Here's a really interesting problem about pushing a wheelchair over a curb from the book that we teach from. So I'm going to steal it. I mean, I'm going to do a tribute to it here. And I really thought it was interesting. So let's go through it. It's about torque. So the idea, if you know how a wheelchair works, is you've got this big wheel and you're come up against a curb of height h. So h here. And you know the way you do it is you push forward on the wheel and you lift the wheelchair over the curb. So we want to calculate how much force does it take to do that. So to describe it clearly, I'm actually first going to draw the vectors and everything going on. For first, we're going to do static and no push. And we'll label some things on the wheel. So for example, the wheel has a radius, big r. And the wheel and the chair and the person, everything together has a mass, big m. And we're going to pretend it's tilted back and I'll send it over the wheel more or less. So we can just say the force of gravity on everything is right there, mg. So in this static situation where we're not yet pushing over the curb, the curb doesn't have to be there. The only other force is a normal force, the earth pushing back up on the wheelchair so it doesn't fall through the earth. So this static problem is just mg equals the normal force. What we want to know though is something a little bit different. What if we have the wheel sitting here like this and it's on the ground and here's the curb at height h. And now we're going to push forward with force f on the top of the wheel. What that's going to do is it's going to push into the curb and the curb is going to push back for the force I'll call c like that. We still have mg right here pulling it down. But one case here is that we're still going to say static. We're trying to lift it over the edge. We're calculating the moment and moment to poor choice of word, the moment in time that the normal force goes to zero. So what that means is as we push harder with f, we get more of a vertical force from the curb and eventually you reach a point that you've lifted and the normal force here doesn't need to exist anymore. You have enough curb upward curb force to cancel mg. And if you push any harder you're going to start to accelerate up over the curb. So it's right that point where as you push harder and harder right when you've canceled in, that's the magic part. And we're going to say you're not moving yet. So we're going to say it's static. OK. So if one of the forces that takes one approach would be to do statics in x and y and say there's no normal force, the problem is the nature of this force on the curb is a little hard to know because it depends on how the curb edge is curved and how the tire deforms under the curb in which way we'll let go, very complicated. But we can do it without that complication by doing static for the torque for the rotational motion. So we can say the sum of the external torques on the tire has to be zero. And if there's one nasty force we don't want to deal with, how do we get rid of it? We put the rotation axis there. We're going to calculate the torques around this point, which is convenient because that's actually the point around which it is really rotating. OK. OK. So what we got to do then is write the torques for these two forces, set equal to zero, and then see what the force is. So let's do F first. The force you're applying with your hand is tending to rotate the system this way, clockwise, what's negative. So it's minus F. And then we need the distance. And it's not just big R because our actual rotation isn't here. It's here. So we could get on the whole complicated vector thing with the vector R from here to there. We don't want to do all that. Remember, you can just put a straight line along the force vector and go with the moment arm that's perpendicular from that line to the point of rotation. So it's really just F times this distance. Remember, that's what the cross product and the sign ended up doing anyway. So this distance we need, it's not just big R. It's 2R minus H is what's left. It's the diameter minus the height of the curve. So it's 2R minus H. All right. So that's the torque pushing you clockwise. Gravity is pulling you counterclockwise. OK. So plus MG is the force. And now we need this one's distance. Or this one's moment arm, right? So we can draw a line along this force and say there's the moment arm there. So it's not quite the radius of the circle. I could have drawn it a little better. It's not big R. It's a little bit less than big R because we've come off the radius a little bit. So this distance we can actually get by drawing a right triangle where we can draw a right triangle here. So there is big R. From the center to the edge of the circle is always big R. This is the radius minus H, R minus H. So this part that we need, we get from the Pythagorean theorem. All right. So if we called it say D, and we're going to put D here, we could say, OK, minus F to R minus H plus MG. And now D, so it must be that R minus H squared plus D squared equals big R squared like that. So D must be the square root of big R squared minus R minus H squared. So the geometry gets a little hairy, but really nothing worse than that. This all has D equals zero. All right. So let's work on this a little bit. Minus F, this needs to stay the same for now. Plus MG times the square root. And if we foil that out, we're going to get minus R squared plus 2RH minus H squared. And that first R squared and this will cancel. So it's 2RH minus H squared because that negative sign there is what you're left with. OK. And then you could, if you're daring, you could pull an H out of here. And I know it's not squared, but you could still do it. And we could also go ahead and move this to the other side and set them equal each other since it equals zero. So let's do that. MG, the square root of H, times the square root of 2R minus H equals F times 2R minus H. Now you can see what I'm doing that. I want to get 2R minus H all by its own, some self there. So then I can take this and put it under there and that becomes a square root. So I get MG. Now we're just doing algebra for fun. F square root of 2R minus H. And eventually I got to solve for the force. So we have to stop having fun here and talk about the answer. So if the force you have to apply is what? It's this factor, the square root of H over 2R minus H times MG. So let's see if the big wheel helps us. So you can see the force you apply is some fraction of your weight. The whole system is a force you have to apply forward. And if your chair has a huge radius, you can see a big number here against a small number here, a small curb, brings the force way down. I had a small number say 1 over 2 times minus 20 minus just 1, 19. 1 over 19, it drops it down to 5%. But the curb gets really big, that becomes a problem because then the numerator gets big and then you don't take too much off the denominator. With a big wheel and a small curb, you can push yourself right over.