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(B) The contact force $F_c$ between the block and the plane is the resultant of the normal force $N$ and the frictional force $f$. $F_c = \sqrt{N^2 +

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first remains constant then decreases

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(B) The contact force $F_c$ between the block and the plane is the resultant of the normal force $N$ and the frictional force $f$. $F_c = \sqrt{N^2 + f^2}$. For a block on an inclined plane,the normal force is $N = mg \cos 	heta$. Case $1$: When the block is at rest (static friction),$f = mg \sin 	heta$. Then $F_c = \sqrt{(mg \cos 	heta)^2 + (mg \sin 	heta)^2} = \sqrt{m^2g^2(\cos^2 	heta + \sin^2 	heta)} = mg$. This holds until the angle of repose $	heta_r$ is reached,where $	an 	heta_r = \mu$. Case $2$: When the block starts sliding (kinetic friction),$f = \mu N = \mu mg \cos 	heta$. Then $F_c = \sqrt{N^2 + f^2} = \sqrt{(mg \cos 	heta)^2 + (\mu mg \cos 	heta)^2} = mg \cos 	heta \sqrt{1 + \mu^2}$. As $	heta$ increases from $0^o$ to $	heta_r$,$F_c$ remains constant at $mg$. As $	heta$ increases from $	heta_r$ to $90^o$,$F_c = mg \cos 	heta \sqrt{1 + \mu^2}$ decreases because $\cos 	heta$ decreases. Therefore,the contact force first remains constant and then decreases.

$A$ body is placed on a rough inclined plane of inclination $\theta$. As the angle $\theta$ is increased from $0^o$ to $90^o$,the contact force between the block and the plane

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