# Car - Traction Force

## Adhesion and tractive effort

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The tractive effort or force between a car wheel and the surface can be expressed as

*F = μ _{t} W *

* = μ _{t} m a_{g} (1)*

*where *

*F = traction effort or force acting on the wheel from the surface (N, lb _{f})*

*μ _{t }= adhesion coefficient between the wheel and the surface*

*W = weight or vertical force between wheel and surface (N, lb_{f}))*

*m = mass on the wheel (kg, slugs)*

*a _{g} = acceleration of gravity (9.81 m/s^{2}, 32.17405 ft/s^{2})*

### Adhesion Coefficients

Surface | Adhesion Coefficient - μ_{t} - |
---|---|

Wet Ice | 0.1 |

Dry Ice | 0.2 |

Loose Sand | 0.3 - 0.4 |

Dry Clay | 0.5 - 0.6 |

Wet rolled Gravel | 0.3 - 0.5 |

Dry rolled Gravel | 0.6 - 0.7 |

Wet Asphalt | 0.4 - 0.7 |

Wet Concrete | 0.4 - 0.7 |

Dry Asphalt | 0.4 - 0.7 |

Dry Concrete | 0.4 - 0.7 |

### Example - Traction Force on an Accelerating Car

The **maximum traction force** available from one of the two rear wheels on a rear wheel driven car - with mass *2000 kg* equally distributed on all four wheels - on wet asphalt with adhesion coefficient *0.5* - can be calculated as

* F _{one_wheel} = 0.5 ((2000 kg) (9.81 m/s^{2}) / 4)*

* = 2452 N*

The traction force from both rear wheels

*F _{both_wheels} = 2 (2452 N)*

* = 4904 N*

Note that during acceleration the force from the engine creates a moment that tries to rotate the vehicle around the driven wheels. For a rear drive car this is beneficial by increased vertical force and increased traction on the driven wheels. For a front wheel driven car the traction force will be reduced during acceleration.

The **maximum acceleration** of the car under these conditions can be calculated with Newton's Second Law as

*a _{car} = F / m *

* = (4904 N) / (2000 kg)*

* = 2.4 m/s ^{2}*

*where *

*a _{car} = acceleration of car (m/s^{2})*

The minimum time to **accelerate from 0 km/h to 100 km/h** can be calculated as

*dt = dv / a _{car} *

* = ((100 km/h) - (0 km/h)) (1000 m/km) (1/3600 h/s) / (2.4 m/s ^{2})*

* = 11.6 s*

*where *

*dt = time used (s)*

*dv = change in velocity (m/s)*

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