A block kept on a rough inclined plane,as sho | vedclass

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Question

(A) Let $m$ be the mass of the block and $	heta = 30^\circ$ be the angle of inclination. The forces acting on the block are gravity ($mg \sin 	heta$

Vedclass · Accepted Answer

$\frac{2\sqrt{3}}{9}$

Vedclass · Answer

(A) Let $m$ be the mass of the block and $	heta = 30^\circ$ be the angle of inclination. The forces acting on the block are gravity ($mg \sin 	heta$ down the plane),normal force $(N = mg \cos 	heta)$,and friction $(f_{max} = \mu N = \mu mg \cos 	heta)$.
Case $1$: When the block is about to move down,the external force $F_1 = 2 \ N$ is applied down the plane. The friction acts up the plane.
$mg \sin 	heta = F_1 + f_{max} \implies mg \sin 30^\circ = 2 + \mu mg \cos 30^\circ \implies \frac{mg}{2} = 2 + \mu mg \frac{\sqrt{3}}{2} \quad ... (1)$
Case $2$: When the block is about to move up,the external force $F_2 = 10 \ N$ is applied up the plane. The friction acts down the plane.
$F_2 = mg \sin 	heta + f_{max} \implies 10 = mg \sin 30^\circ + \mu mg \cos 30^\circ \implies 10 = \frac{mg}{2} + \mu mg \frac{\sqrt{3}}{2} \quad ... (2)$
Adding $(1)$ and $(2)$:
$10 + 2 = 2 \left( \frac{mg}{2} ight) \implies 12 = mg$
Substituting $mg = 12$ in $(2)$:
$10 = \frac{12}{2} + \mu (12) \frac{\sqrt{3}}{2} \implies 10 = 6 + 6\sqrt{3} \mu \implies 4 = 6\sqrt{3} \mu$
$\mu = \frac{4}{6\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$

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