Laws Of Motion Question 417

Question: Three blocks are placed at rest on a smooth inclined plane with force acting on $ m _{1} $ parallel to the inclined plane. Find the contact force between $ m _{2} $ and $ m _{3} $ :

Options:

A) $ \frac{(m _{1}+m _{2}+m _{3})F}{m _{3}} $

B) $ \frac{m _{3}F}{m _{1}+m _{2}+m _{3}} $

C) $ F-(m _{1}+m _{2})g $

D) F

Show Answer

Answer:

Correct Answer: B

Solution:

[b]

$ F=(m _{1}+m _{2}+m _{3})g\sin \theta $ …(i)

$ N=m _{3}g\sin \theta $ ..(ii)

Dividing Eq. (ii) by Eq. (i), we get

$ N=\frac{m _{3}F}{m _{1}+m _{2}+m _{3}} $



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