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The Complete Equilibrium

Two weights w...

Question

Two weights $w_{1}$ and $w_{2}$ are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up at an acceleration $g$ , the tension in the string will be

A

\frac{2 w_{1} w_{2}}{w_{1} + w_{2}}

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B

\frac{w_{1} w_{2}}{w_{1} + w_{2}}

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C

\frac{w_{1} w_{2}}{2 (w_{1} + w_{2})}

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D

\frac{4 w_{1} w_{2}}{w_{1} + w_{2}}

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Solution

The correct option is D $\frac{4 w_{1} w_{2}}{w_{1} + w_{2}}$
Given: Acceleration of pulley, $a_{p} = g$
Let us consider the weight $w_{2}$ is going downward with an acceleration $a$ .
The effective acceleration due to gravity acting is given by,

$g_{e f f} = g + g = 2 g$

The weight of both the block changes to $2 w_{1}$ and $2 w_{2}$
Equating all the forces for $m_{1}$ vertically, we get,

$T - 2 w_{1} = \frac{w_{1}}{g} a$ ..... $(1)$

Similarly for $m_{2},$ Resolving all the forces,we get,

$2 w_{2} - T = \frac{w_{2}}{g} a$ ..... $(2)$

Dividing $(1)$ by $(2)$ , we get,
$\frac{w_{1}}{w_{2}} = \frac{T - 2 w_{1}}{2 w_{2} - T}$

$\Rightarrow 2 w_{1} w_{2} - w_{1} T = w_{2} T - 2 w_{1} w_{2}$

$\Rightarrow 4 w_{1} w_{2} = T (w_{1} + w_{2})$

$\Rightarrow T = \frac{4 w_{1} w_{2}}{w_{1} + w_{2}}$
Hence, option $(A)$ is correct

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