Two blocks A and B of mass 6kg and 3g respectively are placed on a smooth surface of an elevator in such a way that B is placed above A.The coefficient of friction between the two blocks is 1/4 and the elevator is ascending up with acceleration g/3.A horizontal force 10N is applied on A. Find the accelerations of the blocks w.r.t. the elevator.

							First draw the diagram by yourself according to the the given conditions and draw FBD.
Mass of A= 6Kg, Mass of B= 3Kg  
As elevator is accelerating upwards with acceleration of g/3,value of Normal Reaction(N) will change
N-Mg=Mg/3
N=4Mg/3
Hence value of  N between 2 blocks will be = 4*3*9.8/3= 39.2Newton
Now common acceleration of 2 blocks in horizontal direction = 10/(3+6) =10/9 m/s2

*Now check whether there will be slipping between 2 blocks or not
Force required by B to move with common acceleration in horizontal direction = 3*10/9 = 10/3 Newton
Maximum friction between blocks = µ*N = 1/4*39.2 = 9.8 Newton
As 9.8>10/3,
There will be no slipping between the blocks.
They will move with common acceleration in horizontal direction that is 10/9 m/s2


Hence acceleration of both blocks w.r.t. the elevator is = (10/9 i cap + g/3 j cap) m/s2....
						

							Sorry my final answer is w.r.t Earth.

Hence acceleration of both blocks w.r.t. the elevator is = (10/9 i cap ) m/s2....
						

							Is it 3g or 3kg.....