Question

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. If 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

Answer 1

Answer: (D)

The situation is as shown in the figure

The equation of motion for body $B$
$2\, mg \,sin\,45^{\circ}-\mu_{1}R_{1}-T_{2}=2ma$
$2\,mg\,sin\,45^{\circ}-\frac{1}{3}2\,mg\,cos\,45^{\circ}-T=2ma$
$\Rightarrow 2\,mg\times\frac{1}{\sqrt{2}}-\frac{1}{3}2\,mg \times\frac{1}{\sqrt{2}}-T=2ma \dots\left(i\right)$
In this problem as $\left(m_{B}-m_{A}\right)$ $g\,sin\,\theta=\left(mg \sqrt{2}\right)$
is lesser than
$\left(\mu_{B}m_{B}+\mu_{A}m_{A}\right)g\, cos\,\theta=\left(4 mg 3\sqrt{2}\right)$ the masses will not move and hence,
Acceleration of $B$ = acceleration of $A = 0$