Anderson Video - Work Lowering an Object

Professor Anderson
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Hello class, Professor Anderson here. Let's take a look at a problem of lowering a mass at constant speed. So, pretend you have a crane and you're lifting some box, right And then after a while you're going to lower the box at constant speed. Okay, here's our mass M. If it is heading downward at constant speed, there must be some forces that are acting on the box, right. The forces that are acting on the box are what? Well, we have some cable here that's attached to it it that has a tension T, but we know that there's got to be some force that's pulling it down if we're on the Earth. That is force due to gravity: mg and now we're going to pretend that this box moves this distance D. Okay, it starts there and it heads down to some lower position. Let's see if we can calculate the work due to these various forces. Let's look at the work due to gravity. All right we know that work is always F d cosine theta. The work due to gravity is therefore the force of gravity mg. The distance that we go– let's call this H because we're used to doing vertical distances as H. So we put an H right there. These are magnitudes, so when you're putting in F and D those are always positive numbers, okay. The direction is taken into account with the cosine theta. So in the case of gravity, we have a force that's down. We have a displacement that is down. So what is the angle between those two? What do you guys think? okay if they are both What do you guys think? Okay, if they are both heading down they're in the same heading down, they're in the same direction and so the cosine is zero degrees, right. It's the angle that we want to use cosine at zero degrees like we said is of course one. So, what's the work due to gravity in this case? mgh. Okay, what about the work due to tension in the cable? WT What's the work due to the tension in the cable? Well, it's the force, which is tension T. It's the distance which is H. And now it is the cosine of the other angle that you guys said, which is a hundred and eighty degrees. Why? Because tension is up. Displacement is down. That angle between those two is 180 degrees. And so this is where you pick up the minus sign right there. We get minus th and if it's moving at constant speed, we know that tension has to be equal to mg. And so we get minus mgh Okay, anytime you have a force and the thing moves, you have done work on it. That work could be positive or it could be negative. Okay. Or it could be zero, right. If they're at If they're a right angle to each other then it could be zero. Questions about this one? >> When we have the tests and everything, will we have to make... We can't assume that up is positive or whatever, we're gonna have to go with the direction of movement? >> Great question. So Brent's question was about this equation right here, okay. What is important in this equation is this theta right here. And remember this theta is the angle between two vectors. The force and the displacement D. So in some sense it doesn't matter what direction you're moving in, it's really the relative direction between D and F. In this case they were both moving down, okay. And so the angle between them was 0 degrees. These guys right here, these are magnitudes. So those are always positive numbers. F and D are always positive numbers. Theta is where you take into account the direction of them. Okay? All right, let's go back to this problem for a second. We just looked at the problem where we're gonna lower this mass at constant speed. Let's do a slightly different problem where we raise it at constant speed. If we raise the mass at constant speed instead of lower, how do these answers change here? Well the first one, theta, for gravity versus displacement. What do I put here? >> 90. 180. >> 180 degrees okay. And so this is not And so this is not mgh, it's minus MGH. Tension is in the same direction as the force. And so here we put zero degrees which means this becomes a positive. Okay. So if we change the direction of the displacement we've just swapped the signs on both of those. Everybody more or less okay with that? All right. I'm reminded of one of those Seinfeld episodes where George is saying he's gonna do a complete 360 and change his life and Jerry points out: Isn't that coming back to where you started? I think you mean 180.
Hello class, Professor Anderson here. Let's take a look at a problem of lowering a mass at constant speed. So, pretend you have a crane and you're lifting some box, right And then after a while you're going to lower the box at constant speed. Okay, here's our mass M. If it is heading downward at constant speed, there must be some forces that are acting on the box, right. The forces that are acting on the box are what? Well, we have some cable here that's attached to it it that has a tension T, but we know that there's got to be some force that's pulling it down if we're on the Earth. That is force due to gravity: mg and now we're going to pretend that this box moves this distance D. Okay, it starts there and it heads down to some lower position. Let's see if we can calculate the work due to these various forces. Let's look at the work due to gravity. All right we know that work is always F d cosine theta. The work due to gravity is therefore the force of gravity mg. The distance that we go– let's call this H because we're used to doing vertical distances as H. So we put an H right there. These are magnitudes, so when you're putting in F and D those are always positive numbers, okay. The direction is taken into account with the cosine theta. So in the case of gravity, we have a force that's down. We have a displacement that is down. So what is the angle between those two? What do you guys think? okay if they are both What do you guys think? Okay, if they are both heading down they're in the same heading down, they're in the same direction and so the cosine is zero degrees, right. It's the angle that we want to use cosine at zero degrees like we said is of course one. So, what's the work due to gravity in this case? mgh. Okay, what about the work due to tension in the cable? WT What's the work due to the tension in the cable? Well, it's the force, which is tension T. It's the distance which is H. And now it is the cosine of the other angle that you guys said, which is a hundred and eighty degrees. Why? Because tension is up. Displacement is down. That angle between those two is 180 degrees. And so this is where you pick up the minus sign right there. We get minus th and if it's moving at constant speed, we know that tension has to be equal to mg. And so we get minus mgh Okay, anytime you have a force and the thing moves, you have done work on it. That work could be positive or it could be negative. Okay. Or it could be zero, right. If they're at If they're a right angle to each other then it could be zero. Questions about this one? >> When we have the tests and everything, will we have to make... We can't assume that up is positive or whatever, we're gonna have to go with the direction of movement? >> Great question. So Brent's question was about this equation right here, okay. What is important in this equation is this theta right here. And remember this theta is the angle between two vectors. The force and the displacement D. So in some sense it doesn't matter what direction you're moving in, it's really the relative direction between D and F. In this case they were both moving down, okay. And so the angle between them was 0 degrees. These guys right here, these are magnitudes. So those are always positive numbers. F and D are always positive numbers. Theta is where you take into account the direction of them. Okay? All right, let's go back to this problem for a second. We just looked at the problem where we're gonna lower this mass at constant speed. Let's do a slightly different problem where we raise it at constant speed. If we raise the mass at constant speed instead of lower, how do these answers change here? Well the first one, theta, for gravity versus displacement. What do I put here? >> 90. 180. >> 180 degrees okay. And so this is not And so this is not mgh, it's minus MGH. Tension is in the same direction as the force. And so here we put zero degrees which means this becomes a positive. Okay. So if we change the direction of the displacement we've just swapped the signs on both of those. Everybody more or less okay with that? All right. I'm reminded of one of those Seinfeld episodes where George is saying he's gonna do a complete 360 and change his life and Jerry points out: Isn't that coming back to where you started? I think you mean 180.