This is module 29 of applications in engineering mechanics. We are going to use the information that we learned about belts, in the last module and apply it to another static equilibrium problem. Here's the situation, we have a person that weighs a 150 pounds they're holding a cable. That's wrapped around and attached to a bar, which is 60 pounds and has a pin on the right. A pin reaction. And this cable is over, going over two fixed pegs. The peg on the left has a coefficient of friction of 0.1 on the, on the right a coefficient of 0.2. And we want to find the smallest force the person can exert on the cable on the left, to hold the bar in a vertical position. So, you should be able to solve this problem on your own using the tools you've learned in the course, but if you get stuck come on back and we will walk through it together. So, as always the first thing you want to do is to draw the Free-body diagrams ah,I'd like you to do that go ahead and do it for all the bodies in the problem. Let's do the left poly, the right poly the man and the bar. And so first of all do the left poly. If you look here overall in the cable, on the right hand side will be your largest tension. On the left hand side will be your smallest tension. For the left cable, however, the tension in the center here will be the largest for the left cable, and the smallest for the right cable. So, when I draw the free body diagram of the left pulley, I'm going to use just the generic T for the largest tension over here on the right. Ts is the smallest tension, which we're trying to solve for. Now, do the right pulley. In this case the generic T ends up being the smallest tension for this Free-body diagram and TL is the largest tension that attached back to the bar. Now do a Free-body diagram of the man. For the man we just have his, his weight we have the T sub S, acting equal and opposite T sub S up here. And we have the normal force acting up through point A from the beam. And then the last free body diagram I want you to draw is the beam itself. There it is, with the normal force from the man, T sub L from the cable, 60 pounds, it's weight. And the two force reactions from the pin on the right-hand side. So now that we have the free-body diagrams, we're going to go ahead and apply the equations of equilibrium. I'll do the equation of equilibrium for the left pulley to start. The largest tension in this case is that generic T. So I have T equals T sub s, e to the mew times beta or T equals T sub S, in this case mew was given as 0.1 and beta is 90 degrees or in radians, that's pi over 2. So, I would like you to do the same thing for the right pulley, and once you have that finished come on back. Okay, so for the right pulley the generic T becomes the smallest tension in this situation and T sub L is your largest tension. New is point 2 and your angle of surface contact is again 90 degrees or pie over 2n radius. And so we're going to combine these two equations now to get an overall expression for the overall T sub L and T sub S in the cable. And the way I'm going to do that is I am going to substitute this T into this equation. And so I've got T sub L equals T sub S e to the 0.1 pie over 2 times e to the 0.2 pie over 2. And you should remember from mathematics, when I multiple these two terms I'm going to add their exponents and so I've got T sub L equals T sub S e to the 0.3 times pi over 2. And that ends up being, if you multiply out e to the 0.6 times pi over 2, that ends up being T sub L equals 1.6 T sub s. Okay so, do the same thing for the man. Do the equation of equilibrium for the man. And when you get that done, you should have summed forces in the Y direction and arrived at this equation. And so now I have two equations. I've got the equation that relates. the cable in the left side, tens, tension in the cable on the left side to the right-hand side, that's equation one. And I've got the relationship between the normal force and the, the tension in the, the small tension in the cable. And so I have two equations and three unknowns, T sub L, T sub s, and N. I'll need one more equation of equilibrium think about it, which, which equation of equilibrium do you want to use, to finish this problem up? And what you should have said is, we're going to sum moments about r and let's go ahead and do that together. Why did I choose r? Think that about that and come on back and I'll explain my rationale. Normal force here is going to have a tendency to cause a clock, counter clockwise rotation, which is positive in accordance with my sine conventions. So I've got N times it's moment arm, which is 12. And I've got T sub L, which is going to tend to cause a clockwise rotation, negative in accordance with my sine convention. So minus T sub L times its moment arm, which is 10. And then I have plus 60 times its moment arm, which is 8, equals 0. And so now I can substitute in I want to solve for T sub S. So I'll substitute in T sub L in terms of T sub S, and N in terms of T sub S. And so if I do that I've got, N is 150 minus T sub S times 12, minus T sub L is 1.6 T sub S. Times 10 plus 480 equals 0. And if you do that mathematics, you'll get 2280 equals 28 T sub S. And T sub S is equal to 81.4 pounds. And that's our answer. So, in this situation, the person weighs 150 pounds but given this cable wrapped around these fixed pegs, they only need to exert 81.4 pounds to hold this. This situation in static equilibrium. So, that's it for this module.