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(D) The forces acting on the block are: $1$. Weight $mg$ acting downwards.$2$. Applied force $F$ at an angle of $60^{\circ}$ with the horizontal,actin

Vedclass · Accepted Answer

$20$

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(D) The forces acting on the block are: $1$. Weight $mg$ acting downwards.$2$. Applied force $F$ at an angle of $60^{\circ}$ with the horizontal,acting downwards.$3$. Normal reaction $N$ acting upwards.$4$. Frictional force $f$ acting horizontally.Resolving the force $F$ into components:Horizontal component: $F \cos 60^{\circ} = \frac{F}{2}$Vertical component: $F \sin 60^{\circ} = \frac{F \sqrt{3}}{2}$For vertical equilibrium:$N = mg + F \sin 60^{\circ} = \sqrt{3} 	imes 10 + \frac{F \sqrt{3}}{2} = 10\sqrt{3} + \frac{F \sqrt{3}}{2}$The limiting friction is $f_{max} = \mu N = \frac{1}{2\sqrt{3}} 	imes (10\sqrt{3} + \frac{F \sqrt{3}}{2}) = 5 + \frac{F}{4}$For the block not to move,the horizontal component of the applied force must be less than or equal to the limiting friction:$F \cos 60^{\circ} \le f_{max}$$\frac{F}{2} \le 5 + \frac{F}{4}$$\frac{F}{2} - \frac{F}{4} \le 5$$\frac{F}{4} \le 5$$F \le 20 \ N$Thus,the maximum value of $F$ for the block not to move is $20 \ N$.

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