Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

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
Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App