Welcome to module 16 of An Introduction to Engineering Mechanics. Today, we're going

to learn how to calculate this single force result. Instead of a force and a

couple, a single force result for a coplanar 4 system. So a coplanar force

system is one in which all forces lie in the plane, and the moment vectors are

normal to the plane. So you can see I have several forces here that lie in the x, y

plane or the i, j plane. And then I have several moments, which are either into the

board or out of the board. And so, this is a coplanar force system. Now, the last two

modules, we've been able to come up with a single equivalent system with a single

force and a, a moment at a particular point in this case, point P. What we're

going to do today is to reduce it even further to a single force resultant. And

the way we're going to do that is we're going to change the moment to a couple

equivalent. So the moment about P is going to be equal to this couple here, with two

forces equal and opposite at a perpendicular distance apart of d. And

then I'm going to add all those forces, and come up with a resultant of a single

force. And so the easiest way to do this and to learn this is to go through a

problem. So let's look at a problem. We're going to reduce the coplanar system shown

to a single force resulta. And so in this case, I've come up with a single force and

moment. Resulting, which I, again, we've done in the last 2 modules. Now, I want to

reduce that to a single force and the first thing we're going to do is we're

going to replace this 60 foot bound couple with a force couple with the same

magnitude as but the forces are going to have a magnitude of 15 pounds so, we're

going to replace the 60 foot pound couple with a force couple, having a magnitude of

15 pounds. Where the forces have a magnitude of 15 pounds. Okay? So if I draw

this visually, I've got my 15 pound force here which is on a 4 on 3 slope. And I

want to replace the equivalent the, the couple system, or the 60 foot pound couple

syste m with an equivalent force couple, so this is going to be minus. 15 pounds.

This is going to be 15 pounds. And we're going to have some distance, D, in between

them, perpendicular distance, D, here. And again, this force is also going to be on

the same 4 on 3 slope. So, to find that distance, D I say that the 60 foot pound

force is equal to the magnitude of the force couple force couple, times the

distance in between them, and so we find out that this distance d is equal to 4

feet. And I'll put the therefore symbol here. Three dots means therefore. d equals

4 feet. Okay? So I have now changed the couple into two forces so I have a, all

forces on my, my equivalent system. And so here it is. 15 pounds.

15 pounds. Equal and opposite 60 pound foot pound

couple. And I have my original 15 pound force here. If I add those vectorially

now. This force and this force cancel each other. And so I end up with just a single

15 pound force. It's on a 4 on 3 slope, so it's parallel to this. The line of action

is parallel to this original line of action, which is also on a 4 on 3 slope.

And the distance from o out to that force is that perpendicular distance between

this force couple which was four feet. And so that's a single force resultant for

this system, and that's the end of this module.