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Physics

The Simple Pulley

A smooth ligh...

Question

A smooth light pulley fixed to the ceiling of a lift carries a light thread whose ends are attached to the loads to masses $m_{1}$ and $m_{2}$ . The lift starts going up with acceleration a. The forceexerted by the pulley on the ceiling is.( $m_{1} > m_{2}$ )

\frac{4 m_{1} m_{2}}{(m_{1} + m_{2})} (g + a)

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\frac{2 m_{1} m_{2}}{(m_{1} + m_{2})} (g - a)

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\frac{4 m_{1} m_{2}}{(m_{1} + m_{2})} (g)

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\frac{3 m_{1} m_{2}}{(m_{1} + m_{2})} (g + a)

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Solution

The correct option is A $\frac{4 m_{1} m_{2}}{(m_{1} + m_{2})} (g + a)$
For the motion as per above figure,
Let's assume (m $_{1}$ >m $_{2}$ ) so that acceleration of m $_{2}$ is A downwards and m $_{1}$ is A upwards,

Then we have,
m $_{2}$ (g+a) - T=m $_{2}$ A
T-m $_{1}$ (g+a) =m $_{1}$ A

Solving we get T= $\frac{2 m_{2} m_{1}}{(m_{1} + m_{2})} (g + a)$

Thus force exerted by pulley on ceiling =2T= $\frac{4 m_{2} m_{1}}{(m_{1} + m_{2})} (g + a)$

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A pulley which is fixed to the ceiling of an elevator car carries a thread whose ends are attached to the masses $m_{1}$ and $m_{2}$ . The car starts going up with an acceleration $a_{0}$ . Assuming the masses of the pulley and the thread as well as the friction to be negligible, find:

Match the given columns according to the options where,

is the acceleration of the load $m_{1}$ relative to the ground.

Is the acceleration of the load $m_{1}$ relative to the elevator car.

is the force exerted by the pulley on the ceiling of the car. Given, $m_{1} > m_{2}$

(i) $a_{\frac{m_{1}}{g}}$ (u) $\frac{4 m_{1} m_{2} (a_{0} + g)}{(m_{1} + m_{2})}$

(ii) $a_{\frac{m_{1}}{e l e v a t o r}}$ (v) $\frac{2 a_{1} m_{1} - g m_{2}}{(m_{1} + m_{2})}$

(iii) $F_{p u l l e y o n c i e l i n g}$ (w) $\frac{2 a_{0} m_{2} - g (m_{1} - m_{2})}{(m_{1} + m_{2})}$

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A smooth light pulley fixed to the ceiling of a lift carries a light thread whose ends are attached to the loads to masses m1 and m2. The lift starts going up with acceleration a. The forceexerted by the pulley on the ceiling is.(m1>m2)

The correct option is A 4m1m2(m1+m2)(g+a)For the motion as per above figure, Let's assume (m1>m2) so that acceleration of m2 is A downwards and m1 is A upwards, Then we have, m2(g+a) - T=m2AT-m1(g+a) =m1ASolving we get T=2m2m1(m1+m2)(g+a)Thus force exerted by pulley on ceiling =2T=4m2m1(m1+m2)(g+a)

A smooth light pulley fixed to the ceiling of a lift carries a light thread whose ends are attached to the loads to masses $m_{1}$ and $m_{2}$ . The lift starts going up with acceleration a. The forceexerted by the pulley on the ceiling is.( $m_{1} > m_{2}$ )