How Do Forces Interact in a Frictionless Hinged V-Beam System?

[djMan]

Homework Statement

Two identical, uniform beams weighing 260 N each are connected at one end by a frictionless hinge. A light horizontal crossbar attached at the midpoints of the beams maintains an angle of 53.0 ∘ between the beams. The beams are suspended from the ceiling by vertical wires such that they form a "V", as shown in the figureWhat force does the crossbar exert on each beam?What is the magnitude of the force that the hinge at point A exerts on each beam?What is the direction of the force that the hinge at point A exerts on the right-hand beam?What is the direction of the force that the hinge at point A exerts on the left-hand beam?

Homework Equations

Ʃτ = 0
ƩF = 0

The Attempt at a Solution

I have tried to sum the torques and solve for the force of the bar however I end up getting a fraction with a zero numerator. I don't really know how to "see" this problem even after drawing it.

 

[tiny-tim]

hi djMan! welcome to pf!

I have tried to sum the torques and solve for the force of the bar however I end up getting a fraction with a zero numerator.

shouldn't do

show us your full calculations ​

 

[djMan]

Ok,

Let L = length
T = tension of each wire
Fbar = force exerted by bar


Well what I have so far is:

Net torque about hinge = Tsin(153.5)L - Fbar * sin(63.5)L/2 - Tsin(153.5)L + L/2*Fbar*sin(63.5) = 0

And I get a zero each time I try to solve for Fbar

 

[djMan]

Hi everyone, I actually found out how to do this problem. Given the fact that everything is in equilibrium I can cut the "V" in half and look at each side using torque. Then I can solve for Fbar without getting a zero for an answer. Lol mastering physics...

 

[tiny-tim]

hi djMan!

(just got up :zzz:)

Given the fact that everything is in equilibrium I can cut the "V" in half …

yes …

as you've probably realized, if you're finding a tension, you have to "cut in half" the thing with the tension before you do your free body diagram, otherwise the tension occurs twice, as a pair of internal forces, which of course add to 0