- Introduction to torque
- Moments (part 2)
- Finding torque for angled forces
- Rotational version of Newton's second law
- More on moment of inertia
- Rotational inertia
- Rotational kinetic energy
- Rolling without slipping problems
- Angular momentum
- Constant angular momentum when no net torque
- Angular momentum of an extended object
- Ball hits rod angular momentum example
- Cross product and torque
Introduction to moments. Created by Sal Khan.
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- why is F capitalized and D is not?(29 votes)
- Its just the symbols for force it is 'F' and for displacement it is 's' or 'd'(6 votes)
- what is the difference between moment and torque? are the both same ?(31 votes)
- Moment refers to inertia, ie. rotational mass. Depending on the axis of rotation, you have different moments for the same mass. Thus inertia is a tensor quantity.(20 votes)
- At09:55, he said counter-clockwise, drawing an arrow depicting clockwise. Shouldn't it point the other way?(19 votes)
- It's also ok because if he consistently gets all of them backwards, he'll get the right answer again. :)(16 votes)
- Can somebody tell me what exactly is Axis of Rotation?
Thanks in advance.(5 votes)
- It is an imaginary line around which the object is rotating. For instance in a door, the axis of rotation is the line that passes through the hinges. When we speak about the rotation of the Earth, is the line that passes through the Poles.(25 votes)
- does the Clockwise and the Counter clockwise matter that much? I mean, as long as the question or problem doesn't ask the direction, in the end the equation is F1D1=F2D2
now if we put subtracted the D2F2 from both sides, we will come to the original thing...i am kinda confused on this.
can someone help
- Hey Nnnamuujin,
I'm not sure exactly what you are asking, but I do believe that the direction of torque matters. As you watch in this video and Moments (Part 2), you can have CW and CCW forces on both sides of the fulcrum. For the system to be in equilibrium, the clockwise forces must equal the counter-clockwise forces.(7 votes)
- Why did you subract 180 from 48?(0 votes)
- He subtracted 48 from 180 because in the equation 180 = 48 + 10F you want 10F alone on one side, so he did 180 - 48 = 48 + 10F - 48, i.e minus 48 from both sides. This leaves you with 132 = 10F, then you just divide by 10 on both sides to get F = 13.2(21 votes)
- ok, this is kind of a stupid question, but he said that F1d1=F2d2. But that's only when it is in equilibrium, right?(5 votes)
- That's really the same thing as saying that they're still in equilibrium with all three forces. If the system isn't in equilibrium, you can solve for the net torque if you know the angular acceleration it causes as well as the rotational inertia of the object (T=Ia (not the same "a")). Think of this as the direct equivalent of typical translational forces, where F=ma.(2 votes)
- how to calculate the reaction force if you have many forces acting on one object(4 votes)
- You have to use vector analysis to figure out what the net force is.(5 votes)
- A SPHERE cannot roll on a SMOOTH INCLINED SURFACE . Why ?
Please reply soon .
