Hi and welcome to module five of applications in engineering mechanics. Today's learning outcome is to apply the equilibrium equations to again, a real world problem. We're looking at multi-force members. We talked about multi-force members as being frames and machines. And we've already done frames, so today we're going to look at an example of a machine. And the machine I have is a pair of pliers that has moving parts. It's designed to transmit and alter the affect of forces, and so you can see on my example problem here, I have my pair of pliers. Ten pound force applied at each handle, which makes sense, and I'm asked to find the forces acting on the nut, and the pin. So your question should be, how should I approach this problem? And hopefully by now you know you should sketch a free body diagram And since the problem is asking for forces on the nut and the pin, let's go ahead and proceed with sketching those free body diagrams. And so I'd like you to do that on your own and come on back and then we'll check and see how you did. Okay, here's the free body diagrams for the nut and the pin that I've drawn. You'll note that I've just drawn the, the directions arbitrarily, I, I've just assumed the directions, the mathematics will tell me if they're in the opposite direction by having negative numbers come out of my mathematics. you will note on, on, on the free body diagram of the pin I have the forces acting from one half of the pliers, body B1 and body B2, and so those are equal and opposite acting on my pin to keep it in equilibrium. So the question becomes, how many equations of equilibrium do you have available for each of these five free body diagrams, and how many unknowns are there? And I'd like you to do that real quick. Okay, so what you should have found is that, over here you're going to have, as always, three equations of equilibrium, and we have four unknowns. The unknowns are N1, N2, F1, and F2. And over here we have three equations again and we have four unknowns, Rx1, whoops Rx1, Ry1, Rx2, and Ry2. Now, the question is what do you do next, and you should say, okay, I need to do another free body diagram, because I have too many unknowns for the number of equations. But before we go ahead and do that, which we will have to do, let's try this out for as many unknowns from these free body diagrams as we can. And so I'd like you to try that on your own, and then we'll come back and we'll do it together. Okay, so, let's try it here on member free body diagram of member which is the nut. We'll go ahead and, let's assume that this distance is l. And I'm going to call this point P down here. And I'm going to start by summing moments about point P, setting it equal to zero. I'll call counterclockwise positive. And so I've, I only have one force that's going to cause a rotation about point P. That's F1. It's causing a clockwise rotation. So that's negative. So I have minus F1, times its moment arm, which is l, equals zero. All of the other forces line of action go through point P, so they don't exert any kind of a moment. And so, from that result I find that F1 equals 0. And so let's go ahead and write that on our diagram. I can apply another equation of equilibrium. I'll go ahead and say the sum of the forces in the X direction equals zero. I'll choose to the right positive. I get F1 plus F2 equals 0. But we found F1 itself was equal to 0, and so that tells me that F1 and F2 are both zero. And I have one more equation of equilibrium. Let's go ahead and sum forces in the Y direction, set it equal to zero. And I'll get N2 which is up for positive, minus N1 equals 0 and so I can't solve for N1 and N2 explicitly but, I can find the relationship between the two and so I know that N2 is equal to N1. Let's do the same thing on the pin. And so, on the pin, let's start by summing forces in the X direction, setting it equal to 0. I'll choose to the right positive, and I've got Rx1 minus Rx2 equals 0. And so, what I find is Rx1 equals Rx2 and so I'll just go ahead and call those Rx. So I have Rx2 equals Rx1 equals, we'll just say Rx. I can sum forces in the Y direction. Set it equal to zero. I'll choose up as positive for my sign convention, and I get Ry1 minus Ry2 equals 0, and so that states that Ry2 equals Ry1 and I'll just call those Ry. And so Ry2 equals Ry1 equals Ry. we could also sum moments but, If you try it, I want you to try it on your own. You'll find out no matter what point you sum moments around on the body of the pin, you'll get no more information. So sum of the moments in this case, and you can try it on your own, doesn't add any information. So at this point, we've gone as far as we can. we still haven't ex, solved explicitly for N1, N2, Ry or Rx and so we're going to have to draw one more free body diagram. Let's choose body B1 with a pin included. So I'd like you to draw a free body diagram of B1 with a pin included yourself and then we'll do it. Together. Okay, this is the result you should have come up with. you will notice that I've drawn Rx and Ry, as being from body B2. So they're opposite the direction that they were on the pin acting on body B1 and the pin together. So, how many equations and how many unknowns do I have now? Well, I'm going to have three equations. And I only have three unknowns. This being N1, Rx, and Ry. So we can solve for those directly. You can try that on your own. But, I'm also going to do it together with you. We have some of the forces in the X equals zero. I'll choose to the right positive, and I've got minus F1 plus Rx equals 0. Since we know that F1 equals 0, that results in Rx equaling 0, which means the forces on both side of the pins equal 0 in the horizontal direction. We've got sum of the forces in the Y direction equals 0. I'll choose up as positive, and I've got N1 plus Ry plus 10 equals 0. that's two equations, one unknown. I'm going to need one more equation. I'm going to go ahead and solve er, sum the moments about point A so I can solve N1 directly. So I'll sum moments about point A, set it equal to zero. And, I'm going to get N1 causes a clockwise rotation that's opposite my sign convention. So it's N, minus N1 times its moment arm, 1.5. And then, I've got ten causing a counter clockwise rotation. So that's 10 times its moment arm of 3.75. Equals 0. And so if I solve that, I'll find that N1 equals 25. So I'll put N1 of 25 up here, which will lead me to Ry equals minus 35. And so Rx was equal to 0, Ry was equal to minus 35, and therefore, my final answer for the pin's force is nothing in the horizontal direction. Since the value is negative, I've got Ry opposite the direction I've drawn, 35 pounds. And that's my answer there. Do the same thing for the nut. No forces in the X direction. But we have two compressive forces, woops, 25 pounds and 25 pounds for my other answer. So you can see I exerted ten pounds of force on the handle, my mechanical advantage gave me 25 pounds on the nut and that's the end of the problem, so I'd like you to do one more problem on your own. There's a work sheet here. And I've provided the worksheet solution in the in the module handouts so that you can practice on your own. See you next time.