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Homework Help: Rotational Equilibrium of metal bar

  1. Oct 8, 2007 #1
    Rotational Equilibrium - Help!

    1. The problem statement, all variables and given/known data
    http://photos-a.ak.facebook.com/photos-ak-sf2p/v142/24/75/1238100168/n1238100168_30095452_8017.jpg [Broken]

    A uniform metal bar of mass 100 kg and length L (2.5 m) extends horizontally from a wall and connected to the wall by a pivot (F is exerted on rod by pivot). As seen in the drawing, a 100 kg mass hangs from a massless rope that is L from the building. T, a cable, is attached 0.25 L from the wall. The entire system is in a state of equilibrium.

    - Using rotational equilibrium, find tension T

    2. Relevant equations

    Net Sum of Torques = 0
    g = 9.81 m/s^2

    3. The attempt at a solution

    So, if you use F as the point to take the torques about, then the bar becomes the moment arm. Therefore, the mass has a force of 981 (M*g) and its torque is 2.5 * 981, or 2452.5. T's torque is 0.25*2.5 * T, right? So, obviously, you would set the two equal to each other. But I think that the bar mass has Mg as well - do i put its torque as 0.5L * 981, or where would I put it?

    And are the sigfigs for this problem 1 or 2?

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 8, 2007 #2
    You are on the right track. Torque is a vector; hence has a sign. By convention, counterclockwise torque is taken as +ve and clockwise -ve. Torque due to the weight of the bar and 100 kg mass is -ve and that due to T is +ve. Put algebraic sum of the torques equal to zero and solve for T.
  4. Oct 8, 2007 #3
    But I still don't know where to place the weight of the bar on the bar itself, and therefore, I can't find the torque for it. If I assume that it's 0.5L, though, then T comes out to be 5886 N. Is this correct?
  5. Oct 8, 2007 #4
    Weight of the bar acts at its centre of mass which is the centre of the rod. Assume whole mass of the rod is located at the centre. You have rightly taken torque due to it as 0.5L x 981.
  6. Oct 8, 2007 #5
    Thanks! The problem also asks for the force the pivot exerts on the bar at F, using the translational equilibrium, so the sum of the forces must = 0. But if the force of T is already "pulling" it up so that the system is in rotational equilibrium, what forces do I use when finding the sum?
  7. Oct 8, 2007 #6
    5886 N is correct.
  8. Oct 8, 2007 #7
    The bar is in rotational equilibrium as well as in translational equilibrium. Using rotational equilibrium T has been determined. Using translational equilibrium determine F. Put algebraic sum of the forces equal to zero (take upward or downward force as +ve then the other direction is -ve) and determine F.
  9. Oct 8, 2007 #8
    Oh, okay, so it's just the force, right? not the torque.

    So that's T + F = Mg + mg
    5886 + F = 981 + 981
    F = -3924 N

    So it's a negative force?
  10. Oct 8, 2007 #9
    Or is it a positive one, because of where F is placed?
  11. Oct 9, 2007 #10
    5886 + F - 981 - 981 = 0

    But is F positive or negative? I'm not sure, because of the placement.
  12. Nov 1, 2007 #11
    I need help? i dont know where to begin

    I dont know where to begin
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