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 $B$
$2mgsin45^{\circ }-{\mu }_1{R}_1-T_2=2ma$
$2mgsin45^{\circ }-\frac{1}{3}2mgcos45^{\circ }-T=2ma$
$\Rightarrow 2mg\times \frac{1}{\sqrt{2}}-\frac{1}{3}2mg\times \frac{1}{\sqrt{2}}-T=2ma\dots \left(i\right)$
In this problem as $\left(m_B-m_A\right)$ $gsin\theta =\left(mg\sqrt{2}\right)$
is lesser than
$\left({\mu }_B{m}_B+{\mu }_A{m}_A\right)gcos\theta =\left(4mg3\sqrt{2}\right)$ the masses will not move and hence,
Acceleration of $B$ = acceleration of $A=0$