How to calculate tractive force
Tractive force is the propulsive force that drive wheels or tracks apply to the ground to move a vehicle along a path, whether it is a road, a railroad track, or soft earth. Tractive force for hard, smooth surfaces is a proportional function of weight and the coefficient of friction of the two materials in contact.
Calculation complexity increases when rubber or deep treads come into play. A deeper understanding of tractive force principles can be gained by considering several varied examples.
Tractor image by Victor M. from Fotolia.com
Calculate the tractive force of an accelerating 1,360 kg (3,000-pound) auto-mobile with rubber tires on a level asphalt roadway with a 0.68 coefficient of friction. Assume that the propulsive force of the engine through the transmission and rear-drive axle gearing is 567 kg (1,250 pounds) and that the weight over the rear-drive wheels during acceleration is 726 kg (1,600 pounds). First, multiply the 726 kg (1,600 pounds) by the 0.68 coefficient of friction to determine the point where the tires would start to slip and lose traction (726 kg x 0.68 = 494 kg [1,088 pounds]). If the car is applying 567 kg (1,250 pounds) of force to the wheels, the traction would be broken, and the car would "burn rubber" with an accompanying shrieking sound. Any force less than 494 kg would smoothly propel the car forward.
Calculate the different tractive forces of a 113,400 kg (250,000-pound) railway locomotive attempting to come to an emergency stop and then once it is stopped. The static coefficient of friction for mild steel (from which both the wheels and rails are made) is 0.74; the kinetic (moving) coefficient is 0.57. While the locomotive is braking to a stop, the tractive force is only the 113,400 kg (250,000-pound) weight x 0.57, the kinetic coefficient of friction. Maximum tractive force is therefore 64,637 kg (142,500 pounds). Once the locomotive stops, however, the static coefficient must be applied, and it would take a force of 113,400 kg (250,000 pounds) x 0.74 = 83,915 kg (185,000 pounds) to force the locomotive's wheels to start to skid on the tracks from a dead stop. Once they are sliding, though, the force would again be 64,637 kg (142,500 pounds). The difference is the "breakaway" force.
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Calculate the effective coefficient of traction (friction) for a 29,500 kg (65,000-pound) bulldozer that is able to impart 68,039 kg (150,000 pounds) of force to its deeply cleated steel tracks from its diesel-hydraulic drivetrain. Experimental data from the bulldozer manufacturer shows that the bulldozer has demonstrated a drawbar pull of 50,621 kg (111,600) on average hard earthen ground during design testing. Dividing the 50,621 kg of tractive force by the bulldozer's weight would yield the coefficient of traction (as opposed to friction since the bulldozer's tracks have steel cleats that penetrate into the ground to increase grip). Taking 50,621/29,500 kg (111,600/65,000-pound) weight results in a coefficient of traction for the bulldozer of about 1.72, which is vastly greater than the coefficient of friction that would be developed by smooth, un-cleated tracks on hard ground. It is the reason why bulldozers are so effective in moving huge amounts of earth (or anything else their blades might come in contact with).
- Consult coefficient of friction data comparing different road surfaces, such as concrete and asphalt, for stopping distances.
- Ice near or at 0 degrees Celsius (32 Fahrenheit) is far more slippery and treacherous than the same ice at minus 17.8 degrees Celsius (0 degrees Fahrenheit) because of the film of water on the surface of the warmer ice (near-zero friction).
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