Hi, welcome to module 19 of, introduction to engineering mechanics. Today we're going to identify the centroid for areas and volumes, and we're going to calculate again the magnitude of a resultant force, and its location for a force distributed along a straight line. So first of all, I want to talk real briefly about centroids of areas and volumes. And I said last time that the centroid is the geometric center. it's important to note that the geometric center or the centroid does not have to lie on the body and so I've got a couple of examples here. Let's say that we have like a, this is like a washer, okay? Or a, a disc with the hole in the center. The geometric center would be at the point C, which is not on the body itself. here's half a washer. And again, this would be not down here but a little bit higher because. This would be the center, geometric center in this direction and this direction. And so, they do not have to necessarily lie in a body. The centroid will lie on an axis of symmetry however. And you can see if I run a line down here, that this is symmetric on both sides. If I put a mirror there, the left hand side looks same as the right hand side, so that's symmetric. for homogenous bodies, it's important to note that the centroid is the same as the center of mass. So let's now do a, again another example. We did one last, time, last time. We're going to do another one a little bit more complicated of a force distributed along a straight line and we're going to look for the magnitude of the resultant force and its location. So the force in this case is is a function 50x squared Newtons per meter, and so this is its shape. And it's acting three meters along the x-axis. And so, just so this is not so abstract, we can think of this as, as, maybe as snow load on a roof or something like that. So these, this has real-world engineering applications. And so we've got our definitions of finding the magnitude and the location of the resultant forces. Let's go ahead first of all, and find the location. So, X sub r is equal to the integral from zero to L of s, x f dx over the integral from zero to L of f dx. And so we've got integral from zero to L. F in this case is 50X squared, so we have x times 50x squared, dx. And that's going to be over let's put in L. L we know what that is now, it's three meters. And, so we have that over integral from 0 to 3 of f, which is 50 x squared dx. And if you do that integral, with that basic calculus background, you're going to get 50 x to the 4th over 4. Evaluated from zero to 3, divided by 50x cubed over 3. Evaluated from zero to 3, and that ends up being 1012.5 over 450 or a distance of 2.25 meters. So that's the location, that'll be the location of the resultant force. It's going to be a distance 2.25. The origin. Meters. Now let's find the magnitude of that result. And so the magnitude is this equation here. The area under the loading curve. so that's the integral from 0 to L of F of x dx. f dx that's [COUGH] integral from 0 to 3, of 50 x squared dx. Which is 50 x cubed over 3, evaluated from 0 to 3, and so the magnitude of the resultant force is 450 newtons. And so on my picture here, final result will be a resultant force down of 450 newtons. That's equal to the area under this curve, and it acts at a distance of 2.25 meters from the left side. So that's another example of finding a resultant for a force distributed along a straight line. And we'll see you next time.