A block rests on a rough inclined plane makin | vedclass

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(A) The angle of repose is given by $\alpha = 	an^{-1}(\mu) = 	an^{-1}(0.8) \approx 38.6^{\circ}$.Since the angle of the inclined plane $	heta = 30

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$2$

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(A) The angle of repose is given by $\alpha = 	an^{-1}(\mu) = 	an^{-1}(0.8) \approx 38.6^{\circ}$.Since the angle of the inclined plane $	heta = 30^{\circ}$ is less than the angle of repose $\alpha$,the block remains at rest.For a block at rest on an inclined plane,the static frictional force $f_s$ balances the component of the gravitational force acting down the plane.Thus,$f_s = mg \sin(	heta)$.Given $f_s = 10 \, N$,$g = 10 \, m/s^2$,and $	heta = 30^{\circ}$,we have:$10 = m 	imes 10 	imes \sin(30^{\circ})$$10 = m 	imes 10 	imes 0.5$$10 = 5m$$m = \frac{10}{5} = 2 \, kg$.

$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \, N$,the mass of the block (in $kg$) is (take $g = 10 \, m/s^2$)

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