In an arrangement of a doubly inclined plane, block $M$ is placed on the face inclined at $53^\circ$ and block $m$ on the face inclined at $37^\circ$. The blocks are connected by a light string passing over an ideal pulley at the top. The coefficient of friction between the surface of the plane and the blocks is $0.25$. The value of $m$ for which $M = 10\ \text{kg}$ will move down with an acceleration of $2\ \text{m s}^{-2}$ is: (take $g = 10\ \text{m s}^{-2}$ and $\tan37^\circ = \frac34$)
Answer: (B) $4.5\ \text{kg}$
$\sin53^\circ = \cos37^\circ = 0.8$, $\cos53^\circ = \sin37^\circ = 0.6$.
Block $M$ (moving down): $Mg\sin53^\circ - \mu Mg\cos53^\circ - T = Ma$
$$80 - 15 - T = 20 \Rightarrow T = 45\ \text{N}$$
Block $m$ (moving up): $T - mg\sin37^\circ - \mu mg\cos37^\circ = ma$
$$45 = m(6 + 2 + 2) \Rightarrow m = 4.5\ \text{kg}$$
Solution by Sreeraj P, M.Sc Physics