A wooden box lying at rest on an inclined sur | vedclass

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(D) To keep the box in static equilibrium, the downward force along the incline $(mg \sin 	heta)$ must be balanced by the maximum static friction for

Vedclass · Accepted Answer

$\geq 16.3 \ N$

Vedclass · Answer

(D) To keep the box in static equilibrium, the downward force along the incline $(mg \sin 	heta)$ must be balanced by the maximum static friction force $(f_{max})$.The normal force $N$ acting on the box is the sum of the component of weight perpendicular to the incline and the applied force $F$:$N = mg \cos 	heta + F$The maximum static friction force is given by:$f_{max} = \mu N = \mu(mg \cos 	heta + F)$For equilibrium, $mg \sin 	heta \leq f_{max}$, so the minimum force $F$ required is when $mg \sin 	heta = \mu(mg \cos 	heta + F)$.Rearranging for $F$:$F = \frac{mg \sin 	heta}{\mu} - mg \cos 	heta = mg \left( \frac{\sin 	heta}{\mu} - \cos 	heta ight)$Given $m = 1 \ kg$, $g = 10 \ m/s^2$, $	heta = 30^{\circ}$, and $\mu = 0.2$:$F = 1 	imes 10 \left( \frac{\sin 30^{\circ}}{0.2} - \cos 30^{\circ} ight)$$F = 10 \left( \frac{0.5}{0.2} - \frac{\sqrt{3}}{2} ight) = 10 \left( 2.5 - 0.866 ight)$$F = 10 	imes 1.634 = 16.34 \ N$Thus, the minimum magnitude of $F$ is $\geq 16.3 \ N$.

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