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 theblock B remains stationary with respect to block C.

Taking blocks B and C as a system, Nsin⁡30∘=mB+mCa0(leftward).Here, N is normal reaction between A and C.From free body diagram of block B in the frame of block C.    mBgsin⁡30∘=mBa0cos⁡30∘∴     a0=103m/s2From free body diagram of block A in the frame of block CN+mAa0sin⁡30∘=mAgcos⁡30∘or  mB+mCa0sin⁡30∘+mAa0sin⁡30∘=mAgcos⁡30∘or  3×103×2+mA×1023=mA×10×32∴mA=6kg