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Static Equilibrium and tension of a cord

  1. Nov 19, 2007 #1
    A non-uniform bar of weight W = 12.5 N is suspended at rest in a horizontal position by two massless cords. The angle φ between cord 2 and the vertical is equal to 27.5°. The bar has a length L = 6.0 m and the distance of its center of gravity from the left-hand end of the bar is 4.4 m. What is the tension in cord 2?

    My lab partner and I set up this equation:
    Fx: T2*cos (27.5)-T1*cos(theta)=0
    Fy: T2*sin(27.5)+T1*sin(theta)-mg=0

    Torque: -mgL-T2*cos(27.5)=0

    I'm not sure if that L is 4.4 or 6, nor do I know if I am starting this properly.
    Last edited: Nov 19, 2007
  2. jcsd
  3. Nov 19, 2007 #2
    Eqn for Fx and Fy are correct, but torque is incorrect; firstly, dimensionally its wrong and secondly, which is cord 1? left side one or right side one?
  4. Nov 19, 2007 #3
    Cord 1 is on the left
  5. Nov 19, 2007 #4
    k. then L should be 4.4
  6. Nov 19, 2007 #5
    so is the torque equation correct if i use 4.4
  7. Nov 19, 2007 #6
  8. Nov 19, 2007 #7
    I'm afraid I don't understand
  9. Nov 19, 2007 #8
    T2*cos(..) has dimensions of force and mgL has dimensions of work.
  10. Nov 19, 2007 #9
    so would torque be the length of the bar*-mg
  11. Nov 19, 2007 #10
    torque is : -mgL + T2cos(27.5)*6 (6 is the perpendicular distance of force vector from the point at which you are taking torque)
    Last edited: Nov 19, 2007
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