Hi and welcome the module 21 of applications of engineering mechanics. We're going to continue on with the problem we started last time for cable support system with concentrated loads. And so far we found the reactions at A and B. We found the tension in this cable. And we found the Sag at P2. And so this is where we left off last time. what do you do next? And so what you should do at this point is do another drawing cut or section cut to continue to solve. we'll do a joint cut at point P2. And if I draw that free body diagram, I've got my 1,400 pound force up and to the right. I have the 900 pound force down. I have two components of my cut between P1 and P2, I've got an x component and a y component. I know that the x component, again, is the same throughout. Because there's only, there's no external forces in the x direction. So, I know right away. That T12 in the x direction is a 1109.9 pounds, and so I can take that joint cut now, and all I have to do is find the T12 in the y direction, its y component. So, to do that what should you do? Okay and so what, what you should have said is, okay I can just sum forces in the Y direction so that it will be equal to 0. I will choose up as positive and so I get the T12 y component minus 900 plus the y component of the 1,400 force, which is 1,400 times sine of 37.56 degrees equals 0. And if you solve for T12 Y now, you find it's equal to 46.67 pounds. So now I have T12 x component and y component. I can use Pythagorean's theorem to sum those up together. So T12 will be equal to the square root of T12 X squared plus T12 Y squared, T12 X is 1109.9, T12y we just found to be 46.67. So T12 overall is equal to 1111 pounds in tension. Listen, now I've got this section, and this section, I'v got Y2, I still need the tension in that segment and the sag at P1. And so, think about what you do next and then come on back. Okay, there's, there's where we stand that this point Let's go ahead and do a joint cut at,at A, to try to find the tension in the cable segment from A to P1, I've drawn a free body diagram of the cut at joint A, I know AX and I know AY. I know the tension in the A1 section has to be 1,109 pounds again. 1,109.9 pounds because of tension in the x directions, the x components are the same throughout. I don't know the tension in the in the y direction and so you can do that. do that on your own, come on back. And then add it up, and find the overall tension in the segment from point A to point P1. Okay? What you should have done, sum, sum forces in the y direction, you get TA1 in the y direction, is 546.7 using Pythagorean's theorem to add those x and y components together. You get the overall tension in segment A1 to be 1,237 pounds of tension. So, all we have left now is to find the sag at point P1, and so here's again a look at joint A. I'd like you to now go ahead and solve for Y1 using geometry and similar triangles, I did it before in the last module and so you should be able to do that on your own this time, but if you do it then come on back, we'll do it together, okay. So, let's look at these similar triangles here. this 546.7 eh, relates to the y1 side, and 1,109 relates to the eight side, so I've got Y1 is to 546.7 as eight is to 1,109.9. And so y1 ends up equaling 3.94 feet which equals the sag at P1. And so we finished all the parts of the problem. And so that's the end of the section on cable support systems supporting concentrated loads, we'll come back next module and a look at suspension type systems like this suspension bridge. See you then.