11) Fup = mgsinα +μgcosα
Fdo = mgsinα-μgcosα
therfore sin α = 3μ cos α thus tan α = 3μ
2) F-f= ma
therefore F-2μm1g = m1a
Similarly F-μm2g = m2a
therefore simply subtracting them we get a>μg
Q1.The force required to just move a body up in the inclined plane is double the force required to just prevent the body from sliding down the plane.The coefficient of friction is u.
If @ is the angle of inclination then tan@ is equal to
(a)u
(b)3u
(c)2u
(d)0.5u
Q2.Two blocks of masses m1 & m2 are placed in contact with each other on a horizontal platform.The coefficient of fricn betwn m1 and platform is 2u and the block m2 & the platform is u.The platform moves with an accn a.The normal rxn betwn the blocks is
(a)zero in all cases
(b)zero only if m1=m2
(c)non zero only if a>2ug
(d)non zero only if a>ug
ANS-Q1-(b)
Q2-(d)
Please give the answer with proper explanation.
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3 Answers
1
621st is correct.
2nd there will be cases..
simply substracting is not correct
I told you in that class na
it is a sin to write f=μmg
You have committed the sin :P
