This is Module 10 of an Introduction to Engineering Mechanics. Let's start by looking at an overview of the course. Again, I refer to this as the force as opposed to the trees. You'll see that we've completed the first major topic and, Showing forces in 2d and 3d components. We've done particle equilibrium or a balance of forces on a particle and now we're in the third major topic which is moments, and we're calculating moments of a force about a point. In last module, we calculated the moment. In a 2D problem. This time we're going to solve a 3 dimensional moment problem. And so this is the same problem that we worked on in module number 4 and if you don't recall, you should go back and review module 4. But we found that we could do a 3D force representation of the force F and we found that it was minus 80i plus 240j plus 60k. Newtons and so now we want to find the moment of that force about point O and with 3D problems it's almost always easier to do the vector method. It's very difficult, many times, to find the perpendicular distance from the point about which you're rotating to the line of action of the force and so we just go ahead and use the. vector method. And so the moment about O is equal to the position vector from O to a line of action on, a point on the line of action of the force. In this case, we have a couple of obvious choices. Either point A on the line of action of the force, or point B on the line of action of the force. We'll actually do both of them, to show you that you get the same result. But let's start with R from O to A. And then we're going to cross it with the force itself. And so the moment, about O, is equal to R from O to A is 4 meters in the i direction, or 4i. And then we cross it with the force vector minus 80 i plus 240 j plus 60 k and if you do that cross product operation, you'll find that the moment about o is equal to minus 240. J plus 960K. And the units are force and distance so that's going to be Newtons-meters. And so that's our answer. So w e could have found the same result by using a position vector from O to B instead, and so let's go ahead and do that. So moment about O. Is equal to now we'll say r from O to B instead of r from O to A, crossed with the force vector. So the moment about O is going from O to B. We go 12 units in the y direction, or j direction, and 3 units in the k direction. So that's going to be. Plus 12 j, plus 3 k, crossed with, minus 80 i plus 240 j, plue 60 k. And again, you should. Do the cross product on your own, make sure you can calculate that, but you'll find that the moment about o again ends up being -240 j + 960 k Newton meters. So we'll get the same result. Okay. So, now that we've completed determining the force, or the moment of the force f about point o, what I'd like you to do on your own is to determine the moment of force p about point o in this same problem. And I've included a pdf of the solution in the module handouts. So, please work it on your own, and then check yourself out to make sure that you can acheive this learning outcome. And we'll see you next time.