Welcome to the PF.What would the torque on a driveshaft be for a 8 ton vehicle in park on a 16% grade? Tire radius is 19". Axle ratio it 7.1.
Trying to determine the loads on an output shaft of a transmission when parked on a grade with different axle ratios. Not for school work. Just general knowledge.Welcome to the PF.
Is this question for schoolwork? What is the context of the question?
Shouldn't this include the acceleration of gravity?If the rear wheels support the entire load caused by the slope, the force acting at the rear tires is ##F_w = W\sin\theta##, where ##W## is the weight of the vehicle and ##\theta## is the angle of the slope (reference). ##100\tan\theta = \%slope## to find the angle of the slope (reference).
The wheel torque produced is ##T_w = F_w r##, where ##r## is the tire radius.
The gear ratio reduces the torque seen by the driveshaft, so ##T_d = \frac{T_w}{GR}##, where ##GR## is the axle gear ratio.
##\theta = \arctan\frac{16}{100} = 9.1°##
##F_w = (16000\ lb)\sin9.1° = 2530\ lb##
##T_w = (2530.5\ lb) * (1.583\ ft) = 4006\ lb.ft##
##T_d = \frac{4006\ lb.ft}{7} = 572\ lb.ft##
Shouldn't this include the acceleration of gravity?
F_w = 16000 Lbs* 32.174 ft/s2
That is because we don't use the same numbers:I used SI units.
Mass = 8000 kg.
Force due to gravity = 8000 * 9.8 = 78400 newton.
Wheel radius is 39” = conveniently 1 metre.
16% grade = 9.09 deg; Sin(9.09°) = 0.158
7:1 axle ratio.
Drive shaft torque = 1m * 78400N * 0.158 / 7 = 1769.5 Nm
1769.5 Nm = 1305.1 ft.lbs
This is quite different to jack action's 572. ft.lb
https://en.wikipedia.org/wiki/Short_tonAre vehicle weights in the USA always specified in Short Tons = 907.18 kg = 2000 lbs ?