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Another forces problem

  1. Mar 11, 2007 #1

    1.A chair attached to a vertical rotating pole by two cables, is spun in a horizontal circle at a constant speed. The speed is sufficient to create tension in both the upper and lower cables. The tension in the upper cable is 3500N, L=10 m, and the mass of the chair is 134kg.

    c. Predict which tension force is greater, T1(tension in the upper cable) or T2 (tension in the lower cable).

    d. What is the tension in the lower cable?

    2. Relevant equations
    Mathematical equations for equilateral triangle
    like height =(((3)^0.5)/2)x side
    and F=ma equation

    3. The attempt at a solution
    for c, I think upper cable has more tension
    and for d, I just found the vertical component of the upper cable, and also the weight of the chair, and assumed that the vertical component of the lower cable is vertical comp of the upper cable minus the Weight

    wondering if i am right in my assumption?
  2. jcsd
  3. Mar 12, 2007 #2


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    Science Advisor
    Homework Helper

    That looks right.

    For part (c), the forces on the chair are its weight (downwards) and a horizontal force because it is rotating.

    You can consider the force in the cables caused by each of those separately and add up the result.

    By symmetry the horizontal force will produce an equal tension F1 in both cables.

    The vertical force will produce a tension F2 in the top cable and a compression -F2 in the bottom cable.

    So the total force in the top cable is F1+F2 and in the bottom cable is F1-F2.

    (d) That's a good way to do it.
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