# Thread: Mechanics Of A Pulley

1. ## Mechanics Of A Pulley

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

$\displaystyle (P): T - 3g = 3a$ ------------$\displaystyle (1)$
$\displaystyle (Q): 6g - T = 6a$ ------------$\displaystyle (2)$
$\displaystyle (1) + (2) \implies a=\frac{1}{3} ms^{-2}$
$\displaystyle \therefore \text{Using }s=ut + \frac{1}{2} a t^2.$ $\displaystyle \text{Where }s=2, u=0, a=\frac{1}{3}$
$\displaystyle t= \sqrt \frac{2s}{a}$
$\displaystyle t=3.46s$
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But...The real answer is $\displaystyle t=1.11s$ so where have I gone wrong? 2. Originally Posted by Air $\displaystyle (P): T - 3g = 3a$ ------------$\displaystyle (1)$
$\displaystyle (Q): 6g - T = 6a$ ------------$\displaystyle (2)$
$\displaystyle (1) + (2) \implies a=\frac{1}{3} ms^{-2}$
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

3. Originally Posted by topsquark 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.

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