You should've said, let's go ahead and sum moments abut point D.

we could sum forces in the x direction, but then we

would have the unknown FGH, and the unknown component of FDE.

If we sum forces in the y direction, it wouldn't have helped us with FGH, and so

we can just one equation of equilibrium, by summing moments about point D.

We'll solve directly

for the force in member GH.

And that's because FDG and FDE, the other two unknowns,

have their line of action that go through point D.

And one other thing I'd like you to note is, we can

sum moments anywhere on the body or what we call the extended body.

So it could be off the body.

And so in this case our section is down here.

Point D is off the body but we're perfectly fine in summing moments.

It still has

to be equal to zero for the body to

be in equilibrium or this section to be in equilibrium.

So we'll go ahead and do this together.

If I sum moments about D, I've chosen counter-clockwise as positive.

I've got FGH times its moment arm. and it's going to be negative

in accordance with my sign convention. So I've got minus FGH times 4 and then

I've got the 3 kilo newton force which is also going to

cause a clockwise rotation about G, so that's going to

be minus 3 times its moment arm which is 1.5.

And the only other force that's going to cause a

rotation about point D is the 3.92 force reaction

at at F and so if I look at that, its going to

cause a counter clockwise rotation that's positive

in accordance with my sign conventions, so it's

plus 3.92 times its moment arm which is 4.5

equals 0. If you calculate that math you'll

find FGH equals 3.29. It's positive, I've assumed that this

member is in tension, so therefore since it's positive, it is indeed in tension.

So I get FGH equals 3.29, the

units are kilo-Newtons. And it's

in tension, so I'll put T in, in brackets.