Question

Four blocks are arranged on a smooth horizontal surface as shown in the figure. The maximum value of horizontal force $F$, applied to one of the bottom blocks that makes all four blocks move with the same acceleration, is -

• A. $\mu mg[\frac{m+M}{2m+M}]$
• B. $2\mu mg[\frac{m+M}{2m+M}]$
• C. $3\mu mg[\frac{m+M}{2m+M}]$
• D. $0.5\mu mg[\frac{m+M}{2m+M}]$

Answer 1

The correct option is B $2\mu mg[\frac{m+M}{2m+M}]$

FBD of block $B$ is given by -


$F-f_1=Ma$

$F-\mu mg=Ma....\left(1\right)$

FBD of block $A$ is given by -


$f_1-T=ma$

$\Rightarrow \mu mg-T=ma....\left(2\right)$

FBD of block $C$ is given by -


$T-f_2=ma....\left(3\right)$

FBD of block $D$ is given by -



$f_2=Ma....\left(4\right)$

From Eq. $\left(3\right)$ and $\left(4\right)$

$T=(m+M)a....\left(5\right)$

Putting Eq. $\left(5\right)$ in Eq. $\left(2\right)$ we get,

$\mu mg-(m+M)a=ma$

$\Rightarrow a=\frac{\mu mg}{2m+M}....\left(6\right)$

Putting Eq. $\left(6\right)$ in Eq. $\left(1\right)$,

$F-\mu mg=\frac{\mu mMg}{2m+M}$

$F=\mu mg[1+\frac{M}{2m+M}]$

$F=\mu mg[\frac{2m+M+M}{2m+M}]$

$\therefore F=2\mu mg[\frac{m+M}{2m+M}]$

Hence, option $\left(B\right)$ is correct.