A block of weight W rests on a horizontal flo | vedclass

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(A) Let the force $F$ be applied at an angle $	heta$ with the horizontal.For horizontal equilibrium,$F \cos 	heta = \mu R$ ... $(i)$For vertical equ

Vedclass · Accepted Answer

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

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(A) Let the force $F$ be applied at an angle $	heta$ with the horizontal.For horizontal equilibrium,$F \cos 	heta = \mu R$ ... $(i)$For vertical equilibrium,$R + F \sin 	heta = W$,so $R = W - F \sin 	heta$ ... $(ii)$Substituting $R$ from $(ii)$ into $(i)$:$F \cos 	heta = \mu (W - F \sin 	heta)$$F \cos 	heta = \mu W - \mu F \sin 	heta$$F (\cos 	heta + \mu \sin 	heta) = \mu W$$F = \frac{\mu W}{\cos 	heta + \mu \sin 	heta}$ ... $(iii)$For $F$ to be minimum,the denominator $(\cos 	heta + \mu \sin 	heta)$ must be maximum.Taking the derivative with respect to $	heta$ and setting it to $0$:$\frac{d}{d	heta} (\cos 	heta + \mu \sin 	heta) = 0$$-\sin 	heta + \mu \cos 	heta = 0$$	an 	heta = \mu \implies 	heta = 	an^{-1}(\mu)$Using $	an 	heta = \mu$,we have $\sin 	heta = \frac{\mu}{\sqrt{1 + \mu^2}}$ and $\cos 	heta = \frac{1}{\sqrt{1 + \mu^2}}$.Substituting these into the expression for $F$:$F_{\min} = \frac{\mu W}{\frac{1}{\sqrt{1 + \mu^2}} + \mu \left(\frac{\mu}{\sqrt{1 + \mu^2}}ight)} = \frac{\mu W}{\frac{1 + \mu^2}{\sqrt{1 + \mu^2}}} = \frac{\mu W}{\sqrt{1 + \mu^2}}$

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

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