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 $
F
Show Answer
Answer:
Correct Answer: B
Solution:
[b]
$ F=(m _{1}+m _{2})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}} $
BETA
