Next, we turn our attention to node F, this one right here.

And here is our free body diagram for

the node F with the three forces, which again, I'm assuming are in tension.

The arrows are pointing away from the node point.

And a similar procedure.

First, I sum the forces in the horizontal, the x direction, so

I have FE minus, because it's pointing to the left,

I have cosine theta is equal to 0, and in the vertical direction,

sum Fy, we have FB and the vertical component of AF.

So we have minus AF sine theta minus FB is equal to 0.

However, we've already solved for the force in the member AF.

We already know this.

So again, we have two equations, and we can solve for the two unknowns.

So we just progress in this way progressing to the next member,

the next node, until we've solved for all of them.

And that's the basic method of joints.