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(C) The angle of friction is given by $	heta = 	an^{-1}(\mu)$.$	heta = 	an^{-1}\left(\frac{1}{\sqrt{3}}ight) = 30^{\circ}$.Suppose the body is d

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(C) The angle of friction is given by $	heta = 	an^{-1}(\mu)$.$	heta = 	an^{-1}\left(\frac{1}{\sqrt{3}}ight) = 30^{\circ}$.Suppose the body is dragged by a force $F$ acting at an angle $\alpha$ with the horizontal. The normal reaction $N$ is given by $N = mg - F \sin \alpha$.The horizontal force required to overcome friction is $F \cos \alpha = \mu N = \mu(mg - F \sin \alpha)$.Rearranging for $F$,we get $F = \frac{\mu mg}{\cos \alpha + \mu \sin \alpha}$.For the force $F$ to be minimum,the denominator $(\cos \alpha + \mu \sin \alpha)$ must be maximum.Differentiating with respect to $\alpha$ and setting to zero: $-\sin \alpha + \mu \cos \alpha = 0$,which gives $	an \alpha = \mu = \frac{1}{\sqrt{3}}$.Thus,$\alpha = 30^{\circ}$.Substituting $\alpha = 30^{\circ}$ and $\mu = \frac{1}{\sqrt{3}}$ into the expression for $F$:$F_{\min} = \frac{\left(\frac{1}{\sqrt{3}}ight) \cdot 25 \cdot g}{\cos 30^{\circ} + \frac{1}{\sqrt{3}} \sin 30^{\circ}} = \frac{\frac{25g}{\sqrt{3}}}{\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{3}} \cdot \frac{1}{2}} = \frac{\frac{25g}{\sqrt{3}}}{\frac{3+1}{2\sqrt{3}}} = \frac{25g}{\sqrt{3}} \cdot \frac{2\sqrt{3}}{4} = \frac{50g}{4} = 12.5g$.Therefore,the least force required is $12.5 \ kgf$.

$A$ heavy body of mass $25 \ kg$ is to be dragged along a horizontal plane $\left( \mu = \frac{1}{\sqrt{3}} \right)$. The least force required is ........ $kgf$.

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