Question

Consider the diagram shown below in which two masses of $m$ and $2 m$ are placed on a fixed triangular wedge.

The coefficient of friction between block $A$ and the wedge is $2 / 3$, while that for block $B$ and the wedge is $1 / 3$.
If the whole system is released from rest, then acceleration of block $A$ is

• A. Zero
• B. $ \frac{2m^2}{3} g$
• C. $ \frac{4 m^2}{3} g$
• D. $ \frac{m^2}{\sqrt{2}} g$

Answer 1

Answer: (A)

Redrawing the diagram.

For block $B$ of mass $2 m$
$2\, m g \sin 45^{\circ}-f_{2}-T =2 m a $
$ \Rightarrow \frac{2 m g}{\sqrt{2}}-\mu_{1} R_{1}-T_{2} =2 m a$
$\Rightarrow \frac{2 m g}{\sqrt{2}}-\frac{1}{3} \cdot 2 m g \cos 45^{\circ}-T =2 m a$
As $ \left(m_{B}-m_{A}\right) g \sin \theta=\frac{m g}{\sqrt{2}} $ is lesser
then $\left(\mu_{B} m_{B}+\mu_{A} m_{A}\right) g \cos \theta$
$=\frac{4 m g}{3 \sqrt{2}}, $ the masses will not move.
So, acceleration of the system will be zero.