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 ${37}^o$.

Answer 1

Let us draw free body diagram of each, i.e. of rod A and of wedge B and also draw the kinematical diagram for acceleration, after analyzing the direction of motion of A and B. kinematical relationship of acceleration.

$tan\alpha =\frac{W_A}{W_B}$

Let us write Newton's second law of both bodies in terms of a projection having taken the positive direction of y and x-axes as shown in the figure

$m_{A}g-Ncos\alpha =m_A{W}_A$

$Nsin\alpha =M_B{W}_B$

$w_A=\frac{m_{A}gsin\alpha }{m_{A}sin\alpha +m_{B}cot\alpha cos\alpha }=\frac{g}{(1+nco{t}^2\alpha )}$

$w_b=\frac{w_A}{tan\alpha }=\frac{g}{(tan\alpha +ncot\alpha )}$