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Equilibrium Uniform Beam Problem

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Two identical, uniform beams weighing 260 N each are connected at one end by a fric-
    tionless 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 Fig. 11.56(see attached file) (a) What force does the crossbar exert on each beam?


    2. Relevant equations

    [tex]\[\sum {{\tau _z} = 0} \][/tex]


    3. The attempt at a solution

    If i apply the above formula for the system shown in the attached fig, then the forces - i) Tensions in the attached string ii)weight of the each rod iii)Force exerted by the crossbar on the beams iv)Force exerted by the hinges on the beam

    All the forces mentioned above cancel out with each other because all the forces are equal in magnitude and produces torque in opposite directions.
    Kindly give suggestions to solve this problem.
     

    Attached Files:

    • 76_a.bmp
      76_a.bmp
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  2. jcsd
  3. Sep 1, 2009 #2
    I can't see your pic yet, but I'll try to help.

    The sum of the forces is zero (because the system is at equilibrium) but this doesn't mean that the forces cancel. Rather, each force is being perfectly balanced by another force. I'm not sure where the strings are attached, so I can't help much more than that, but basically:

    But, in general, the force exerted by the cross bar should be the torque of the bars minus the tension in the strings.

    You say the bars weight 260 Newtons, but I think that's a typo. N is a force, not a mass.
     
  4. Sep 1, 2009 #3
    I can download the pic.... kindly chk again...

    For example let us consider the Tensions of the vertical wires attached to the ceiling. If i take torque about the point A ie at the Hinges then the magnitude of the Tensions of the string are equal in magnitude and opposite in direction. Hence they cancel with each other. Same thing with other forces??

    They have given weight not the mass... so 260N is correct....
     
  5. Sep 1, 2009 #4
    Oh, I understand the weight thing now. You're right.

    Your pic says "Pending approval" and I can't see it.

    As for the forces, they "cancel" but only because they sum to zero. They still exist and you can calculate them. It's like saying that force A equals negative force B. They sum to zero, but they still exist.

    If you had no cross bar, then it sounds like the strings would hold the entire mass. Since you do have a cross bar, the strings will not balance the entire mass. It sounds like you need to calculate the center of mass of the bar and find the distance of the crossbar attachments from the center of mass of each bar. This will give you the proportion of mass that is supported by the cross bar as opposed to the string.
     
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