Laws Of Motion Question 150
Question: A particle of weight $ W $ rests on a rough inclined plane which makes an angle $ \alpha $ with the horizontal. If the coefficient of static friction $ \mu =2 $ tan a, find the horizontal force $ H $ acting transverse to the slope of the plane when the particle is about to slip.
Options:
A) $ 2W\sin \alpha $
B) $ W\sin \alpha $
C) $ \frac{\sqrt{3}}{2}W\sin \alpha $
D) $ W\sqrt{3}\sin \alpha $
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Answer:
Correct Answer: D
Solution:
[d] Net force: $ F=\sqrt{{{(Wsin\alpha )}^{2}}+H^{2}} $ When the particle is about to slip: $ F=\mu N $
$ \Rightarrow \sqrt{{{(Wsin\alpha )}^{2}}+H^{2}}=(2tan\alpha )Wcos\alpha $
$ \Rightarrow H=\sqrt{3}W\sin \alpha $
BETA
