Question

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

Answer 1

Answer: 6

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 $\frac{3\times 10}{\sqrt{3}}\times 2+m_A\times \frac{10}{2\sqrt{3}}$
$=m_A\times 10\times \frac{\sqrt{3}}{2}$
$\therefore m_A=6kg$