Hi. Welcome to Module 17 of an Introduction to Engineering Mechanics. Today's learning outcome is to calculate the single force resultant for a parallel force system. So the last time, we found the single force resultant for a coplanar system. This time, we're going to find the resultant for a parallel force system, and this figure is a, is a rendering, my own rendering of a figure out of the, the McGill King text that I referred to in Module 6. So, we have a number of, excuse me, forces acting parallel to each other with an xyz axis here. If I want to find the resultant of that parallel force system, I add the magnitude of all those forces together, and that's going to give me the magnitude of the resultant force. To find where that resultant force acts, I take a summation of all of the individual forces, and the direct, and the, the distance in the x-direction from the y-axis, out to those forces. Sum it up, divide by the total magnitude of the, the resultant force, and that's going to give me the, the overall distance in the x-direction to the resultant force. Same thing for the y-direction. I take the individual forces, the distance from the origin or the x-axis, in this case, out to the individual forces in the y-direction. Sum all that up, divide it by the result in force. And that's going to give me the overall distance in the y-direction of the resultant force. And the best way to actually use this theory and understand is to do an example. And I'm going to do an example that's very similar to one the examples that are in the McGill King text. So, here is a problem. I've shown my, my definitions over here of how to find the results. to keep you on your toes, I've changed my coordinate system slightly. In my theory here I have x to the right, y out of the page, and z down. in this problem that we're actually going to do, I have x out of the board, y to the right, and z up. And so, the first thing we're going to do is we're going to find the total force resultant. And so, the force resultant is going to be equal to the sum of all of these individual forces, and so I've got it. And all of these forces are parallel to each other. Some of them are up, some of them are down, okay? And so, we have minus 20 in the z-direction. So, minus 20k. And we have a minus 60 in the k-direction. We've got a minus 80 in the k-direction. And we've got a plus 50 in theK k-direction. So, if I add those altogether, I get the resultant force is going to have a magnitude of 110 pounds, and it's going to be in the negative k-direction. So, this is going to be a 110 pounds down. But I'm not going to put it over here because my coordinate system is actually separate. So, we'll leave that off. Our, my coordinate system is different between these two so it would be down 110 pounds k. Alright. So, here's my result in force magnitude 110 down, now I've got to find the distance in x and the y-direction to that resultant force. And so, we'll use these definitions over here. So first of all, I've got x equals the sum of the individual forces times there distances in the x-direction divided by the resultant force. And so, in this case, we have the minus 20 pound force and the distance in the x-direction to that force is 1, 2 feet. This is times 2. And then, I've got my 60-pound force. This is going to be, and it's down so it's going to be negative, so plus minus 60, for down, times its distance, which is 2. And then, I've got a plus 50. 50 pounds is up, so, it stays positive times the distance in the x-direction to its line of action, which is one foot. And finally, I've got minus 80 times the distance in the x-direction, which is 2. I divide all of that by the total magnitude, which is minus 110 and I'll find out that the distance in the x-direction for my resultant force is 2.45 feet. And then, I do the same thing in the y-direction. So, y equals the sum of the individual forces times the distance to their line of actions in the y-direction divided by the resultant force and magnitude. An d so, in this case, the 20-pound force lies on the x-axis so it doesn't go out any distance in the y-direction. We've got the minus 60-pound force and it goes out 2 units or 2 feet in the y-direction, so, and then we have the 50-pound force and it's up, so it's positive. And we go 3 units or 3 feet to it, and then finally, the minus 80-pound force times its distance, which is 4, all divided by 110 minus 110 and that equals 2.64 feet. So, our resultant force will be 2.45 feet in the x-direction and 2.64 feet in the y-direction. So, it's going to be about here. It's going to be down, it's going to be 110 pounds and so again, this distance and this distance, distance in the x-direction is 2.45 feet and the distance in the y-direction is 2.64 feet, and that's the solution. So, you should be able to do any of these single force resultants for a parallel force system. And I'll see you next module.
