If the acceleration of mass $‘ n m^{'} = \frac{7 g}{p}$ , find the value of $p$ if $n = 2$ . Assume all surfaces to be smooth and strings inextensible .

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Solution

The given condition is as shown. Let the acceleration of pulley, mass $m$ and mass $n m$ be $a_{p}, a_{1}$ and $a_{2}$ respectively.

Since strings are inextensible,
$l_{1} + l_{2} = constant$
Double differentiating w.r.t time, we get equation in the form of acceleration i.e
$_{1} +_{2} = 0 \Rightarrow - a_{1} + a_{p} = 0 \Rightarrow a_{p} = a_{1}$
Simillarly for other string,
$_{4} +_{3} = 0 \Rightarrow - a_{p} - a_{p} + a_{2} = 0 \Rightarrow a_{2} = 2 a_{p}$
or $a_{2} = 2 a_{1} - (1)$

From the FBD of block of mass $n m$ ,
$n m g - T_{2} = (n m) a_{2}$
$\Rightarrow n m g - T_{2} = (n m) 2 a_{1}$
$\begin{matrix} \Rightarrow 2 n m g - 2 T_{2} = 2 (n m) 2 a_{1} \\ (1) \end{matrix}$
From the FBD of block of mass $m$ ,
$\begin{matrix} T_{1} - m g s i n θ = m a_{1} \\ (2) \end{matrix}$
From the FBD of moving pulley,
$T_{1} = 2 T_{2}$ (Since the pulley is massless)
Since $2 T_{2} = T_{1}$ , solving $(1)$ and $(2)$ ,
$m g (2 n - sin θ) = m a_{1} (4 n + 1)$
Now for $θ = 30^{\circ}$ , we get the values of accelerations as
$a_{1} = g \frac{(4 n - 1)}{2 (4 n + 1)} & \Rightarrow a_{2} = g [\frac{4 n - 1}{4 n + 1}]$
So putting the value of $n = 2$ we get acceleration of mass $n m$ as
$a_{2} = \frac{7 g}{9}$ , which is given in the question as $\frac{7 g}{p}$
Thus, comparing both we get the value of $p = 9$

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