o Coefficient of Rolling friction is lesser than that of Kinetic friction .
fr < fk < fs
Problem : Calculate the force required for pushing a 30 kg wooden bar over a wooden floor at a constant speed . Coefficient of friction of wood over wood = 0.25
Solution .
M = 30 kg μ = 0.25 ( Fa – f ) = ma For constant speed , a = 0 So , Fa = f = μN
From free body diagram , N = mg Therefore , Fa = f = μN = μmg = 0.25 * 30 * 9.8 = 73.5 N
Problem 2 : A homogenous chain of length L lies on a table . What is the maximum length l of the part of the chain hanging over the table if the coefficient of friction between the chain and the table is u , the chain remaining at rest with the table ?
Solution .