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Anderson Video - Torque and Tipping Point

Professor Anderson
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Hello class, Professor Anderson here. Let's talk about a very simple torque problem and this is the one where you have a beam and then supporting that beam are two posts. There's one right at the end and then there's one located near the other end but not all the way at the end. And now you take a man, And he starts walking towards the end of the beam. Now, depending on the relative masses and so forth, you could get to a point where the man has walked so far to the end of that beam that the beam just starts to lift up off that back post. So when he gets out there, this side of the beam could lift up and the whole thing will topple over. And we want to figure out what that location is. How far does he have to walk along the beam for that to happen? So, to visualize this let's give some parameters first off. Let's say that the mass of the beam is M sub B. The mass of the person is M sub P. And we'll give some distances. So let's let the left side of our beam be at x equals zero. We'll say that this position right here is x1 and this position of the person is X. So as x increases eventually we'll get to a point of unstable equilibrium and the thing will tip. Alright, so let's analyze the forces that are acting on our beam. So we have the force due to this beam on the left, and that gives a normal force N zero. There's also a force due to this beam on the right, and that is a normal force N one. There is the mass of the beam acting down: MB times gravity. And let's say the man has made it all the way out to here, therefore there is the mass of the man– or the mass of the person times gravity acting down. So those are the forces that are acting on that beam, and now we need to analyze both the forces and the torques. So, forces in the x-direction are obviously zero. There's nothing pointing right or left. What about the forces in the y-direction? Well, we have N naught going up. We have N one going up. We have M B G going down and we have M P G going down. And all that has to add up to zero. Alright, now let's analyze the torques. So the torques are what? Well remember torque is equal to force times lever arm. And in this case, all the forces are at right angles to beam, so the lever arm is just how far is that force from the axis of rotation. So let's pick an axis of rotation. Let's say it's the left end of the beam, so post N naught, that will be our axis of rotation. And now let's analyze the torques. All right, some of the torques equals what? Well, N naught is the first force but if that's going right through the axis of rotation, then there's no lever arm. So that one is zero And now we have N one going up. And remember we said if it's trying to rotate it counterclockwise, that's a positive torque. So what is the x value for N one? It is X one. We have the mass of the beam pulling down and that's and that's trying to rotate it clockwise and so we give it a minus sign. So this is minus M sub B times G. And we have to figure out what the lever arm is for that and if it's a uniform beam then that center of mass is halfway down the beam. So this becomes L over 2. And then finally we have to analyze the person and we said that person is out a distance X from the axis, also trying to rotate the beam clockwise so we give it a minus sign and so this becomes M sub P times G times X. Okay, and now we have a few equations. The X equation doesn't really help us. The Y equation certainly does, and the torque equation certainly does. But let's see how many unknowns we have. We have N naught, N one and X. So that's not very good, right? We have three equations or two equations really and we have three unknowns, so we need another equation to help us and it is this– we have to think about the physics of it a little bit and so there is a condition for tipping. Namely, when this beam tips up, it loses contact with the post on the left– with post N naught. And so if the beam loses contact with the post, the post can't push up on that beam at all and so the condition is N naught equals zero. Okay, so let's see if we can solve this thing for X now. So, some of the forces in the y-direction told us we had this, but now we're saying N naught is equal to zero. So it's really N one minus M B G minus M P G is all equal to zero. And now we can solve this very quickly for N one. We move the other stuff to the other side and we can just lump sum terms together. And it looks like that: N one is equal to M B plus M P times G. And now let's go back to our torque equation. So torque was N zero was multiplying zero– that one's gone. We have N one times X 1 but N one is now this. So we have MB plus M P times G all times x1. Then we have minus M B G L over 2 and minus M P G times X. All of that equals 0. And now we can solve this equation for x. So if we move that last term over to the other side, we get x equals– we're gonna have to divide by M P G. And the rest of the terms are all this other stuff: MB plus MP GX1 minus... Ok so you can work out the torque equation and get a solution for X as I've shown here and depending on those values you'll get a real number for X. When you're all done, make sure that your value of X makes sense in terms of its location on the beam. All right, hopefully that's clear. If you have any questions, come see me in my office. Cheers.