Four block are arranged on a smooth horizontal surface as shown in figure .The masses of the blocks are given (see the fig ) The coefficient of static friction between the top and the bottom blocks is `mu_(s)` What is the maximum value of the horizontal force `F` applied to one of the bottom blocks as shown that makes all four block with the same acceleration ?

Answer

`2mu_(s) mg((m + M)/(2m + M))`

Explanation

Draw free body diagram of all the block

Try to identify the case of motion of block on which force is not applied
Block `A` move due to static friction when sliding static it is `f_(max) = mu_(s) mg` This force must be greater than tension `T` only then if will accelerate forward block `C` moves due to tension `T` which must be greater than `f` the static friction nbetween `C` and `D` block `FBD` of block due to `f`
From `FBD` of block `B`
`F - mu_(s) mg = Ma`...(i)
From `FBD` of block `B`
`F - mu_(s) mg = Ma`...(i)
From `FBD` of block `A`
` mu_(s) mg - T= ma`...(ii)
From `FBD` of block `C`
`T - f = ma`...(iii)
From FBD of block `D,f = Ma`....(iv)
From Eq(iii) and (iv) ` T = (m + M)a`...(v)
`a = ((T)/(m + M))`
putting `T` in Eq (i) `F - mu mg = (mu_(s) Mmg)/((M + 2m))`
`F = 2mu_(s) mg((m + M)/(2m + M))`