A pulley fixed to the ceiling of an elevator car carries a thread whose ends are attached to the loads $m_{1}$ and $m_{2}$ ( $m_{1}$ $>$ $m_{2}$ ). The car starts going up wih an acceleration $a$ . Assuming the masses of the pulley and the thread, as well as the friction, to be negligible find the acceleration of $m_{1}$ relative to the elevator shaft and relative to car.

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$a = a - a_{0} = \frac{(m_{1} - m_{2}) g - 2 m_{2} a_{0}}{(m_{1} - m_{2})}$

$a = a - a_{0} = \frac{(m_{1} - m_{2}) g - m_{2}}{(m_{1} - m_{2})}$

$a = a - a_{0} = \frac{(m_{1} - m_{2})}{(m_{1} + m_{2})}$

$a = \frac{(m_{1} - m_{2}) g - 2 m_{2} a_{0}}{(m_{1} - m_{2})}$

answer is A.

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Detailed Solution

Suppose the acceleration of load $m_{1}$ with respect to car is a downward, then acceleration of $m_{2}$ will be $a$ upward.

If their accelerations are $a_{1}$ and $a_{2}$ respectively w.r.t. an observer on the ground, then

$a_{1} = (a - a_{0})$ downward

and $a_{2} = (a + a_{0})$ upward

By Newton’s second law, we have

$m_{1} g - T = m_{1} (a - a_{0})$ …(i)

and $T - m_{2} g = m_{2} (a - a_{0})$ …(ii)

Solving equations (i) and (ii), we get

$a = \frac{(m_{1} - m_{2}) (g + a_{0})}{(m_{1} + m_{2})}$

$T = \frac{2 m_{1} m_{2} (g + a_{0})}{m_{1} + m_{2}}$

$∴ a = a - a_{0} = \frac{(m_{1} - m_{2}) g - 2 m_{2} a_{0}}{(m_{1} - m_{2})}$

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