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So these are the things you should be able to achieve by the end of this module.

Â First of all distributed force systems are an application of resultants, and we've

Â looked at several modules on resultants, and we've done a, a number of, of

Â problems. So now we're going to get into an application of that where we're going

Â to have distributed force systems. And so the first thing I want you to do, is to

Â give me some examples of body forces. So I'd like you to take a few minutes, and on

Â a piece of paper, write down some examples of body forces. So, now that you've

Â answered that question, here's a, a couple typical examples of body forces. gravity

Â is a body force, and so, if I have a body over here Gravity acts on each piece of

Â mass in the body, and so, or each little part of it, mass in the body and so it

Â acts throughout the body. electro-magnetism is another example of a

Â body force Let's. Also give examples of surface forces, and

Â so, take a few minutes and write down some surface forces and then we'll come back.

Â Now that you've answered that question. Here are some examples of surface forces.

Â One would be, if I was pushing on this book. Okay? This notebook with my finger.

Â That's a force acting externally to the to the, to the body, and so that's a surface

Â force acting of the surface of the body. if there was a huge gust of wind that came

Â through this, this room and the wind was in this direction that wind force would

Â act on the surface of the body, or the surface of the book. And that would be an

Â example of an external force. And you can think of others as well, I'm sure. Okay,

Â let's talk about how to find the result for a force distributed along a straight

Â line. And so here, I hav e, An x, y coordinate system. I've got a force

Â distributed along the x-axis, and it has a varying magnitude as it goes along. And

Â that magnitude is defined by the function f of x, where f is a force per unit

Â length. And what we're going to look for is what the resultant is? What the

Â magnitude of the resultant is and where it acts? And so what we're going to do is

Â we're going to take this total force and we're going to split it up into little

Â differential elements. And so I've written a little differential element here. I've

Â exaggerated the size of it. It would be infinitely small but it has a distance in

Â the x direction or a width of dx and it has A magnitude of f, and so the total

Â force acting down from that little element would be fdx. The force per unit length

Â times its unit length, which is dx. And so, if I want to find the resultant for

Â the entire load here, I take the, the force of that individual differential

Â element fdx, and then I integrate it over the entire force acting from zero to l.

Â And that's going to give me the area under the loading curve, or the magnitude of the

Â resultant force. If I want to find where that resultant force acts. So I found the

Â magnitude here. If I want to find where it acts, this distance x of r. What I do is,

Â I look at the moment due to the re, the, to the distributed force. And that's got

Â to equal the moment Due to the resultant force. So, for the distributive force,

Â each of these little differential elements are acting down at a distance x, so x

Â times fdx, that's going to be the moment. We integrate that from zero to l for the

Â entire distributive force. That's got to equal to magnitude of the Resultant force,

Â which is the integal from zero to l, fdx, which we looked at up here, times it's

Â moment arm, which is xr. And so xr, the distance to the resultant force Its just

Â the integral from zero to l of xfdx and then we divide by the area under the

Â loading curve in the mom the result not the load which is the integral of 0 to l

Â fdx. And then xr becomes the x cordinate of what we call the centroid of this load

Â or the Geometric Center of the Load. So that's the definition of a centroid.

Â Geometric Center. So let's do an example again. Here, I

Â have, up in the right hand corner of my, my definitions. xr is the distance to the

Â resultant force. We just talked about that. F magnitude of fr is the magnitude

Â of the result in force load. And so, here's my example. I've got a simple ramp

Â load here. It goes from zero to b. The function is just a straight line. So the.

Â The force per unit length is B over L, the slope times x. And so I can use my

Â definition x sub r is equal to the integral from 0 to l of xfdx over the

Â integral from zero to l of fdx and so in this case we've got the integral from zero

Â to L, f is bl over x and we multiply that by x so that's going to be B L x squared.

Â DX divided by the integral from zero to L of F which is B over L times X, DX. And so

Â that's going to equal B over L is a constant, B's constant, L's constant. I

Â can pull it out of the integral, so I've got B. Over L times the integral from zero

Â to L of x squared dx. Divided by B over L times the integral from zero to L of xdx.

Â If I integrate those I get X sub r equals B over L times the interval of x squared

Â is x cubed evaluated from zero to L and then B over L times the integral of xdx is

Â x squared over 2 evaluated from zero to L and so that's simply going to be 2 3rds L.

Â So, that's the answer for my resultant location. So, if I have my resultant here,

Â it's going to be at a distance, this is F of R, it's going to be at a distance 2

Â 3rds L. And we still need to find the magnitude of that resultant. So to find

Â the magnitude I have it written here. Magnitude of Fr is equal to the integral

Â from zero to L of fdx that's equal to the integral from zero to L, F is B over L x

Â and that's going to be dx, that's just going to be equal to the area under the

Â loading curve which is the triangle in this case. So if I do the integral, I get

Â the magnitude of the resultant force is e qual to B over L is a constant. Pull it

Â out. Integrate xdx. That's x squared over 2 again. I evaluate

Â it from zero to L. And so I find out that the magnitude of my resultant force is BL

Â over 2. And that should make sense to you because that's the same as the area,

Â that's the area under a curve for a triangle. One half base times height. The

Â base in this case is L the height is B. And so I have my total resultant force, or

Â a single resultant force for this distributed load. It's acting down. It's

Â magnitude is BL over 2 and the distance to that results is 2 3rds L. So, a nice

Â application for a distributed load surface load acting on a body and it's a nice

Â application of results. I'll see you next time.

Â