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Handing beams

  1. Nov 14, 2005 #1
    I am having trouble solving this problem from Young and Freedman: University Physics. 11.76.
    Here's a picture of it:
    [​IMG]

    I can't even get started. Hope someone can help. Thanks.
     
  2. jcsd
  3. Nov 14, 2005 #2
    I'd think about the free-body diagram for a single beam in that position and angle hanging idly in the air just with the thread.
    Then the forces would be the gravity and the tension of the thread
    Next, at the hinge the two beams are in a symmetrical configuration and the forces they exert on each other must be symmetrical and at the same time equal and opposite.
    Thus follows that the forces are horizontal.

    And in reference to the c.m. of one of the beams, the tension of the thread gives a torque in ccw.
    Since the crossbar is attached to the midpoint, the only other force that balances the torque is the force at the hinge.
    And it follows that the force at the hinge for a beam must act inward.

    If you are supposed to neglect the weight of the crossbar , it cannot pull the beams up or down.(think of the symmetrical configuration here again). Thus the direction of the force by the crossbars are also horizontal. And the sign and magnitude are to be determined from the combined equations from all of these.

    [edit]
    One more thing you can use is that the horizontal forces canel out each other and the crossbar must exert outward force on the beam, equal in magnitude to the force at the hinge. So it must be under compression.

    I hope this helps.
     
    Last edited: Nov 14, 2005
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