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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

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

a) remains constant b) first remains constant then decreases c) first decreases then increases d) first increases then decreases

1 Answer

Till $\theta < \tan ^{-1} \mu$ contact force is constant and gravitational force are equal and opposite.
Once the body starts sliding
$F_c=\sqrt {(mg \cos \theta)^2+(\mu \;mg\; \cos \theta)^2}$
$\quad= mg \cos \theta \sqrt {1+\mu ^2}$
So as $\theta $ increases $\cos \theta$ decreases.
So contact force decreases.
Hence b is the correct answer.

 

answered Jul 15, 2013 by meena.p
edited Mar 21, 2014 by thagee.vedartham
 

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