Torque & Equilibrium - Online Tutor, Practice Problems & Exam Prep

In mechanics, equilibrium involves both linear and rotational aspects. Linear equilibrium occurs when the net force is zero, resulting in no acceleration. Rotational equilibrium requires the sum of all torques to also be zero, preventing angular acceleration. For complete equilibrium, both conditions must be satisfied. Understanding these principles is crucial for analyzing rigid bodies, as various force and torque configurations can lead to different states of equilibrium, influencing stability and motion.

Torque & Equilibrium

Hey guys. So you may remember back when we talked about forces that we mentioned how if the forces on 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 going to call rotational equilibrium. Let’s check it out. So remember, if the net force on an object or the sum of all forces on an object is 0, then the acceleration on that object is 0. And that’s because of Newton’s second law, 𝐏 = 𝐉𝐰. If the net force is 0, it means the sum of all forces equals 0. And therefore, the acceleration has to be 0. This situation is called equilibrium. But now that we know about torques, things are not going to be that simple.

Okay? Sometimes this condition here of the net force being 0 is not sufficient for equilibrium. It was before we knew about torques. So here’s an example. You can imagine you have a continuous bar that has a uniform mass distribution. So the bar’s center of mass happens right here. And at this point is where the weight force acts. There’s a little fulcrum here that holds the bar up, and this is going to push back with a force of the normal force. And this normal force may even be equal to 𝐱𝐫. The problem is that even though the forces will cancel on the vertical, this 𝐱𝐫 produces a torque here, torque of 𝐱𝐫. 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, 𝐻 = 0. It acts on the axis of rotation. So the net torque will be this.

So the net torque will be equal to this. And because I have a net torque, I will have an 𝐱𝐻. So even though the forces cancel, the torques don’t cancel. So what happens is that this thing would tip over this way and fall. Okay? So we’re going to have a situation where we don’t have equilibrium. Alright? This brings us to the fact that there has to be actually, the first condition is that the sum of all forces must be 0. And this gives us an acceleration of 0. And this is good old equilibrium, but now we’re going to call it linear equilibrium, because there are going to be 2 types.

Okay. The second condition is that the sum of all torques also has to be 0. And this will give us an 𝐱𝐻 that is 0. And we’re going to call this rotational equilibrium. And if we have both of them, we’re going to have what I call complete equilibrium. Okay. This topic is in most textbooks referred to as static equilibrium. And static refers to the fact that you’re going to be in a situation where there’s no velocity, no linear velocity, no angular velocity. So not only do you have complete equilibrium, but also the object is not moving. Okay?

This is sometimes called also the equilibrium of rigid bodies because we’re going to deal with rigid bodies exclusively. So we’re always going to have these extended objects rather than a point mass. Okay? Let’s do I want to do a sort of an introductory example here to talk about different situations where you may have one type of equilibrium but not the other or both or neither. Alright? 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 axis through its center. So the bar, here’s the axis of rotation. The bar can spin either this way or that way, but the middle is fixed. Right? 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, the 2 forces are pushing this thing up. So the bar if the bar has 2 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 only rotate in the middle, but somehow it could actually move up and down. Alright.

All these little arrows have the same magnitude and if you see a double arrow, which we see here, this just means double the magnitude. Okay. We want to know, do we have linear equilibrium and rotational 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 are both pushing up so the bar has to move up. However, we do have rotational equilibrium and that’s because this force here causes a torque this way. I’m going to call this torque 1. And this force causes a torque this way, torque 2. Those two torques are going the opposite direction. This one is counterclockwise positive. This one is clockwise negative, and they have the same magnitude. The reason they have the same magnitude, let me bring back the torque equation, is torque equals 𝐱𝐷 sin of Ƹ͸. The forces are the same on both sides. Notice that they’re both the same distance from the axis of rotation, 𝐻1 and 𝐻2. Okay. They’re both two little sticks away and they both make an angle of 90 degrees. So same force, same distance, same angle, the torques are the same. They will cancel each other out.

What about here? Here you have two forces, one pushing up, the other one pushing down. They will cancel each other out and we will have linear equilibrium. However, we’re not going to have rotational equilibrium because both of these cause a torque about the middle that has the same direction. So this is torque 1 which is counterclockwise negative and torque 2 which is counterclockwise 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. Alright.

So what about here? Here again, we have two forces canceling each other. So we have linear equilibrium, but we’re not going to have rotational equilibrium. We didn’t have rotational equilibrium here. We’re not going to have rotational equilibrium here. Why? Well, because even though these forces are going in different directions, the torques are different in magnitude. Check it out. So this guy is going this way, torque 1. This is counterclockwise, so it’s positive. And this guy is producing a torque that is, that is I’m sorry, this guy is actually to the left of the dot so the torque will be this way. Torque 2 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 𝐻1 is farther away, so 𝐻1 is greater than 𝐻2. Here’s 𝐻1 and here’s 𝐻2. Therefore, torque 1 has a greater magnitude than torque 2. So torque 1 wins and the bar ends up spinning this way. So no rotational equilibrium. What about here? So I can say that there are 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 1, 1, 2, 3, 4. I hope you see that 1 and 4 will cancel. Torque 1:right_align: goes this way which is negative. Torque 4 goes this way which is positive. They’re both the same distance, from the axis, same angles, everything so these two cancel. These two guys also are opposite to each other but they’re going to have the same magnitude because it’s the same distance. So torque 2 looks like this. Again, you’re to the left of the axis pushing it down so it’s going to cause it to spin like this and torque 3 is going to go the other way. So I hope you see how 1 and 4 cancel and 2 and 3 cancel each other as well. So here, I actually have both rotational and linear equilibrium.

