Section 5 8 Friction Friction Friction We must

Section 5. 8: Friction

Friction • Friction: We must account for it to be realistic! – Exists between any 2 sliding surfaces. – Two types of friction: Static (no motion) friction Kinetic (motion) friction – The size of the friction force: Depends on the microscopic details of 2 sliding surfaces. • The materials they are made of • Are the surfaces smooth or rough? • Are they wet or dry? • Etc. , etc.

• Kinetic Friction: Experiments determine the relation used to compute friction forces. • Friction force fk is proportional to the magnitude of the normal force n between 2 sliding surfaces. • DIRECTIONS of fk & n are each other!! fk n fk n a FA (applied) mg • Write relation as fk k n (magnitudes) k Coefficient of kinetic friction k depends on the surfaces & their conditions k is dimensionless & < 1

• Static Friction: Experiments are used again. • The friction force fs exists || 2 surfaces, even if there is no motion. Consider the applied force FA: n fk FA (applied) mg ∑F = ma = 0 & also v = 0 There must be a friction force fs to oppose FA FA – fs = 0 or fs = FA

• Experiments find that the maximum static friction force fs(max) is proportional to the magnitude (size) of the normal force n between the 2 surfaces. • DIRECTIONS of fk & n are each other!! fk n • Write the relation as fs(max) = sn (magnitudes) s Coefficient of static friction – Depends on the surfaces & their conditions – Dimensionless & < 1 – Always find s > k Static friction force: fs s n

Coefficients of Friction μs > μk fs(max, static) > fk(kinetic)

Static & Kinetic Friction

Fig. 5 -16, p. 119

Example n f

Example ∑F = ma n fs y direction: ∑Fy = 0; n – mg – Fy = 0 n = mg + Fy ; fs (max) = μsn x direction: ∑Fx = ma n fs y direction: ∑Fy = 0; n - mg + Fy = 0 n = mg - Fy; fs(max) = μsn x direction: ∑Fx = ma

Example n a fk a ∑F = ma For EACH mass separately! x & y components plus fk = μkn

Example 5. 11 Place a block, mass m, on an inclined plane with static friction coefficient μs. Increase incline angle θ until the block just starts to slide. Calculate the critical angle θc at which the sliding starts. Solution: Newton’s 2 nd Law (static) y direction: ∑Fy = 0 n – mg cosθ = 0; n = mg cosθ x direction: ∑Fx = 0 mg sinθ – fs = 0; fs= mg sinθ Also: fs = μ s n Put (1) into (3): fs = μsn mg cosθ Equate (2) & (4) & solve for μs = tanθc (1) (2) (3) (4)

Example 5. 12

Example 5. 13