A block of weight W rests on a… - JEE Main 2012 Physics PYQ

JEE Main 2012 — Physics Mechanics

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

Official previous-year question

Held on 19 May 2012 · Verified 6 Jul 2026.

Options

A
$\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}$
B
$\theta=\tan ^{-1}\left(\frac{1}{\mu}\right), F=\frac{\mu W}{\sqrt{1+\mu^2}}$
C
$\theta=0, F=\mu W$
D
$\theta=\tan ^{-1}\left(\frac{\mu}{1+\mu}\right), F=\frac{\mu W}{1+\mu}$

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Solution

Let the force $F$ is applied at an angle $\theta$ with the horizontal. JEE Main 2012 Physics, Laws of Motion — solution figure

For horizontal equilibrium, $F \cos \theta=\mu R$ For vertical equilibrium, $R+F \sin \theta=\mathrm{mg}$ or, $R=m g-F \sin \theta$ Substituting this value of $R$ in eq. (i), we get $\begin{aligned} & F \cos \theta=\mu(\mathrm{mg}-F \sin \theta) \\ & =\mu \mathrm{mg}-\mu F \sin \theta \\ & \text { or, } F(\cos \theta+\mu \sin \theta)=\mu \mathrm{mg} \\ & \text { or, } \mathrm{F}=\frac{\mu \mathrm{mg}}{\cos \theta+\mu \sin \theta} \end{aligned}$ For $F$ to be minimum, the denominator $(\cos \theta+\mu \sin \theta)$ should be maximum. $\therefore \frac{d}{d \theta}(\cos \theta+\mu \sin \theta)=0$ or, $-\sin \theta+\mu \cos \theta=0$ or, $\tan \theta=\mu$ or, $\theta=\tan ^{-1}(\mu)$ Then, $\sin \theta=\frac{\mu}{\sqrt{1+\mu^2}}$ and $\cos \theta=\frac{1}{\sqrt{1+\mu^2}}$ Hence, $F_{\text {min }}$ $=\frac{\mu w}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu w}{\sqrt{1+\mu^2}}$

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Options

$\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}$

$\theta=\tan ^{-1}\left(\frac{1}{\mu}\right), F=\frac{\mu W}{\sqrt{1+\mu^2}}$

$\theta=0, F=\mu W$

$\theta=\tan ^{-1}\left(\frac{\mu}{1+\mu}\right), F=\frac{\mu W}{1+\mu}$

Solution

Let the force $F$ is applied at an angle $\theta$ with the horizontal. JEE Main 2012 Physics, Laws of Motion — solution figure

Options

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Solution

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Options

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Solution

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JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

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Keep practicing Mechanics