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Question

(A) Let the mass of the block be $m$.The block is at rest on the inclined plane.The component of the gravitational force acting down the plane is $mg

Vedclass · Accepted Answer

$2$

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(A) Let the mass of the block be $m$.The block is at rest on the inclined plane.The component of the gravitational force acting down the plane is $mg \sin 30^{\circ}$.Since the block is in equilibrium,the static frictional force $F$ must balance this component:$F = mg \sin 30^{\circ}$Given $F = 10 \, N$,$g = 10 \, m/s^2$,and $\sin 30^{\circ} = 0.5$:$10 = m 	imes 10 	imes 0.5$$10 = 5m$$m = \frac{10}{5} = 2 \, kg$.Note: We should check if this state is possible. The maximum static friction is $f_{max} = \mu_s N = \mu_s mg \cos 30^{\circ} = 0.8 	imes 2 	imes 10 	imes \frac{\sqrt{3}}{2} \approx 13.86 \, N$. Since $F = 10 \, N < 13.86 \, N$,the block remains at rest.

$A$ block rests on a rough inclined plane making an angle of $30^o$ 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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