 Hi everyone. So in this video I want to talk about band breaks. So band breaks are, well, effectively what they sound like. You have your rotating thing, which is a pulley or a drum or something like that. And you have a band wrapped around it. When you apply tension to that band, it increases the friction between the rotating drum and the stationary band and brings it to a stop or at least slows it down, hopefully. So we want to look at how we might analyze band breaks and kind of what is going into that situation. So when we talk about band breaks, there's a couple of different versions that we might think about. Each involve a rotating member. The first one that I'm going to show is just a basic band break. And the format is that we have a fixed pivot with a handle attached to it. And then we have the band, which wraps, obviously in contact. I'm just kind of drawing this around as rough as I, or as good as I'm able, actually. So a band that wraps around it. And when we apply a force to the handle, it increases the tension in the band and applies that friction to the rotating drum, causing it to stop. We can also have another form, which is a slight variation. And so I have my same pivot point here. But now my handle, which is out here, also extends down this way and has something like that. So now I've got my band wrapped around my drum and attached to my handle. I apply a force, right? And this may may look like kind of a weird situation, right? Because if I'm applying a force down here and pushing the handle down, I'm increasing the tension here, but presumably decreasing the tension on this side, right? Because I'm allowing more slack into the band as I go. And therefore, you know, it requires a little bit of careful consideration of the two lengths here, one between the pivot point and the contact on the right hand side, and the distance between the pivot point and the contact on the left hand side. So I've got the first labeled A and the second labeled B. So as long as A is greater than B, then I'm net increasing the tension in the band, right? As a net result. But what this does for me, this configuration, is it makes this break what we would call self-energizing. And really what it means is that, so as the tension and therefore the friction increases between the rotating drum and the band, the drum is trying to pull the band clockwise in this case around it. And as it pulls, it's increasing the tension on the handle itself down here on the left hand side. So it's actually lending assistance to the breaking process through that friction. So it's self-energizing in that it's using some of the friction from that interface to actually apply more breaking force. So in that case, it's kind of providing, I guess, assistance. Alright, so let's take a look at what it would look like then to analyze this. And I'm going to use the simpler version of the band break for this analysis example. So kind of zooming in, I guess, on the drum, I have my drum and it rotates, right? And I have my band wrapped around it. And on either side, I have tensions. I'm going to call them P1 and P2. So these tension forces are just the tension in the band. Now, you know, you might say, well, shouldn't the tension be the same in the band, right? Because it's all just one thing. And that's generally true when we talk about things like pulleys, and we're assuming negligible friction in like the pulley bearing and things like that. But in this case, we're not talking about a pulley. We're talking about something that has friction as this band passes over the drum. And therefore, these two tensions are not likely to be the same. Because of the rotation, tension one is likely to be greater than tension two. So as this drum rotates, it's going to be pulling and increasing the tension in the band at P1 while decreasing the tension P2. And therefore, we would expect that P1 to be larger. So let's take a look at this by isolating a small segment of the band and pulling it out for a free body diagram. So we're going to zoom in on a small segment, a differential segment again, of the band in order to better analyze it. So we're going to say that this is a small differential piece where its length is an arc length, which you free remember from other courses, other places, arc length is r theta. But in this case, we're going to say it's r d theta because again, we're using a differentially sized piece of material. So very, very small for our analysis. Now on either side of this small chunk of the band, we have tension. So we'll say we have on this side tension P. And it's on the right hand side, which means it should be smaller than the tension on the left hand side. So we'll say that at this side, it's tension P plus whatever change in tension delta or DP occurs across that differential piece. Now the orientation of this tension vector is that d theta, which is our segment, how much of a chunk we've separated out is d theta, but only half of that, right? Because we're changing angle on both sides. So we have another d theta over two over here on the right hand side. Okay. So this theta again comes from whichever, however big the differential chunk of material is. So if I wanted the full d theta, I