Laws Of Motion Question 324
Question: The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is$ \mu $ . The inclination $ \theta $ of the plane is
Options:
A) $ {{\tan }^{-1}}\mu $
B) $ {{\tan }^{-1}}( \mu /2 ) $
C) $ {{\tan }^{-1}}2\mu $
D) $ {{\tan }^{-1}}3\mu $
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Answer:
Correct Answer: D
Solution:
[d] In case [a] In case [b] $ mg\sin \theta ={{F} _{1}}-\mu N$
$\text{N=mg sin}\theta \text{ }\therefore \text{mg sin}\theta \text{+}\mu \text{mg cos}\theta \text{=}{{\text{F}} _{1}}$
$ \text{In second case }\left( b \right)$
$\mu N+{{F} _{2}}=\text{mg sin}\theta$
$\Rightarrow \mu \text{mg cos}\theta -{{F} _{2}}=mg\sin \theta $
$\text{or }{{F} _{2}}=mg\sin \theta -\mu mg\cos \theta \text{ but }{{F} _{1}}=2{{F} _{2}}$
$\text{therefore mg sin}\theta +\mu mg\cos \theta $
$=2\left( mg\sin \theta -\mu mg\cos \theta \right)$
$mg\sin \theta =3\mu mg\cos \theta $
$ \text{or }\tan \theta =3\mu \text{ or }\theta \text{=ta}{{\text{n}}^{-1}}\left( 3\mu \right)$
[d]
BETA
