A block placed on a rough inclined plane of i | vedclass

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(C) In the first case,the block is on the verge of moving upwards,so the friction force acts downwards. The force balance equation is: $F = mg \sin 30

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$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

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(C) In the first case,the block is on the verge of moving upwards,so the friction force acts downwards. The force balance equation is: $F = mg \sin 30^{\circ} + \mu mg \cos 30^{\circ} \dots (1)$In the second case,the block is on the verge of sliding downwards,so the friction force acts upwards. The force balance equation is: $F + \mu mg \cos 60^{\circ} = mg \sin 60^{\circ} \implies F = mg \sin 60^{\circ} - \mu mg \cos 60^{\circ} \dots (2)$Since the magnitude of $F$ is the same in both cases,equate $(1)$ and $(2)$:$mg \sin 30^{\circ} + \mu mg \cos 30^{\circ} = mg \sin 60^{\circ} - \mu mg \cos 60^{\circ}$$\mu (\cos 30^{\circ} + \cos 60^{\circ}) = \sin 60^{\circ} - \sin 30^{\circ}$$\mu (\frac{\sqrt{3}}{2} + \frac{1}{2}) = \frac{\sqrt{3}}{2} - \frac{1}{2}$$\mu = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

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