What is the maximum value of the force F such | vedclass

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(A) The forces acting on the block are the applied force $F$ at an angle of $60^\circ$ with the horizontal,the weight $W = mg$,the normal reaction $R$

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The forces acting on the block are the applied force $F$ at an angle of $60^\circ$ with the horizontal,the weight $W = mg$,the normal reaction $R$,and the frictional force $f$.Resolving the force $F$ into components:Horizontal component: $F_x = F \cos 60^\circ$Vertical component: $F_y = F \sin 60^\circ$For vertical equilibrium:$R = W + F \sin 60^\circ$Given $m = \sqrt{3} \ kg$ and $g = 10 \ m/s^2$,$W = mg = 10\sqrt{3} \ N$.So,$R = 10\sqrt{3} + F \sin 60^\circ = 10\sqrt{3} + F \frac{\sqrt{3}}{2}$.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 \leq \mu R$$F \cos 60^\circ \leq \mu (W + F \sin 60^\circ)$Substituting the values $\mu = \frac{1}{2\sqrt{3}}$,$\cos 60^\circ = \frac{1}{2}$,and $\sin 60^\circ = \frac{\sqrt{3}}{2}$:$F \cdot \frac{1}{2} = \frac{1}{2\sqrt{3}} (10\sqrt{3} + F \cdot \frac{\sqrt{3}}{2})$$F \cdot \frac{1}{2} = \frac{10\sqrt{3}}{2\sqrt{3}} + \frac{F \sqrt{3}}{4\sqrt{3}}$$F \cdot \frac{1}{2} = 5 + \frac{F}{4}$$F \cdot \frac{1}{2} - \frac{F}{4} = 5$$\frac{F}{4} = 5$$F = 20 \ N$.

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move (in $N$)? (Given $m = \sqrt{3} \ kg$,$\mu = \frac{1}{2\sqrt{3}}$,$g = 10 \ m/s^2$)

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