A block of mass m is at rest on an inclined p | vedclass

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(B) The block is at rest on the inclined plane. This means the net force acting on the block along the inclined plane must be zero.The gravitational f

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$mg \sin 	heta$

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(B) The block is at rest on the inclined plane. This means the net force acting on the block along the inclined plane must be zero.The gravitational force acting on the block is $mg$,which can be resolved into two components:$1$. $A$ component perpendicular to the inclined plane: $mg \cos 	heta$.$2$. $A$ component parallel to the inclined plane: $mg \sin 	heta$.The component $mg \sin 	heta$ acts downwards along the plane,trying to slide the block down. Since the block is at rest,the static frictional force $f$ must balance this component to prevent motion.Therefore,the frictional force $f = mg \sin 	heta$.

$A$ block of mass $m$ is at rest on an inclined plane with an angle of inclination $\theta$ and coefficient of friction $\mu$,as shown in the figure. The frictional force acting on the block is:

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