Anderson Video - Two Masses, a Pulley and an Incline

Professor Anderson
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>> Hello, class, Professor Anderson here. Let's take a look at an inclined plane problem and let's make it a little more complicated. Let's add a pulley to this incline plane. And we will hang a mass from this pulley, call that M1. And we will have the pulley tied to another mass right there, M2. And let's add some friction on the plane with coefficient Mu K. All right, we don't know exactly which way this thing is going to accelerate, but let's just pick a direction. And, if we get a negative sign when we're all done, then you know you had your picture wrong and it's actually accelerating the other way. So let's say that M2 is big enough that it's going to pull down that way and the entire thing is going to accelerate down the plane. If block M2 is accelerating down the plane, then block M1 is accelerating up with the exact same acceleration A because they're tied together. All right. So, this is what the picture looks like. Let's say that we are trying to figure out what that acceleration A is, in fact, equal to. So how do we do this? Well, we've got our picture. We have our givens, M1, M2, Mu K and theta. The next step is to draw a free body diagram. So let's draw a free body diagram for particle number 1, M1, what are the forces on it? Well, of course, gravity is pulling down and we have tension T in the cable going up. And that's it for particle 1, okay? What about number 2? Number 2, it's on an incline so, again, we want to use our rotated coordinate system. And in that rotated coordinate system, there is going to be a force down the plane and we know that that force is M2G sine of theta. There is another component of gravity which is into the plane and that is M2G cosine of theta. Again, if you can't remember which is which, look at the limits as theta goes to 0 and that should tell you if you have it right or you have it backwards. What other forces are acting on M2? Well, there is the normal force from the plane. There is also tension, T, because it's tied to the cable. And there is also friction, F sub K. And that's all the forces that are acting on M2. And now, with this picture, we can write down Newton's Second Law and see if we can solve for the acceleration A. So let's make a little room here. And we will attack it. All right, let's look at particle number 1. Sum of the forces, it's all in the Y direction for that guy. And it's equal to tension going up minus M1G going down and all that is going to be equal to the mass times the acceleration. And so now you can write the tension T. Tension T, according to this equation, is just M1A plus M1G. All right we like that one and we'll put a box around it. What about number 2. For number 2, this is number 1, number 2, we have sum of the forces in the X direction. Remember we're in this rotated coordinate system. So the X direction means positive X is down to the right. So we have M2G sine theta, okay. And then we have 2 that are opposing that. We have tension T and we also have frictional force FK. And all of that is equal to the mass times its acceleration. We said it's going to accelerate down the plane, okay? That looks pretty good. What about the sum of the forces in the Y direction for that box? We've got N2 in the positive Y. Remember we're in this rotated coordinate system so N2 minus M2G cosine theta and all of that is equal to 0 because it's staying on the incline. It's not jumping up off the incline or falling through the incline. And so now we immediately know what the normal force is on that box. It's M2G cosine theta. All right so now we're getting somewhere. Let's take a look at what is left. Well, it looks like this equation right here is a good place to start. So let's write that up here. M2G sine theta minus T, but we know what T is, it's right there. M1A plus M1G. And now we have to subtract FK, but we know exactly what FK is. Fk is equal to Mu K times N, the normal force which is M2G cosine theta. And all of that is equal to M2A. So now look what we have, we have 1 equation and we can solve this for acceleration A and I'm not going to bore you with all the details, you can double check it on your own. But you do a little bit of math and what you should end up with is the following. A is equal to G times the quantity M2 sine theta minus Mu K M2 cosine theta minus M1. All of that over M1 plus M2. Okay? Double check the math, make sure you get to the same answer. Hopefully that is clear and, if you have any trouble with that one, come see me in my office. Cheers.
>> Hello, class, Professor Anderson here. Let's take a look at an inclined plane problem and let's make it a little more complicated. Let's add a pulley to this incline plane. And we will hang a mass from this pulley, call that M1. And we will have the pulley tied to another mass right there, M2. And let's add some friction on the plane with coefficient Mu K. All right, we don't know exactly which way this thing is going to accelerate, but let's just pick a direction. And, if we get a negative sign when we're all done, then you know you had your picture wrong and it's actually accelerating the other way. So let's say that M2 is big enough that it's going to pull down that way and the entire thing is going to accelerate down the plane. If block M2 is accelerating down the plane, then block M1 is accelerating up with the exact same acceleration A because they're tied together. All right. So, this is what the picture looks like. Let's say that we are trying to figure out what that acceleration A is, in fact, equal to. So how do we do this? Well, we've got our picture. We have our givens, M1, M2, Mu K and theta. The next step is to draw a free body diagram. So let's draw a free body diagram for particle number 1, M1, what are the forces on it? Well, of course, gravity is pulling down and we have tension T in the cable going up. And that's it for particle 1, okay? What about number 2? Number 2, it's on an incline so, again, we want to use our rotated coordinate system. And in that rotated coordinate system, there is going to be a force down the plane and we know that that force is M2G sine of theta. There is another component of gravity which is into the plane and that is M2G cosine of theta. Again, if you can't remember which is which, look at the limits as theta goes to 0 and that should tell you if you have it right or you have it backwards. What other forces are acting on M2? Well, there is the normal force from the plane. There is also tension, T, because it's tied to the cable. And there is also friction, F sub K. And that's all the forces that are acting on M2. And now, with this picture, we can write down Newton's Second Law and see if we can solve for the acceleration A. So let's make a little room here. And we will attack it. All right, let's look at particle number 1. Sum of the forces, it's all in the Y direction for that guy. And it's equal to tension going up minus M1G going down and all that is going to be equal to the mass times the acceleration. And so now you can write the tension T. Tension T, according to this equation, is just M1A plus M1G. All right we like that one and we'll put a box around it. What about number 2. For number 2, this is number 1, number 2, we have sum of the forces in the X direction. Remember we're in this rotated coordinate system. So the X direction means positive X is down to the right. So we have M2G sine theta, okay. And then we have 2 that are opposing that. We have tension T and we also have frictional force FK. And all of that is equal to the mass times its acceleration. We said it's going to accelerate down the plane, okay? That looks pretty good. What about the sum of the forces in the Y direction for that box? We've got N2 in the positive Y. Remember we're in this rotated coordinate system so N2 minus M2G cosine theta and all of that is equal to 0 because it's staying on the incline. It's not jumping up off the incline or falling through the incline. And so now we immediately know what the normal force is on that box. It's M2G cosine theta. All right so now we're getting somewhere. Let's take a look at what is left. Well, it looks like this equation right here is a good place to start. So let's write that up here. M2G sine theta minus T, but we know what T is, it's right there. M1A plus M1G. And now we have to subtract FK, but we know exactly what FK is. Fk is equal to Mu K times N, the normal force which is M2G cosine theta. And all of that is equal to M2A. So now look what we have, we have 1 equation and we can solve this for acceleration A and I'm not going to bore you with all the details, you can double check it on your own. But you do a little bit of math and what you should end up with is the following. A is equal to G times the quantity M2 sine theta minus Mu K M2 cosine theta minus M1. All of that over M1 plus M2. Okay? Double check the math, make sure you get to the same answer. Hopefully that is clear and, if you have any trouble with that one, come see me in my office. Cheers.