Okay. Welcome to Module 7 of an Introduction to Engineering Mechanics. Today, we're going to take many of the concept that we, concepts that we've learned in previous modules and we're going to go ahead and solve the two-dimensional or 2D equilibrium problem. This is the problem we're going to look at or examine and solve. We have a point or a particle O here that has a tension and a cable OB pulling to the right. We have a tension up into the left and cable O, A AO. We have a 200 pound force acting down, and down and to the left we have a 100 pound force. And that's 100 pounds because throughout this cable is the same tension. Remember, we, we assumed that this pulley was a frictionless bearing, and we neglected its weight. And so, the tension in the cable throughout is the same, or 100 pounds. And so, the first thing we do in approaching this problem, these problems, is to draw a free-body diagram. And, we need to identify our body of interest. In this case, the body of interest is point or particle O, because that's where all the forces are acting. And, I now apply my external forces. And so, I have a, a tension in OB, that's cable OB to the right. And you might have, may have asked yourself in, in, in previous modules, or this module, why am I always drawing the tension, it, in tension? Why wouldn't I draw a force acting in compression back to the left? I have a simple demonstration here. I've got a my mouse cable. And you can see that I can apply tension to this cable but I can't apply compression. I can't apply a force back towards my, my hand. And so, there's a saying that goes, you can't push on a rope, and you can't push on a cable or a belt either. A cable or a belt or a rope can only be in tension, so that's why I'm always drawing them in tension. we also have our tension in, in cable OA or AO, that's up and to the left. And so, this is TOA and it's on a 5 on 12 slopes, that's a 5, 12, 13 triangle. Also acting at point O, external forces, there's a 200 pound force down. A nd I've got a tension down and to the left that's going to be equal to 100 pounds. And that is on a four on three slope. Once I have that good graphical tool, that free body diagram, I can now apply my equations of equilibrium. I can apply them in either order, they're, they're independent equations, so I can do some of the forces in the x first or some of the forces in the y first. This time I'll go ahead and choose to do some of the forces in the y direction first equal 0. And, I will, for assembling my equation, I'll assume up is positive. I could assume down is positive. You may want to do that on, on your own to see that you get the same answer. And so, first of all, I have a tension in cable OA. I have the y component is the 5 side of the 5, 12th, 13th triangle. And so, by similar triangles, that's going to be plus 5 13ths TOA, and it's positive because it's pulling up. TOB does not have a y component. I have my 200 pound force, which is down, so that's going to be minus 200. all in the y direction. And then, the 100 pound force, its y component is only the 4 5ths component. So, we have minus 4 5ths times 100 equals 0. And so, I can solve for TOA. TOA is equal to, it'll come out to be 728 pounds. If I want to express it as a vector, I have to show the direction as well. And so, it's going to be up and to the left on a 5 on 12 slope. And so, that's my complete answer for TOA. So, I'm halfway there. And, you should ask your, yourself, how can I find now the tension in OB? And what you should determine is, well, I've used the equation for equilibrium in the y direction. Let's try to use that other independent equation to see if we can find TOB or, and so if I do that here is a sketch again of my free body diagram. And, here's my results from the last slide where I summed forces in the y direction and found TOA was equal to 728. Now, let's sum forces in the x direction. I'll choose right as positive for assembling my equation. And, first I have TOA. It's pulling to the left. Its x component is the 12 13ths component of TOA, so that's minus 13 13ths TOA. And then, I have TOB to the right, so that's positive according to my sign convention. It's all in the right, all to the right, so it's just plus TOB. The 200 pound force does not have an x component, it's all in the y direction. And then, finally, the 100 pound force has only the 3 5ths component in the x direction, and it's going to be negative because it's pulling back to the left. So that's minus 3 5ths times 100 equals 0. So, I have one equation now, two unknowns. But I've already solved for TOA. And so, I can substitute that in. That's 728. And that'll allow us to get our final solution for TOB which is 732 pounds. And again, if I want to express it as a vector, I should express it as a vector. And so, I have to put a direction here, and so I'll call it to the right. And that's my solution. That concludes today's module. Thank you.
