A body of mass 1 , kg rests on a horizontal f | vedclass

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(C) For the body to move,the applied force $F$ must overcome the limiting friction $f_L = \mu N$.Resolving the forces:Horizontal component: $F \cos

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$5$

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(C) For the body to move,the applied force $F$ must overcome the limiting friction $f_L = \mu N$.Resolving the forces:Horizontal component: $F \cos 	heta = f_L = \mu N$Vertical component: $F \sin 	heta + N = mg \Rightarrow N = mg - F \sin 	heta$Substituting $N$ into the friction equation:$F \cos 	heta = \mu (mg - F \sin 	heta)$$F (\cos 	heta + \mu \sin 	heta) = \mu mg$$F = \frac{\mu mg}{\cos 	heta + \mu \sin 	heta}$To minimize $F$,we maximize the denominator $D = \cos 	heta + \mu \sin 	heta$. The maximum value of $a \cos 	heta + b \sin 	heta$ is $\sqrt{a^2 + b^2}$.Here,$a = 1$ and $b = \mu = \frac{1}{\sqrt{3}}$.$D_{\max} = \sqrt{1^2 + (\frac{1}{\sqrt{3}})^2} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.Thus,$F_{\min} = \frac{\mu mg}{D_{\max}} = \frac{(\frac{1}{\sqrt{3}}) 	imes 1 	imes 10}{\frac{2}{\sqrt{3}}} = \frac{10}{2} = 5 \, N$.

$A$ body of mass $1 \, kg$ rests on a horizontal floor with which it has a coefficient of static friction $\frac{1}{\sqrt{3}}$. It is desired to make the body move by applying the minimum possible force $F \, N$. The value of $F$ will be (Nearest Integer). [Take $g = 10 \, m s^{-2}$]

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