Four blocks are arranged on a smooth horizontal surface as shown. The masses of the blocks are given (see the diagram). The coefficient of static friction between the top and the bottom block is $μ_{s}$ . 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?

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$F_{\max} = μ_{s} m g (\frac{2 m + M}{m + M})$

$F_{\max} = μ_{s} m g (\frac{m + M}{2 m + M})$

$F_{\max} = 2 μ_{s} m g (\frac{2 m + M}{m + M})$

$F_{\max} = 2 μ_{s} m g (\frac{m + M}{2 m + M})$

answer is C.

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Detailed Solution

Let all the four blocks move with common acceleration a, then $a = \frac{F}{2 (m + M)}$
Friction between C and D will be more than friction between A and B. Let this friction is f, then

$\begin{array}{l} f = (m + m + M) a = \frac{(2 m + M) F}{2 (m + M)} \\ F o r n o s l i p p i n g f \leq f_{1} \\ \frac{(2 m + M) F}{2 (m + M)} \leq μ_{s} m g \Rightarrow F \leq \frac{2 μ_{s} m g (m + M)}{2 m + M} \end{array}$