So today I want to talk about the difference between rolling friction and sliding friction. One of the challenging things about that is to understand that once you get to a free body diagram, like we're going to draw in a second, what's F? And it's so tempting always, like they taught you in physics, F is always mu times theta, except it's not always mu times theta. So how is it different between sliding friction and rolling friction? So I have my 250 gram car, there's a 10.3 distance between the two wheels. The height is about 4.4 centimeters and what I want to do is look at when it rolls and when it slides. Now I'm going to make one assumption at the beginning. I'm going to assume that these plastic wheels have approximately the same sliding coefficient on this aluminum track as the top plastic surface is all hard plastic and aluminum. So it's probably very close to the same. At that point, what I'm going to do is turn this over. I can't lock the wheels on my car. So I'm going to turn it over and I'm going to lift this up until it slides. And then we're going to see if we can find out what the coefficient of static friction is from that sliding angle, from the angle of static repose. So here's my car, here's my angle. I'm going to lift this up until it moves. I'm going to see how far I can get. Lifting. I'm lifting. I'm still lifting. How far can I get? 17, 18, something degrees like that. It's a pretty big angle. So if I take that pretty big angle, my phi s, that's the angle of static repose, is about 18 degrees. Now on my free body diagram, what am I going to have? I'm going to have the normal force. That's going to act at some distance x away from the design because remember the normal force has to move to take care of where the equations of equilibrium need it to be. I have my weight of the car acting at some angle theta. And I'm going to have my friction force along the surface. That's what I want to look at. I want to know what that friction value is for these two cases. I'm going to add an axis system here so I have x along the track and y perpendicular to the track so that I can sum the forces and x and y. So some of the forces and x just gives me along the track, w times sine theta equals f. And the sum of the forces perpendicular to the track tells me that w cosine theta is n. Now once you have those things, what you want to say is when does it slip? So I want to know what the slipping condition is. Slipping condition is always the same. f is mu times n. That's all the ever slipping condition you ever get. So given that we have f is w sine theta, what I really have is w sine theta equals mu times w cosine theta. That allows you to solve mu is the arc tan of this angle that we just found. That's the same thing as the definition of the angle of static repose. It's very nice when you can derive a formula that you've seen before. This gives you mu is about 0.3, which is about what we would always expect it to be. Now these aren't exact numbers because my 18 degrees is something that I'm reading on this indicator. So it's not exactly precise, but it's very close. And if you take that and plug it back into f, my f value here is about 0.7 newtons, which seems very reasonable for this car. So that's what it takes to make this thing slide. Now what does it take to make it roll? Presumably less, right? Things roll before they slide off. So I take my car, I'm going to turn it right side up again. And I'm going to see how high can I lift this track before it actually rolls off? Took me 18 degrees before I could get it to slide off. What's it going to take to get it to roll off? So when I start lifting this track, I can get to, ah, maybe a degree, maybe a degree and a half before it goes. That's about it. I mean, you can barely see that. That's all it takes is about a degree or a degree and a half to make it go. So theta at this point is about 1.5 degrees, not very big. So what do I do with my equations of equilibrium when I'm looking at that? Because I need my free-buddyed ira. So I will now have normal forces at both wheels. I will have friction forces at both wheels. Now remember, I can't solve for both of them together. So what I'm going to do is I'm going to come down here and I'm going to say F is F A plus F B because you cannot solve for two collinear forces. Whatever they are, they're collinear and I can't deal with that. I also need the weight, of course, acting at its angle theta. So that's all I have on my free-buddy diagram. That free-buddy diagram, when you come over here, you could do the sum of the forces. Instead of F, I now have F A plus F B. Instead of N, I now have N A plus N B. But notice that the free-buddy diagram did not change very much. The free-buddy diagram can change. The equations of equilibrium can change very much. So that part is pretty much the same. But when I'm dealing with rolling friction, I now have to look at what's happening at the wheel itself. So at the wheel itself, I now have some weight, but I don't have a whole weight. So that weight's going to be about a quarter of the weight because I have four wheels. So a quarter of the weight acting at an angle theta. And instead of having F and N, what I would like to do instead is come back and say I'm going to replace those with the resultant force R. Nice thing about that is that this free-buddy diagram only has two forces. The force is on a free-buddy diagram that's in equilibrium. You know they have to be both equal and upset and along the same line of action. So I know that this R has to be right there. A word. We're modeling a system. That R doesn't act on that wheel, but that's okay. We're modeling a system. In the real world, these do shift a little bit. The wheel compresses, the track compresses. In fact, what you have is an area, a force, a distributed load. All of these things are ways of modeling this. But to model this, what we're going to do is we're going to say that this acts at a distance B away from the center line of the wheel. Now I can come back and I can say what is the sum of the moments at the center of the wheel? So now I have W over 4 times cosine of theta times B. R is not going to show up because I'm taking the sum of the moments at this point. And I'm going to have W over 4 sine theta. And I'll put it down here. W over 4 sine theta times R. So once I've got this, that's the sum of the moments taken at this spot. And R does not appear because there you have it. That allows me, in this case, W is going to cancel. The force or cancel. Everything else is going to cancel and you can solve for B. Given that your radius is 1 centimeter, B is .03 centimeters. Remember that your coefficient of rolling friction has units of length because it's actually this distance that we had to shift that R value. Once you have those bits, that's the coefficient of rolling friction. Take that and plug it back into the equations of equilibrium. What do I get for F? My F value ends up being .05 newtons. .05 newtons. Now a couple things are different. This one is very different from this one. And you know that. You know that if something rolls, it rolls much faster than it's going to slide. I have to get it way up here before it'll slide. So you know that that's different. You also are going to need to know that this friction force and this friction force are very different. Those are the reaction from the track onto the car along the track. That's what that reaction force is. It changes from .7 newtons to .05. It's a lot smaller. That's the difference between sliding friction and rolling friction.