Laws of MotionMedium
Question
It the figure show a ring of mass M and a block of mass m are in equilibrium. The string is light and pulley P does not offer any friction and coefficient of friction between pole and M is Vμ. The frictional force offered by the pole on M is
Options
A.Mg directed up
B.mg μ directed up
C.(M-m) g directed down
D.μ mg direction down
Solution
Here ′M′ is in equilibrium.
So net force on ′M′ must be zero.
∴ f = Mg (upwards)
So net force on ′M′ must be zero.
∴ f = Mg (upwards)
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