Welcome to module 24 of an Introduction to Engineering Mechanics. For today's learning outcome we're going to continue to employ the rules for drawing a Free Body Diagram and Sketch Free Body Diagrams for some bodies. It's a really important tool so we're going to spend some time on it to cement that concept, and make sure that you understand how to do this. So the first thing that we're going to do is draw the free body diagram for this homogeneous bar. The weight of the bar is 2 pounds. It's a little bit different than what we saw last time. So let's go ahead and look at the model. This time we have our pin reaction, on the on, on the right hand side. We still have our 5 pound weight acting down on the left hand side. But now instead of a roller reaction, or a roller constraint, I have a cable that's hooked from my beam up to the wall here, keeping it from, from going down. And so let's go ahead and, and, and draw that free body diagram. Okay, so the first thing I want to do again, as always is to sketch the body free of constraints. So here's my body. This time again I want to label some points. And so I'm going to call the point where the cable attaches point A. I'll call the roller reaction, or roller constraint point B. And, you can even call the place where the 5 pound weight axe is point c, so let's put those on our free body diagram. And we did last time if you, if you forgot, you should go back and review, but we did last time the roller or spin, excuse me, the pin constraint and the five pound weight, and the weight of the beam itself. And so let's draw those on. So we had a 5 pound weight. No different than last time. We've got a, two force reactions due to the pin Bx, and By. And we've got a weight down, of the beam itself. Which in this case is 2 pounds just given and the only constraint that we have to replace now with, with a first reaction is the cable itself. So lets see what that cable does to my beam. If I take this weight off the right hand side. I'm going to leave the roller reaction on there for a second. But, we can see that this or excuse me, the pin restraint. We can see that, cable, what it does is it Puts a tension force on my beam along the line of action of that cable. So there's one force reaction just in the line of action of the cable itself. And so I'm going to remove that, and I'm going to remove my pin constraint, and on my free body diagram I'm going to draw that. Draw that cable we placed with a force reaction. Now just call that a tension T and we got to put it. The dimensions on there, so we're going to put it on a 5 on 12 slope since it's only in one direction. Last thing we do is add other dimensions to the body. So we've got from point, a to the left hand side is nine inches. The distance to the weight will be the middle of the beams, that's 11.5 inches. And, we have 14 inches from the point a to the right hand side. Okay, so that's a good free body diagram. Let's go ahead and do another one. I'm going to change models here. In this case I've got what's called cantilever beam. Okay, and we're just, the body of interest in this case is this beam up here which is like a diving board if you will. And so on the right hand side or excuse me on the left hand side here, we're going to say that there's a 50 pound force acting down and the bar itself weighs 15 pounds. So that's the physical model scale model of, of this system. And over here on my, my slide. I've got a representation of that canti, what we call a cantilever beam. So first thing I did was identify the body of interest. We said it was this, this beam here. We're going to go ahead and draw it free of constraints. I'm going to put on any external forces or couples. In this case lets label some points here again, are called the, left hand side, o already have label A and B so I'm in good shape. So on the, on the left hand side here I've got a 50 pound force acting down. In the middle it says, if it's a homogeneous beam, which is it says our bar, we have a 15 pound weight on the beam, acting down. And let's label those points. This point B, this is point A. The only other thing we need to put on there as far as force's or moments are concerned is due to the constraint. The only constraint we have is that this right hand wall where the can lever is attached. And so let's look at beam model our bar model again. And this case over here where the can lever is, is, is attached to the wall, it does not allow motion left or right. It is not, does not allow motion up or down, so there's going to be a force reaction in the x direction and in the y direction. And it also doesn't allow rotation and so unlike the pin where I could rotate around and there was no force, moment reaction, in this case there's a constraint for, from rotation, and so I'm also going to have, a moment reaction for a cam lever beam like this. And so let's go ahead and draw that, on, my free body diagram. So I've got, as before two force reactions, A sub y and A sub x but in addition, it's preventing rotation. So there's also a moment reaction I have to include so I'm going to put moment sub A, and that. The last thing we have to do is add the dimensions. Go to the center, it would be 5 inches to with the weight of the bar axe, in the overall dimension is 10 inches. And that's a complete and accurate free body diagram. So you should be getting pretty good at this by, by now or understanding it quite well. So at this point what I want you to do is, I would like you to draw the free body diagram for this homogeneous slender bar AB that's shown here. I want you to go ahead and neglect the thickness, okay. So we're going to say that the thickness is, is negligible compared to the lengths. And I want you to also neglect the weight of the beam. We'll say that these, this force, external force being applied, and this external couple being applied is much more than the weight of the beam. And so we can call the weight of the beam insignificant. And so, that's what Id like you to do for homework, additional homework. And then that's the free body diagram of this structure is what I'm going to pick up with next time, and we're going to actually solve for the reactions on this beam or this bar. So I'll see you next time.
