 20, gravity 9.8, r is the distance from the axis of rotation to the point where the force happens. We have a uniform mass distribution which means mg happens in the middle of the bar, the bar is 3 meters long so the r vector is 1.5, put a 1.5 here. The angle that these two guys make is 90 degrees. Sine of 90 is 1. If we multiply this whole thing, we get 294 Nm and that's it. That's the answer here. Very straightforward question for us to calculate one torque based on some other information. That's it for this one. Let me know if you have any questions and let's keep going.