A rod AB of weight W1 is placed over a sphere of weight as shown in figure. Ground is rough and there is no friction between rod and sphere and sphere and wall. Draw free body diagrams of sphere and rod separately.
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Solution
Given,
Lac < Lcb
Also, Weight of rod of length L is W1
Let weight of part ac be Wac and of part cb be Wcb.
therefore,
Wac = W1 ( Lac / L )
and, Wcb = W1 ( Lcb / L)
clearly, Wac < Wcb (since, Lac < Lcb)
Therefore, rod will tend to move anticlockwise,
Thus friction at contact point of rod and ground will act in backward direction.