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JEE Main 2021 (17 Mar, Shift 2) · Friction Basics

Friction Basics question from JEE Main 2021 (17 Mar, Shift 2)

Q17JEE Main 2021 · 17 MAR · SHIFT 2Medium

Friction Basics

A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction 13\frac{1}{\sqrt{3}}. It is desired to make the body move by applying the minimum possible force F N. The value of F will be ____________. (Round off to the Nearest Integer) [Take g = 10 ms−{}^{-}2 ]

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. Pull at angle θ\theta above horizontal: Fcos⁡θ=μ(mg−Fsin⁡θ)⇒F=μmgcos⁡θ+μsin⁡θF\cos\theta = \mu(mg - F\sin\theta) \Rightarrow F = \dfrac{\mu mg}{\cos\theta + \mu\sin\theta}.
  2. Minimum when tan⁡θ=μ\tan\theta = \mu: Fmin=μmg1+μ2F_{min} = \dfrac{\mu mg}{\sqrt{1+\mu^2}}.
  3. Fmin=(1/3)(10)4/3=103⋅32F_{min} = \dfrac{(1/\sqrt{3})(10)}{\sqrt{4/3}} = \dfrac{10}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}
  4. So F=5 NF = 5\text{ N}.
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