Thank you .(8 votes)
- What is a Pseudo-vector?It is mentioned by Sal Sir at4:16.(4 votes)
- A vector-like object which is invariant under inversion is called a pseudovector, also called an axial vector (as a result of such vectors frequently arising as vectors describing rotation. Hope This Works!(2 votes)
Welcome to the presentation on moments. So just if you were wondering, I have already covered moments. You just may not have recognized it, because I covered it in mechanical advantage and torque. But I do realize that when I covered it in mechanical advantage and torque, I think I maybe over-complicated it. And if anything, I didn't cover some of the most basic moment to force problems that you see in your standards physics class, especially physics classes that aren't focused on calculus or going to make you a mechanical engineer the very next year. So we did that with-- why did I write down the word "mechanical?" Oh yeah, mechanical advantage. If you do a search for mechanical advantage, I cover some things on moments and also on torque. So what is moment of force? Well, it essentially is the same thing as torque. It's just another word for it. And it's essentially force times the distance to your axis of rotation. What do I mean by that? Let me take a simple example. Let's say that I have a pivot point here. Let's say I have some type of seesaw or whatever. There's a seesaw. And let's say that I were to apply some force here and the forces that we care about-- this was the exact same case with torque, because there's essentially the same thing. The forces we care about are the forces that are perpendicular to the distance from our axis of rotation. So, in this case, if we're here, the distance from our axis of rotation is this. That's our distance from our axis of rotation. So we care about a perpendicular force, either a force going up like that or a force going down like that. Let's say I have a force going up like that. Let's call that F, F1, d1. So essentially, the moment of force created by this force is equal to F1 times d1, or the perpendicular force times the moment arm distance. This is the moment arm distance. That's also often called the lever arm, if you're talking about a simple machine, and I think that's the term I used when I did a video on torque: moment arm. And why is this interesting? Well, first of all, this force times distance, or this moment of force, or this torque, if it has nothing balancing it or no offsetting moment or torque, it's going to cause this seesaw in this example to rotate clockwise, right? This whole thing, since it's pivoting here, is going to rotate clockwise. The only way that it's not going to rotate clockwise is if I have something keep-- so right now, this end is going to want go down like that, and the only way that I can keep it from happening is if I exert some upward force here. So let's say that I exert some upward force here that perfectly counterbalances, that keeps this whole seesaw from rotating. F2, and it is a distance d2 away from our axis of rotation, but it's going in a counterclockwise direction, so it wants to go like that. So the Law of Moments essentially tells us, and we learned this when we talked about the net torque, that this force times this distance is equal to this force times this distance. So F1 d1 is equal to F2 d2, or if you subtract this from both sides, you could get F2 d2 minus F1 d1 is equal to 0. And actually, this is how we dealt with it when we talked about torque. Because just the convention with torque is if we have a counterclockwise rotation, it's positive, and this is a counterclockwise rotation in the example that I've drawn here. And if we have a clockwise rotation, it has a negative torque, and that's just the convention we did, and that's because torque is a pseudovector, but I don't want to confuse right now. What you'll see is that these moment problems are actually quite, quite straightforward. So let's do a couple. It always becomes a lot easier when you do a problem, except when you try to erase things with green. So let's say that -- let me plug in real numbers for these values. Let me erase all of this. Let me just erase everything. There you go. All right, let me draw a lever arm again. So what we learned when we learned about torque is that an object won't rotate if the net torque, the sum of all the torques around it, are zero, and we're going to apply essentially that same principle here. So let's do it with masses, because I think that helps explain a lot of things and makes this seesaw example a little bit more tangible. Let's say I have a 5-kilogram mass here, and let's say that gravity is 10 meters per second squared. So what is the downward force here? What is the downward force? It's going to be the mass times acceleration, so it's going to be 50 Newtons. And let's say that the distance, the moment arm distance or the lever arm distance here, let's say that this distance right here is 10 meters. Let's say that I have another mass. Let's say it's a 25 kilogram-- no, that's too much. Let's say it's 10 kilograms. Let's say I have a 10-kilogram mass. And I want to place it some distance d from the fulcrum or from the axis of rotation so that it completely balances this 5-kilogram mass. So how far from the axis of rotation do I put this 10-kilogram mass? This is the distance, right? Because we actually carry the distance to the center of the mass. Well, how much force is this 10-kilogram mass exerting downwards? Well, it's 10 kilograms times 10 meters per second squared, it's 100 Newtons. This is acting what? This is acting clockwise, right? This one's acting clockwise and this one's acting counterclockwise, right? So they are offsetting each other. So we could do it a couple of ways. We could say that 50 Newtons, the moment in the counterclockwise direction, 50 Newtons times 10 meters, in order for this thing to not rotate has to be equal to the moment in the clockwise direction. And so the moment in the clockwise direction is equal to 100 Newtons times some distance, let's call that d, 100 Newtons times d, and then we could just solve for d, right? We get 50 times 10 is 500. 