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Torque, problem with FBD

  1. Oct 21, 2007 #1
    Here's an image of the problem I trying to do.

    [​IMG]

    I'm wondering if there is a force exerted by the wall on the beam, and if so in which direction? I think it's pointed upward to the right, but I'm not too sure.

    Any help would be appreciated.
     
  2. jcsd
  3. Oct 21, 2007 #2

    PhanthomJay

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    There is a force and a torque exerted by the wall on the beam. The magnitude and direction of each is determined from equilibrium considerations......sum of forces = 0 and sum of torques =0. Don't confuse torques wuth forces. The force of the wall on the beam does not have a rightward component.
     
  4. Oct 21, 2007 #3
    I don't undestand what you mean. The beam is not in equilibrium, and I know torque caused by the wall is zero, but I was wondering what's the direction of the force exerted by the wall.
     
  5. Oct 21, 2007 #4

    PhanthomJay

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    What makes you say the beam is not in equilibrium? The beam is supported into the wall which exerts a torque about point P, which is equal and opposite to the torque caused by the beam's weight about point B. Now you wish to go one step further and determine the direction of the force on the wall on the beam. What does Newton's first law tell you about in which direction the net force of the wall on the beam must be??
     
  6. Oct 21, 2007 #5
    My teacher told me it wasn't in equilibrium since there is a torque.
     
  7. Oct 22, 2007 #6

    PhanthomJay

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    There are two torques...one caused by the weight (10(9.8)(2) = 196n.m), and one caused by the wall (-196n.m). Bt this is really an engineering statics question, and i suppose your teacher might argue that the beam is rotating about a pin at joint P , in which case, no torque would be supplied by the wall. But nevertheless, there still exists a force at the wall acting on the beam, and since the beams weight always acts vertically down, then the reaction force of the wall on the beam must act --------(fill in the blank).
     
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