First slide

Static friction

Question

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)

Moderate

A

100N
B

250N
C

80N
D

50N

Solution

If N be the normal reaction of each wall, then total frictional force = 2(µN) $\therefore$ 2(µN) = mg
$N = \frac{{mg}}{{2\mu }} = \frac{{40 \times 10}}{{2 \times 0.8}} = \frac{{400 \times 10}}{{2 \times 8}}$
N = 250 newtons.

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