Our next friction application is for belts. This is a situation line all the rest of our equations of our equilibrium problems for rigid bodies. We're going to read the problem, draw the free-body diagrams, and then consider what other equations we have. In this case, we'll get one more equation for belts. So if we look at some vocabulary terms for any pulley with a belt or a rope going over it, we've assumed up till now that that is a frictionless situation, so the tension on both sides the same. As soon as you actually have friction on your belts, they're not the same anymore. So if we have R as the radius of pulley, I'm going to have T2 and T1 are the two tensions, and they're different. So the belt is coming from over here, it's going in this direction. Beta is the angle subtended by contact. When I'll subtend it, is it just a fancy word that says, where is it touching? So it starts touching here, it goes around, and then it ends touching over there. The angle in here is called beta. That can be greater than 2 pi if the rope goes around a couple times. If there is no friction, T1 is equal to T2, but if there is friction, then what I have is here, the belt is tending to go this way. But if it slips, that means the slip is this way. So your friction is going to oppose that motion, so friction is going this way. With friction, then, if you just think about this, I've got T2 coming this way, and friction and T1 going that way. So T2 has to be bigger than T1. But how much bigger is the question? So if you take just a piece of this belt, so one little bit of belt, and you draw the free-body diagram of that, you'll have some tension over here, a little bit more tension over there. You're normal in friction forces. Now, we can't do this as just some of the forces in X and Y at a large level because friction goes all the way around. And this is not a moment. This is friction acting here, tangent, here, tangent, here, tangent. It's a force acting in all kinds of different directions. So what do you do with that? You add it up using calculus. So we're going to take this as this delta theta goes all the way around, which is where you get to beta. You have a situation where you're going to integrate from zero to beta. When you do that integral, and it is available for you on our learning management site, all of that calculus, you get this nice little formula. The ratio of the two tensions is E times E2V, mu tends to beta, where mu is the coefficient of static friction or kinetic friction, bending on whether it's actually slipping or vaptic slip. And beta is that angle. Now, in the process of doing all this, you have to deal with the fact that that beta has to be in radians, because one of the things we use in the formulas here is the arc length formula. So beta has to be in radians. And there's, of course, this extra way of doing the same formula, take the natural logable sites, and then you have new beta is the natural log of T2VRT1. Now, we look at a quick example of one of those. These is two belts, one belt, two pulleys, and this is going all the way around. I have a 5 centimeter pulley, the tensinium merc pulley. I have the coefficient of static and kinetic friction, and this belt goes around. The maximum amount of all the tension on that belt is 500 newtons. I want to know what kind of torque I can put here before I break the belt. Start with the free-bited diagram of just the smaller one. The smaller one is touching less beta would be smaller. So that's where it's going to slip first. If I look at that, because some of the forces in X and Y would give me AX and AY, these, that set the bin, but I don't really care about those. I don't really care about this, but the ratio is between T2 and T1. So that's our extra formula for belts. Notice I changed my 150 degrees into radians. So once you've got that, you can solve for T1. Look at the free-bited diagram of the larger pulley. Some of the forces in X and Y would give me the pin at the middle, but I don't care. So if I'm looking at this, I can take some of the moments, and I have M and the two tensions, the two tensions I don't even have to resolve into component form, because they're acting tangent to the circle. So the perpendicular distance is just going to be 10, and I can plug that in and I get a value for M.