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 ?

Options

A. `F_("max")=2mu_(s)mg((2m+M)/(m+M))`
B. `F_("max")=mu_(s)mg((m+M)/(2m+M))`
C. `F_("max")=2mu_(s)mg((m+M)/(2m+M))`
D. `F_("max")=musmg((2m+M)/(m+M))`

Answer

C

Explanation

Let all the four blocks move with a common acceleration acceleration `a`,
then `a=F/(2(m+M))`

Frictionn between C and D will be more than friction between A and B.
Let this friction be `f` then
`f=(m+m+M)a=((2m+M)F)/(2(m+M))`
For no sliping `flef_1`
`((2m+M)F)/(2(m+M))lemu_smgimpliesFle(2mu_smg(m+M))/(2m+M)`