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Math Help - Another pulley problem/resolving

  1. #1
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    Another pulley problem/resolving

    Hey, another question whereby i don't get how my method gets the wrong answer.... solution is herehttp://clip2net.com/s/VKcK, question is here:part 1 and question i need help with
    tension in the string is 6g, where g=9.8, sin theta is 3/5.

    I resolve tension from the left so that i get R= T(0.6) + T , because one tension is sloped (therefore needs verticall resolved) and one is already completely downward, right?

    I don't understand why the mark scheme suggests that the vertically resolved downward compenent X2 is the force acting down, can someone please explain this to me?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Where did you get 0.6T?

    If you want to resolve the tension on the left, you have a vertical component of T\cos\alpha and a horizontal component of T\sin\alpha

    Then, teh net force downwards is T\cos\alpha + T while the net left force is T\sin\alpha. The net force on the pulley is then the vector sum of those two forces.
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