A block of mass m rests on a rough inclined p | vedclass

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(A) The forces acting on the block are the gravitational force $mg$ acting downwards,the normal reaction $N$ perpendicular to the plane,and the static

Vedclass · Accepted Answer

$	heta = 	an^{-1} \mu$

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(A) The forces acting on the block are the gravitational force $mg$ acting downwards,the normal reaction $N$ perpendicular to the plane,and the static frictional force $f$ acting up the plane.When the block is about to slide,the component of weight down the plane must equal the maximum static friction.Equating the forces parallel to the plane: $mg \sin 	heta = f$ $...(i)$Equating the forces perpendicular to the plane: $mg \cos 	heta = N$ $...(ii)$Since the block is about to slide,the frictional force is at its maximum value,$f = \mu N$.Dividing equation $(i)$ by equation $(ii)$:$\frac{mg \sin 	heta}{mg \cos 	heta} = \frac{f}{N} = \frac{\mu N}{N}$$	an 	heta = \mu$Therefore,$	heta = 	an^{-1} (\mu)$.

$A$ block of mass $m$ rests on a rough inclined plane. The coefficient of friction between the surface and the block is $\mu$. At what angle of inclination $\theta$ of the plane to the horizontal will the block just start to slide down the plane?

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