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JEE Main 2022 · Physics · Kinematics and Laws of Motion · medium

A block of mass 5 kg rests on a rough plane inclined at 53^° to the horizontal; the coefficient of static friction is 0.5. The minimum horizontal…

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A block of mass 5 kg5\ \mathrm{kg} rests on a rough plane inclined at 53∘53^\circ to the horizontal; the coefficient of static friction is 0.50.5. The minimum horizontal force that must be applied to the block (pushing it towards the incline) to keep it from sliding down is (g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, sin⁡53∘=0.8\sin53^\circ=0.8)
  1. A. 15 N15\ \mathrm{N}
  2. B. 25 N25\ \mathrm{N}
  3. C. 30 N30\ \mathrm{N}
  4. D. 40 N40\ \mathrm{N}
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Answer: (B) 25 N25\ \mathrm{N}

Resolve along and perpendicular to the incline. The horizontal force FF has components Fcos⁡θF\cos\theta up the incline and Fsin⁡θF\sin\theta into the plane. Normal reaction N=mgcos⁡θ+Fsin⁡θN=mg\cos\theta+F\sin\theta. At the verge of sliding down: Fcos⁡θ+μN=mgsin⁡θF\cos\theta+\mu N=mg\sin\theta, so F(cos⁡θ+μsin⁡θ)=mg(sin⁡θ−μcos⁡θ)F(\cos\theta+\mu\sin\theta)=mg(\sin\theta-\mu\cos\theta). With cos⁡53∘=0.6\cos53^\circ=0.6: F(0.6+0.4)=50(0.8−0.3)F(0.6+0.4)=50(0.8-0.3), giving F=25 NF=25\ \mathrm{N}. Forgetting the extra normal force from Fsin⁡θF\sin\theta gives 41.7 N41.7\ \mathrm{N}.
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