Hi, and welcome to module twelve of applications in engineering mechanics. Today, we're going to continue that space trust problem that we started last time. Last time, we found the reactions at pins A points A, B, and C. Now we're going to go to the next part which says, using the method of sections, with a section cut plane par- [COUGH] by a plane parallel to the xz plane, find the force in members AD and CD. So we're going to want to cut member AD, member CD, and we want our entire cut to be parallel to the x z plane. And so there is my section cut. And each section of this space trust needs to be in equilibrium, so I'm going to go ahead and draw a good free-body diagram of that section. I'll assume all the members in tension. So I'm going to have F from B to C, F from B to A. F from D to C. And F from D to A. Okay. I also shown the results on our space trust for the reactions that we solved for last time, and you can see that in the section that we're looking at or exploring, we have the force of B in the x direction. And so that's the free body diagram. Let's go ahead and look it here again. And so, since this is a three-dimensional problem, if you recall back to my course, Introduction Engineering Mechanics, we're going to have to, to work on this problem using vectors instead of scalar form. And so we're going to have to express all of these forces in vector form, or the forces that we want in vector form. So I'm going to start by doing the force DC in vector form. And so, we need a magnitude and we need a direction. To do the direction, we're going to take the unit vector from point D to point C. And so, the unit vector we'll give the symbol e. From D to C, and if you recall back to that earlier course, you want to walk from the tail of that vector to the head of the vector. And so we're going to find the position vector from D to C, and then divide it by the magnitude of that position vector, in order to find the unit vector in that direction. So, in going from D to C. We have to go 3 units in the negative x direction, 4 units in the negative y direction and 12 units in the positive z direction. So we are going to have minus 3i minus 4j plus 12k over the magnitude of that vector that position vector which is the square root of the sum of the squares which in this case ends up being 13. And then FDC. The force expressed as a vector is equal to the magnitude of the force, times it's direction. We just found the direction to be the unit vector eDC. And so, we have FDC's magnitude times minus 3 over 13i. Minus 4 over 13 j plus 12 over 13 k. So that's, that's F DC expressed as a vector. What I'd like you to do now, is to express FDA as a vector. The other force that we want, want to find in member DA, or AD. Okay, here is the position vector eDA, it's all in the j direction, negative j direction, and so the force DA is minus for, FDA in the j direction. So now that we have both of those forces, I've shown them up here. those are the only two forces we want to find, or need to find in this problem. So what do you do next? Think about what you would do next, and then let's come back and talk about it. Okay, well what you should have said was I, I'm going to have to apply an equation of equilibrium. To this section that needs to be in equilibrium. And my next question to you is, which equation of equilibrium might you use to find FDC and FDA? Okay, what you should have said was probably the easiest. Or the quickest equation of equilibrium to use, would be the sum of the moments about B. Because we don't need the forces FBA, and FBC, they go, their line of action goes through point B, so they won't be involved in this equation. And we can solve, we should be able to solve for FDA and FDC directly. So, let's go ahead and do that. [COUGH] So I've got the sum of the moments, and again, I've got to use the, the vector form of the moments here, so I'm going to go r cross F like we learned in our earlier course. We've got r from B to D crossed with FDA. And then we have plus r from B to D, crossed with, the supplied force, 100i minus100k. Plus I have one other, force that's going to cause a tendency for rotation about point B, and that's FDC, so again, it's position vector is r from B to D crossed with FDC. And all of that has to equal 0. We'll go ahead and put our values in. actually I wrote this incorrectly here, this should be not r from D to A but the position vector r from B to D. I think I set it correctly and wrote the wrong term. So, all of these position vectors are to a point along the line of action of these three forces, and that's going to be three feet in the i direction, so I have got. RBD is 3i crossed with FDA we found to be minus FDA in the j direction, plus RBD again is 3i crossed with F which is 100i minus 100k. And then I have plus rBD, which is 3i, crossed with FDC, which is minus 3 13ths FDCi minus 4 13ths, FDCj and then plus 12 13ths FDCk. I am running a little bit out of room here but hopefully you can follow along, and all of that has to be equal to 0, and if you do that mathematics if you do those cross products, you should come up with minus 3FDAK plus 300j minus 12 13ths, FDCk. Minus 36 over 13 FDCj. So there's my equation of equilibrium. And so the next question is, where would you go from here? What do you think you should do? Try it, go as far as you can. And so what you should have found is, now that I have this equation of equilibrium, you should say well, what I'm going to do is I'm going to need to match components in the i, j, and k direction, to solve for the unknowns. In this case, I don't have any i components, so I'll go ahead and do the j components. And so I get, let's see, 300 with this j component minus 36 over 13 FDC equals 0. And so FDC then. Is equal to 108.3, since it's a positive value and we assume FDC was in tension. That means that it is actually in tension, so FDC is equal to 108.3 pounds in tension. And so that's one of our answers. And what I'd like you'd to do to finish up the problem is, match components in the k direction to find F, from D to A, or the force in member AD, and once you have that, come on back. So what you should have found was FDA, ends up being a negative value since we assumed it to be intention its negative, it's compression and the value is 33.3 pounds and that finishes up this part of the problem. See you next time.
