# Mechanics Of A Pulley

• December 11th 2007, 04:27 AM
Simplicity
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 $3kg$ and $6kg$ 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:

$(P): T - 3g = 3a$ ------------ $(1)$
$(Q): 6g - T = 6a$ ------------ $(2)$
$(1) + (2) \implies a=\frac{1}{3} ms^{-2}$
$\therefore \text{Using }s=ut + \frac{1}{2} a t^2.$ $\text{Where }s=2, u=0, a=\frac{1}{3}$
$t= \sqrt \frac{2s}{a}$
$t=3.46s$
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But...The real answer is $t=1.11s$ so where have I gone wrong? :confused:
• December 11th 2007, 06:33 AM
topsquark
Quote:

Originally Posted by Air
$(P): T - 3g = 3a$ ------------ $(1)$
$(Q): 6g - T = 6a$ ------------ $(2)$
$(1) + (2) \implies a=\frac{1}{3} ms^{-2}$

Slow down! Where's the fire? Add these two equations again a bit more carefully:
$(T - 3g) + (6g - T) = (3a) + (6a)$

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

$-3g + 6g = 9a$

$3g = 9a$

Do you see your mistake now?

-Dan
• December 11th 2007, 06:37 AM
Simplicity
Quote:

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

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

$-3g + 6g = 9a$

$3g = 9a$

Do you see your mistake now?

-Dan

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

Thank you. :D