Laws Of Motion Question 391
Question: A weight w is to be moved from the bottom to the top of an inclined plane of inclination 9 to the horizontal. If a smaller force is to be applied to drag it along the plane in comparison to lift it vertically up, the coefficient of friction should be such that:
Options:
A) $ \mu >\tan ( \frac{\pi }{4}-\frac{\theta }{2} ) $
B) $ \mu <\tan ( \frac{\pi }{4}-\frac{\theta }{2} ) $
C) $ \mu <\tan \theta $
D) $ \mu >\tan \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ F _{1}=w\sin \theta +\mu w\cos \theta $
$ F _{2}=w $
As,$ F _{1}<F _{2} $
$ \therefore \mu <\tan ( \frac{\pi }{4}=\frac{\theta }{2} ) $
