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.