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Question

Find the acceleration of mass $m_{1}$ and $m_{2}$ in the arrangement shown in figure, if mass $m_{2}$ is twice that of mass $m_{1}$ , and the angle that the inclined plane forms with the horizontal is equal to $30^{\circ}$ . Mass of the pulley and string are negligible and friction is absent everywhere.

(a)_{m_{2}} = \frac{2 g}{18}

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(a)_{m_{2}} = \frac{7 g}{9}

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(a)_{m_{1}} = \frac{7 g}{18}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(a)_{m_{1}} = \frac{7 g}{9}

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Solution

The correct options are
B $(a)_{m_{2}} = \frac{7 g}{9}$
C $(a)_{m_{1}} = \frac{7 g}{18}$
Here pulley $1$ is movable and if it moves by $x$ downwards, then total length of string slackened is $2 x$ .
However, one end of the string is fixed so in order to maintain string length constant, block $m_{2}$ will move downwards by $2 x$ and block $m_{1}$ will move $x$ upwards along the incline.

Differentiating the displacement twice w.r.t time:
$a_{1} = a$ for $m_{1}$ , and $a_{2} = 2 a$ for $m_{2}$ respectively.

Applying Newton's $2^{nd}$ law for $m_{1}$ and $m_{2}$ along the direction of acceleration:
$2 T - m_{1} g sin θ = m_{1} a . . . (i)$
$m_{2} g - T = m_{2} (2 a) . . . (i i)$

Here $m_{2} = 2 m_{1}, θ = 30^{\circ} . . . (i i i)$

Now, adding $(i i) \times 2 + (i)$ ,
$\Rightarrow 2 m_{2} g - m_{1} g sin 30^{\circ} = a (m_{1} + 4 m_{2})$
$\Rightarrow 4 m_{1} g - \frac{m_{1} g}{2} = a (m_{1} + 8 m_{1})$
$[∵ m_{2} = 2 m_{1}]$
$\Rightarrow \frac{7 m_{1} g}{2} = a (9 m_{1})$
$∴ a = \frac{7 g}{18}$
Therefore,
Acceleration of $m_{1} = a = \frac{7 g}{18}$
Acceleration of $m_{2} = 2 a = 2 \times \frac{7 g}{18} = \frac{7 g}{9}$

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Find the acceleration of mass m1 and m2 in the arrangement shown in figure, if mass m2 is twice that of mass m1, and the angle that the inclined plane forms with the horizontal is equal to 30∘. Mass of the pulley and string are negligible and friction is absent everywhere.