>> Hi and welcome to ma, Module 20 of An Introduction to Engineering Mechanics. So, the learning outcome for today's module is to apply something which we call the Method of Composite Parts to calculate the magnitude of a resultant force and it's location distributed along a straight line. So to calculate the result and then its location for a, along a straight line, we're going to break our force into what we'll call standard shapes. And then instead of doing the integrations like we did in the previous modules What we'll do is we'll find a reference, you can find a reference text or a book, or online, for these standard shapes to find out what the magnitude of the resultant force will be. And you'll be able to find the centriod for each of those standard shapes, and so you should be able to find a, a reference like this in any of your standard statics textbooks etc. And so, I've got a real world application here. This is a picture of a sign outside of my office here at Georgia Tech. And, so, we're saying that there is a wind loading on the sign, that can be modeled as shown. In this diagram here. So at the top of the sign I've got a wind load of 8 Newtons per meter. And at the bottom of the sign I've got a load of 3.5, or 3 Newtons per meter, and it acts over an the height of the sign,which is 1.5 meters. And I want to find the magnitude, of the result in force and its location. And so, what I do here is I take my load, and instead of doing an integration, I'm going to break it into standard shapes that are easy to work with. And so I'm going to break this into a rectangle and a triangle. And so for the rectangle, I've got the magnitude is 3 newtons per a meter, across the top. So we've got this is a rectangular shape, and the magnitude is 3 Newtons per meter, and an x over 1.5 meters. And so, first of all, I want to find the magnitude of that force. That's going to be the area under the curve. The area for a, a rectangle is just base times height. So I've got the magnitude of the resultant force for the rectangle is equal to 3, which is the base here, times the height, which is 1.5, or 4.5. Newtons. And the location of the centroid for a, a rectangle is just right in the center here, so I'm going to draw that as well. So I've got my resultant force, rectangular force, that's going to be f. It's an r rectangle, x at this distance from the bottom, which is 0.75, 0.75 meters. Half of 1.5. And so, I can do the same thing for the triangle now. I notice I left out an equal sign here, this should be equal, so 3 times 1.5 is 4.5 Newtons for the overall magnitude of the force for the rectangular. Portion of the, the load. Now, I'm going to look at the triangular portion of the load so pictorially, it looks like this and if I subtract 3 from 8, I see that the height over here is 5 Newtons per meter, and it goes down to zero Newtons per meter at the lower end. And so, define the resultant force there, the magnitude of the resultant force, I've got magnitude of f, r. Triangle, equals the magnitude or the area inside a triangle is one half base times height so that's going to be one half 5 times 1.5, or 3.75 newtons. And if you look in a standard, reference for, centroids of a triangle. The centroid occurs, 2 3rds of the way up from the bottom of this triangle. So I'll draw that picture, or that result in force. So this is. F's of r rectangle, and, not rectangle, I'm sorry. Triangle, and the distance here would be 2 3rds of the total height which is 1.5 or 1m. Okay, and so now the total result for this overall load is the sum of the rectangular portion and the triangular portion. So, the magnitude of the result in force total, is equal to 4.5 plus 3.75 is 8.25 Newtons. And so the last thing I need to do is find the location where this result force acts over all and the way they act I can develop composite centroid by summing moments. So we are going to develop a composite centroid for both shapes together and the method I'm going to use is to sum moments. And so each, over on my total diagram here, I'm going to have a total force resultant, which I'm trying to look for the location of that. Magnitude of resultant total. The moment due to this total resultant has to be equal to the moment due to each of these individual. Results on the triangular and on the rectangular. So, I have to have some reference, I use the bottom of the sine here as my reference where to sum the moments about and so when I take the moment do at this point of the bottom, due to the total resultant force that's going to be, let's call this distance. X sub r, or x resultant. And so, we're going to have the moment is x sub r, the perpendicular distance times the magnitude, and so we're going to have x sub r times the magnitude. Of the resultant total force. And, that's got to equal the moment due to each of these individual shapes, the composite parts. So, for the rectangle, it's going to be the moment arm, or the perpendicular distance to its line of action, which is .75. Meters times the magnitude of that force which we found to be 8.25 of sta, I'm sorry, that was 4.5. 8.25 was the overall magnitude of the resultant force total. And then, the moment due to the resultant force for the triangle is the magnitude 3.75 times its moment arm, which is one. So, this is going to be plus one times 3.75. And so what we get is x of r if you solve for it you know that f total here is 8.25, f's a result and total of the magnitude is 8.25. And so if you multiply this divide by 8.25. You'll get, well that's a messy 8.25, lets try it this way, 8.25, okay, if you do that math, you will get x bar equals 0.864 meters. And so now I have both my magnitude of my resultant force which is 8 point 25 newtons and the distance up from the. Bottom of the low, which is 0.864 meters. And so, that's my solution. Now, there's one thing I want to mention before we finish up the module, usually when we're doing examples for practical problem solving, you're going to find that we seldom have an interest in finding that. Single force result, we'll be able to use the individual resultants and their distances and that would be fine for actual problem solving, and you'll see that as we go through that some examples as we go through the course. So that's it for today's module and I'll see you next time. Thanks.
