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Q.

A man of mass 40 kg is at rest between the walls as shown in the figure. If $μ$ between the man and the walls is 0.8, find the normal reaction exerted by the walls on the man. (g=10ms^–2)

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a

50N

b

80N

c

100N

d

250N

answer is B.

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Detailed Solution

The frictional force from the two sides will be balancing the weight of man.

$2 μN = mg \Rightarrow N = \frac{mg}{2 μ} = \frac{400}{2 (0.8)} = 250 N$

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