Q.
Block $A$ of mass $m$ and block $B$ of mass $2m$ are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure. The wedge is inclined at $45{^\circ}$ to the horizontal on both the sides. The coefficient of friction between the block $A$ and the wedge is 2/3 and that between the block $B$ and the wedge is 1/3 and both the blocks $A$ and $B$ are released from rest, the acceleration of $A$ will be
A) $- 1$
B) 1.2
C) 0.2
D) Zero
Explanation
(d) The situation is as shown in the figure
[image:5_image211.jpeg]
The equation of motion for body B.
$2mg\sin{45{^\circ} - \mu_{1}R_{1} - T_{1} - T_{2} = 2ma}$
$2mg\sin{45{^\circ} - \frac{1}{3}2mg\cos{45{^\circ} - T = 2ma}}$
$\Longrightarrow \ \ \ 2mg \times \frac{1}{\sqrt{2}} - \frac{1}{3}2mg\ \times \frac{1}{\sqrt{2}} - T = 2ma\ \ \ldots.(i)$
In the problem as $\left( m_{B} - m_{A} \right)g\cos{\theta = (mg/\sqrt{2}})$ is lesser than
$\left( \mu_{B}m_{B} + \mu_{A}m_{A} \right)g\cos{\theta = (4mg/3\sqrt{2})}\ $ the masses will not move and hence.
Acceleration of B = acceleration of A = 0.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.