Consider the situation shown in figure. The horizontal surface below the bigger block is smooth. The coefficient of friction between the block is u(mu). Find the minimum and the maximum force F can be applied in order to keep the smaller blocks at rest with respect to the bigger block.

							
If no force is applied, the block A will slip on C towards right and the block B will move downward. Suppose the minimum force needed to prevent slipping is F, taking A+B+C as the system, the only external horizontal force on the system is F. Hence the acceleration of the system is
a = F/2m+M
take the block A as the system. The force on A are
 1)      Tension T by the string towards right
 2)      Friction f by the block C towards left
 3)      Weight mg downward and
 4)      Normal force N upward.
For vertical equilibrium N = mg.
As the minimum force needed to prevent slipping is applied, the friction is limiting thus
f = mN = mmg
As the block moves towards right with an acceleration a,
T –f = ma
Or T - mmg = ma                                                               ----------------- (2)
Now take the block B as the system the forces are as shown in fig (b).      
 1)      Tension T upward.
 2)      Weight mg downward
 3)      Normal force N1 towards right
 4)      Friction f1 upward.
As the block moves towards right with an acceleration a, N1 = ma.
As the friction is limiting, f1= mN1 = mma
For vertical equilibrium
T + f1 = mg
T + mma = mg 
Eliminating T from (1) and (3)
amin = (1-)*g/(1+ )
When a large force is applied the block A slips on C-towards left and the block B slips on C in the upward direction. The friction on A is towards right and that on B is downwards solving as above, the acceleration in this case is
Thus the force should be between
(1-)(M+2m)*g/(1+) and
(1+ )(M+2m)*g/(1- )