As shown in the figure,a block of mass sqrt{3 | vedclass

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(B) Let the applied force be $F$. The horizontal component of the force is $F \cos 60^{\circ}$ and the vertical component is $F \sin 60^{\circ}$.The n

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$10$

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(B) Let the applied force be $F$. The horizontal component of the force is $F \cos 60^{\circ}$ and the vertical component is $F \sin 60^{\circ}$.The normal reaction $N$ on the block from the surface is given by balancing vertical forces:$N = mg + F \sin 60^{\circ}$Given $m = \sqrt{3} 	ext{ kg}$ and $g = 10 	ext{ m/s}^2$,so $mg = 10\sqrt{3} 	ext{ N}$.$N = 10\sqrt{3} + F \left( \frac{\sqrt{3}}{2} ight)$The limiting friction force is $f_L = \mu N = \frac{1}{3\sqrt{3}} \left( 10\sqrt{3} + \frac{F\sqrt{3}}{2} ight) = \frac{10}{3} + \frac{F}{6}$.For the block to just move,the horizontal component of the applied force must equal the limiting friction:$F \cos 60^{\circ} = f_L$$F \left( \frac{1}{2} ight) = \frac{10}{3} + \frac{F}{6}$Multiply by $6$ to simplify:$3F = 20 + F$$2F = 20 \Rightarrow F = 10 	ext{ N}$.The question asks for the value of $3x$,where $F = 3x$.$3x = 10$.

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