Professor Anderson

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>> And it's the old box with pulley problem. So the sort of classic physics problem dealing with pulleys is the following. Let's take a pulley and we'll hang it off the end of the table. We'll put a box on the table, M1. We'll put a box over here, M2. And we'll tie those two together with a rope that goes over the pulley. And let's assume that this surface here is frictionless. And we'll assume that the pulley is lightweight and frictionless as well. And the question is, what is the acceleration, A? So, to figure out the acceleration, A, we have to think a little bit more about the system. If this box is accelerating to the right, with A -- What is this block doing? [ Inaudible Answer ] Accelerating downward with A. OK? And this is called an acceleration constraint. They are tied together. They both have the same acceleration -- different directions, of course, but they both have the same magnitude of acceleration. OK? If M1 accelerates to the right, that rope tells us that M2 is going to accelerate down with the same acceleration, A. Alright, so now what do I do? I've got my block, M1. I've got my block M2. Nice looking picture. I now have to draw a free body diagram. So let's identify all the forces that are acting on M1. What's acting on M1? What's the first thing we always draw? Gravity. OK? What else is acting on M1? Normal force. What else? OK, there's something pulling on it, namely this rope. And if we say that the rope has tension T, tension T is a force to the right. And that's it, right? That's no friction, so that's all of the forces that are acting on M1. So now you can write down Newton's second law for M1. The sum of the forces in the X direction equals the mass times acceleration in the X direction. We only have one force. It's the tension. The acceleration, we said, is just A. OK? So what's the tension in the rope? It's MA, and this is M1 times A. What about the vertical components? Is that really going to help us here? Well maybe not, but let's write it down anyway. The sum of the forces is equal to the mass times acceleration in the vertical direction. This is normal force up, M1G, down. And that equals zero because that block is not accelerating up or down. And so we just get normal force equals M1G. Alright, what about M2? M2, it looks like we only have vertical forces to worry about. And we know that M2G is going down. What's going up? Tension T. How do we know? Newton's third law tells us that that tension is the action-reaction pair with that tension. The rope transmits tensions everywhere throughout it. And so we have tension T going up. And now we just have one simple equation for this. We have the sum of the forces in the Y direction equals the mass times acceleration in the Y direction. But we know that it's going to accelerate down. So let's say down is our positive direction. Pick down as our positive direction. If you picked up as the positive direction, you would just have to put a minus sign on this A. And this is what we mean by the acceleration constraint. If we're going to solve for the same A, we had better get them going in the same direction. OK. So what do we get? We get T minus M2G equals mass number two times acceleration. But we need to reverse those because we said down was positive. And so you make it minus T plus M2G. G and A had better be in the same direction, so they better have the same sign. Alright, so now can we solve this thing for A? Yeah, it doesn't look too bad. Alright, we've got a nice little equation here that says minus T plus M2G equals M2A. And I can take T from that equation and put it right in here. So I get minus M1A plus M2G. All that equals M2A. Pink on pink is kind of hard to see, isn't it? Alright, step to the side and do it like that. OK. And now we just have to solve for A. So if I move this to the other side, I get M2G equals -- what, that becomes a positive when I move it to the other side. I can combine some terms. And now I get an answer for A. A is equal to M2G divided by M1 plus M2. Alright. Picture. Free body diagram. Math. Sorry. F equals MA. Math. And now what was step five on our list? Anybody remember? Step five was double check. And what do we have to double check? We have to double check the units and make sure they work out. This is why we keep all the variables for the end. M on the top is going to cancel with M on the bottom. Kilograms goes away, and we're just going to end up with units of G, which is of course meters per second squared. And we like that for acceleration, so that's good. What are the limits? What limits should we look at in this answer? What do you guys think? Limits always mean let something go to zero or let something go to infinity. What should we let go to zero? How about M2? If M2 is zero, should this block move at all? If I just have a block sitting on a table and I don't hang anything from it, M1 shouldn't move at all. Is that what happens? Let's see. M2, we've got one in the denominator. But we still have an M1 in the denominator. And then we have an M2 up in the numerator. And so, yes. Acceleration will go to zero if M2 equals zero. If M1 equals zero, there is no mass up here. How fast should this block fall towards the ground? Gravity, right? G. Is that what we get? M1 goes away. I've got M2 over M2, which is one. Multiply that by G and I get gravity. Why isn't it negative G? Because we already said that for this block, positive is down. OK, questions on that one?

