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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

equation, and find out what T3 is.