Torque & Equilibrium - Video Tutorials & Practice Problems

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1

concept

Torque & Equilibrium

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Hey, guys. So you may remember back when we talked about forces that we mentioned how if the forces of an object cancel, the object is in equilibrium. Well, now that we're talking about torques, there's a similar situation where you may have what we're gonna call rotational equilibrium. Let's check it out. So remember, if the Net force on an object if the Net force or the sum of all forces an object is zero, then the acceleration on that object is zero. And that's because of Newton's second law. F equals M A. If the Net force zero means that some of all forces equals zero and therefore the acceleration has to be zero. This situation is called equilibrium. But now that we know about torques, things are not gonna be that simple. It's gonna be a little more complicated. Okay, sometimes this this condition here of the Net force being zero is not sufficient for equilibrium. It waas before we knew about torques. So here's an example you imagine have a continuous bar that has uniformed mass distribution. So the bars center of mass happens right here, and at this point is where the weight force acts There's a little folk room here that holds the bar up and this is going to push back with a force of Newton normal. And this normal may even be equal to M. G. The problem is that even though the forces will cancel on the vertical, um, this MG produces a torque here torque of M. G. This is clockwise, so it's negative. But there is no torque due to the normal force. And that's because the normal force acts on the axis of rotation r equals zero. It acts on the axis of rotation. So the net torque will be this. So the network will be equal to this. And because I have a network, I will have a new Alfa. So even though the forces canceled, the torques don't cancel. So what happens is that this thing would tip over this way and fall. Okay, so we're gonna have a situation where we don't have equilibrium. Alright, this brings us to the fact that there has to be. Actually, there are two conditions that are necessary for a knob jek toe. Have what I'm gonna call complete equilibrium. It's not enough that the forces canceled the first condition is that the sum of all forces must be zero. And this gives us an acceleration of zero. And this is good old equilibrium. But now we're gonna call it linear equilibrium because they're going to be two types. Okay, the second condition is that the sum of all torques also has to be zero. And this will give us an Alfa that is zero. And we're gonna call this rotational a political. And if we have, both of them were gonna have what I call complete equilibrium. Okay, um, this topic is in most most textbooks refer to its static equilibrium and static refers to the fact that you're gonna be in a situation where there's no velocity and no linear velocity, no angular velocity. So not only do you have, um, complete equilibrium, but also, the object is not moving. Okay, Um, this is sometimes called also the equilibrium of rigid bodies, because we're going to deal with rigid bodies exclusively. So we're always gonna have these extended objects rather than a point mass. Okay, let's do I want to do a sort of introductory example here to talk about different situations where you may have one type of equilibrium, but not the other or both. Or neither. All right, so let's check it out. So here it says, a light bar, which is this gray horizontal bar, is free to rotate about a perpendicular access through its center. So the bar here's the axis of rotation. The bar can spend either this way or this way, but the middle is fixed, right? Um and then it says the bar is not attached, so it is free to move horizontally, vertically. What that means is, for example, in this first situation, 32 forces air pushing this thing up. So the bar, if the bar has two forces pulling it up and it's not fixed in the middle, the bar would actually do this. Okay, It's only fixed in that it could Onley rotate in the middle, but somehow it could actually move up and down. All right, All these little arrows have the same magnitude. And if you see a double arrow, which we see here, um, this just means double the magnitude. Okay, we wanna know. Do we have linear equilibrium in rotation equilibrium? So check this out here. We don't have linear equilibrium because both of these forces mean that the net force will be going up. Okay, The forces air both pushing up, so the bar has to move up. However, we do have rotational equilibrium. And that's because this force here causes the torque. This way. I'm gonna call this Torque one in this force caused the torque this way. Torque to, um, those two torques air going opposite direction. This one is counter clockwise. Positive. This one is clockwise negative. And they have the same magnitude. The reason we have the same magnitude. Let me bring back the torque equation. Is torque equals f R sine of data? The forces are the same in both sides. Notice that they're both the same distance from the axis of rotation R one and R two. Okay, they're both too little sticks away. And they both make an angle of 90 degrees. So same for same distance. Same angle. The torques are the same. They will cancel each other house. What about here? Here you have two forces. One pushing up, the other one pushing down. They will cancel each other and we will have linear equilibrium. However, we're not going to have rotation equilibrium. Because both of these cause a talk about the middle that has the same direction. So this is torque one, which is counterclockwise negative and torque to which is counter clockwise, negative as well. So there will be a net torque. There will be a net torque. That is negative. Here. There was a net force that is positive going up. All right, so what about here? Here again? We have two forces cancel each other, so we have linear equilibrium, But we're not going to have rotation of equilibrium. We didn't have rotation equilibrium here. We're not gonna have rotation equilibrium here. Why? Well, because even though these forces air going in different directions, the torques are difference in magnitude. Check it out. So this guy is going this way. Torque one. This is counterclockwise, so it's positive. And this guy is producing a torque. That is, um that is I'm sorry. This guy's actually to the left of the dot so the torque will be this way. Talk of two is negative. Okay, so if you imagine a bar, right, if you push this way, it's going to be positive. And if you push this way, it's going to be negative rotation, but t one is farther away. So our one is greater than our two. Here's our one. And here's are to therefore Tark. One has a greater magnitude than torque to so talk one winds in the bar ends up spinning this way. So no rotational equilibrium. What about here? So I can say that there's two forces up and two forces down, so the forces will cancel each other out. And I have a linear equilibrium. What about rotational equilibrium? Well, guy, let's call this 11234 I hope you'll see that one in four will cancel. Um, Torque one goes this way, which is negative. Torque four goes this way, which is positive. They're both the same distance from the axis, Same angles, everything. So these to cancel these two guys also are opposite to each other, but they're gonna have the same magnitude because the same distance so torque to looks like this again, you're to the left of the axis pushing it down. So it's gonna cause it to spin like this, and Torque three is gonna go the other way. So I hope you see how three one in four canceled, and two and three cancel each other as well. So here, actually both rotational in linear equilibrium. What about this problem here? Here, the two forces cancel each other. I have linear equilibrium. Um, what about rotation equilibrium? I only have one on each side. They're both causing torques in different directions. One to torque one Talk to torque. One is clockwise negative. Talking to his counterclockwise. Positive. They perfectly cancel each other out because the ours are the same. They're pushing at the same distance from the axis of rotation. I have both. Equilibrium is here is Well, what about here? Here, All my forces air going up. So I will have a net force that is going up. Let's call that positive. So there is no linear equilibrium. But I do have rotational equilibrium because here I have Let's call this distance here two. And let's call this distance one. You can see how this distance is double. So the first talk over here torque one, which is these two arrows talk One would be two f are and talk to would be, um f to our right. So the guy on the left has double the force. But the guy on the right has doubled the distance. So these two end up canceling each other. Okay? They end up canceling each other, and I have rotation equilibrium as well. All right, so this is just a bunch of Let me get out of the way here. Um, this is a bunch of difference. Uh, situations for you, toe. Sort of get an understanding of when you would have linear and when you have rotational equilibrium. Alright. So that's it for this one. Let's keep going.