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>> And it's the old box with pulley problem. So the sort of classic physics problem dealing with pulleys is the following. Let's take a pulley and we'll hang it off the end of the table. We'll put a box on the table, M1. We'll put a box over here, M2. And we'll tie those two together with a rope that goes over the pulley. And let's assume that this surface here is frictionless. And we'll assume that the pulley is lightweight and frictionless as well. And the question is, what is the acceleration, A? So, to figure out the acceleration, A, we have to think a little bit more about the system. If this box is accelerating to the right, with A -- What is this block doing? [ Inaudible Answer ] Accelerating downward with A. OK? And this is called an acceleration constraint. They are tied together. They both have the same acceleration -- different directions, of course, but they both have the same magnitude of acceleration. OK? If M1 accelerates to the right, that rope tells us that M2 is going to accelerate down with the same acceleration, A. Alright, so now what do I do? I've got my block, M1. I've got my block M2. Nice looking picture. I now have to draw a free body diagram. So let's identify all the forces that are acting on M1. What's acting on M1? What's the first thing we always draw? Gravity. OK? What else is acting on M1? Normal force. What else? OK, there's something pulling on it, namely this rope. And if we say that the rope has tension T, tension T is a force to the right. And that's it, right? That's no friction, so that's all of the forces that are acting on M1. So now you can write down Newton's second law for M1. The sum of the forces in the X direction equals the mass times acceleration in the X direction. We only have one force. It's the tension. The acceleration, we said, is just A. OK? So what's the tension in the rope? It's MA, and this is M1 times A. What about the vertical components? Is that really going to help us here? Well maybe not, but let's write it down anyway. The sum of the forces is equal to the mass times acceleration in the vertical direction. This is normal force up, M1G, down. And that equals zero because that block is not accelerating up or down. And so we just get normal force equals M1G. Alright, what about M2? M2, it looks like we only have vertical forces to worry about. And we know that M2G is going down. What's going up? Tension T. How do we know? Newton's third law tells us that that tension is the action-reaction pair with that tension. The rope transmits tensions everywhere throughout it. And so we have tension T going up. And now we just have one simple equation for this. We have the sum of the forces in the Y direction equals the mass times acceleration in the Y direction. But we know that it's going to accelerate down. So let's say down is our positive direction. Pick down as our positive direction. If you picked up as the positive direction, you would just have to put a minus sign on this A. And this is what we mean by the acceleration constraint. If we're going to solve for the same A, we had better get them going in the same direction. OK. So what do we get? We get T minus M2G equals mass number two times acceleration. But we need to reverse those because we said down was positive. And so you make it minus T plus M2G. G and A had better be in the same direction, so they better have the same sign. Alright, so now can we solve this thing for A? Yeah, it doesn't look too bad. Alright, we've got a nice little equation here that says minus T plus M2G equals M2A. And I can take T from that equation and put it right in here. So I get minus M1A plus M2G. All that equals M2A. Pink on pink is kind of hard to see, isn't it? Alright, step to the side and do it like that. OK. And now we just have to solve for A. So if I move this to the other side, I get M2G equals -- what, that becomes a positive when I move it to the other side. I can combine some terms. And now I get an answer for A. A is equal to M2G divided by M1 plus M2. Alright. Picture. Free body diagram. Math. Sorry. F equals MA. Math. And now what was step five on our list? Anybody remember? Step five was double check. And what do we have to double check? We have to double check the units and make sure they work out. This is why we keep all the variables for the end. M on the top is going to cancel with M on the bottom. Kilograms goes away, and we're just going to end up with units of G, which is of course meters per second squared. And we like that for acceleration, so that's good. What are the limits? What limits should we look at in this answer? What do you guys think? Limits always mean let something go to zero or let something go to infinity. What should we let go to zero? How about M2? If M2 is zero, should this block move at all? If I just have a block sitting on a table and I don't hang anything from it, M1 shouldn't move at all. Is that what happens? Let's see. M2, we've got one in the denominator. But we still have an M1 in the denominator. And then we have an M2 up in the numerator. And so, yes. Acceleration will go to zero if M2 equals zero. If M1 equals zero, there is no mass up here. How fast should this block fall towards the ground? Gravity, right? G. Is that what we get? M1 goes away. I've got M2 over M2, which is one. Multiply that by G and I get gravity. Why isn't it negative G? Because we already said that for this block, positive is down. OK, questions on that one?