Laws Of Motion Question 145
Question: The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is 0.25, the angle of inclination of the plane is
Options:
A) $ 37{}^\circ $
B) $ 45{}^\circ $
C) $ 30{}^\circ $
D) $ ~53{}^\circ $
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Answer:
Correct Answer: A
Solution:
[a] Retardation in upward motion $ =g(sin\theta +\mu cos\theta ) $
$ \therefore $ force required just to move up $ F _{up}=mg(\sin \theta +\mu \cos \theta ) $ Similarly for downward motion$ a=g(\sin \theta +\mu \cos \theta ) $
$ \therefore $ Force required just to prevent the body sliding down $ F _{dn}=mg(\sin \theta -\mu \cos \theta ) $ According to problem $ F _{up}=2F _{dn} $
$ \Rightarrow mg(sin\theta +\mu cos\theta )=2mg(sin\theta -\mu cos\theta ) $
$ \Rightarrow \sin \theta +\mu cos\theta =2sin\theta -2\mu cos\theta $
$ \Rightarrow 3\mu \cos \theta =\sin \theta \Rightarrow \tan \theta =3\mu $
$ \Rightarrow \theta ={{\tan }^{-1}}(3\mu )=ta{{n}^{-1}}(3\times 0.25)=ta{{n}^{-1}}(0.75)=37{}^\circ $
BETA
