Okay.

So, FDG, or excuse me FDE is what

we want to find, we're going to sum moments about G.

I've chosen counterclockwise as positive. And so I've got, okay, I'm going to have,

[UNKNOWN],

I don't know the perpendicular distance between the

line of action of FDE and point G.

And so what I'm going to use is Varignon's theorem

back from our, my last course, Introduction to Engineering Mechanics.

And I'm going to break this FDE force into its two components,

an x component and a y component. And so let's

first look at the moment due to the x component.

And we see that this FDE is on a one on three slope.

This, this distance is one, this distance is 3 meters.

And so the hypotenuse is the square root 10.

And so we can see this is a breakout of, of the slope.

And so the x component then will be

3 over the square root of 10 times FDE. So we've got 3

over the square root of 10 times FDE, times its momental arm.

And its momental arm will be 3 because that's the

perpendicular distance from the x component down to point G.