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

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Question

(A) Case $1$: When the block is pushed upwards on an inclined plane with $	heta=30^{\circ}$,the force $F$ must overcome both the component of gravity

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Case $1$: When the block is pushed upwards on an inclined plane with $	heta=30^{\circ}$,the force $F$ must overcome both the component of gravity and the maximum static friction force acting downwards.$F = mg \sin 30^{\circ} + \mu mg \cos 30^{\circ} = \frac{mg}{2} + \mu mg \frac{\sqrt{3}}{2} = \frac{mg}{2}(1 + \mu \sqrt{3})$  --- $(1)$Case $2$: When the block is prevented from sliding down on an inclined plane with $	heta=60^{\circ}$,the force $F$ acts upwards along with the maximum static friction force to balance the component of gravity acting downwards.$F + \mu mg \cos 60^{\circ} = mg \sin 60^{\circ}$$F = mg \sin 60^{\circ} - \mu mg \cos 60^{\circ} = mg \frac{\sqrt{3}}{2} - \mu mg \frac{1}{2} = \frac{mg}{2}(\sqrt{3} - \mu)$  --- $(2)$Equating $(1)$ and $(2)$ since the magnitude of $F$ is the same:$\frac{mg}{2}(1 + \mu \sqrt{3}) = \frac{mg}{2}(\sqrt{3} - \mu)$$1 + \mu \sqrt{3} = \sqrt{3} - \mu$$\mu(1 + \sqrt{3}) = \sqrt{3} - 1$$\mu = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

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