Question

Four blocks are arranged on a smooth horizontal surface as shown in Fig. The masses of the blocks are given (see in figure). 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 blocks move with the same acceleration?

• A. $2{\mu }_{s}mg(\frac{m+M}{2m})$
• B. $2{\mu }_{s}mg(\frac{m+M}{2m+M})$
• C. $2{\mu }_{s}mg(\frac{m+M}{m+M})$
• D. $2{\mu }_{s}mg(\frac{m+M}{M})$

Answer 1

The correct option is B $2{\mu }_{s}mg(\frac{m+M}{2m+M})$

From F.B D of block B.

$F-{\mu }_{s}mq=Ma-\left(i\right)$

From $F\cdot B\cdot D$ of block $A$,

${\mu }_{s}mg-T=ma$ - (ii)

From F.B.D of block C,

$T-f=ma$ - (iii)

From F.B. D of block D, $f=ma-\left(iv\right)$

From (iii) \& (iv) $T=(m+M)a$ - (v)

$\Rightarrow a=\left(\frac{T}{m+M}\right)$

putting $T$ in eq (i)

$F-\mu mg=\frac{{\mu }_{s}Mmg}{M+2m}$

$F=2{\mu }_s\left[\frac{m+M}{M+2m}\right]$