Question

The two pulley arrangements shown in the figure are identical. The mass of the rope is negligible. In (a) the mass $m$ is lifted up by attaching a mass $2m$ to the other end of the rope. In (b) $m$ is lifted up by pulling the order end of the rope with a constant downward force of $2mg$. The ratio of acceleration in two cases will be

Answer 1

Let the accelerations in both cases are $a$ and $a^{\prime }$ respectively.

The tension is given as,

$T-mg=ma$ (1)

$2mg-T=2ma$

By adding equations (1) and (2), we get

$mg=3ma$

$a=\frac{g}{3}$

In the second case the tension is given as,

$T^{\prime }-mg=m{a}^{\prime }$ (3)

$2mg-T^{\prime }=0$ (4)

By adding equations (3) and (4), we get

$mg=m{a}^{\prime }$

$a^{\prime }=g$

The ratio of acceleration is given as,

$\frac{a}{a^{\prime }}=\frac{\frac{g}{3}}{g}$

$\frac{a}{a^{\prime }}=\frac{1}{3}$