A block of mass $\sqrt{3}$ kg resting on a horizontal surface. A force $F$ is applied on the block as shown in figure. If coefficient of friction between the block be $\ \frac{1}{2\sqrt{3}}$, what can be the maximum value of force $F$ so that block does not start moving? (Take $g = 10\ ms^{- 2}$) [image:5_image52.emf]

Question

Dr. Abubacker · Accepted Answer

(a) From acting on block are shown in adjoining figure
As the block does not move, hence
$F\cos{60{^\circ} = f = \mu N = \mu\left( Mg + F\sin 60{^\circ} ight)}$
[image:5_image197.emf]
$	herefore F\frac{1}{2} = \frac{1}{2\sqrt{3}}\left( \sqrt{3} 	imes 10 + F \bullet \frac{\sqrt{3}}{2} ight)$
On simplification, we get $F = 20$ N

Q. A block of mass $\sqrt{3}$ kg resting on a horizontal surface. A force $F$ is applied on the block as shown in figure. If coefficient of friction between the block be $\ \frac{1}{2\sqrt{3}}$, what can be the maximum value of force $F$ so that block does not start moving? (Take $g = 10\ ms^{- 2}$)

Explanation

Reviewed by Dr. Abubacker

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