# Force required to move stationary vehicle?

1. Dec 3, 2005

### BirchUK

$#force required to move stationary vehicle??$#

I am undergoing a product development as part of my full time business. I have forgotten a lot of the Physics I learnt back in my school days and have been having trouble getting help with this issue, so I hope someone here can help.

The device requires the movement of a vehicle from a stationary position and so I therefore need a formula to calculate what kind of force is needed to move varying weights of cars from stationary position.

I found the coefficient of rubber on dry concrete to be 1.0 (0.8 on wet concrete) and I will also have the various weights of cars.

How do I calculate the force needed to get the car moving?

Does the calculation differ if all four wheels are touching the floor in comparison to having the front two wheels on a platform (and then the platform is pulled)?

Any help much appreciated.

Thanks,

Steve.

2. Dec 3, 2005

### BirchUK

I have found a formula for 'Rolling Friction':

F = f x W/R

f= coefficient of ROLLING friction (which is 0.01-0.02 for rubber on concrete)
W=weight of cylinder (car in this case)

PROBLEM: how does having four 'cylinders' effect this calculation? The equation as it stands is for calculating force on a single rolling cylinder. Also, how can it be applied if the front two wheels of the car are on the platform that is doing the pulling??

3. Dec 3, 2005

### FredGarvin

Honestly, you'll have to do a lot of hand waving. What you need to calculate is the force to overcome friction. Not just friction between the tires and the ground, but the friction in the drivetrain, the friction in the wheel hubs, etc...That's a lot of variation in one calculation especially across so many different cars.

What you might do is start with the age od $$F_{friction}=\mu N$$ and then braanch out and do some actual pull tests. See if there is a deccent correlation between your calculations and the real world. You may simply add a multiplying fudge factor to overcome unforseen things and be done with it that way.

4. Dec 3, 2005

### BirchUK

Yes, you have a good point...and I don't fancy wasting a box of pencils working all that out.

You suggest that I pop out with a newton meter and a bit of rope and pull a few cars so I have some figures for force required to pull certain weights. How would this then be applied to a formula so I can estimate the force needed for other weights? And would there be a way of predicting what impact there would be in having the two wheels on the pulling platform? The wheels of this platform would also have to be considered in the calculations surely?

Is this all looking extremely complex? Problem is, I could really do with coming up with a fairly reliable formula. This could also help decide what size wheels would be best on the platform.

5. Dec 3, 2005

### dicerandom

I don't think the wheels will come into it much, you're more concerned with things like the bearings in the wheel hubs here. If the car is rolling along, and not skidding, then the frictional force between the wheels and the road serves to rotate the wheels, not slow the car down.

I think Fred gave you a good first principles formula in his post. You can measure both the mass of the car and the force which is required to get it rolling, you can therefore solve for the equivilant coefficient of friction. Try running some experiments with the same car and different weights, add in some sand bags or something, and see if you get the same coefficient of friction, which you should if this simple model holds. Once you've verified the model you can try measuring the coefficient of friction for other makes of cars, perhaps they will all fall within some general range and you can determine a value to use for any vehicle so that your answer will be correct to within some degree of precision.

6. Dec 3, 2005

### Danger

Aside from the math part, of which I know nothing, your design will be highly dependent upon how far and how fast you have to move the cars. There's a big difference between winching one across the garage floor and towing it to another city.

7. Dec 3, 2005

### BirchUK

Some good advice again, much appreciated.

Thanks guys, I'll get testing it then and see what I come up with.

Regards,

Steve.

8. Dec 3, 2005

### Staff: Mentor

friction between ground and tires is key

I'd say that the only force that needs to be overcome to get the car moving is the friction of the ground on the tires. I agree that it won't be a simple calculation, because--as you note--that force depends on the internal friction sources that you mentioned (in the drive train and wheel hubs). Nonetheless, only an external force can directly affect the motion of the car. Perhaps this is what you meant, Fred?

If there's friction between the wheels and road, it will both slow the car down and rotate the wheels. (It can't do one without the other!)

9. Dec 3, 2005

### Staff: Mentor

Ideally, you would find a direct linear relationship between the weight of the car and the force required to get it moving.

10. Dec 4, 2005

### brewnog

I don't think experiment will provide anything like the ideal linear relationship you are seeking. I would expect 'little' things like tyre pressures and sticking brakes to give so much additional variation in your results that you won't be able to fit your empirical results to any model with much success at all.

For design purposes, I'd just find a worst case scenario, - (perhaps a really heavy car with rusted brake discs and flat, fat tyres), get the newton meter out, and use that as your worst case design parameter.

Good luck, I'd be interested to hear how this one goes!

11. Dec 4, 2005

### FredGarvin

That is correct Sir.

12. Dec 4, 2005

### Danger

I have to agree with Brewski on this. My first reaction to your question was to do what I always do... take a quess at what is required in an un-ideal situation and then double it. Since I have no education, I didn't think that it was appropriate to post it. Now that the guy with the degree has brought it up, I feel confident in stating it.

13. Dec 4, 2005

### bkelly

sticktion

There is a formal term called sticktion. I am not sure about the spelling and it may not be directly applicable, but the concept is. The force to begin movement will be greater than the force to continue movment. Examples include avelanches and cerial pouring out of a box.

I believe there is significant rolling resistance due to tires along. When a bicycle coasts by me, I can clearly hear the tire noise. Noise requires energy and that energy must come from the effort reuired to continue motion. The tire noise from cars is much greater as there is more weight and the tires are bigger.

Here is one way to measure. Get some kind of vehicle such as a trailer with no drive train. Measure the rolling resistance, starting and continuing, then add weight and measure.