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Equilibrium Tension Torque

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Note Picture...

    A uniform beam, 5.00 m long with a mass of 2.00 x 103 kg, is held against a wall in the position shown by a hinge and a horizontal steel wire attached to its end.



    a) Find the tension in the wire.
    b) Find the magnitude of the total force supplied by the hinge.




    3. The attempt at a solution

    the only forces for the torque is the mass of the beam perpendicular to the beam as well as the tension of the cable

    beam torque = mgdcosQ = 2000(9.8)(2.5)(cos20)
    tension torque = TxsinQ this is where im a littel confused because the tension is in the x direction but the torque should be perpindicular to the beam so its confusing me.

    2000(9.8)(2.5)(cos20) = TxsinQ.....Tx = 135426.28

    is this correct
    T
     

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  3. Nov 29, 2009 #2

    rl.bhat

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    Torque due to the tension is T*sinθ*L where L is the length of the beam.
     
  4. Nov 29, 2009 #3
    oh right right...

    2000(9.8)(2.5)(cos20) = Tx(5)sin20.....Tx = 26926.87 N

    is this correct
     
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