# Car - Traction Force

## Adhesion and tractive effort

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

F = μt W

= μt m ag   (1)

where

F = traction effort or force acting on the wheel from the surface (N, lbf)

μt = adhesion coefficient between the wheel and the surface

W = weight or vertical force between wheel and surface (N, lbf))

m = mass on the wheel (kg, slugs)

agacceleration of gravity (9.81 m/s2, 32.17405 ft/s2)

### Adhesion Coefficients

SurfaceAdhesion Coefficient
- μt -
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

Fone_wheel = 0.5 ((2000 kg)  (9.81 m/s2) / 4)

= 2452 N

The traction force from both rear wheels

Fboth_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

acar = F / m

= (4904 N) / (2000 kg)

= 2.4 m/s2

where

acar = acceleration of car (m/s2)

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

dt = dv / acar

= ((100 km/h) - (0 km/h)) (1000 m/km) (1/3600 h/s) / (2.4 m/s2)

= 11.6 s

where

dt = time used (s)

dv = change in velocity (m/s)

## Related Topics

• Dynamics - Motion - velocity and acceleration, forces and torques

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