Continuing our discussion of statics, now I want to look at Centroids and moments of inertia. So, these are the overall topics we'll be looking at here. And in particular in this segment, we'll look at why these are important, how we compute center of gravity and related properties,. Centroids of lines and areas and how we computer properties of composite areas. Now, firstly, Centroids and moments of inertia and moments of inertia are important especially when we have distributed forces, forces which are distributed over a line or an area or a volume. And it's important for example to get the location of the, of a resultant force. For example, the weight of a body is the resultant force due to gravity on that body. And distributed forces can arise in a number of ways. For example, the first case here. A hanging cable with some weight, the force is distributed along the line in that case. Or the second example, force distributed over an area arises when we have a hydrostatic pressure distribution. For example, the water being held back by a dam. And the last one where we have some large volume, where the resultant force is distributed over the volume of that body, that's a volume distribution. So, all three cases are important. There a number of topics that are presented in the reference handbook. Which are computation of Centroids of Mass, area, length, and volume. Moments of inertia, the Parallel axis theorem, Radius of gyration, Product of inertia. But we won't be covering all of these, just the most important ones of these. And these are wise it's profitable to think first in terms of the center of gravity of an object. So let's suppose that we have some object which is hanging here. And let's suppose I have this hanging from a cable and it's suspended at some point A. So, the weight of the object, w, is hanging straight downwards, and the line of action, of the, of the cable here, will pass through some point. Now, let's suppose I drill another hole in the object, and suspend it from a new point, B. Again the line of action through the cable will pass through some point here which I'll call G. And if I do it yet again supported from some new point C. Again the line of axis will pass through some point which I'll call G. So the lines of axis or the, through the cable will all pass through some common point here G, which is of course the Center of gravity, which is the location of the resultant force, or where the resultant force of the body acts, the way it acts. And the weight of course is equal to the mass of the object multiplied by the acceleration due to gravity, or in USCS units m times acceleration due to gravity, divided by the universal acceleration due to gravity. Now. Generally speaking, we can compute the location of the Center of gravity g by taking moments and for example, if I have an object of arbitrary volume and I divide it up into elemental volumes, each of weight dW, then by taking moments about the x y and z axis for equilibrium. I find these equations for the x y and z coordinates of the central gravity. X bar is integral of xdW divided by the total weight, et cetera. Now, if this material is of uniform density, like a block of wood of constant density or iron or steel,. Then this equation becomes this. Xc is equal to integral xdV where dV is the volume of an element divided by V where V is the total volume of the element. So, my equations for Xc, Yc and Zc are as given here, so these coordinates depend only on the shape of the object. And that location, those coordinates is known as the Centroid. So, the Centroid is purely, depends on the shape of the object. It's a purely geometrical thing. And if the, object is of uniform density, then the location of the Centroid and the location of the Center of gravity are Coinstant. They're in the same point. However, if it's not uniform density, then, for example, in my, here. Let's suppose that this was wood, say, on top, but something very heavy on the bottom, like lead. So now, it's not of uniform density. So, the Centroid would still be the same place, because all that depends on is the shape, but the center of gravity would then be some point below that. They would not be coincident. Now, we will be concerned with competing the Centroids of Lines, areas, and volumes, although mostly we'll just look at lines and areas, and here's the corresponding general section from the reference handbook with their definition of the Centroid given in vector form like this. Now, let's do a couple of examples. Firstly, the Centroid of a line. So here we have a long line source, which is arbitrarily given in space, generally in three dimensions. And the Centroid of that line. Is given by this expression, this, li-, of the, of the general case that we had on the, on the slide before. So rC, the vector coordinate of the Centroid of that line is given by that expression, where r is the radius vector to. To some small line, line element L dL. And the total length of the line is capital L. So expressing that vector equation in terms of its Cartesian coordinates. We see that the x coordinate of the Centroid is integral xdL divided by the total length of the line. And y bar and z bar are similar equations. If we're dealing with sections of lines, for example multiple sections like this, then if each section is denoted by i, i equals one, two, three et cetera. Then, the x coordinate of the Centroid of the line is given by the summation of xi, Li divided by the total length and the y coordinate, summation yi, Li divided by the total length. This is most easily exam, shown as usual by means of an example. And here we have a wire segment ABC consisting of two straight sections. AB is 1 meter, at an angle of 45 degrees to the horizontal, and BC is 2 meters long and is horizontal. Where is the combined Centroid of that wire segment? These are the possible x, y coordinates. So, to answer this then, we use our equation in its finite segment form, xiLi over L. So the Centroid presumably is going to be possibly somewhere out here. So, the x coordinate here, xC. Is equal to. We have two segments. I'll call this segment 1 and 2. X1L1 plus x2L2 over the total length, which is L1 + L2. So that is equal to. First of all, the segment 1. The Centroid of this section. My