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, I really thought was interesting, so let's go through it. It's about torque, so the idea if you know how wheelchair works as you got this big wheel and your come up against a curb, of height H. Okay so. H here. Either 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 will label some things on the wheel okay. 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 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, we're not yet pushing over the curb. The current have to be there, the only other forces normal force the earth pushing back up the wheelchair so it doesn't fall to the earth, right? So this static problem is just MG equals normal force. 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 going to push forward F, top of the wheel. Well, that's going to do is it's going to push into the curb and the curb is going to push back for the for sale 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, okay, we're trying to lift it over the edge were calculating the moment, and moment support choice award. The moment in time. That the normal force goes. To 0 ,so that means is as we push harder with F, we get more 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, if you push any harder, you're going to start to accelerate up over the curb. So it's right that point where you push harder and harder right when you cancelled in. That's the magic part and we going to we're not moving yet, so we going to say it's static, okay? So if one of the forces that takes one approach would be to do statics and 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 edges curved and how the tire deforms under the curb and which way will it 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 some of the external torques, on the tire has to be 0. And if there is one nasty forest we don't want to deal with, how do we get rid of it? We put the rotation axis there, going to calculate that works around this point, which is convenient. Because that's actually the point around which it is really rotating okay? Okay, so we gotta do then is right the torques for these two forces said equal to 0 and then see what the forces. So let's do first right before she replying with your hand is tending to rotate the system this way clockwise, so it's negative, so it's- F. And then we need their distance ,and it's not just big are cause we're not r access rotation isn't here, it's here. So we could get on the whole complicated vector thing with the vector are from here to there we want to do all that. Remember, you can just put a straight line along the force vector an go with the moment arm. It's perpendicular from that line to the point of rotation, so it's really just F. Times this distance ,that's what the cross product of the sign ended up doing anyway. So this distance we need it's not just big r its 2R- H is what's left right? It's the diameter minus the height of the curb so it's, -F(2R- h). Right, so that's the torque pushing you clockwise. Gravity is pulling you counterclockwise. Okay, so that's +mg is the force, and now we need this ones distance or this one 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 bigger, 'cause we've come off the radius a little bit. So, this distance we can actually get by drawing a right triangle where we can drive right triangle here. Right, so there is big are right from the center to the edge of the circle is always big R. This is the radius minus H, R- h ,so this part that we need we get from the pythagorean theorem, all right? So if we called it d and we going to put d here, we could say okay,- F(2R-h) +mg and now. So, d must be that ( R- h) squared + d squared equals big R squared like that. So d must be the square root of big R-squared- R -h squared. So the geometry gets a little hairy, but really nothing worse than that. This all has to equal 0. All right, so let's work on this a little bit,- F this needs to stay the same for now plus mg times the square root. And if we foil that out, we going to get minus R-squared +(2R- h )squared, and that first R-squared and this will cancel. Right, so it's a 2Rh- h squared cause that- sign there. Is what you're left with us okay? 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 can still do it. And we could also go ahead and move this to the other side, and Stephanie equal each other since it equals 0. So let's do that,let's mg the square root of h times the square root of 2R-h equals F( 2R -h). And now you can see what I'm doing that I want to get 2R- h all by its Lonesome 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 2 -h. And eventually got to solve for the 4th. 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 ,okay? So you can see the force you apply some fraction of your weight or of the weight of the whole system is of course 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 at a small number, say one over 2 * 10 is 20- 119, one over 19 ,It drops it down to 5%. But the curve gets really big that becomes a problem because in the numerator gets big. And then you don't take us any take too much off the dominator, but a big wheel in a small curb you could push yourself right over.
