Sections | |||
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Torque & Equilibrium | 23 mins | 0 completed | Learn |

Review: Center of Mass | 14 mins | 0 completed | Learn |

Equilibrium with Multiple Objects | 30 mins | 0 completed | Learn |

Equilibrium with Multiple Supports | 31 mins | 0 completed | Learn |

Center of Mass & Simple Balance | 31 mins | 0 completed | Learn |

Equilibrium in 2D - Ladder Problems | 40 mins | 0 completed | Learn |

Beam / Shelf Against a Wall | 53 mins | 0 completed | Learn |

More 2D Equilibrium Problems | 29 mins | 0 completed | Learn |

Concept #1: Torque & Equilibrium

**Transcript**

Hey guys! 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. 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. 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. That's because of Newton's second law, F = ma. If the net force is zero, it means the sum 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 going to be that simple. It's going to be a little more complicated. Sometimes this condition here of the net force being zero is not sufficient for equilibrium. It was before we knew about torques. Here's an example. Imagine you have a bar that has uniform mass distribution, so the barÕs center of mass happens right here. 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 normal. This normal may even be equal to mg. The problem is that even though the forces will cancel, the vertical, this mg produces a torque here, torque of mg. This is clockwise so it's negative but there is no torque due to the normal force. That's because the normal force acts on the axis of rotation, r = 0. It acts on the axis of rotation. The net torque will be this. The net torque will be equal to this. Because I have a net torque, I will have an _. Even though the forces cancel, the torques don't cancel so what happens is that this thing would tip over this way and fall. We're going to have a situation where we don't have equilibrium. This brings us to the fact that there are two conditions that are necessary for an object to have what I'm going to call complete equilibrium. It's not enough that the forces cancel. The first condition is that the sum of all forces must be 0. 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 two types. The second condition is that the sum of all torques also has to be 0, and this will give us an _ that is 0 and we're going to call this rotational equilibrium. If we have both of them, we're going to have what I call complete equilibrium. This topic is most textbooks refer to as static equilibrium. Static refers to the fact that you're going to be in a situation where there's no linear velocity and no angular velocity. Not only do you have complete equilibrium, but also the object is not moving. This is sometimes called also the equilibrium of rigid bodies because we're going to deal with rigid bodies exclusively. We're always going to have these extended objects rather than a point mass. I want to do 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. Let's check it out. Here it says a light bar which is this gray horizontal bar is free to rotate about a perpendicular axis through its center. Here's the axis of rotation. The bar can spin either this way or this way, but the middle is fixed. 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 two forces are pushing this thing up. If the bar has two forces pulling it up and it's not fixed in the middle, the bar would actually do this. It's only fixed in that it could only rotate in the middle but somehow it could actually move up and down. All these little arrows have the same magnitude. If you see a double arrow which we see here, this just means double the magnitude. We want to know do we have linear equilibrium and rotational equilibrium. Check this out. Here we don't have linear equilibrium because both of these forces mean that the net force will be going up. 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 the torque this way, Torque 2. Those two torques are going in 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 = Frsin_. The forces are the same in both sides. Notice that they're both the same distance from the axis of rotation, r1 and r2. They're both two little sticks away and they both make an angle of 90 degrees. Same force, same distance, same angle Ð 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 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. This is Torque 1 which is counterclockwise negative and Torque 2 which is counterclockwise negative as well. There will be a net torque that is negative. Here there was a net force that is positive going up. What about here? Here, again, we have two forces cancel 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? Because even though these forces are going in different directions, the torques are different in magnitude. Check it out. This guy is going this way, Torque 1. This is counterclockwise so it's positive. This guy is producing a torque actually to the left of the dot so the torque will be this way. Torque of 2 is negative. If you imagine a bar, if you push this way, it's going to be positive. If you push this way, it's going to be negative rotation. But T1 is farther away, so r1 is greater than r2. Here's r1 and here's r2, therefore Torque 1 has a greater magnitude than Torque 2. Torque 1 wins and the bar ends up spinning this way, so no rotational equilibrium. What about here? I can say that there's two forces up and two forces down. The forces will cancel each other out and I have a linear equilibrium. What about rotational equilibrium? Let's call this one 1, 2, 3, 4. I hope you see that 1 and 4 will cancel. Torque 1 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 of the same distance. 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. 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. Here actually we 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 one is clockwise negative. Torque 2 is counterclockwise positive. They perfectly canceled each other out because the rÕs 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. Let's call that positive. 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. You can see how this distance is double. The first torque over here, Torque 1 which is these two arrows. Torque 1 would be (2F)r and Torque 2 would be F(2r). The guy on the left has double the force, but the guy on the right has doubled the distance so these two end up cancelling each other. They end up cancelling each other and I have rotational equilibrium as well alright. This is just 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. That's it for this one. Let's keep going.

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.

0 of 4 completed

Two objects of masses 28 kg and 21 kg are hung from the ends of a stick that is 70 cm long and has marks every 10 cm, as shown.
If the mass of the stick is negligible, at which of the points indicated should a cord be attached if the stick is to remain horizontal when suspended from the cord?
1. B
2. D
3. G
4. F
5. C

A ball of radius r with a wire glued to one spot on its surface can be pulled along the floor and will slide without any tendency to roll only if the wire is horizontal and is a distance h above the floor. In terms of F, the magnitude of the applied force, fk, the magnitude of the force of kinetic friction, and the radius r of the ball, what is h?
1. h = r (f k − F)
2. h = 0; the wire must run along the floor.
3. h = r (f k/F)
4. h = r
5. h = r (1 + F/f k)
6. h = r (F − f k)
7. h = r (F/f k)
8. h = r (1 − f k/F)
9. h = r (1 + f k/F)
10. h = r (1 −F/f k)

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