Find the tension (in $N$ ) acting on $2 m$ mass when the system is released from rest. Assume that the pulleys are ideal and strings are inextensible.

\frac{24 m g}{29}

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\frac{23 m g}{24}

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\frac{25 m g}{23}

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\frac{23 m g}{25}

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Solution

The correct option is A $\frac{24 m g}{29}$
The given system can be reframed as shown

From constrained condition and using method of intercept we get,
a) ${. . L}_{1} + {. . L}_{2} = 0 \Rightarrow - a_{1} + a_{p} = 0$
$\Rightarrow a_{1} = a_{p}$

b) ${. . L}_{3} + {. . L}_{4} = 0$
$\Rightarrow - a_{2} - a_{p} - a_{p} + a_{3} = 0$
$\Rightarrow 2 a_{p} = a_{3} - a_{2}$
$\Rightarrow 2 a_{1} = a_{3} - a_{2}$ .............(1)

From FBD of $2 m, 3 m$ and $m$
$3 m : 3 m g - T_{2} = 3 m a_{3} \dots \dots (2)$ and $T_{1} = 2 T_{2} \dots \dots (3)$
$2 m : T_{2} - 2 m g = 2 m a_{2} \dots \dots (4)$
$m : T_{1} - m g = m a_{1} \dots \dots (5)$

Putting the values of $a_{1}$ , $a_{2}$ and $a_{3}$ into eq (1) from (2), (3), (4) and (5) we get

$2 [\frac{2 T_{2}}{m} - g] = [g - \frac{T_{2}}{3 m}] - [\frac{T_{2}}{2 m} - g]$
$\Rightarrow \frac{4 T_{2}}{m} + \frac{T_{2}}{3 m} + \frac{T_{2}}{2 m} = 4 g \Rightarrow T_{2} = \frac{24 m g}{29}$
Thus, the tension acting on the mass $2 m$ is $\frac{24 m g}{29} N$

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