Question

Find the acceleration of rod $A$ and wedge $B$ in the arrangement shown in figure if the mass of rod equal that of the wedge and the friction between all contact surfaces is negligible and rod $A$ is free to move downwards only. Take angle of wedge as ${37}^{\circ }$. (Take the masses of both rod and wedge to be "m")

Answer 1

Consider the acceleration of the wedge be $a$ towards right and the acceleration of rod be $b$ downwards

$b\cos 37=a\cos 53$ from the contraction of the motion

$\begin{array}{c}\frac{4b}{5}=\frac{3a}{5}\\ \Rightarrow 4b=3a------\left(1\right)\end{array}$

Now consider the contact force between them is $N$

Then,

For rod

$mg-N\cos 37=mb--------\left(2\right)$

For wedge

$\begin{array}{c}N\sin 37=ma\\ \Rightarrow N=\frac{5ma}{3}--------\left(3\right)\end{array}$

Now

$\begin{array}{c}mg-\frac{5ma}{3}\times \frac{4}{5}=mb\\ \Rightarrow g-\frac{4a}{3}=b\\ \Rightarrow g-\frac{4a}{3}=\frac{3a}{4}\\ \Rightarrow g=\frac{16a+9a}{12}\\ \Rightarrow 25a=120\\ a=\frac{120}{25}=4.8m/s^2\\ b=\frac{3a}{4}=3.6m/s^2\end{array}$