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Three blocks ...

Question

Three blocks of masses $m_{1}$ , $m_{2}$ and $m_{3}$ are connected as shown in the figure. All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of mass $m_{1}$ .

\frac{2 g}{1 + \frac{m_{1}}{4} (\frac{4}{m_{2}} + \frac{3}{2 m_{3}})}

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\frac{g}{1 + \frac{2 m_{1}}{4} (\frac{3}{m_{2}} + \frac{2}{m_{3}})}

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\frac{g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}

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\frac{2 g}{1 + \frac{2 m_{1}}{7} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}

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Solution

The correct option is B $\frac{g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}$
Let tension in string attached to $m_{1}$ be $T_{1}$ , and that in the other be $T_{2}$ . Let the acceleration of block of mass $m_{1}$ $b e$ A $, a n d t h e r e l a t i v e a c c e l e r a t i o n o f t h e o t h e r t w o b l o c k s b e$ a_r$.
From the freebody diagram of the blocks,
$m_{1} : T_{1} = m_{1} A$
and $m_{2} : T_{2} - m_{2} g = m_{2} (a_{r} - A)$
and $m_{3} : m_{3} g - T_{2} = m_{3} (a_{r} + A)$
Also the forces on the suspended pulley must balance each other. Hence
$2 T_{2} = T_{1}$
Solving the above equations gives:
$A = \frac{g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}$

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