Question

Consider the diagram shown below in which two masses of $m$ and $2m$ 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. $\frac{4m^2}{3}g$
• D. $\frac{m^2}{\sqrt{2}}g$

Answer 1

Answer: (A)

Redrawing the diagram.

For block $B$ of mass $2m$
$2mg\sin 45^{\circ }-f_2-T=2ma$
$\Rightarrow \frac{2mg}{\sqrt{2}}-{\mu }_1{R}_1-T_2=2ma$
$\Rightarrow \frac{2mg}{\sqrt{2}}-\frac{1}{3}\cdot 2mg\cos 45^{\circ }-T=2ma$
As $\left(m_B-m_A\right)g\sin \theta =\frac{mg}{\sqrt{2}}$ is lesser
then $\left({\mu }_B{m}_B+{\mu }_A{m}_A\right)g\cos \theta$
$=\frac{4mg}{3\sqrt{2}},$ the masses will not move.
So, acceleration of the system will be zero.