Question

The situation shown in the figure, all surfaces are frictionless. Masses of blocks $B$ and $C$ are $1\, kg$ and $2 kg$, respectively. Find the mass of the blocks $A$ for which the block $B$ remains stationary with respect to block $C$.

Answer 1

Taking blocks $B$ and $C$ as a system,
$N \sin 30^{\circ}=\left(m_{B}+m_{C}\right) a_{0}$ (leftward).
Here, $N$ is normal reaction between $A$ and $C$.

From free body diagram of block $B$ in the frame of block $C$.
$m_{B} g \sin 30^{\circ}=m_{B} a_{0} \cos 30^{\circ}$
$\therefore a_{0}=\frac{10}{\sqrt{3}} m / s ^{2}$
From free body diagram of block $A$ in the frame of block $C$
$N +m_{A} a_{0} \sin 30^{\circ}=m_{A} g \cos 30^{\circ}$
Or $\frac{\left(m_{B}+m_{C}\right) a_{0}}{\sin 30^{\circ}}+m_{A} a_{0} \sin 30^{\circ}$
$=m_{A} g \cos 30^{\circ}$
Or $\frac{3 \times 10}{\sqrt{3}} \times 2+m_{A} \times \frac{10}{2 \sqrt{3}}$
$=m_{A} \times 10 \times \frac{\sqrt{3}}{2}$
$\therefore m_{A}=6\, kg$