Welcome to module four of Applications in Engineering mechanics. Today's learning outcome is to apply the 2D equilibrium equations to solve a real world problem. We're going to do a frame problem. This is a frame, it's not exactly the same but it's, it's very much like a frame structure of my, stepladder that I've used several times as an example in this course. [SOUND]. So, in this problem we have 150 Newton force, or applied at pin B, we have the, the weight of the structure to be considered negligible with no, with no weight. And we have the question of finding the force exerted at point B by. On to member ABD. And so what do you, what should you do first to solve this problem? I'd like you to think about that. What you should do is sketch the free body diagram. And so the question become, what free body diagram of, of what body should you sketch and once you've decided that, go ahead and sketch that free body diagram. I chose to go ahead and, and, and draw a free body diagram of member ABD. And the reason I did that, was because the problem statement said, what's the force of the pin at point B on member ABD? To decide, Or to, to, to draw that free body diagram, let's first look at a top view of the structure. You can see we have the 150 Newton p-, force applied at the pin. And we have member B, C, E, C connected here. And we also have member A, D, B connected to the pin. So at the pin itself. We have two force reactions. Bx and By, I'll just assume them to the right and up. We also have a pin reaction at point A, so I'll draw Ax and Ay. Again I have just assumed the direction. If you look at the entire structure, you'll see intuitively that if the 150 newton force is pulling to the right that Ay should actually be down. I've drawn it up, no problem there. If I end up with a negative number for Ay, which we probably should, that'll indeed confirm that the Ay force is down. The last force we have on member ADB is from the two-force member DE. I don't know whether it's in tension or compression. I'll just go ahead and assume that D is in tension. So I've got Dx to the right. Okay, how many equations do we have now, and, and how many unknowns are there? Go ahead and do that yourself. What you should say is, okay there are three equations of equilibrium, three independent equations of equilibrium. And we have five unknowns, Ax, Ay, Bx, By, and Dx, and so that kind of gives us a sad face here because we can't solve the problem yet. So the question becomes, what do you do next? Well, what you should do next is, draw another free body diagram. And I'm going to choose to draw a free body diagram of the entire structure. the reason I'm going to do that is, if I do the entire structure, I'll actually be able to solve for A x and A y, and we'll be able to use them back in the free body diagram of A B D. So I have my A x. Excuse me. That's Ay. And my Ax. We have a force reaction at C as well. That's a roller. You can see from my demonstration here that a roller allows motion perpendicular to the plane. In this case, the plane is at 60 degrees, so it doesn't allow motion perpendicular to the plane, but it does allow motion along the, the, the direction of the plane. And so here on my diagram, I'm going to draw A as shown. and now, the question becomes, what do you do next? And what you should say is, I need to solve for the unknown. So I'd like you to take some time to apply the equations of equilibrium. And you should be able to solve for the unknowns in this particular case. Alright, let's go ahead and do that together. Alright, if you got stumped as we, you went along let's begin by summing moments about point C. Applying equation for summing the moments about point C. I'll choose counterclockwise as positive, for assembling my equations. One of the reasons I chose the moments about point C is the line of actions of AX and C both go through point C and won't cause a moment. Therefore the only unknown I'll be able to solve for is AY directly, that's a good thing. So I have AY causes a clockwise moment about point C. My sign convention is counterclockwise positive, so that's going to be minus Ay times it's moment arm, the perpendicular distance between the line of Ay and C is three. And then I have this 150 Newton force, which is also going to cause a clockwise rotation, so it's negative. 150 times its moment arm, which is 4, equals zero. And so, I end up with Ay equals minus 200. That checks, because we had drawn it up. We said, phsysically, it looks like it has to be down. And it does come out to be down, because Ay ends up being negative. Alright, let's do another equation of equilibrium. At this point, let's go ahead and sum forces in the Y-direction. And you don't have to do these in any particular order, I'm, I'm just going through the process that I used. But you'll get the same answer no matter how you go through the sequence. In this case, I've got Ay is up. As shown with the arrow. It's going to be a negative value, but it's up, so that's positive. And then I have plus C times its component in the Y direction, which is sine of 30 degrees equals 0. I'll go ahead and put in my value of Ay of minus 200. And I get C equals 200 over sine of 30 degrees or that's a value of 400. And so C has been solved for as 400. And then finally, to get Ax, we'll sum forces in the X direction. I'll choose to the right, positive. And so we have Ax is to the right, so it's positive. 150 is to the right, so it's positive. And then we've got minus C. It's X component is cosine of 30 degrees equals 0. I'll put in my value of C of 400 and I'll get AX equals 196. Okay, so we've solved for those reactions. What do you do next? Take some time, figure it out, and then come on back. Okay, what you have, should have figured out that you do next is to go back to member ABD where we had five unknowns. We've now solved for two of those unknowns. And so we'll apply the equations of equilibrium and solve the, the last two or three unknowns. And so I'll start, I'll go ahead and sum moments about B to get us started. I'll choose again, counterclockwise positive. And I'm going to get Ay. Ay as drawn is going to cause a clockwise motion, so that's going to be negative. Ay times its moment arm, which from point B is 1.5, and then I've got. Ax which is going to cause a clock, a counterclockwise rotation about these, so that's going to be positive in accordance with my sign convention. So I've got plus Ax times its moment arm which is four, and then I also have Dx, so plus Dx times its moment arm, which is 2, equals 0. Go ahead and put in the values of Ay, which is minus 200. And Ax which is 196. And you can then solve for Dx which ends up being minus 542. And so, although we assumed member DE, the two force member in tension, it actually ends up being in compressions because we get a negative value. At this point I want you to go ahead and pick up and and do the rest of the problem. How do you solve for BX and BY? Go ahead and do it and then come on back. Solving for Bx and By should be simple. Just sum forces in the X-direction. You get Bx equals 346. Sum forces in the Y-direction, you get By equals 200. And so we now know the complete vector forces on point. At point B on member ADB exerted by the pin. It's 346 in the i direction and 200 in the j direction Newtons. That's a great problem. Why don't you take some time and do another problem to cement these concepts. This is a worksheet I'd like to do on your own. And I've included the solution in the module handouts, and after you've completed that we'll see you next time.