A force F is applied on block A as shown in figure. The contact force between the blocks A and B and between the blocks B and C respectively are (Assume frictionless surface)

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F, $\frac{F}{2}$

$\frac{4 F}{7}, \frac{6 F}{7}$

$\frac{F}{7}, \frac{2 F}{7}$

$\frac{6 F}{7}, \frac{4 F}{7}$

answer is B.

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Detailed Solution

$a = \frac{F}{m + 2 m + 4 m} = \frac{F}{7 m}$

Let normal force between A and B is $N_{1}, then$

$N_{1} = (2 m + 4 m) a = \frac{6 F}{7}$

and $between B and C is N_{2}, then$

$N_{2} = 4 ma = \frac{4 F}{7}$

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