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 $

Show Answer

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 $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक