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>> Hi and welcome to ma, Module 20 of An Introduction to Engineering Mechanics.

Â So, the learning outcome for today's module is to apply something which we call

Â the Method of Composite Parts to calculate the magnitude of a resultant force and

Â it's location distributed along a straight line.

Â So to calculate the result and then its location for a, along a straight line,

Â we're going to break our force into what we'll call standard shapes.

Â And then instead of doing the integrations like we did in the previous modules What

Â we'll do is we'll find a reference, you can find a reference text or a book, or

Â online, for these standard shapes to find out what the magnitude of the resultant

Â force will be. And you'll be able to find the centriod

Â for each of those standard shapes, and so you should be able to find a, a reference

Â like this in any of your standard statics textbooks etc.

Â And so, I've got a real world application here.

Â This is a picture of a sign outside of my office here at Georgia Tech.

Â And, so, we're saying that there is a wind loading on the sign, that can be modeled

Â as shown. In this diagram here.

Â So at the top of the sign I've got a wind load of 8 Newtons per meter.

Â And at the bottom of the sign I've got a load of 3.5, or 3 Newtons per meter, and

Â it acts over an the height of the sign,which is 1.5 meters.

Â And I want to find the magnitude, of the result in force and its location.

Â And so, what I do here is I take my load, and instead of doing an integration, I'm

Â going to break it into standard shapes that are easy to work with.

Â And so I'm going to break this into a rectangle and a triangle.

Â And so for the rectangle, I've got the magnitude is 3 newtons per a meter, across

Â the top. So we've got this is a rectangular shape,

Â and the magnitude is 3 Newtons per meter, and an x over 1.5 meters.

Â And so, first of all, I want to find the magnitude of that force.

Â That's going to be the area under the curve.

Â The area for a, a rectangle is just base times height.

Â So I've got the magnitude of the resultant force for the rectangle is equal to 3,

Â which is the base here, times the height, which is 1.5, or 4.5.

Â Newtons. And the location of the centroid for a, a

Â rectangle is just right in the center here, so I'm going to draw that as well.

Â So I've got my resultant force, rectangular force, that's going to be f.

Â It's an r rectangle, x at this distance from the bottom, which is 0.75, 0.75

Â meters. Half of 1.5.

Â And so, I can do the same thing for the triangle now.

Â I notice I left out an equal sign here, this should be equal, so 3 times 1.5 is

Â 4.5 Newtons for the overall magnitude of the force for the rectangular.

Â Portion of the, the load. Now, I'm going to look at the triangular

Â portion of the load so pictorially, it looks like this and if I subtract 3 from

Â 8, I see that the height over here is 5 Newtons per meter, and it goes down to

Â zero Newtons per meter at the lower end. And so, define the resultant force there,

Â the magnitude of the resultant force, I've got magnitude of f, r.

Â Triangle, equals the magnitude or the area inside a triangle is one half base times

Â height so that's going to be one half 5 times 1.5, or 3.75 newtons.

Â And if you look in a standard, reference for, centroids of a triangle.

Â The centroid occurs, 2 3rds of the way up from the bottom of this triangle.

Â So I'll draw that picture, or that result in force.

Â So this is. F's of r rectangle, and, not rectangle,

Â I'm sorry. Triangle, and the distance here would be 2

Â 3rds of the total height which is 1.5 or 1m.

Â Okay, and so now the total result for this overall load is the sum of the rectangular

Â portion and the triangular portion. So, the magnitude of the result in force

Â total, is equal to 4.5 plus 3.75 is 8.25 Newtons.

Â And so the last thing I need to do is find the location where this result force acts

Â over all and the way they act I can develop composite centroid by summing

Â moments. So we are going to develop a composite

Â centroid for both shapes together and the method I'm going to use is to sum moments.

Â And so each, over on my total diagram here, I'm going to have a total force

Â resultant, which I'm trying to look for the location of that.

Â Magnitude of resultant total. The moment due to this total resultant has

Â to be equal to the moment due to each of these individual.

Â Results on the triangular and on the rectangular.

Â So, I have to have some reference, I use the bottom of the sine here as my

Â reference where to sum the moments about and so when I take the moment do at this

Â point of the bottom, due to the total resultant force that's going to be, let's

Â call this distance. X sub r, or x resultant.

Â And so, we're going to have the moment is x sub r, the perpendicular distance times

Â the magnitude, and so we're going to have x sub r times the magnitude.

Â Of the resultant total force. And, that's got to equal the moment due to

Â each of these individual shapes, the composite parts.

Â So, for the rectangle, it's going to be the moment arm, or the perpendicular

Â distance to its line of action, which is .75.

Â Meters times the magnitude of that force which we found to be 8.25 of sta, I'm

Â sorry, that was 4.5. 8.25 was the overall magnitude of the

Â resultant force total. And then, the moment due to the resultant

Â force for the triangle is the magnitude 3.75 times its moment arm, which is one.

Â So, this is going to be plus one times 3.75.

Â And so what we get is x of r if you solve for it you know that f total here is 8.25,

Â f's a result and total of the magnitude is 8.25.

Â And so if you multiply this divide by 8.25.

Â You'll get, well that's a messy 8.25, lets try it this way, 8.25, okay, if you do

Â that math, you will get x bar equals 0.864 meters.

Â And so now I have both my magnitude of my resultant force which is 8 point 25

Â newtons and the distance up from the. Bottom of the low, which is 0.864 meters.

Â And so, that's my solution. Now, there's one thing I want to mention

Â before we finish up the module, usually when we're doing examples for practical

Â problem solving, you're going to find that we seldom have an interest in finding

Â that. Single force result, we'll be able to use

Â the individual resultants and their distances and that would be fine for

Â actual problem solving, and you'll see that as we go through that some examples

Â as we go through the course. So that's it for today's module and I'll

Â see you next time. Thanks.

Â