First slide

Friction on inclined plane of angle more than angle of repose

Question

A box of 100 kg is sliding down along a rough inclined plane of height 2m and length 4m. If coefficient of friction is 0.5 then the force required to be applied so that it comes down with a constant velocity is

Easy

A

250 N
B

500 N
C

933 N
D

67 N

Solution

The force required to make it fall with a constant velocity is the same as the force due to which it is coming down but in opposite direction. i:e. $m g \sin θ - μ m g \cos θ$ = $m g (\sin θ - μ \cos θ) h e r e \sin θ = \frac{2}{4} = 0.5 θ = 30^{0}$
$100 \times 10 (0.5 - 0.5 \frac{\sqrt{3}}{2}) = 500 \times 0.134 = 67$

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