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Question

Three masses $m_{1}, m_{2}$ and $m_{3}$ are attached to a string pulley system as shown. All three masses are held at rest and then released. To keep $m_{3}$ at rest, the condition is:

\frac{1}{m_{3}} = \frac{1}{m_{1}} + \frac{1}{m_{2}}

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m_{1} + m_{2} = m_{3}

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\frac{4}{m_{3}} = \frac{1}{m_{1}} + \frac{1}{m_{2}}

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\frac{1}{m_{1}} + \frac{2}{m_{2}} + \frac{3}{m_{3}}

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Solution

The correct option is C $\frac{4}{m_{3}} = \frac{1}{m_{1}} + \frac{1}{m_{2}}$
Let tension produced string consisting of $m_{3}$ be $T$ .Since $m_{3}$ has to be at rest, $T$ should be equal to $m_{3} g$ .Now consider the lower pulley.Forces acting on this pulley are $T_{1}$ on one string and $T_{1}$ o the other string.Since tis pulley is at rest so
$2 T_{1} = T$ and $T_{1} = \frac{T}{2}$ .
Now take acceleration of $m_{2}$ as $a$ downwards and $m_{3}$ as $a$ upwards.
Equation of motion of these two blocks are given,
$\frac{T}{2} - m_{1} g = m_{1} a ⟶ 1$
$- \frac{T}{2} + m_{2} g = m_{2} a ⟶ 2$
divide $e q .1$ by $e q .2$ and rearrange the divided equations to get
${\frac{4}{m}}_{3} = {\frac{1}{m}}_{1} + {\frac{1}{m}}_{2}$ , $i . e .$ option $c$ .

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Three masses m1,m2 and m3 are attached to a string pulley system as shown. All three masses are held at rest and then released. To keep m3 at rest, the condition is:

Three masses $m_{1}, m_{2}$ and $m_{3}$ are attached to a string pulley system as shown. All three masses are held at rest and then released. To keep $m_{3}$ at rest, the condition is: