Q.

A block of mass m = 2 kg is resting on a rough inclined plane of inclination 370. The coefficient of friction between the block and the plane is μ = 0.5. What minimum force F (in newton) should be applied perpendicular to the plane on the block, so that the block does not slip on the plane?

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answer is 8.

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Detailed Solution

Since mg sin 37°> μmg cos 37°, the block has a tendency to slip downwards.Let F be the minimum force applied on it, so that it does not slip. Then,N=F+mgcos⁡37∘    ∴     mgsin⁡37∘=μN=μF+mgcos⁡37∘or  F=mgsin⁡37∘μ−mgcos⁡37∘         =(2)(10)(3/5)0.5−(2)(10)45=8 N
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