IIT JEE-AIEEE 2012, Entrance Notifications, Study Tips, Engineering Preparation, Practice Test Papers, Experts Advice

Ask iit jee aieee pet cbse icse state board community

Discussion Response Post to: plz solve

Forum Index -> Mechanics -> View Full Question

Author

Message

BALGANESH (1654)

Blazing goIITian

total posts: 1025
Offline

the centre of mass remains the same if we place particles of mass 3/2mg at A and B
now let the force be applied and the particle looses cantact with ground
the fbd will be as shown in the figure
Fcos@ = mgsin@
F = mgsin@/cos@ = 3/2mgtan@
for F to be minimum
dF/d@ = 3/2mg (sec^2@)
@ is almost 0
therefore sec@ approaches 1

therefore minimum force rqd = 3/2mg

NOT SURE
PLZ CORRECT ME IF I'M WRONG

Like 0 people liked this

Free Sign Up!