symmetry is in the middle. The length is 1, so the horizontal distance is -1/2)sin 45 degrees, multiplied by its length, which is 1 meter. Then we have the Centroid of the horizontal section BC. By symmetry that is in, in the middle, a distance of 1 meter to the right of the y-axis. So x2 is 1 meter, and multiplied by. I'm sorry, x1 is 1 meter, 2 divided by 2. Multiplied by it's length, which is 2. Divided by the total length which is 3 meters so xC is equal to 0.55 meters. The vertical coordinate yc is summation yiLi over L, is equal to y1L1 plus y2 L2 over L1 plus L2. So similarly. We have the coordinates here of 0.1 is plus one half sine 45 degrees multiplied by its length, which is 1. Plus the elevation of the Centroid of BC which is obviously 0, because it's lying on the x-axis. Divided by the total length is three, is equal to 0.12 meters. So the correct answer is B. Now, one thing to note about this is that the Centroid of this segment is here. The Centroid of this segment is here. If I draw a straight line between those two Centroids, the Centroid of the combined segment will lie on that line. Next we'll look at Centroids of areas. And areas can be arbitrary shape in a, x y z space, as shown here. And the definition of the Centroid of a ,of an area like that. The Centroid is here is given by this expression. Integral xdA divided by the total area, et cetera. Where dA is the area of an elemental, area of the surface. And generally speaking, if this is a, a, curved section, the central ar, of that area might not lie actually on that area. The quantity integral xdA is known as the "First moment of area", although in the handbook they just call it the moment of area. More normally we call it the first moment of the area. And if we divide this instead into finite segments. Then we get the equation that they suggest in reference handbook, that the x coordinate of the Centroid is equal to summation x and am over A, where x and Am refer to each section or composite or each area of a composite area. And then, the summation xn an is the first moment of the area. You can compute this for all different types of areas, and this is an extract from the handbook. So the, the reference handbook contains several pages with properties of various shapes, triangles, circles, rectangles, etc. And this is just an extract from them. So, generally speaking, you probably won't have to compute these things. You can just look them up in these, in these handbooks. And, the ones that we'll probably most commonly be dealing with will be rectangles, cir, circles, and possibly triangles, will be the most common ones. What if we have a composite area, such as this one here where we have two areas or a composite area comprised of two rectangles, 1 and 2? And let's suppose that the Centroid of each of these separate sections, 1 and 2, are shown by the crosses here. Then the formula in that case is that the x coordinate of the Centroid of the composite area is equal to summation xi,ai where xi is the x coordinate of the Centroid of the area 1i and ai is it's area. In summation, Ai is the total area. Similarly, the y-coordinate, is given by this expression. So, in this example, xc is equal to x1a1, where x1 is the horizontal coordinate of the Centroid of area 1, times a1, plus x2a2,. Divided by a1 plus a2 for that particular example. Now, if we have holes. For example here we have a rectangle with a hole in it. We can consu-, we can do this the same thing except the hole is considered as a negative area. So, the same formula applies, xc is equal to summation xA, xiAi, et cetera. Only now the whole area is negative. So in this case, we have x1A1 minus, because this area 2. As a whole, it's negative minus x2 A2 divided by A1 minus A2. An example. We have a Tee-shaped area comprised of two rectangles, each 4 meters by 1 meter. The x and y coordinates of the, of the Tee-shape, dissent, of the dissent of the Centroid of the Tee-shape are most nearly which of these alternatives? So here is our relevant formula, xc is summation xiAi over A. So in this case, that's equal to x1A1 plus x2A2 divided by A1 plus A2. But I think you can see that in this case the, this is symmetrical. So the x coordinate of the Centroid is going to lie on the line between those two. In other words we expect the Centroid to be somewhere here. So it, by symmetry here, it's going to be equal to 2 meters from the y-axis. And you can substitute in the numbers and find that that's true. The y coordinate yc is summation yiAi over A. Which is equal to y1A1, et cetera. So, in this case, we have y1 is the height of the centroid of area 1. This height is 2 meters and the area is 4 times 1 plus the height of the Centroid of 2. This height. From simple geometry is 5 meters by symmetry, 5 meters multiplied by its area which is 4 times 1. And the area on the bottom is 4 times 1 plus 4 times 1 which gives me a y coordinate of 3.5 meters. So the correct answer here is D. Now an example with a hole in it. So here we have a rectangle which is 4 meters tall by 1 meter wide with a hole here which is 0.5 meters in diameter. The x and y coordinates of its Centroid, in other words, the rectangle with a hole in it,. Are mostly nearly which of these alternatives? So the same formulas apply, xz is summation x1A1, et cetera. So in this case we have x1A1 minus because this is a whole here. x2A2, and A1 minus A2, again, because it's a hole. So, again, I think you can see just by symmetry here, that the x coordinate of the combined Centroid is going to be on the center line there, therefore is equal to 0,5 meters. And you can substitute in numbers and confirm that. The vertical coordinate Yc is equal to summation yiAi over A. So in this case we have y1A1 minus y2A2 over A1 minus A2 which is equal to. The area of the rectangle is 1 times 4. And the height of the Centroid of the rectangle here is 2 meters. Minus the height of the Centroid of the circle here is 3. And its area is pi by 4 times the diameter squared. Or pi by 4 times 0.5 squared. And in the denominator, the area of the rectangle is 1 times 4, and the area of the whole, pi by 4 times 0.5 squared, is negative. So, calculating the numbers, the height is 2.57 meters, and the correct answer is C. So this concludes my discussion of Centroids.