A block of mass m rests on a horizontal table | vedclass

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(A) Let a force $F$ be applied at an angle $	heta$ with the horizontal.Resolving the forces,the vertical equilibrium gives: $N + F \sin 	heta = mg$,

Vedclass · Accepted Answer

$\frac{\mu}{\sqrt{1+\mu^{2}}} mg$

Vedclass · Answer

(A) Let a force $F$ be applied at an angle $	heta$ with the horizontal.Resolving the forces,the vertical equilibrium gives: $N + F \sin 	heta = mg$,so $N = mg - F \sin 	heta$.The horizontal force required to overcome friction is: $F \cos 	heta = f_s = \mu N$.Substituting $N$: $F \cos 	heta = \mu (mg - F \sin 	heta)$.Rearranging for $F$: $F (\cos 	heta + \mu \sin 	heta) = \mu mg$,which gives $F = \frac{\mu mg}{\cos 	heta + \mu \sin 	heta}$.To find the minimum force $F$,we maximize the denominator $D = \cos 	heta + \mu \sin 	heta$.Setting the derivative with respect to $	heta$ to zero: $\frac{dD}{d	heta} = -\sin 	heta + \mu \cos 	heta = 0$.This implies $	an 	heta = \mu$.Using the identity $\cos 	heta + \mu \sin 	heta = \sqrt{1+\mu^2} \sin(	heta + \alpha)$ where $	an \alpha = 1/\mu$,or simply substituting $\sin 	heta = \frac{\mu}{\sqrt{1+\mu^2}}$ and $\cos 	heta = \frac{1}{\sqrt{1+\mu^2}}$ into the expression for $F$:$F_{\min} = \frac{\mu mg}{\frac{1}{\sqrt{1+\mu^2}} + \mu \frac{\mu}{\sqrt{1+\mu^2}}} = \frac{\mu mg}{\frac{1+\mu^2}{\sqrt{1+\mu^2}}} = \frac{\mu mg}{\sqrt{1+\mu^2}}$.

$A$ block of mass $m$ rests on a horizontal table with a coefficient of static friction $\mu$. What minimum force must be applied on the block to drag it on the table?

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