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Hi and welcome to Module 6 of An Introduction to Engineering Mechanics.

Â Today's learning objective is to calculate the forces acting on systems of particles

Â or multibody. Problems. This is a typical multibody problem. this

Â is an example from a, a textbook by Doctors Dave McGill and Wilton King.

Â Although I said in this Coursera course you didn't, there was no required textbook

Â I am going to use several figures and examples From, a textbook called

Â engineering mechanics, statics by, doctors McGill and King, and, I will give, I, I'll

Â put a little note on, below those figures, to sh, to show when I'm using those, but,

Â I know doctors, McGill and King, and they were kind enough. To allow me permission

Â to use material from their textbooks in this course. both of those gentlemen are

Â professors emeritus at Georgia Tech and so it's great that they're supporting this

Â Corsera course. Okay so here's the situation. We have this, this, this person

Â who is exerting a force in this rope. And holding up these systems of pulleys where

Â there's a weight, which would be known, weight 1 here on the right and a weight 2

Â here which is greater than weight 1. So if we go over here to a model. Got a model of

Â this situation so I'd be the, the giant person if you will, over here on the, on

Â the left hand side. That's holding this cable, as it's attached, and holding this

Â system up with its 2 weights. So, let's go back over here, and look at the

Â assumptions that we would make in solving this problem. Typically, for these

Â problems, we, we, we make several simplifying assumptions, these, pulley

Â problems, to help us to. Getting a, a, a, a simple but an, a fairly accurate

Â solution. If we find that these assumptions are, are not good, and we need

Â more accurate an, a, more accurate answers, we may need to go back, and, and

Â relook and include some of this some of these forces and things that we're

Â neglecting. So, one of them is that we assume that there's frictionless bearing,

Â bearings in these problems. We assume that there's a, a constant tension throughout

Â the cable. Or the belt or the rope, and we neglect the pulley weights. And so let's

Â look at these assumptions again over here on, on, on my model. And so again, if I'm

Â the, the, the man here the giant person over here on the left. I'm holding this

Â cable and we're saying that the, or this string if you will and we're saying that

Â the tension in this string is, is the same throughout, over to here, and the tension

Â in this String or cable is the same throughout. We're assuming that these

Â bearings are frictionless, okay. We're also neglecting the pulley weights. We're

Â saying that the pulley weights are, are much, insignificant when compared to the

Â weights themselves. And so those, those are some, some good assumptions to

Â simplify the problem and still give us a fairly accurate answer. So in solving this

Â problem, we take our system which has to be completely in static equilibrium and we

Â apply the equations of equilibrium, and we talked about these last module. we must

Â have a balance of forces in the x direction and a balance of forces in the y

Â direction and so let me draw my x and y directions here. Here.

Â Lets call the left x, and to the y up. And so this whole system has to be an

Â equilibrium, and each piece of the system must be each, an, an equilibrium, to

Â satisfy static equilibrium. And so, what I'm going to do, is I'm going to separate

Â this thing out, and I'm going to look at just 1 little piece here. I'm going to

Â look at this Pulley P1, and I'm going to draw a free body diagram of that. So this

Â is a free body diagram of pulley 1, or pulley P1 if you will. And so I separate

Â that pulley from the rest of the world here, so I'm drawing it Unconstrained.

Â Here is the center. And now I apply the external forces that are acting on that

Â pulley. And so I see that on the r-, on the right hand side here I have this, this

Â chain if you will and so if I cut that chain there's a tension in that chain

Â which is equal to The weight one, and that's going to be pulling down on the

Â right-hand side of this pulley. So, I've got, w1 here. One of our simplifying

Â assumptions was that the tension in that cable is the same throughout, so if I cut

Â it over on this left-hand side and apply a force reaction there, I get w1 down. I

Â have this a connection of the set of the pulley to the ceiling and so I have to cut

Â that and there'd be tension in that, I called that tension 3 and then I have a

Â tension in this other cable which is coming down from. the center of the pulley

Â and if I cut that, that tension, I'll call t2. And I want you to note on t2 that if I

Â follow the tension in t2 all the way around here, that it's actually the

Â tension or the force that the person must exert to hold the system in equilibrium,

