Ramesh V
Last Activity: 15 Years ago
(1)
here OM = R , to find= AM
on balancing forces we have
normal rxn. N = mg.cosx
f = mg.sinx
where f = friction coeff. * N
i.e.,
f = mgcox /31/2
so tan x = 1/31/2
so height at which particle remains stationary is Am = OM - OA = ( 1 - 31/2 /2 )R
(2)
Its solved in my previous posts asked by you
http://askiitians.com/forums/Mechanics/10/3377/Laws-of-Motion.htm
(3)
on balancing moments about centre of gravity gives:
mg*a/2 + mga/3 = mg (a/2 + x)
on solving for x gives
x=a/3
Hence,the distance b/w the line of action of mg and normal reaction is a/3.
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