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Thread: Another pulley problem/resolving

  1. #1
    Oct 2010

    Another pulley problem/resolving

    Hey, another question whereby i don't get how my method gets the wrong answer.... solution is here, 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
    May 2010
    Where did you get 0.6T?

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

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