Anderson Video - Pulley with Two Masses

Professor Anderson
17 views
Was this helpful ?
0
>> This is the pulley problem, and the idea is the following. We're going to have a pulley hanging from the roof, and we're going to take a block, M1, and a block, M2, and we're going to tie them together with a rope. And let's ask the following question; if block M2 is H above the ground, how long will it take for M2 to hit the ground? >> In other words, we're looking for a time. How long will it take for M2 to hit the ground? So when we approach these problems, it's always good to just really think physically about the problem first. So if M2 is equal to M1, is M2 ever going to hit the ground? No, it's just going to hang there, right? If these things are starting from rest, they're just going to hang there. M2 is never going to hit the ground. So when we're all done, hopefully, that's one of the limits of the solution. The other is if M1 is basically 0, then M2 should be in freefall. Right? It should be accelerating at G. So let's make sure that that is one of the limits of our solution, as well. Okay, how do we figure this out? Well, we've got the free-body diagram for M1, and we know what forces are acting on it. M1G, tension T in the rope. We've got the free-body diagram for M2. We know what forces are acting on it, M2G, tension T in the rope. In this problem, the action/reaction pair is that tension T. Okay, now there's a little bit of a caveat, which is in these problems that we are attacking right now, the pulley is massless and frictionless, and so the tension, T, is everywhere the same throughout the rope, but later on, we're going to get to pulleys that have mass and can grip the rope, and in that case, the tension, T, will not be the same on either side of the rope. Okay, you can have very different tensions on either side of the rope if the pulley itself is gripping the rope. Okay? All right, how do we attack this? Well, first off, we should probably pick an acceleration constraint, which looks like this. A going up there corresponds to A going down there. We know that M2 had better be bigger than M1 if it's going to fall, and so the whole system will do that. So now we just go back to Newton's Second Law. We only have to worry about the Y direction. So we have some of the forces in the Y direction for M1. We get T minus M1G equals M1 times A. For number two, we also have the vertical direction, but remember, we're going to flip it over and we're going to say that positive is down for that one, and so the left side of this equation becomes M2G minus T equals M2A. And this is what we mean by this acceleration constraint. If this one is falling, that one is going up. So pick the forces that are in the same direction as a and make those positive. The ones that are opposite, make them negative. When you're faced with this now, how do we solve this for A? Well, it doesn't look too bad, and one clever little trick that you can do is you can just add the two equations. Take this block and add it to this block, and look what happens if I do that. I get T minus M1G plus M2G minus T. That's the left side of those equations, is going to equal M1A plus M2A. Anytime you have equations on top of each other, you can always just add them together, or you can always just subtract them together, if you just keep both sides separate. Add the left sides. Add the right sides. And now you see what happens, the T there drops out with the T there. And so we get M2 minus M1 times G equals M1 plus M2 times A. And now we can solve this for A. A is equal to M2 minus M1 over M1 plus M2, all of that times G. When we get an answer like this, we've got to go back to the double check part. That was our fifth step. Do the units work out? Well, kilograms on top is going to cancel with kilograms on bottom, and so we're going to end up with units of G, which is, of course, meters per second square, which is what we want for A. So that makes sense. What about the limits? Initially, we said if M2 equals M1, then this thing should not move. And if M2 equals M1, you see that we do get A is equal to 0. Good. We also said that if we don't hang anything on M1, then this thing should be in freefall. Is that what happens? We get 0 up there, and we get 0 down there, and so we get M2 over M2, which is 1, and we get acceleration is G. This thing will accelerate at G. So both of those limits make sense. Now the question was, the original question was, how long does it take to hit the ground? So to do that, we have to go back to the kinematic equations. I know, you thought you could forget about them, right? Awful. What kinematic equation should we use to solve for this? One of my favorites, hopefully it's your favorite, is this. What does this become? It becomes H equals 1/2 AT squared. We can solve this now for T. T is 2h over A, and we're going to take the square root of that whole thing, and this is A. So now if you have the masses, you have the height, can calculate A. You can plug it into here, and you can calculate exactly how long it takes to hit the ground, and you can just verify that if those masses were equal, it would never hit the ground, right? Because A would be 0, and this thing will blow up to infinity. If mass 1 was 0, then we put a G down there, and this becomes good ol' free fall. How long does it take to hit the ground? Square root of 2H over G. All right, questions about that one?
