This is module 24 of Applications in Engineering Mechanics. We're going to continue where we, pick left off last time with this concept of Coulomb friction to actually solve some real world engineering problems. So this is the first problem I want to look at. what I have here is a bar that can rotate, bar B2 and then I have a body on that bar B1. And I want to find the relationship between this angle theta and the coefficient of static friction between the bodies B1 and B2 that will prevent the body B1 from slipping or sliding down the plane. So if you slide over here here's the situation. As I increase the angle theta, at some point the coefficient of friction is not going to be great enough to keep the body from sliding down the plane and I want to find that boundary. I want to find that condition where fric, sliding will not occur. And so my question to you is, how might you, what, what are you going to do to start to analyze this problem? So think about that, come on back. Okay. Like always, what you should do is draw a free body diagram, and so my question to you is, which free body diagram are you going to draw? My subs, my suggestion is to go ahead and draw a, a full body diagram of B1 since that's the body that we want to prevent form sliding down the plane. And so, do that, come on back and let's see how you did. Okay, so here's our free body diagram of my body. you'll notice that I have my, my weight force acting straight down. I've got a normal force acting tangent to this surface, perpendicular to this surface. And then I have a friction force opposing the pending motion, or the impending motion. And so, I've actually rotated my X and Y. frame so that it's along the, the incline. And what I want you to do now is to use the equation of the equilibrium for some of the forces in the y direction to solve for the normal force. Okay, and what you should have done is found that the normal force Is just equal to mg times the cosine of the angle. I'm going to use the equilibrium equation of sum of the forces in the x direction to come up with the next equation. And so in this case, I've chosen up and to the right positive. So I've got f minus mg sine of theta equals 0. I want the condition here just before slipping a curve. And so if I look at my graph of the slipping versus non slip condition that we talked about last module, I'm at the point where I'm going to get the very max friction available just before it starts sliding. And so f has to be f max or mu sub s times N minus mg sine theta. But we know that m from the equation where you found up here, is equal to mg cosine theta. So we get U sub s times mg cosine theta equals mg sin theta. And I can cancel the mgs now from both sides and I find out that the condition from USIV S is that USIV S is sin theta over cosine theta which is the same as tangent of theta. And so that's the intending condition just before the, the, the body would slip, body B1 would slip and so what we would have to have to keep it from slipping is mu sub s would have to be greater than or equal to that value. So we've got U sub s is greater than or equal to tangent theta to prevent slipping. And that's the answer to our problem. [BLANK_AUDIO] ' Kay, let's go ahead and do another problem. In this case changed just slightly. I've got a crate. It's on a fixed plane of 30 degrees. But I have the ability to have an applied force here P. I know now that the crate weighs 200 pounds, and I'm given that the static coefficient of friction between the crate and the plane is 0.4. And so, what I want you to first determine is, if that force P were not present okay, if it was gone, would the crate slip down the plane? So go ahead and do that on your own, and then come on back after you've, you've completed that analysis. So to do that, that is exactly like the, the problem we just looked at previous to this where we were raising the bar to see what, what condition was just before, slipping would occur if there was no force P here, and so all you have to do is check the condition we came up with to prevent slipping. Is u sub s greater than or equal to tangent theta, well you put your values in. U sub s is equal to 0.4, theta is 30 degrees, so 0.4 has to be greater than or equal to 0.577. And so you see that no, we don't have enough friction, and so unless we put some sort of forces on this crate, it is going to slide down, down the, the incline. And so my next question is, what is the smallest force P that we can use to prevent the crate from sliding or slipping down the plane. And again I'd like you to go ahead and try to do this analysis on your own. once you've given it a good shot come on back and we'll do it together. So again, what I'm going to do, what I did is, I've drawn my free body diagram. The only difference in what we had before is, that we now have this force P acting up and to the right. you can again sum forces in the y direction. I'd ask you to do that. And when you do You'll find that N equals 173 pounds. I'll go ahead together with you and we'll sum forces in the x direction and so we've got P plus F minus 200 times the sine of theta, which is 30 degrees, equals 0. So F we know is going to be on the verge of slipping. So, we want the smallest force p that will prevent slipping. So, that's going to be the maxify your friction I can have which is U sub s times N, and I know that U sub s is equal to 0.4. I found N to be 173 pounds. 200 times the sine of 30, well, sine of 30 is 0.5, so this term over here is 100. So I've got minus 100 equals 0. And you find out that the smallest value of P that'll prevent slipping is 30.8 pounds. And so that's our answer. If I have anything less than 30.8 pounds, the crate will slide down the plane. As I get, to, to the point 30.8 and a little bit higher, that's enough to prevent sliding down the plane. So my next question to you in a worksheet for you to do on your own is, okay, what is the largest force now, P, that can be applied before the crane, crate will start sliding up the hill? And so, we know if we have zero force P, Crate slides down as we increase P. Once P gets to 30.8 pounds, the crate will not slide. We can increase it, we can increase it more and more until there's some point where P is going to get so large that The friction is not going to be able to prevent it from sliding up the plane and that's what I'd like you to do in this worksheet. I've got the solution in your module handouts and we'll see you next time.