ME 201 Engineering Mechanics: Statics Chapter 8 Friction

Friction A retarding force that resists the relative movement of two bodies in contact with each other Always acts in oppose direction of motion – actual motion or impending motion Always acts parallel or tangent to surface(s) in contact Magnitude is mainly dependent on surface roughness (other contributing factors: temperature, molecular adhesion, electrostatic attraction, lubrication, relative velocities)

Coefficient of Static Friction on Dry Surfaces Materials Coefficient of Static Friction or μ Metal on metal 0.15 – 0.60 Metal on wood 0.20 – 0.60 Metal on stone 0.30 – 0.70 Metal on leather 0.30 – 0.60 Wood on wood 0.25 – 0.50 Wood on leather Stone on stone 0.40 – 0.70 Earth on earth 0.20 – 1.00 Rubber on concrete 0.60 – 0.90

Friction Two types of friction Dry or Coulomb friction Static – surfaces are at rest with respect to each other Kinetic/dynamic – surfaces are moving with respect to each other Fluid Friction Viscosity, friction developed between layers of a fluid moving at different velocities We’ll limit our discussion to dry friction

Friction Theory Consider a block on a horizontal surface No Frictional resistance N is referred to as the Normal force, perpendicular to the contact surface W N

Friction Theory Next an external force is applied If no friction, block would immediately move If force is applied, F = P, for equilibrium At some point, P > F, motion occurs W P F N

Friction Theory Ratio of F to N is called the Coefficient of Static Friction or μ Where μs = coefficient of static friction (unitless) F – Max or limiting frictional force resistance (lb, k, N) N – Normal force, perpendicular to contact surface (lb, k, N)

Friction Theory Stated another way: Fs = μs N (for impending motion) Fk = μk N (for sustaining motion, kinetic) In general: F ≤ μ N

Example Problem Given: W = 400 lb. μ = 0.4 Find: Force P required to cause motion to impend W P

Example Problem Solution W P Given: W = 400 lb. μ = 0.4 Find: P for impending motion Solution: 1-FBD 2-Eqn of Equil F=μN 400 lb P N F