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Tension in a pulley System

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    2. Relevant equations

    F=mg=(6kg)(9.81m/s^2)=58.86N
    Fx=(cos30)(58.86N)

    3. The attempt at a solution

    Obviously the triangle somehow interferes with the tension between the top of the pulley and the rope attached to the wall. What really confuses me is how the tension in the rope attached to the wall is greater than the gravitational force of the weight itself. By my thinking the Fx at the top of the pulley = 50.97N which is wrong. How is the triangle increasing the tension?
     
  2. jcsd
  3. Dec 15, 2008 #2

    LowlyPion

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    Homework Helper

    You need to consider the sum of the Torques about the pivot C is 0.

    If the tension is greater, the lever arm over which the AD segment acts about C must be shorter than the lever arm that the weight acts at.
     
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