Interactive Lecture
Coulomb Friction, Impending Motion, and Friction Angles
by Dr Arnaldo Delli Carri
Friction is a force that opposes the relative motion between two surfaces in contact. Walking, driving, and holding objects all depend on it — remove friction and each becomes impossible.
Friction arises from the microscopic interactions between surfaces:
The friction force is approximately proportional to the normal force between surfaces, not to the apparent contact area. This occurs because increasing the normal force increases the real contact area proportionally.
The classical model of friction, developed by Charles-Augustin de Coulomb, distinguishes between two types:
Static Friction: \(F_s \leq \mu_s N\)
Where:
The actual friction force adjusts to match the applied force up to the maximum value \(\mu_s N\)
Kinetic Friction: \(F_k = \mu_k N\)
Where:
| Material Pair | Static Coefficient (μₛ) | Kinetic Coefficient (μₖ) | Applications |
|---|---|---|---|
| Steel on Steel (dry) | 0.6 - 0.8 | 0.4 - 0.6 | Machine components, bearings |
| Steel on Steel (lubricated) | 0.1 - 0.15 | 0.05 - 0.1 | Lubricated bearings, engines |
| Aluminium on Steel | 0.61 | 0.47 | Structural connections |
| Copper on Steel | 0.53 | 0.36 | Electrical contacts, bushings |
| Rubber on Dry Concrete | 1.0 - 1.2 | 0.8 - 1.0 | Tyres, footwear |
| Rubber on Wet Concrete | 0.5 - 0.7 | 0.4 - 0.6 | Wet road conditions |
| Wood on Wood (dry) | 0.4 - 0.6 | 0.2 - 0.4 | Furniture, construction |
| Ice on Ice | 0.1 | 0.03 | Ice skating, winter sports |
| Teflon on Teflon | 0.04 | 0.04 | Non-stick coatings |
| Glass on Glass | 0.9 - 1.0 | 0.4 | Optical equipment |
Friction coefficients vary significantly with surface conditions, temperature, humidity, contamination, and wear. Engineers typically use conservative values with appropriate safety factors in design.
Apply force to the block and observe the transition from static to kinetic friction. The graph shows the friction force response.
Pure rolling occurs when an object rolls without slipping. This is possible due to static friction at the contact point.
For pure rolling to occur:
Pure Rolling Condition:
\(v_{cm} = \omega R\)
Where:
The velocity of any point on the wheel: \(\vec{v}_{point} = \vec{v}_{cm} + \vec{\omega} \times \vec{r}\)
In pure rolling, the bottom point of the wheel is instantaneously at rest (zero velocity), whilst the top point moves at twice the centre's velocity. This creates the characteristic velocity distribution shown in the simulation below.
Observe the velocity vectors showing the combination of translation and rotation in pure rolling motion.
Whilst pure rolling theoretically requires no energy to maintain, real objects experience rolling resistance due to material deformation.
Rolling Resistance Force:
\(F_{rr} = C_{rr} N\)
Where:
Alternatively: \(F_{rr} = \frac{b}{R} N\)
Where \(b\) is the rolling resistance arm (distance)
| System | C_rr Value | Notes |
|---|---|---|
| Steel wheel on steel rail | 0.0010 - 0.0025 | Trains - very efficient |
| Car tyre on concrete (optimal pressure) | 0.010 - 0.015 | Standard road conditions |
| Car tyre (underinflated) | 0.020 - 0.030 | Increased deformation |
| Truck tyre on road | 0.006 - 0.010 | High pressure tyres |
| Bicycle tyre on road | 0.002 - 0.005 | High pressure, narrow contact |
Rolling resistance is why a car in neutral will eventually stop on a level road. For typical car tyres, it accounts for about 20% of fuel consumption at highway speeds, which is why tyre pressure has a measurable effect on fuel consumption.
The power required to overcome rolling resistance:
\(P = F_{rr} \cdot v = C_{rr} N v\)
This power is dissipated as heat in the deformed material, primarily in the tyres for vehicles.
Friction is at work in nearly every engineering system that involves contact or motion:
| Concept | Equation | Description |
|---|---|---|
| Normal Force | \(N\) | Perpendicular contact force between surfaces |
| Friction Force | \(F_s \leq \mu_s N\) | Static friction before sliding |
| Kinetic Friction | \(F_k = \mu_k N\) | Friction force during sliding motion |
| Angle of Repose | \(\theta_{max} = \tan^{-1}(\mu_s)\) | Maximum incline angle before sliding begins |
| Rolling Resistance | \(F_{rr} = C_{rr} N\) | Resistance force for rolling objects |
| Power Dissipation | \(P = F_{rr} \cdot v\) | Power lost to rolling resistance |
A 5 kg wooden block rests on a wooden incline. If μₛ = 0.5 and μₖ = 0.3:
Show Solution
Part a:
At the critical angle, static friction equals the component of weight down the slope:
\(\mu_s N = mg\sin\theta_{max}\)
\(\mu_s mg\cos\theta_{max} = mg\sin\theta_{max}\)
\(\tan\theta_{max} = \mu_s = 0.5\)
\(\theta_{max} = \arctan(0.5) = 26.6°\)
Part b:
Forces down the incline: \(F_{down} = mg\sin(35°) - \mu_k mg\cos(35°)\)
\(ma = mg(\sin(35°) - \mu_k\cos(35°))\)
\(a = g(\sin(35°) - 0.3\cos(35°))\)
\(a = 9.81(0.574 - 0.3 \times 0.819) = 9.81(0.328) = 3.22 \text{ m/s}^2\)
A 100 kg crate on a concrete floor has μₛ = 0.7 and μₖ = 0.5. If you push horizontally:
Show Solution
Part a:
Normal force: \(N = mg = 100 \times 9.81 = 981 \text{ N}\)
Force to overcome static friction: \(F = \mu_s N = 0.7 \times 981 = 687 \text{ N}\)
Part b:
Force to maintain constant velocity: \(F = \mu_k N = 0.5 \times 981 = 490.5 \text{ N}\)
Part c:
Net force: \(F_{net} = 800 - \mu_k N = 800 - 490.5 = 309.5 \text{ N}\)
Acceleration: \(a = \frac{F_{net}}{m} = \frac{309.5}{100} = 3.1 \text{ m/s}^2\)
A 1500 kg car with C_rr = 0.015 travels at 100 km/h on level ground. Calculate:
Show Solution
Part a:
Normal force: \(N = mg = 1500 \times 9.81 = 14,715 \text{ N}\)
Rolling resistance: \(F_{rr} = C_{rr} N = 0.015 \times 14,715 = 220.7 \text{ N}\)
Part b:
Velocity: \(v = 100 \text{ km/h} = 27.78 \text{ m/s}\)
Power: \(P = F_{rr} \times v = 220.7 \times 27.78 = 6,131 \text{ W} = 6.13 \text{ kW}\)
Part c:
New rolling resistance: \(F_{rr} = 0.025 \times 14,715 = 367.9 \text{ N}\)
New power: \(P = 367.9 \times 27.78 = 10,218 \text{ W} = 10.22 \text{ kW}\)
Extra power needed: \(10.22 - 6.13 = 4.09 \text{ kW}\) (67% increase!)