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JEE Main 201219 MayOnline
Exam
JEE Main
Year
2012
Date
19 May
Mode
Online
Subject
Physics
Chapter
Laws of Motion

JEE Main 2012 · 19 May · Online · Physics · Laws of Motion

A block of weight W rests on a horizontal floor with coefficient of...

A block of weight WW rests on a horizontal floor with coefficient of static friction μ\mu. It is desired to make the block move by applying minimum amount of force. The angle θ\theta from the horizontal at which the force should be applied and magnitude of the force FF are respectively.

Options

  1. Aθ=tan⁡−1(μ),F=μW1+μ2\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}
  2. Bθ=tan⁡−1(1μ),F=μW1+μ2\theta=\tan ^{-1}\left(\frac{1}{\mu}\right), F=\frac{\mu W}{\sqrt{1+\mu^2}}
  3. Cθ=0,F=μW\theta=0, F=\mu W
  4. Dθ=tan⁡−1(μ1+μ),F=μW1+μ\theta=\tan ^{-1}\left(\frac{\mu}{1+\mu}\right), F=\frac{\mu W}{1+\mu}

Correct answer

A. θ=tan⁡−1(μ),F=μW1+μ2\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}

Step-by-step solution

Let the force FF is applied at an angle θ\theta with the horizontal. For horizontal equilibrium, Fcos⁡θ=μRF \cos \theta=\mu R For vertical equilibrium, R+Fsin⁡θ=mgor,R+F \sin \theta=\mathrm{mg} or,R=m g-F \sin \thetaSubstitutingthisvalueofSubstituting this value ofR in eq.(i),wegeteq. (i), we getFcos⁡θ=μ(mg−Fsin⁡θ)=μmg−μFsin⁡θ or, F(cos⁡θ+μsin⁡θ)=μmg or, F=μmgcos⁡θ+μsin⁡θ\begin{aligned} & F \cos \theta=\mu(\mathrm{mg}-F \sin \theta) \\ & =\mu \mathrm{mg}-\mu F \sin \theta \\ & \text{ or, } F(\cos \theta+\mu \sin \theta)=\mu \mathrm{mg} \\ & \text{ or, } \mathrm{F}=\frac{\mu \mathrm{mg}}{\cos \theta+\mu \sin \theta} \end{aligned}ForForFtobeminimum,thedenominatorto be minimum, the denominator(\cos \theta+\mu \sin \theta)shouldbemaximum.∴ddθ(cos⁡θ+μsin⁡θ)=0or,should be maximum. \therefore \frac{d}{d \theta}(\cos \theta+\mu \sin \theta)=0 or,-\sin \theta+\mu \cos \theta=0or,or,\tan \theta=\muor,or,\theta=\tan ^{-1}(\mu)Then,sin⁡θ=μ1+μ2Then, \sin \theta=\frac{\mu}{\sqrt{1+\mu^2}} and cos⁡θ=11+μ2Hence,\cos \theta=\frac{1}{\sqrt{1+\mu^2}} Hence,F_min =μw11+μ2+μ21+μ2=μw1+μ2{\text{min }} =\frac{\mu w}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu w}{\sqrt{1+\mu^2}}
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