Question

Block A of mass $m$ and block $B$ of mass $2m$ are placed on a fixed tringular 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 horizointal on both sides. If the coefficient of friction between the block $A$ and the wedge is $\frac{2}{3}$ and that between block B and the wedge is $\frac{1}{3}$ 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. $\text{zero}$

Answer 1

The correct option is D $\text{zero}$

FBD of the system:

Then, common acceleration of the system
$a_c=\frac{\text{Supporting force}-\text{Opposing force}}{m_{total}}$

Here, total supporting force
$=2mg\sin {45}^{\circ }-mg\sin {45}^{\circ }=\frac{2mg}{\sqrt{2}}-\frac{mg}{\sqrt{2}}=\frac{mg}{\sqrt{2}}$

Opposing force is provided by friction
Maximum available friction force $=f_A+f_B={\mu }_A{N}_A+{\mu }_B{N}_B=\frac{2}{3}\times \frac{mg}{\sqrt{2}}+\frac{1}{3}\times \frac{2mg}{\sqrt{2}}=\frac{4}{3}\frac{mg}{\sqrt{2}}$
$>\frac{mg}{\sqrt{2}}$

Since the maximum opposing force is more, the mass system will not move
$\Rightarrow$ acceleration of system is zero