Two beams hanging in equilibrium

AI Thread Summary
Two identical beams weighing 260 N each are connected at one end by a hinge and form a "V" shape with a 53-degree angle maintained by a crossbar. The problem involves calculating the force exerted by the crossbar on each beam, using equilibrium equations for forces and torques. The initial approach included breaking the problem into parts but led to confusion regarding angles and force directions. A suggestion was made to reconsider the force direction in relation to the crossbar, emphasizing that it acts along the length of the bar rather than at right angles. Ultimately, the poster resolved the issue after clarifying the force assumptions.
p0ps1c1e
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Homework Statement


Two identical, uniform beams weighing 260 N each are connected at one end by a friction-less hinge. A light horizontal crossbar attached at the midpoints of the beams maintains an angle of 53 degrees between the beams. The beams are suspended from the ceiling by vertical wires such that they form a "V".

What force does the crossbar exert on each beam?
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Here is the free body diagram I drew with the symbols I used as well.
http://imgur.com/aFndxKR

Homework Equations


Στ = 0
ΣF = 0

3.
So I thought to break this problem into two and just focus on one of the beams pivoting about point A and then solve for N1. I called the length of the beam L. I feel like I messed up the angles though

Doing that I came up with the equations...
T - mg - N1*sin(53) = 0

T * L cos(37)-mg*(L/2)*cos(37) - N1*(L/2) = 0



Then I thought to solve for N1 but I don't know the length L so I got stuck there


T = mg + N1sin(53) and T = (N1*(L/2) + mg*(L/2)*cos(37)) / L*cos(37)Can somebody give me a hint?
Thanks

 
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p0ps1c1e said:
53 degrees between the beams
This does not match your diagram.

Your diagram also appears to assume the force exerted on a beam by the crossbeam is at right angles to the beam. Any other ideas?
 
The cross bar is a two force member which means that the forces in it act along the length of the bar, not in the directions you have assumed. Try again with this understanding. (Note what happens to the sum of vertical forces on the cross bar with the forces as you have them!)
 
Yeah that sounds right... I originally had it that way but my TA drew it this way. It really confused me haha I'll try it that way and let you all know how it works out
 
I figured it out. Thanks a lot!
 
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