Motor torque required to move a trolley

Question

This is a real situation.

The trolley is mounted with a rack underneath it and multiple steel wheels on a steel track. The trolley is driven by the rack and pinion mechanism from a geared motor.

The actual specification:

• Motor: 1.5 kW, 460V 3 phase
• Trolley weight (including rack, wheels, and load):20000 kg (20 tonne）
• Pinion size:0.2 meter
• Gearbox: gear ratio 88, output torque 660 N.m

Now, based on F = mg × friction + ma

If I ignore the acceleration part, and assume the static friction is 0.1.
F = 20000 &times 9.81 × 0.1 = 19620 N
Torque needed = F × r = 19620 × 0.1 = 1962 N.m

I imagine the trolley would not be able to move at all and the motor will overload. But I have seen the trolley run without any problem.

What is wrong in my calculation? Is the wheel against steel friction much lower than I thought?

Answer 1

Your friction coefficient is simply too large for a good bearing. It appears to be consistent with lubricated metal to metal contact, but bearings do much better than this. One could model bearings by applying lubricated metal coefficients throughout the bearing geometries. The effect is that a small radius such as at the balls will considerably reduce the torque from a force. Metal dragging friction has a place where the balls do not roll - because the outer and inner race are different diameters, balls must slide without rolling for a portion of their movement.

Consider a simpler example where a needle bearing that takes a large force and turns it into a tiny friction torque by making the force apply to a tiny radius.

Solving backwards and assuming your gearbox outputs the full 660Nm, to move F*19620Nm, F <= .03. Bearings are more along the lines of F=.02 than F=.1 so your device working is plausible.