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A Bar Suspended by Two Vertical Strings

  1. Feb 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A rigid uniform horizontal bar of mass m1 = 85.00 kg and length L = 5.300 m is supported by two vertical massless strings. String A is attached at a distance d = 1.500 m from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floor. An object of mass m2 = 2000 kg is supported by the crane at a distance x = 5.100 m from the left end of the bar.

    Throughout this problem, positive torque is counterclockwise. Use 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

    (1) Find T(A) the tension in string A.
    (2) Find T(B) the magnitude of the tension in string B.

    [​IMG]

    2. Relevant equations

    Ta = (Torque due to weight of bar + Torque due to weight of object) / (l/2)
    Tb = Ta*d - m1g(L/2) - m2gx


    3. The attempt at a solution

    I got Ta = 68111.63, It says that I am close, rounding error, and it has to be in 4 sig figs.

    I got Tb = 34056.45, and no dice with this one.
     
    Last edited: Feb 8, 2008
  2. jcsd
  3. Feb 8, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I don't understand how you arrived at these equations.

    To find Ta, set up a torque equation about the point where string B connects to the bar; To find Tb, set up a torque equation about the point where string A connects to the bar.
     
  4. Feb 8, 2008 #3
    Are you using T for both torque and tension? Some units would be a big help to you.
     
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