>> Hi, this is Module 21 of An Introduction to Engineering Mechanics. So let's take a look at the forest here, what I've called the forest before. This is the overview of the course, as opposed to the trees as we go through each lesson. We've gotten quite a ways along in the course. We've, we've done the representation of forces in 2 and 3D. We've went through the syst-, the part of the course on particle equilibrium, we've learned about the concept of moments and how to follow the moment of force about a point in a line or an act a line or an axis, we've also discuss couples we're in the section about equilibrium equations and equivalent systems, we're working on resultants and, where they act looking at distributed forces finding centroids and then method we all use last time in what I'm going to do again today is another example. Who have using the method of composite parts. Because it's a really important skill for engineers to know. And then we'll go into the last block of instruction, starting with the next module. So, the learning outcome for today is to solve another problem using the method of composite parts to find a resultant force and couple on a, on a, on a body. And so here, here's the problem we're going to work with we want to, we have a, a, cantilever beam here, it's fixed at the left hand end, it's like a diving board if you will, here's the beam itself and I've got some distances on there and I've got some sort of a load, distributed load, starting with a 100 pounds per foot up here, wrapping down to 60 pounds per foot, leveling off, and then going down to zero at the end. And so, we want to find an equivalent resultant force, and a couple at point o over here on the left. And so, what I'd first like you to do for practice is to break this load into. Using the techniques that we learned last module, breaking it into standard shapes, and finding the resultant force for those standard shapes, and the location of those resultant force for each of those standard shapes. And so take, take some time to do that and then come on back. Okay. Just as a hint, in case you couldn't get started, what I want to do is I want to show you the standard shapes that you should break this in to. So, if we draw a line here, and a line here, and a line here. Those are the standard shapes you should work with. You should work with this triangle and this rectangle, this rectangle and this triangle. And so, if you didn't finish up now is a good time to go ahead and finish and find the result in forces and their locations for each of these standard shapes. And then come on back. Okay, these are the results you should have come up with. For the upper triangle here, its magnitude is the area for a triangle one half the base times the height or 100 pounds. Its location is 1 3rd from the left-hand side, or 2 3rds from the right-hand side, so it's 1.67 feet from point o, 1 3rd of five feet. This rectangle has an area of 300 pounds. It's resultant force is acting in the middle of this rectangular section, which is. 2.5 feet. This rectangle is a, a base of 60 and a height of 4, for 240 pounds. Its location is halfway along this four feet section so it's 5 plus 2, or 7 feet from the left, left hand edge. And finally the rectangle here is one half base times height, or 120 pounds, going from the left, you go 5 feet, 4 feet. And then one-third of four feet. And so, if that's the result you got, you're in pretty good shape for breaking the system into composite parts, and finding those resultants, and their locations. So now what I want to do is I want to come up with a single equivalent force and couple at point o. So this will be system two. So I'll draw my cantilever beam again. And we know from earlier lessons, when we found equivalent systems where we work with equivalent systems that the sum of the forces on system 1 has to be equal to the sum of the forces on system 2, and so in this case. We sum the forces on system 1. That's going to be, and I'll, I'll call down positive, so I'm going to have 100 pounds, plus 300 pounds, plus 240 pounds, plus 120 pounds equals the Total sum of the forces on system, two. And so that ends up equaling, 760 pounds, and so that's one of our answers. That's the result in force. At point O, and so that would act down over here at point O on system 2 and will be equal to 760 pounds. Okay, lets now keep on going and find or satisfy the condition for sum of the moments being equal on both system 1 and system 2 So let's start out by drawing system 21 here again. What we've done so far, this is system two And this is point O. We found that the result in force at point o is 760 pounds down. And now we have to satisfy the other condition, which is the sum of the moments About a common point on both systems, so this is point o in this case, have to be equal. And so let's go ahead and sum moments about o on system one. And so we've get, 100 is and the, and the distance to that perpendicular distance to that line of action of that force is 1.67, so I've got 100 times 1.67 and then I've got 300 times 2.5 and then I've got plus 240. Perpendicular distance is 5 plus 2, or 7. And plus 120 times the perpendicular distance to the line of action at that force, which is 5, plus 4 is 9 plus 1.33, is 10.33. And, all of that has to equal the moment about o on our second system. So, the sum of the moments about O on system 2, if you do that math, is 3,836. The units are. Foot, pounds. And these individual resultants cause a clockwise rotation, so the moment of o or the couple at o has to also be clockwise. And so that's my equivalent couple. And I can draw it on the figure. Clockwise array, arrow and it's going to be 3836 foot pounds. And so we've completed the problem, we've replace the loading shown to the right with an equivalent resultant force and couple at point O, and I'll see you next module.
