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?

Question

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?

Infinity Learn · Accepted Answer

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⁡$$37$$∘$$=$$μ$$N=$$μ$$F+mgcos$$⁡$$37$$∘or  $$F=mgsin$$⁡$$37$$∘μ−mgcos⁡$$37$$∘         $$=(2)(10)(3/5)0.5$$−$$(2)(10)45=8$$ N

Friction

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A block of mass m = 2 kg is resting on a rough inclined plane of inclination $37^{0}$ . 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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Friction

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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?

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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