# Torque to move a loaded cart

SevenToFive
I would like to see if I am on the correct path here and didn't miss anything for a project at work. If I am trying to move a cart that weighs 10 tons, this includes the weight of the cart and its load, the wheels of the cart are 10 inches in diameter, and with a 40:1 ratio gearbox I need to figure out the torque needed to move this cart.

The cart is steel wheels on a steel rail so I figured 0.5 coefficient of friction.
The total tractive effort (TTE) = Rolling Resistance(RR) + Grade Force(GF) + Force for Accelerate(FA)

For RR=weight * surface friction = 20,000lbs * 0.5
RR=10000lbs

GF= Weight * Sin(Degrees) = 20000lbs * Sin(0)
GF=0

FA = Weight * max speed(ft/s) /(32.2ft/s^2 * Time in Seconds)
Max speed would be radius * rpm * 0.10472 = 5" * 44rpm * 0.10472
Max speed (ft/s) = 1.92ft/s
20000lbs * 1.92ft/s /(32.2ft/x^2 * 1sec)
Force Acceleration (FA) = 1193lbs

TTE=10000lbs+0+1193lbs
TTE=11193lbs

Wheel torque = TTE * Wheel Radius
Wheel torque = 11193lbs * 5inches
Wheel torque needed to move the 10 ton cart is 15,965in-lbs.

If I am close I can size the gearbox and then only have to worry about the operating temperature that the gearbox will be in since it is being used to move a cart in and out of a heat treating furnace.

Homework Helper
Gold Member
The way that I recall, the rolling resistance comes from the tire slightly deforming while in contact with the surface. I think the resistance that you want is from the wheel bearings, not the friction between the wheel and track. I think it will be much less than 0.5
I base this on the fact that I can put my 3000 pound vehicle in neutral on a level surface, and be able to push it. I'm guessing that I'm not exerting a lot more than 100 pounds force.

If the wheel starts sliding in relation to the track, then you are loosing power. The goal is to have a high enough static friction (between wheel and track) so that it does not slide.

There is more than one coefficient of friction here. The 0.5 coefficient of friction relates to spinning or skidding the wheels. It does not apply to your situation. The coefficient of rolling friction is the force to make the cart move divided by the weight of the cart. That can be from 0.0003 to 0.005 (based on a quick internet search). It will vary widely depending on smoothness of track and wheels, bearings, alignment, slope, etc.

I realize this is physicsforum, where we try to calculate everything, but this is a case where you would be well advised to run a test. Get a spring scale, a rope, and a few people. Pull on the rope and measure the force. Do this in both directions in case there is a slight slope.

Walmart has a deer scale that measures up to 550 lbs for 25. https://www.walmart.com/ip/Allen-550-lb-Deer-Scale/17403736. You don't need a super accurate scale because you will put a safety factor of about 3 on to whatever you measure. • scottdave and Tom.G SevenToFive There is more than one coefficient of friction here. The 0.5 coefficient of friction relates to spinning or skidding the wheels. It does not apply to your situation. The coefficient of rolling friction is the force to make the cart move divided by the weight of the cart. That can be from 0.0003 to 0.005 (based on a quick internet search). It will vary widely depending on smoothness of track and wheels, bearings, alignment, slope, etc. I realize this is physicsforum, where we try to calculate everything, but this is a case where you would be well advised to run a test. Get a spring scale, a rope, and a few people. Pull on the rope and measure the force. Do this in both directions in case there is a slight slope. Walmart has a deer scale that measures up to 550 lbs for25. https://www.walmart.com/ip/Allen-550-lb-Deer-Scale/17403736. You don't need a super accurate scale because you will put a safety factor of about 3 on to whatever you measure.

JR, I would like to do the simple way of just measuring rather than doing all of the calculations but this unit isn't even built yet and where it is being built is 2500 miles away from me.

Looking through my past work, I think my calculated torque value is way way to high. I am going have to run through the numbers again and as JR mentioned using a 0.005 rather than a 0.5 for friction.

Thanks

• scottdave
tygerdawg
Unless I didn't see it, you seem to have missed the most important part: time required to accelerate to speed.

Websearch for Smart Motion Cheat Sheet, is a nice summary of equations of motion. Other nice sources are gear motor Engineering Guides from a variety of manufacturers.

You need to calculate peak torque, which is mostly the torque needed to accelerate that inertia (rotational & linear) up to your speed in a specified time. Add to that all the friction, gravity, etc., plus safety factor. Torque is cheap, be sure to use plenty of it.

T = J x α (Torque = mass moment of inertia X angular acceleration)
α ≈ Δvelocity / Δtime