This is module 13 of Applications in Engineering Mechanics. Today we're going to continue on that space truss we've been working on for the last couple of modules and finish up that problem. So, we're up to the point where we want to find the forces in members AB, BC, and BD using the method of joints, and so we want to find the force in members AB, BC and BD, so I want to cut through all those members. I'll join that cut together and come up with a joint cut around joint B. the whole structure has to be in equilibrium. Each piece of it has to be in equilibrium. So that joint has to be in equilibrium, therefore I'm going to use my graphical tool for equilibrium, which is the free body diagram. I'll assume these members are intentions, so I've got FBC, FBA and FBD. Plus I've got this reaction force FBX, which is negative 75 at point B as well. And so that's a good free body diagram. Here it is again. And so what should you do now? And what you should say that you have to do is well, this is a three dimensional problem, we're going to have to work in vector form. So we're going to have to put these forces in vector notation. And so I'll, I'll, I'll do one of those with you, I'll take EBD or FBD. I need the unit vector in the BD direction. So that's going to be BD unit vector is equal to the position vector from B to D divided by the magnitude of that position vector. So that's 3i. And going from B to D over the magnitude which is 3. So the unit vector is completely in the i direction. That means that FBD is equal to the magnitude of FBD, times its direction. And so that takes care of FBD in vector form. Let's to BA next. So, we've got the position vector from B to A, or the unit vector from B to A will be the position vector from B to A divided by its magnitude again. And so the position vector from B to A is, we have to go 3 in the i direction and minus 4 in the j direction. So 3i minus 4j over the magnitude of that position vector, which is 5 square root of the sum of the squares. And so now again I can find [COUGH] the force FBA as a vector, its equal to its magnitude times the unit vector. 3 5ths i minus 4 5ths j, and that is an expression of FBA as a vector. What I'd like you to do is to now do FBC as a vector on your own. And you get that done, come on back. Okay, this is what you should've co, come up with for FBC, using the same technique I did above. And at that point, I'd like to say what do we do next? What, where, what, what are you going to to do to continue the problem? If you can continue it, try to do it all the way to the end, come on back when you get stumped. And what you should do next, is we have these three forces as vectors, these three unknown forces as vectors. We have this known reaction force, and so we are going to have to apply the an equation of equilibrium. In this case, this is a concurrent force distance, so a moment equation is not going to do us any good. So let's go ahead and sum forces vectorially at point B to come up with our equation of equilibrium. And so in this case I've got minus 75 in the i direction for that force plus each of three forces. So plus FBD i, plus 3 5ths FBA i, minus 4 5ths FBA j. Minus 4 over the square root of 160 FBC j, plus 12 over the square root of 160 FBC k. And all that has to equal 0. And what next? So, what you should say that we do next is solve for these unknowns by matching components. By matching components, I'm going to come up with three scalar equations: one for the i, one for the j and one for the k direction. I have three unknowns, so I should be good to go. So I'll do the, the k components here to start. And for the k components ,I've got no k, no k. The only term I have that has k components is this last sum of terms. So, I have 12 over the square root of 160 FBC has to equal 0. On the right hand side, there is no k components, and so that means that FBC must equal 0. And so that is one of our answers. And now I am going to do the j components. And for the j components, I've got no j component, okay, there's the first j component. Minus 4 5ths FBA, and then I've got minus 4 over the square root of 160 FBC equals 0. But I just found that FBC was equal to 0, so what that leads me to is the conclusion that FBA also has to be equal to 0. And so I've got my second force in member FBA or member AB. Okay, why don't you go ahead and do the last one by matching the i components. Come on back and see how you did. There's the i components matched, and what you should have found was FBD in that case comes out to be a positive value. We assumed it was in tension, so it is in fact in tension. And it's 75 pounds for its magnitude. All right, that takes care of that part of the problem. Let's go ahead and do the last part of the problem, which is find the force in member AC by inspection. If I look at this joint A here and I look in the z direction. I can see if I isolate this that the only force that it could possibly have in the z direction is the z component of FAC. And since that's the case, it must be equal to 0 by inspection. So you should be able to say that member AC is 0 by inspection. But if you can't, you can always do a free body diagram of that joint. And so if I put my force here, FAC, FAB and FAD. If I sum forces now at that point A just in the z direction. So I'm going to sum forces in the z direction at point A. They have to be equal to 0. And in this case, we only have the z component of FAC equal 0, which means the overall force FAZ has to be equal to 0. And so, that is our answer. And so we've completed that problem. I have another space pro space truss problem for you to complete on your own for practice so that you get real good at these problems, and I've put the solution in the handouts for this module. And we'll see you next time.