Â so let's make a note of that. We'll note that T2 is the force the person exerts. So

Â that's going to be our answer. , So now we apply our equations of equilibrium, as I

Â mentioned. here they are on the right. We've done 'em in the last module. We

Â don't have any forces in the x direction in this problem, so we don't have to worry

Â about this first equation of equilibrium, so we'll just sum forces in the y

Â direction. And, we'll make sure they're balanced, so we'll set them equal to 0. I

Â have to choose a sign convention for assembling my equations. This is

Â arbitrary, and I'll show you that it doesn't matter when we do this. So, what

Â I'll do to start here, is I'll just choose up, as being positive. And so when I

Â assemble my equation, I have T3 was, is up, so that's going to be positive. And

Â then I have weight one is down, so that's going to be negative and in accordance

Â with my sign convention. T2 is down so that's negative, and the weight one on the

Â other side is down, so it's negative. And all that has to equal zero. And so I end

Â up with T3 = 2 W1. Plus T2. And so that's my equation that results

Â from that force balance. Just to show you that it wouldn't have mattered If I had

Â chosen my sign convention as being down arbitrarily, let's just redo that. So I

Â have some of the forces in the y = 0, I'll choose down is being positive. And I would

Â get, well in this case T3 is up, so it would be negative. -T3.

Â W sub 1 is down, so that's positive in accordance with this sign convention.

Â That's plus W1. T2 is down, so that's plus T2. W1 is down so that's plus W1 = 0. And

Â you'll note, that if I carry T3 to the other side, add the T1s, W1s together that

Â these two equations are exactly the same. And so you get the same result regardless

Â of the sign conven-, convention that you've arbitrarily chose to assemble the

Â equation. Okay, so, the equation that results though, we have a problem here

Â because we have, how many, equations? Well we have one, equation. And let's count the

Â number of unknowns. Well, we would know W1. We'd know that weight. Given in the

Â problem, but we don't, don't know, tension T2, which is what we want to find. That's

Â the force the person exerts, and we don't know tension T3. So we have 1 equations, 2

Â unknowns, and so we have unhappy face. and so what do we do next? And you may want to

Â think about this for a second and then come back. Try to think about what we

Â would do next. Okay, what you thought about that what you need to do in this

Â case is. Okay, every piece of this system has to be in eq. Let's take another piece

Â and draw another free body diagram, so we're going to try. Another free body

Â diagram, and the one I'll choose next is, is pulley p2. So, at the top here I've

Â written my analysis from pulley p1. I, I arrived at the equation t3 = t. 2W1 + T2,

Â so 2 equations, or 1 equation, 2 unknowns. Now lets do a free body diagram of Pulley

Â P2. ,, , And in fact I'd like you to take a few minutes on your own and draw that

Â free body diagram and then come back we'll do it together. Okay, now that you have

Â completed that here is my body pulley P2, here is the centre, I apply my external

Â forces I have, this was t2, and on Pulley 1 in tension, so it was pulling down on

Â Pulley 1, it's going to be pulling up on Pulley 2. This tension in that cable is

Â the same throu ghout, so it's also, if I cut it on this left-hand side it's going

Â to pull up on the left-hand side t2. In the middle here, if I cut this chain

Â remember the tension in the, in this chain was the same throughout and it was equal

Â to W1. So that's up W1. and then finally I have tension is this rope which is hooked

Â to weight 2 and so that's going to be down. And that's weight too, so that's a

Â good body diagram, and that's what you should have arrived at. Once we've got

Â that, we can assemble our, our equation again. We'll sum, there's no forces in the

Â x, so we'll only sum forces in the y direction. Set it = to 0.

Â I'll choose up as being positive. And so, I've got T2, positive, because it's up,

Â plus W sub 1, positive because it's up, plus T2 - W2 = 0. And if I solve for T2

Â now, I get T2 = W2 - W1 / 2. And so that's my answer. So if I'm given the values for

Â the weight 1 and weight 2, I, I subtract the 2, and divide by 2. That's going to

Â give me the force that the person must exert. And you'll notice that I get a

Â bonus in this, that I can actually, go up here, and put the result for T2, into this

Â equation, and find out what T3 is.

Â