1. Section 5.8: Friction

2. Friction Kinetic (motion) 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. DIRECTIONS of fk & n are  each other!! fk  n
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
Write relation as fk  k n (magnitudes)
k  Coefficient of kinetic friction
k depends on the surfaces & their conditions
k is 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 force fs to oppose FA
FA – fs = 0 or fs = FA n fk
FA (applied) mg

5. s  Coefficient of static friction
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  sn

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

7. Static & Kinetic Friction

8. Figure 5.16: When pulling on a trash can, the direction of the force of friction f between the can and a rough surface is opposite the direction of the applied force F. Because both surfaces are rough, contact is made only at a few points as illustrated in the “magnified” view. (a) For small applied forces, the magnitude of the force of static friction equals the magnitude of the applied force. (b) When the magnitude of the applied force exceeds the magnitude of the maximum force of static friction, the trash can breaks free. The applied force is now larger than the force of kinetic friction, and the trash can accelerates to the right. (c) A graph of friction force versus applied force. Notice that fs, max > fk .
Fig. 5-16, p. 119

9. Example n f

10. Example ∑F = ma y direction: y direction: n = mg + Fy ; fs (max) = μsn
∑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 a  a  ∑F = ma For EACH mass separately! x & y componentsfk
∑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 θc at 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
Figure 5.19: (Example 5.12) After the puck is given an initial velocity to the right, the only external forces acting on it are the gravitational force mg, the normal force n, and the force of kinetic friction fk .

14. Example 5.13
Figure 5.20: (Example 5.13) (a) The external force F applied as shown can cause the block to accelerate to the right. (b, c) The free-body diagrams assuming the block accelerates to the right and the ball accelerates upward. The magnitude of the force of kinetic friction in this case is given by fk = μkn = μk (m1g − F sin θ).