500 Newton-meters is equal to 100 Newtons times d. That's 100. Divide both sides by 100, you get 5 meters is equal to d. So d is equal to 5. That's interesting. And I think this kind of confirms your intuition from playing at the playground that you can put a heavier weight closer to the axis of rotation to offset a light weight that's further away. Or the other way to put it is you could put a light weight further away and you kind of get a mechanical advantage in terms of offsetting the heavier weight. So let's do a more difficult problem. I think the more problems we do here, the more sense everything will make. So let's say that we have a bunch of masses. Actually, let's not do it with masses. Let's just do it with forces because I want to complicate the issue. So this is the pivot. And let's say I have a force here that's 10 Newtons going in the clockwise direction, and let's say it is at-- let's say if this is 0, let's say that this is at minus 8, so this distance is 8, right? Let's say that I have another force going down at 5 Newtons. And let's say that its x-coordinate is minus 6. Let's say I have another force that's going up here, and let's say that it is 50 Newtons. This might get complicated. 50 Newtons, and it's at minus 2, so this distance right here is 2. Let's say that I need to figure out-- and I'm making this up on the fly. Let's say that I have another force here that is 5 Newtons. No, let's make it a weird number, 6 Newtons, and this distance right here is 3 meters. And let's say that I need to figure out what force I need to apply here upwards or downwards-- I actually don't know, because I'm doing this on the fly-- to make sure that this whole thing doesn't rotate. So to make sure this whole thing doesn't rotate, essentially what we have to say is that all of the counterclockwise moments or all of the counter clockwise torques have to offset all of the clockwise torques. And notice, they're not all on the same side. So what are all of the things that are acting in the counterclockwise direction? So counter clockwise is that way, right? So this is acting counterclockwise, this is acting counterclockwise, and that's it, right? So the other ones are clockwise. And we don't know this one. Let's assume for a second. We could assume either way. And if we get a negative, that means it's the opposite. So let's assume that this is a-- all of the clockwise ones I'll do in this dark brown. Let's assume this is clockwise, let's assume that this is clockwise, and let's assume that our mystery force is also clockwise. All of the counterclockwise moments have to offset all the clockwise moments. So what are the counterclockwise moments? Well, this one's counterclockwise, so it's 10 Newtons, 10 times its distance from its moment arm. We said it's 8, because it's at the x-coordinate minus 8 from 0, so it's 10 times 8, plus 50. This is also counterclockwise times 6, 50 times 6, and those are all of our counterclockwise moments and that has to equal the clockwise moments. So clockwise moments, let's see. We have 5 Newtons that's going clockwise times 6. 5 Newtons. Actually, was this 6? No, if this is 6, I must have written some other number here that I can't read now. How far did I say this was? Let's say that this is 2. So that 50, let's say this is 2, it's negative 2, because that's what it looks like. I apologize for confusing you. So what were all the counterclockwise moments? This 10 Newtons times its distance 8, the 50 Newtons times this distance, 2. Don't get confused by the negative. I just kind of said we're in the x-coordinate axis or at minus 8 if this is 0, but it's 8 units away, right? And this 50, its moment arm distance is 2 units. So that has to equal all of the clockwise moments. So the clockwise moments is 5 Newtons times 6. Its distance is 6 and it's 5 Newtons going in the clockwise direction. And then we have plus 6 Newtons times 3, plus 6 times 3. And then we're just assuming, we don't know for sure. Let's say we're applying the force. I should have told you ahead of time so you could do this problem. Let's say that we're applying the force at 10 meters away from our fulcrum arm. So force times 10. So now let's just solve for the force. We get 80 plus 100 is equal to 30 plus 18 plus 10F. We get 180 is equal to 48 plus 10F. What's 180 minus 48? It's 132 is equal to 10F, or we get F is equal to 13.2 Newtons. So we guessed correctly that this is going to be a-- sorry, this is going to be a-- I keep mixing up all of the clockwise and counterclockwise. This is going to be a clockwise force. These were all of the-- sorry, this is going to be a counterclockwise force, right? A clock, this is counterclockwise. Let me label that because I think I said it wrong several times in the video. So these go clockwise. And it's this one and this one. And what were the counterclockwise? These go counterclockwise. So we have to apply a 13.10 Newton force 10 meters away, which will generate 132 Newton-meters moment in the counterclockwise direction, which will perfectly offset all of the other moments, and our lever will not move. Anyway, I might have confused you with all the counterclockwise/clockwise. But just keep in mind that all the moments in one rotational direction have to offset all the moments in the other rotational direction. All a moment is is the force times the distance from the fulcrum arm, so force times distance from fulcrum arm. I'll see you in the next video.