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Thread: Mechanics Of A Pulley

  1. #1
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    Mechanics Of A Pulley

    I'm going wrong somewhere but don't know where. Can someone help? Thanks in advance.
    ______________________________________________

    Q: Two particles P and Q of masses $\displaystyle 3kg$ and $\displaystyle 6kg$ respectively are attached to the ends of a light inextensible string. The string is released from rest with both masses a distance of 2m above a horizontal floor. Find how long it takes for the particle Q to hit the floor.
    ______________________________________________

    My attempt:

    $\displaystyle (P): T - 3g = 3a$ ------------$\displaystyle (1)$
    $\displaystyle (Q): 6g - T = 6a$ ------------$\displaystyle (2)$
    $\displaystyle (1) + (2) \implies a=\frac{1}{3} ms^{-2}$
    $\displaystyle \therefore \text{Using }s=ut + \frac{1}{2} a t^2.$ $\displaystyle \text{Where }s=2, u=0, a=\frac{1}{3}$
    $\displaystyle t= \sqrt \frac{2s}{a}$
    $\displaystyle t=3.46s$
    ______________________________________________

    But...The real answer is $\displaystyle t=1.11s$ so where have I gone wrong?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Air View Post
    $\displaystyle (P): T - 3g = 3a$ ------------$\displaystyle (1)$
    $\displaystyle (Q): 6g - T = 6a$ ------------$\displaystyle (2)$
    $\displaystyle (1) + (2) \implies a=\frac{1}{3} ms^{-2}$
    Slow down! Where's the fire? Add these two equations again a bit more carefully:
    $\displaystyle (T - 3g) + (6g - T) = (3a) + (6a)$

    $\displaystyle T - 3g + 6g - T = 3a + 6a$

    $\displaystyle -3g + 6g = 9a$

    $\displaystyle 3g = 9a$

    Do you see your mistake now?

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Slow down! Where's the fire? Add these two equations again a bit more carefully:
    $\displaystyle (T - 3g) + (6g - T) = (3a) + (6a)$

    $\displaystyle T - 3g + 6g - T = 3a + 6a$

    $\displaystyle -3g + 6g = 9a$

    $\displaystyle 3g = 9a$

    Do you see your mistake now?

    -Dan
    Oh yes. My 'g' vanished in the equation.

    Thank you.
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