Two beams hanging in equilibrium

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SUMMARY

The discussion revolves around the equilibrium of two identical beams, each weighing 260 N, connected by a frictionless hinge and forming a "V" shape with a 53-degree angle between them. The participant initially struggled with the correct application of torque and force equations, specifically Στ = 0 and ΣF = 0, to determine the force exerted by the crossbar on each beam. After receiving clarification on the nature of the crossbar as a two-force member, the participant successfully resolved the problem.

PREREQUISITES
  • Understanding of static equilibrium principles, including torque and force balance.
  • Familiarity with free body diagrams and their construction.
  • Knowledge of trigonometric functions and their application in physics problems.
  • Basic mechanics of materials, particularly regarding two-force members.
NEXT STEPS
  • Study the principles of static equilibrium in greater depth, focusing on ΣF and Στ equations.
  • Learn how to construct and analyze free body diagrams for complex systems.
  • Explore the characteristics and calculations related to two-force members in structural mechanics.
  • Investigate the application of trigonometric functions in resolving forces and moments in physics problems.
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their understanding of static equilibrium and force analysis in structures.

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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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