Truss Prob: finding the force in a member

AI Thread Summary
The discussion focuses on calculating the force in member HG of a truss using the method of sections. The participant has attempted to solve the problem by applying static equilibrium equations but is struggling due to the presence of a pulley system. They seek clarification on how the pulley affects the forces in the truss and whether it can be replaced by equivalent forces. Suggestions include taking moments around specific points and summing forces in both x and y directions to find the necessary values. The importance of accurately identifying horizontal forces in the free body diagram is emphasized for a correct solution.
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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... ?
 

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

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

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