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JEE Main 2019 (9 Jan, Shift 1) · Friction on an Incline

Friction on an Incline question from JEE Main 2019 (9 Jan, Shift 1)

Q22JEE Main 2019 · 9 JAN · SHIFT 1Medium

Friction on an Incline

A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g = 10 ms ${}^{-}$ 2)

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Answer: (a)

Forces down the incline: $3 + mg\sin 45^\circ = 3 + \dfrac{100}{\sqrt{2}} = 73.7\text{ N}$ .
Maximum friction (up the incline): $\mu mg\cos 45^\circ = 0.6 \times 70.7 = 42.4\text{ N}$ .
Minimum $P$ : $P + 42.4 = 73.7 \Rightarrow P \approx 31.3\text{ N}$ .
So the answer is (a) 32 N (nearest option).

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