Truss Prob: finding the force in a member

[lazypast]

Homework Statement

The diagram is attached. I must solve using method of sections the force in member HG.

Homework Equations

ƩF(y) = 0, ƩF(x) = 0, ƩM = 0

The Attempt at a Solution

My hand-drawn pic shows the section.
Solving vertically, ƩF(y) = 0
F(DG)cos45 = 10
F(DG) = 14.1kN

Moments about E
10(2) + 2F(DG)cos45 + 2F(HG) = 0
Dividing by 2 and rearranging:
F(HG) = -10 - 14.1cos45 = -20kN



I don't think I'd have a problem if there wasn't a pulley system and I can only assume the reason I'm not getting the correct answer for F(HG) is because I'm not taking into account the effect the pulley has.

Can anyone advise a nudge in the right direct... ?

 

[SteamKing]

Well, what effect does the pulley have on the truss? Can the pulley be replaced by equivalent forces at the end of the truss?

 

[brinethery]

Your free body diagram looks right. Now take a moment around G to find what member DE is. Next, sum the forces in x and y to find DG and GH.

Was I helpful to you with this? I really struggled with trusses in statics and now have a really good understanding of how to solve these problems.

 

[pongo38]

In your free body diagram you need to identify the value of the horizontal force at the top of the pulley. What do think that is? (in theory, and in reality)

 

[lazypast]

I have attached my workings out to find the reactions at the supports. Originally I was taking the total external forces at face value (i.e 10kN vertically down) but now I can see because the pulley system support is not connected to the structure there are extra forces.



Going back to the original FBD, I'll add and resolve with a horizontal force of 10kN in there, too. I've attached my workings out.

 

[pongo38]

Reactions look good, but meanings of F1 F2 and F3 not well-defined.