# Force Required

What would be the approximate force required on the perimeter of a 30 ft diameter table that weighs 200,000 lbs to move it? The table is sitting on twelve wheels (10" diameter each) that are mounted to ball bearings.

A quick approximation would be:

Force = coefficient of static friction X weight.

The rubber-to-road friction is likely larger than the bearing friction (assuming the bearings are in good condition).

My old physics textbook lists for coefficient of static friction = 1 for rubber on dry concrete.

Edit:

The method above will significantly over-estimate the force since this would only apply if the tires could not roll.

I will look into this a little more...

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The rolling resistance coefficient (RRC) for rubber automobile tires is about 0.01 (less if the tires are inflated more). Thus for a car weighing 3000 pounds, the horizontal force needed to push the car (overcome the RRC) is about 30 pounds.

For your case, I would guess about 0.01 x 200,000 pounds = 2,000 pounds tangential force on the perimeter of the table.

Bob S

AlephZero
Homework Helper
I don't think you are going to support 200,000 lbs on 12 10-inch-diameter rubber tires. The weight on each tire would be about 7.5 tons. Compare "one 10-inch diameter wheel" with the wheels and tires on a typical 7-ton road truck!

The question is impossible to answer unless we know more about the wheels, and what surface they are running on. You would probably need a specially built track to support that weight with such a small contact area. Otherwise, the most of the force will be overcoming the deformation of the ground and/or the wheels, not overcoming rolling resistance or friction in the bearings.

Where the heck do you have a 100 ton table and why would such a table have wheels?

AlephZero