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Challenger Torque question

  1. Nov 5, 2008 #1
    1. The problem statement, all variables and given/known data
    The diagram shows a non uniform bar of mass 59kg suspended by light ropes. Calculate the force of tension in each rope and the distance between the left end of the bar and its center of gravity.


    2. Relevant equations
    ΔT(torque) = 0
    ΔFx = max
    ΔFy = may


    3. The attempt at a solution
    First I wrote down all the variables that were given to me:

    Mg (weight of the bar) = (59)(9.80) = 578.2N
    M (mass of the bar) 59kg
    g = 9.80 m/s(squared)
    l1(lever arm 1) = 8.0m

    I am unsure how to do this question because this question is on a worksheet full of questions where the mass or object is uniform. This question is the only question in the whole worksheet where the mass or object is not uniform thus (I guess) why its a "challenger" question.

    Basically, could someone just show me how to find the distance "d" between the left end of the bar and its center of gravity because I am unsure as to how to even start out finding this.

    Furthermore, to find the tension in each rope do you put the point of rotation at one of the ends where the ropes are, making the tension for that side zero then solve using the torque equation for the tension (of the other side). Then use the translational equilibrium equation to find out what the torque you set to the point of rotation was?
     

    Attached Files:

  2. jcsd
  3. Nov 6, 2008 #2

    Doc Al

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    Staff: Mentor

    Since you're not given the center of gravity, why not try another approach? One that doesn't require calculating torques.
     
  4. Nov 6, 2008 #3
    The only other thing that I can think of to apply to this situation or system would be the translational equilibrium equation because a system in rotational equilibrium has to also be translational equilibrium. However, I am not really sure how I could use this to find the center of gravity for this question.
     
  5. Nov 7, 2008 #4

    Doc Al

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    Staff: Mentor

    Good. Do it.
    Use it to find the tension in the ropes. Then use those tensions (and torque equations) to find the center of gravity.
     
  6. Nov 7, 2008 #5
    For this question is it possible to get a number for the center of gravity (a numerical answer) because I isolated the center of gravity in my torque equations and I got:

    [Tysin(39)]*(8.0)/578.2 = l3

    l3 being the center of gravity
     
  7. Nov 9, 2008 #6

    Doc Al

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    Staff: Mentor

    Of course it's possible to solve for the center of gravity. All I suggest is that you first solve for the tensions.
     
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