>> This is the pulley problem, and the idea is the following. We're going to have a pulley hanging from the roof, and we're going to take a block, M1, and a block, M2, and we're going to tie them together with a rope. And let's ask the following question; if block M2 is H above the ground, how long will it take for M2 to hit the ground? >> In other words, we're looking for a time. How long will it take for M2 to hit the ground? So when we approach these problems, it's always good to just really think physically about the problem first. So if M2 is equal to M1, is M2 ever going to hit the ground? No, it's just going to hang there, right? If these things are starting from rest, they're just going to hang there. M2 is never going to hit the ground. So when we're all done, hopefully, that's one of the limits of the solution. The other is if M1 is basically 0, then M2 should be in freefall. Right? It should be accelerating at G. So let's make sure that that is one of the limits of our solution, as well. Okay, how do we figure this out? Well, we've got the free-body diagram for M1, and we know what forces are acting on it. M1G, tension T in the rope. We've got the free-body diagram for M2. We know what forces are acting on it, M2G, tension T in the rope. In this problem, the action/reaction pair is that tension T. Okay, now there's a little bit of a caveat, which is in these problems that we are attacking right now, the pulley is massless and frictionless, and so the tension, T, is everywhere the same throughout the rope, but later on, we're going to get to pulleys that have mass and can grip the rope, and in that case, the tension, T, will not be the same on either side of the rope. Okay, you can have very different tensions on either side of the rope if the pulley itself is gripping the rope. Okay? All right, how do we attack this? Well, first off, we should probably pick an acceleration constraint, which looks like this. A going up there corresponds to A going down there. We know that M2 had better be bigger than M1 if it's going to fall, and so the whole system will do that. So now we just go back to Newton's Second Law. We only have to worry about the Y direction. So we have some of the forces in the Y direction for M1. We get T minus M1G equals M1 times A. For number two, we also have the vertical direction, but remember, we're going to flip it over and we're going to say that positive is down for that one, and so the left side of this equation becomes M2G minus T equals M2A. And this is what we mean by this acceleration constraint. If this one is falling, that one is going up. So pick the forces that are in the same direction as a and make those positive. The ones that are opposite, make them negative. When you're faced with this now, how do we solve this for A? Well, it doesn't look too bad, and one clever little trick that you can do is you can just add the two equations. Take this block and add it to this block, and look what happens if I do that. I get T minus M1G plus M2G minus T. That's the left side of those equations, is going to equal M1A plus M2A. Anytime you have equations on top of each other, you can always just add them together, or you can always just subtract them together, if you just keep both sides separate. Add the left sides. Add the right sides. And now you see what happens, the T there drops out with the T there. And so we get M2 minus M1 times G equals M1 plus M2 times A. And now we can solve this for A. A is equal to M2 minus M1 over M1 plus M2, all of that times G. When we get an answer like this, we've got to go back to the double check part. That was our fifth step. Do the units work out? Well, kilograms on top is going to cancel with kilograms on bottom, and so we're going to end up with units of G, which is, of course, meters per second square, which is what we want for A. So that makes sense. What about the limits? Initially, we said if M2 equals M1, then this thing should not move. And if M2 equals M1, you see that we do get A is equal to 0. Good. We also said that if we don't hang anything on M1, then this thing should be in freefall. Is that what happens? We get 0 up there, and we get 0 down there, and so we get M2 over M2, which is 1, and we get acceleration is G. This thing will accelerate at G. So both of those limits make sense. Now the question was, the original question was, how long does it take to hit the ground? So to do that, we have to go back to the kinematic equations. I know, you thought you could forget about them, right? Awful. What kinematic equation should we use to solve for this? One of my favorites, hopefully it's your favorite, is this. What does this become? It becomes H equals 1/2 AT squared. We can solve this now for T. T is 2h over A, and we're going to take the square root of that whole thing, and this is A. So now if you have the masses, you have the height, can calculate A. You can plug it into here, and you can calculate exactly how long it takes to hit the ground, and you can just verify that if those masses were equal, it would never hit the ground, right? Because A would be 0, and this thing will blow up to infinity. If mass 1 was 0, then we put a G down there, and this becomes good ol' free fall. How long does it take to hit the ground? Square root of 2H over G. All right, questions about that one?