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. $-1m{s}^{-2}$
• B. $1.2m{s}^{-2}$
• C. $0.2m{s}^{-2}$
• D. Zero

Answer 1

Answer: (D)

Zero

$\text{Total maximum friction force},f_t=f_A+f_B$

${\mu }_{A}mg\cos {45}^{\circ }+{\mu }_{B}2mg\cos {45}^{\circ }$

$\text{or,}{f}_t=$ $mg(\frac{2}{3}\times \frac{1}{\sqrt{2}}+\frac{1}{3}\times 2\times \frac{1}{\sqrt{2}})$

$=\frac{4}{3\sqrt{2}}mg$

The net pulling force, $F=2mg\sin {45}^{\circ }-mg\sin {45}^{\circ }$

$=mg\sin {45}^{\circ }$

$=\frac{mg}{\sqrt{2}}$

$\because$ $F<f_t$

$\therefore$ The blocks will not move.

So, acceleration of block $A$ is $\underset{–}{\text{Zero.}}$

$\therefore$ option (D) is correct.