Hi and welcome to module 20 of Applications in Engineering Mechanics. Today's learning outcomes a are to continue with a cable systems supporting concentrated loads and find the tensile loads in each of the cables, and to determine an unknown sag. So here's the next problem that we're going to look at. I don't know either of the sags at point P1 or P2, but I do know in this case that the maximum tension allowable at any segment of the cable is 1400 pounds. First thing I'd like you to do on your own is to determine the reactions at point A and B. And so you can go through and do that, and once you've completed that come on back, and I'll show you what you should have done together. 'Kay so you should have drawn, whoops, a free body diagram of the entire structure. And if you did that, some moments about A, you could find By, some forces on the y direction, you find Ay. And if there is some forces in the x direction, you can't solve for Ax or Bx explicitly again, but you can solve for their relationships, which is that they're equal. And so having done that. those are the results. We want to find Bx and Ax, and so what I'll do is, I'll go ahead and use the fact that the x components are constant throughout the cable, and so we can therefore find the section with the largest tension. And the one with the largest tension is the one that will have the largest y component. Because if you take the x component which is the same throughout and you use Pythagorean's theorem to find the overall tension in the segment. The one with the y component that's the largest will be the segment with the largest overall tension. And so in this case, that will be section BP2, because it has the, the highest slope. And so in doing that, we find that TB2 then is1400 pounds in tension. So we've solve for one of the ta, cable segments. and then let's do a joint cut at B to find B sub x. And so here is the joint cut a joint B, I know what it, B sub y is, I don't know what B sub x is. I do however now know that the TB2 is 1400 pounds, I just don't know the angle at which it'd, which it's at. And so to do that, here is my joint cut again, what I'm going to do is I'm going to apply the equations of equilibrium to that joint cut. First of all some forces in the y direction, set it equal to zero. I'll choose up as positive. And so I'll get 853.3 minus the 1400 pound tension in B2. But just it's y component which is going to be the sign of theta B2 equals 0. so I'll find sine of theta B2 equals 0.6095, or theta [COUGH] excuse me, B2 is equalled to 37.56 degrees. Now knowing that I can sum forces in the x direction. Set it equal to 0, and I get the x component of the 1400 pound force which is going to be negative according with my sign convention. So this is going to be minus 1400 times the cosine of theta B2, which was 37.56 degrees, plus BX equals 0. And so therefore, BX equals 1109.9 pounds to the right, which means that Ax is going to be equal to 1100.9, 1109.9 pounds to the left. Okay, so now I have the reactions at A and B done. I found the, the tension in that cable segment because it's limited to 1400 pounds. before I find the tensions in the other cables, we'll do that next module, let's go ahead and determine the sag at point P1, excuse me, at P2. And the way I'll do that is again, I'm going to draw this section. So, there's the joint cut at B. Bx is now found, By was found, I know that TB2 was 1400 pounds. And so I have this distance is 6 feet. I don't know what the distance y2 is, but I can use the geometric relationship using the angle here. So I have got opposite over adjacent will be the tension function. So the opposite side is y2 over 6 is equal to the tangent of 37.56 degrees. Or y2 ends up equaling 4.613 feet, which is the sag at point P2. So all we have left to do in the problem, and we'll pick up next time in the module 21, is to find the segment and this segment's tension. See you there.