Section 5.8: Friction

1. Section 5.8: Friction

2. 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., etc.

3. 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. • DIRECTIONSof fk & n areeach other!!fk n • Write relation as fk k n (magnitudes) k  Coefficient of kinetic friction kdepends on the surfaces & their conditions kis dimensionless & < 1 n a fk FA(applied) mg

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

5. Experiments find that the maximum static friction forcefs(max) is proportional to the magnitude (size) of the normal force n between the 2 surfaces. • DIRECTIONSof 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  sn

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

7. Static & Kinetic Friction

8. Fig. 5-16, p. 119

9. Example n f

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

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

12. 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 θcat which the sliding starts. Solution:Newton’s 2nd Law (static) y direction: ∑Fy = 0  n – mg cosθ = 0; n = mg cosθ(1) x direction: ∑Fx = 0  mg sinθ – fs = 0; fs= mg sinθ(2) Also: fs = μsn(3) Put (1) into (3): fs = μsn mg cosθ(4) Equate (2) & (4) & solve for μsμs = tanθc

13. Example 5.12

14. Example 5.13