Anderson Video - Leaning Ladder Problem

Professor Anderson
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>> Hello class, Dr. Anderson, here. Let's talk about the leaning ladder problem. So in this problem we have the following setup. We have a floor and a wall and we have a ladder that's leaning against the wall. Now there is friction on the floor with coefficient of friction mu sub s. The wall is frictionless. Now, let's say we do the following. Let's say we ask the question how far up can a person climb on the ladder before the ladder slips and falls? And you're all sort of familiar with this idea. You know, when you lean a ladder up against the wall, if you're down at the bottom it's stable, but as you go further and further up it can get unstable and eventually fall. So how would we model this problem? Well, let's draw the ladder again, and the ladder has length L. Let's say that the person is a distance x up the ladder, and the angle between the ladder and the floor is theta. All right. So now with all this stuff, let's see if we can think about the forces and the torques that are acting on the ladder. So down at the bottom we of course have the normal force, n. There is a static frictional force, which is trying to keep the bottom of that ladder in place, so that's f sub s. There is the weight of the ladder m sub L times g. There is the weight of the person m sub p times g and there's one more thing, which is the push of the wall on the top of the ladder. All right. So that's our force diagram and now let's think about the forces that are acting on the ladder and also the torques that are acting on the ladder, and let's make a little room. Let's make some room over here. Okay. So we have sum of the forces in the x direction. What are the forces in the x direction? Well f sub s is going to the right. P is going to the left. That's it. Those have to add up to zero. What about forces in the y direction? In the y direction we have normal force n going up, we have the weight of the ladder mLg going down, we have the weight of the person going down, and that is it. All of that has to add up to zero. All right. So that's the easy part of the problem. The hard part of the problem is identifying all the torques. Now when you write down some of the torques you have to immediately pick an axis of rotation. So let's pick this point right here at the ground as our axis of rotation. By doing that we immediately eliminate n and f sub s from our torque equation, because they don't have any lever arm. They go right through the axis of rotation. But what about all these other forces? All right. We have the weight of the ladder. So it is trying to rotate this way clockwise. Clockwise is negative by our convention, so we put a minus sign, the weight is mLg, so that's the force. What is the lever arm? Well, the lever arm is this. Okay. Here's our ladder. There is our force. This is a lever arm for that guy right there. And this side of the triangle is in fact, L over 2, because it's a uniform ladder. So the mass is acting about the center of mass, okay. So this becomes L over 2 times cosine theta, that's what this lever arm is equal to. Good. What about the person? The person is also trying to rotate this thing in the clockwise direction. What is the lever arm for the person? Well, it's the same idea as we just talked about right here except the distance is no longer L over 2, it is x. And so this becomes x cosine of theta. And now there is one more force, which is giving a torque and that is the push p from the wall. And the push p from the wall is up at the top of the ladder. So if we extend that force line that is push p. What is the lever arm in that case, it's that right there. This was Theta. What is that side? Well if that's theta then this is theta. The length here is L and so the lever arm becomes L sine theta. So this last term is positive p times L sine theta. All that has to add up to zero, because it's not rotating. Okay. And now look you have three equations and you have a number of unknowns. There is, in fact, fs, which is unknown, P which is unknown, n is unknown, and x which is unknown. And so we need, in fact, one more equation to make this work. And it is of course the static friction equation. What's the biggest that static friction can be? It's f sub s equals mu s times n, and now with those four equations you can, in fact, solve for x how far up the ladder can you go. All right. Good luck with that one. If you're having a lot of trouble, come see me in my office. Cheers.
>> Hello class, Dr. Anderson, here. Let's talk about the leaning ladder problem. So in this problem we have the following setup. We have a floor and a wall and we have a ladder that's leaning against the wall. Now there is friction on the floor with coefficient of friction mu sub s. The wall is frictionless. Now, let's say we do the following. Let's say we ask the question how far up can a person climb on the ladder before the ladder slips and falls? And you're all sort of familiar with this idea. You know, when you lean a ladder up against the wall, if you're down at the bottom it's stable, but as you go further and further up it can get unstable and eventually fall. So how would we model this problem? Well, let's draw the ladder again, and the ladder has length L. Let's say that the person is a distance x up the ladder, and the angle between the ladder and the floor is theta. All right. So now with all this stuff, let's see if we can think about the forces and the torques that are acting on the ladder. So down at the bottom we of course have the normal force, n. There is a static frictional force, which is trying to keep the bottom of that ladder in place, so that's f sub s. There is the weight of the ladder m sub L times g. There is the weight of the person m sub p times g and there's one more thing, which is the push of the wall on the top of the ladder. All right. So that's our force diagram and now let's think about the forces that are acting on the ladder and also the torques that are acting on the ladder, and let's make a little room. Let's make some room over here. Okay. So we have sum of the forces in the x direction. What are the forces in the x direction? Well f sub s is going to the right. P is going to the left. That's it. Those have to add up to zero. What about forces in the y direction? In the y direction we have normal force n going up, we have the weight of the ladder mLg going down, we have the weight of the person going down, and that is it. All of that has to add up to zero. All right. So that's the easy part of the problem. The hard part of the problem is identifying all the torques. Now when you write down some of the torques you have to immediately pick an axis of rotation. So let's pick this point right here at the ground as our axis of rotation. By doing that we immediately eliminate n and f sub s from our torque equation, because they don't have any lever arm. They go right through the axis of rotation. But what about all these other forces? All right. We have the weight of the ladder. So it is trying to rotate this way clockwise. Clockwise is negative by our convention, so we put a minus sign, the weight is mLg, so that's the force. What is the lever arm? Well, the lever arm is this. Okay. Here's our ladder. There is our force. This is a lever arm for that guy right there. And this side of the triangle is in fact, L over 2, because it's a uniform ladder. So the mass is acting about the center of mass, okay. So this becomes L over 2 times cosine theta, that's what this lever arm is equal to. Good. What about the person? The person is also trying to rotate this thing in the clockwise direction. What is the lever arm for the person? Well, it's the same idea as we just talked about right here except the distance is no longer L over 2, it is x. And so this becomes x cosine of theta. And now there is one more force, which is giving a torque and that is the push p from the wall. And the push p from the wall is up at the top of the ladder. So if we extend that force line that is push p. What is the lever arm in that case, it's that right there. This was Theta. What is that side? Well if that's theta then this is theta. The length here is L and so the lever arm becomes L sine theta. So this last term is positive p times L sine theta. All that has to add up to zero, because it's not rotating. Okay. And now look you have three equations and you have a number of unknowns. There is, in fact, fs, which is unknown, P which is unknown, n is unknown, and x which is unknown. And so we need, in fact, one more equation to make this work. And it is of course the static friction equation. What's the biggest that static friction can be? It's f sub s equals mu s times n, and now with those four equations you can, in fact, solve for x how far up the ladder can you go. All right. Good luck with that one. If you're having a lot of trouble, come see me in my office. Cheers.