What about this problem here? Here, the two forces cancel each other. I have a linear equilibrium. What about rotational equilibrium? I only have one on each side. They’re both causing torques in different directions. 1, 2, torque 1, torque 2. Torque 1 is clockwise negative. Torque 2 is counterclockwise positive. They perfectly cancel each other out because the 𝐻s are the same. They’re pushing at the same distance from the axis of rotation. I have both equilibriums here as well. What about here? Here, all my forces are going up. So I will have a net force that is going up. 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 2 and let’s call this distance 1. Okay. You can see how this distance is double. So the first torque over here, torque 1, which is these two arrows, torque 1 would be 𝐱 𝐷 and torque 2 would be 𝐲𝐷. Right? So the guy on the left has doubled 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 rotational equilibrium as well. All right? So this is just a bunch of let me get out of the way here. This is a bunch of different situations for you to sort of get an understanding of when you would have linear and when you have rotational equilibrium.

Balancing a bar with a force

Hey guys. So 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. And 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. So this here is a distance of 1 meter, to the end. Okay? And you want to push straight down on the left edge, so this is you with a 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 does the fulcrum apply on the bar? Well, if the bar is rested on top of the fulcrum, the fulcrum is going to 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 the force? Well, 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. So we want to have rotational equilibrium, which means the sum of all torques will be 0. Okay. There are 2 torques that are going to act here. 1, 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. So the torque of \( f \) is going to be counterclockwise, positive. The normal force acts at the axis of rotation. Therefore, it produces no torque. The torque of normal would be \( \text{normal} \times r \times \sin(\theta) \) but \( r \) is 0 because the force acts on the axis of rotation, so the whole thing is 0. So really, what you have is these 2 guys. So I can do this. I can say, the torque of \( f \) plus negative torque of \( mg \) equals 0. And if I send this to the other side, I get the torque of \( f \) equals the torque of \( mg \). And this should make a ton of sense. The torques this basically just says that the torques are going in opposite directions or canceling each other out. So the next thing you do is you expand these two sides. So the torque of \( f \) is going to be \( f \times r_f \times \sin(\theta_f) \), and on the right side, I have \( mg \times r_{mg} \times \sin(\theta_{mg}) \). We're looking for \( f \). So let's plug in everything here. The distance, the \( r \) 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 \( r_f \) which is 1. And the angle between \( f \) and \( r \) is 90 degrees. Okay. So this is \( r \), the \( r \) vector for \( f \), and 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. \( \sin(90) \) is 1. \( mg \), 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 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. Okay? So 1 meter from, it's 2 meters from the left to the center of the mass. It's in the middle. But it's 1 meter from the left to the fulcrum, so you got another meter here. And this is the distance for \( r_{mg} \). Okay. So \( r_{mg} \) is this, which is 1 meter, and \( r_f \) is this, which is 1 meter as well. So I'm going to put 1 here and the sine will be \( \sin(\theta) \) 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 \( f = 98 \) newtons. \( f = 98 \) newtons. Cool. 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. And then for part b, very quickly, to find the normal force, we have to use the fact that the sum of all forces is 0 on the y-axis. Right? And if you look at all the forces, all the forces, there are 2 forces going down, which is \( f \) plus \( mg \). They're both going down. And then there's one force going up which is normal and they all equal to 0. So I can say that normal equals \( f + mg \). 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 newtons. Okay? So this is how much force you would need to keep this thing balanced, and this is how much you get, as a result of doing that. That's how much normal force you have as a result. Okay? So that's it for this one. Let me know if you have any questions.

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

Pin holding a horizontal bar

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. We got the red pin here, and we're using it to try to balance the bar that has a mass of 20. So m equals 20 in a length of 3 meters. And it's held horizontally against the wall by the pin right here. Okay? 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, axis. Okay. And 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. Okay. So here, if we're holding it up, we're going to say that the sum of all torques equals 0 because the bar is not going to accelerate. It's not gonna rotate. It's not 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. Okay. So we're going to write torque of mg negative plus torque of pin positive equals 0. This gives us that I'm going to send the negative to the other side. Torque of pin equals torque of mg. This should make sense. All we're doing is getting these 2 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 of the pin is just by calculating the torque of mg. Okay?

Torque of pin will be mg, I'm going to expand the right side, r sine of Theta. Now 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. And the angle that these two guys make is 90 degrees. So sine of 90 is 1. 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. Alright? So it's it for this one. Let me know if you have any questions and let's keep going.

T = m g r ⁢ sin ⁡ ( θ )

Where:

• m = 20 kg (mass of the bar)
• g = 9.8 m/s² (acceleration due to gravity)
• r = 1.5 m (distance from the axis to the point where the force is applied)
• θ = 90° (angle between r and the force)