The maximum value of the applied force F such | vedclass

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(A) For the block to remain at rest, the horizontal component of the applied force must be equal to the limiting friction force.From the free body dia

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$20$

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(A) For the block to remain at rest, the horizontal component of the applied force must be equal to the limiting friction force.From the free body diagram, the normal reaction $R$ is given by $R = mg + F \sin 60^{\circ}$.Given $m = \sqrt{3} \,kg$, $g = 10 \,ms^{-2}$, and $\mu = \frac{1}{2\sqrt{3}}$.$R = \sqrt{3} 	imes 10 + F \sin 60^{\circ} = 10\sqrt{3} + F \frac{\sqrt{3}}{2}$.The limiting friction force is $f = \mu R = \frac{1}{2\sqrt{3}} \left( 10\sqrt{3} + F \frac{\sqrt{3}}{2} ight) = \frac{10}{2} + \frac{F}{4} = 5 + \frac{F}{4}$.The horizontal component of the applied force is $F \cos 60^{\circ} = F 	imes \frac{1}{2} = \frac{F}{2}$.Equating the horizontal force to the limiting friction for the maximum value of $F$:$\frac{F}{2} = 5 + \frac{F}{4}$$\frac{F}{2} - \frac{F}{4} = 5$$\frac{F}{4} = 5$$F = 20 \,N$.

The maximum value of the applied force $F$ such that the block as shown in the arrangement does not move is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,N$)

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