1. ME 201 Engineering Mechanics: Statics
Chapter 8
Friction

2. 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)

3. 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

4. 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

5. 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

6. 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

7. 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)

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

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

10. Example Problem SolutionW 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