Consider the situation shown in Fig. The hirizontal surface below the bigger block is smooth the coefficient of friction between the block is `mu` find the minimum and the maximum force `F` that can be applied in order to keep the smaller blocks at rest with respoct to the bigger block

Explanation

If no force is applied blocks A will slip on C toward right and block B will move downward Suppose the minimum force needed to prevent slipping is `F` Taking `A + B+ C` as the system is `F` hence , the acceleration of the system is
`a = (F)/(M + 2m)` ...(i)
New take the block A as the system . The forces on A are Fig.

a. Tension `T` by the string toward right
b. Friction `f` by the block C towards left
c. Weight `mg` downward and
d. Normal force `N` upward
For vertical equlibrium `N = mg`
As the minimum force needed to prevent all slipping is applied the friction is limiting Thus .
`f = mu N = mu mg`
As the block toward right with an acceleration a
`T - f = ma`
`or T - mu mg = ma ` .....(ii)
Now take block B as the system . The forces are [Fig]
a. Tension `T` upward
b. weight `mg` downward
c. normal force `N` toward right ' , and
d. friction `f` upward.
As the block moves toward right with an acceleration `alpha`
`N = ma`
As the friction is limiting ,`f = mu N '= mu ma`
For vertical equation `T = f = mg`
`T + mu ma = mg`
Eliminating `T` from (ii) and (iii) , we get
`a_("min") = (1 - mu)/(1 + mu) g`.
When a large force is applied block A slip on C toward left and block B slip on C in the upward direction . The friction on A is towards right and that on B is downward solving as above the acceleration in this case is
`a_("max")= (1 +mu)/(1- mu) g`
Thus les between`(1 - mu)/(1+ mu) g` and `(1 + mu)/(1- mu) g`
From (i) , the force `F` should be between `(1 - mu)/(1+ mu) (M + 2m)g` and `(1 + mu)/(1- mu) (M + 2m)g`