Question

In the figure shown $P_1$ and $P_2$ are massless pulleys. $P_1$ is fixed and $P_2$ can move. Masses of A, B and C are $\frac{9m}{64},$ $2m$ and $m$ respectively. All contacts are smooth and the string is massless. If $\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$, then the acceleration of block $C$ in $m/s^2$ is

Answer 1

Let $x,y,z$ be displacements of $A,B,C$ respectively
$a_A,a_B,a_C$ be acceleration of $A,B,C$ respectively

Constraint relation between $a_A$ and $a_B$

$\tan \theta =\frac{y}{x}\Rightarrow y=x\tan \theta \underset{B}{a}=\underset{A}{a}\tan \theta a_B=\frac{3a_A}{4}(\because \theta ={37}^{\circ })if{a}_A=a\Rightarrow a_B=\frac{3a}{4}$
Between B and C, work done by tension must be zero
$\Rightarrow -Ty+2Tz=0y=2z\Rightarrow a_B=2a_C\Rightarrow a_C=\frac{3a}{8}$

$FBD$ of $A$

Considering only the horizontal direction,
$N\sin \theta =\frac{9m}{64}a\frac{3N}{5}=\frac{9ma}{64}(\because \tan \theta =\frac{3}{4}\Rightarrow \sin \theta =\frac{3}{5},\cos \theta =\frac{4}{5})\Rightarrow N=\frac{15ma}{64}......\left(i\right)$

$FBD$ of $B$:




Considering only the vertical direction
$2mg-T-N\cos \theta =2m\left(\frac{3a}{4}\right)2mg-T-\frac{4}{5}\times \frac{15ma}{64}=\frac{3ma}{2}\left(\text{using}\right(i\left)\right)2mg-T=\frac{3ma}{2}+\frac{3ma}{16}2mg-T=\frac{27ma}{16}......\left(ii\right)$
$FBD$ of $C$

$2T-mg=\frac{3ma}{8}........\left(iii\right)4mg-2T=\frac{27ma}{8}........2\left(ii\right)\text{Adding}\left(iii\right)and2\left(ii\right)\Rightarrow 3mg=\frac{30ma}{8}\Rightarrow a=\frac{8g}{10}\Rightarrow a_C=\frac{3a}{8}=\frac{3}{8}\times \frac{8g}{10}=3m{s}^{-2}$