A wedge of mass 2m placed on a rough surface, its part AB is circular of radius R. A small block of mass m is released from rest at A. Find minimum value of friction between wedge and ground so that wedge remains at rest.
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Mass = 2m
net horizontal force
f= mg.cos45°sin45°
=mg/2
since
cos45°=sin45°=1/√2
Normal reaction from the ground N
N = 2mg+mg(cos45°)
=2mg+mg/2
= 5mg/2
f=μN
mg=μ5mg/2=1/5
= 0.20
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