would label it right here like that. And finally we have some normal force dn acting on it like this. And then we have a friction force mu dn acting on it as well. So that's the friction force due to the drum rotating against this small chunk of material. All right. So scroll down a little so we can see some more space. So now we need to take what effectively is a free body diagram and construct some equations. So in the vertical direction, we're going to do an equilibrium. And basically what we have is a force on the left hand side of pdp, p plus dp on the right hand side of p, but only the vertical or y direction if you want to call it components. So on the left we have p plus dp times sine of the angle. Remember the angle is d theta over two p on the right hand side again times sine of d theta over two. And then we have the normal force. Now it might be a little confusing in that I've left these two as positive, which means I've indicated that down is positive. It doesn't really matter. I just need to make sure that I end up with the opposite for my vertical positive or upwards direction component, which is dn. And then this is all in equilibrium. So it's equal to zero. Now this is kind of a big ugly equation. So I'm going to make some simplifying assumptions or corrections to it. This is going to look a little bit like mathematical hand waving. And that's because it kind of is. But we're going to take something called the small angle assumption. And if you're not familiar with the small angle assumption, it's basically that for small angles, which d theta would be a small angle because it's differential, which means it's very very small, we can say that sine of the angle, which in this case is d theta over two, is approximately equal to the angle itself, d theta over two. So that basically just says we're dropping the sine from this because it's it's not affecting things very much for very small angles. We're also going to say that dp is much much smaller than p, which means that over here in this p plus dp, I can basically cross out the dp part because it's so much smaller than p being a differential piece that it doesn't have much influence. So if I do that, I've gotten rid of these signs. So cross that out, cross that out, cross out this dp. What am I actually left with? Well, I've got a p times d theta over two plus a p times d theta over two. So that would be two p times d theta over two. So those twos cancel out. And I'm left with p d theta minus dn equals zero, which then gives me dn equals p d theta. So I've simplified this kind of long, ugly equation down into something, you know, something we can handle. Now we can do the same thing in the horizontal, or if you want to think of it as the x direction, we can write an equilibrium here. So same general idea. But now instead of the sine component, we're using the cosine component of the tension forces. They're in opposite directions. And I have the friction component of my normal force in there as well. Alright, so I'm going to do some similar mathematical hand waving and use the small angle assumption where cosine d theta over two is approximately equal to one for small angles. So if I factor this in, then I've got this cosine d theta over two is approximately equal to one, approximately equal to one. And now I can rewrite my equation. I can say I have p plus dp times one minus p. So that's p plus dp minus p. So I'm left with dp minus mu dn equals zero, which then gives me that dp equals mu dn. So I'm going to take this dn substituted in down here into this equation. And that gives me dp equals mu p d theta. So I've got what's starting to look like a differential equation here. I can collect my terms. So doing that, I'm going to divide p over. So I have one over p dp is equal to mu d theta. And I can integrate both sides of this. So I'm going to integrate this from p2 to p1 because I'm integrating over p. And I'm going to integrate this side from zero to phi, where phi, which I may not have noted before, phi is going to be basically the angle over which the band makes contact with the drum. So it's this angle here, angle phi is going to be that angle. So it's basically the range over which my band makes contact with my drum. All right. So I've got this integration set up. The integral of one over p is going to be the ln or the natural log of p evaluated from p2 to p1. So when you subtract these, you can actually bring them inside the natural log like this. And I'm evaluating mu times the integration of d theta, which would be just theta evaluated from zero to phi, which means it's just phi. Rearranging this equation to solve for p1 over p2, I take e to the power of each side, which means I get something that looks like this as an expression. So that's useful as a relationship between p1 and p2. Now, I also can look at my free body diagram of my drum where I have p2, p1, and some torque that's being applied by that friction, which means that I can solve for t as p1 minus p2 times radius. So now between these two equations, I in theory also have enough information to actually solve this problem and figure out what my torque is for a given known tension or a given known torque, solve for tension and so forth.