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

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.

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{m}_{B}g\sin {30}^{\circ }=m_B{a}_0\cos {30}^{\circ }\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}_0=\frac{10}{\sqrt{3}}\mathrm{m}/{\mathrm{s}}^2\end{array}$

From free body diagram of block A in the frame of block C

$\begin{array}{l}N+m_A{a}_0\sin {30}^{\circ }=m_{A}g\cos {30}^{\circ }\\ \mathrm{or} \frac{\left(m_B+m_C\right)a_0}{\sin {30}^{\circ }}+m_A{a}_0\sin {30}^{\circ }=m_{A}g\cos {30}^{\circ }\\ \mathrm{or} \frac{3\times 10}{\sqrt{3}}\times 2+m_A\times \frac{10}{2\sqrt{3}}=m_A\times 10\times \frac{\sqrt{3}}{2}\\ \therefore m_A=6\mathrm{kg}\end{array}$