Hi, this is Module 26 of Applications in Engineering Mechanics. We're going to finish up our block of instruction on Coulomb friction Coulomb's Law of Dry Friction in static equilibrium problems. And so this is our last problem, a little bit more complex than what we've done before. We know that we have a homogeneous block that weighs 50 pounds. We know that the pulley here is massless and frictionless, and so that means that the tension on either side of the pulley is going to be equal. And we learned that in my earlier course, Introduction to Engineering Mechanics. And we want to find the range of values of the weight of this block A, here, that will maintain static equilibrium of the, of the system. And so again, what do you do first? Well you need to draw a free body diagram and so draw a free body diagram of block B and come on back. Okay so now what you should've found here is you've got the the weight of the, the block over here A, so weight sub A. You also know the normal, the weight is down at the center of the homogeneous block B. you know that the normal force is going to be up and to the right, but you don't know where it acts. And on friction, you know that it opposes motion, but in this case which motion are we talking about? Are we talking about motion up the plane or down the plane and we don't know. And so, what you're going to do here is you're going to have to think about four different cases that could occur. And I want you to think about what those four cases are and then come on back and let's explain em together. So what could happen here is WA could be so small that this block could either slide or tip down the, the plane, which would be two cases or it could be so large that it could either slide or tip up the plane. And so what I want you to do, is to look at all, what we're going to do together actually is to look at all the possible cases of impending motion. And we'll look at impending motion down the plane first. And the first case I'm going to look at is assuming that it is a slide down the plane and that now allows us to draw the Free-body diagram, because now I know that friction will oppose that motion so it's going to be up and to the left and we know that the normal force again is going to be perpendicular to the plane. And so there is a free body diagram of the system. let's go ahead and check, check slipping down the plane. So I've got let's, I've got my X and Y components here. Let's go ahead and sum forces in the Y direction. Set it equal to zero, I'll choose up into the right positive. And so I'm going to get N minus the four-fifths component of 50, equals 0, where N equals 40 pounds. If we're saying or we're assuming for this first case that we have impending slipping down the plane, that means that f has to have reached its maximum possible value. And so for impending slipping, we've got f equals f max which is equal to mu sub N or in this case mu is given, the coefficient of friction was given as 0.3. I didn't specify whether it was static friction or kinetic friction so we'll just assume that it's the static friction, or the static and the kinetic friction are the same. So it's 0.3 times N, N was equal to 40, and so we get friction is equal to 12 pounds. Now I can sum forces in the x direction. And I have got I'll use down and to the right as positive, and I have got minus my weight of block A minus f plus the three-fifth components of the weight which is 50 pounds equals zero. I know that the friction force was 12, and if I solve that, what you should find is that the weight of A equals 18 pounds for impending sliding down the plane. Okay so that's case one. Let's next look for tipping down the plane. What does WA have to be if we assume that the block has tipped down the plane? And in this case, we want to draw the free body diagram for tipping down the plane. And I'd like you to do that on your own. And what you should find is that okay, if we're assuming tipping down the plane, that means that the normal force is going to act at that corner, and the friction force is, again going to oppose that pending motion. We'll call that corner point A. Let's go ahead then and find out what WA has to be for the impending tipping case. And the way to do that is to apply the equation of equilibrium of the moments about point A. So I'm gong to say the sum of the moments about point A equals 0. I'll choose counter-clockwise positive and I'm going to get, let's see here, I'm going to get WA times its moment arm that's the perpendicular distance between the line of action of the force and the point about which we're tendency to rotate which is 10. And then I've got the four-fifths component of the 50 pound force, the weight that's going down here. And its moment arm is going to be three. And then I've got the three-fifths component is going to tend to rotate this body clockwise around point A, so that's going to be negative in accordance with my sine convention. And its moment arm will be one half of 16, or eight. because this total side is 16, equals 0. And once you've done that you can solve for WA, and in this case WA ends up equaling 12 pounds for impending tipping down the plane. 'Kay, so now let's do a little thought experiment, here. Let's say that WA, here, is 20 pounds. And we want to look at impending motion down the plane. And, at 20 pounds, we're going to assume that it's in static equilibrium. As I lower the weight, once I get to 18 pounds, the block B is going to slide down the plane. We'll never get to the level of 12 pounds, where it will tip down the plane, because it will already have slid first. And so therefore, WA must be greater than or equal to 18 pounds to, for static equilibrium, to avoid slipping down the, the, the plane. [NOISE]. So that takes care of all impending motion down the plane. And now we have to check the impending motion cases up the plane, so let's look at impending motion up the plane. For Case 3 let's assume that the block slides up the plane. So go ahead and, and, and do that and what you find is now the friction force will oppose the motion which is assumed to be up the plane. We have our normal force the same, so the normal force actually ends up being the same as we found last time, 40 pounds. And the friction force, although it's the opposite direction, since we're assuming sliding, it has to be, at its max value and so that's 0.3 times 40, the coefficient of friction times 40 or 12 pounds still. And so now I can sum forces in the extra kind of I like you to do that on your own and see what WA would be for the Case 3 where we have impending sliding up the plane. Okay what you should have found that WA was equal to 42 pounds for impending, slipping up the plane. Let's check the last case, Case 4, for tipping up the plane. In that case, we know if it's tipping up the plane that the normal force is going to act at this upper corner, and friction again is going to oppose motion. The equation of equilibrium that you should use to find WA in this case is the summable moments about b. if you do the sum of the moments about B you find that WA is 36 pounds, and so now let's go back to that thought experiment. If WA was 20 pounds and we said it was in equilibrium, if I increase WA more and more and more when I get to 36 pounds The block B is going to tend to tip up the plane. We'll never get to 42 pounds where block B will slide up the plane. So the limiting case is tipping at 36 pounds. And we already found that the limiting case for motion down the plane was that WA had to be greater than 18 pounds. And so our overall condition for static equilibrium is that WA has to be between 18 pounds and 36 pounds. If WA is less than 18 pounds, block B will start sliding down the plane. If it's greater than 36 pounds, block B will tip up the plane. And so this completes this block of instruction on coulomb friction as applied to static equilibrium problems.