Laws Of Motion Question 141
Question: A house is built on the top of a hill with $ 45{}^\circ $ slope. Due to the sliding of material and sand from top to the bottom of hill, the slope angle has been reduced. If the coefficient of static friction Between sand particles is 0.75, what is the final angle attained by hill? $ (ta{{n}^{-1}}0.75\simeq 37^{0}) $
Options:
A) $ 8{}^\circ $
B) $ ~45{}^\circ $
C) $ 37{}^\circ $
D) $ 30{}^\circ $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] As sand particles are sliding down, the slope of the hill gets reduced.
The sand particle stops coming down when component of gravity force along the hill is balanced by limiting friction force.
$ mg\sin \theta ={{\mu } _{s}}mg\cos \theta $
$ \Rightarrow \theta ={{\tan }^{-1}}({{\mu } _{s}})\simeq 37{}^\circ $ Where $ \theta $ is the new slope angle of hill.