2

example

Balancing a bar with a force

Video duration:

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

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?

A

300 N

B

667 N

C

60 N

D

167 N

4

example

Pin holding a horizontal bar

Video duration:

2m

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Hey, guys. So here in this example, we're trying to balance a sort of a shelf on a wall by using a pin. So let's check it out. So that the red pin here Andi were using it to try to balance the bar that has a mass of 20. So am equals 20 in a length of 3 m and it's held horizontally against the wall by the pen right here. Okay. The idea is that there's going to be an mg right in the middle that's pulling the bar down. Um, this is the axis of rotation somewhere over here. Axis. Okay. And this MGI is producing a torque that would cause this to spin. But the pen, 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. Okay, so 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. Um, it's not gonna rotates. Not gonna have angular acceleration. The torque of M. G is this way, which is negative. So you'd want the torque of the pin to be positive so that they cancel. Okay, so we're gonna write torque of M G Negative plus torque of pin positive equals zero. This gives us that I'm going to send the negative to the other side. Torque of pin equals torque of N G. There should make sense. All we're doing is getting these two guys to cancel each other out. And what we're looking for here is what is the torque on the pin that's needed and the way we're going to calculate the torque on the pin is just by calculating the torque of M G. Okay, so torque of pin will be m g gonna expand the right side. Our sign of theta Now the masses 20 gravity 9.8 r is the distance from the axis of rotation to the point where the force happens. We have a uniformed mass distribution, which means mg happens in the middle of the bar. The bars 3 m long. So the our vector is 1.5. Put a 1.5 here and the angle that these two guys make is degrees. So sign of 90 is one. Okay. And if we multiply this whole thing. We get 294 Newton Meter, and that's it. That's the answer here. Okay, So very straightforward question for us to see to calculate one torque based on some other information, Right? So that the finish one. Let me Do you have any questions? Let's keep going.

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