Calculating the Force to Move a 200,000 lb Table on 12 Wheels

In summary, the approximate force required to move a 30 ft diameter table weighing 200,000 lbs that is sitting on twelve 10" diameter wheels with ball bearings would be around 2,000 pounds. However, this estimation may be significantly over-estimated as it assumes the tires cannot roll. The rolling resistance coefficient for rubber tires is about 0.01, so for a car weighing 3000 pounds, the horizontal force needed to push it is about 30 pounds. The question is difficult to answer without more information about the wheels and surface they are on. It is rare to have a 100 ton table with such small wheels, as the weight would need to be evenly distributed to avoid excessive deformation of the ground or wheels
  • #1
mach6
1
0
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.
 
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  • #2
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...
 
Last edited:
  • #3
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
 
  • #4
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.
 
  • #5
Where the heck do you have a 100 ton table and why would such a table have wheels?
 
  • #6
There are plenty of mobile engineering structures that weigh more than 100 tons. That is only the weight of three large "standard" road trucks. But supporting them on twelve 10-inch diameter wheels is not so common.
 
  • #7
I don't mean to get off track here, but I hotly disagree that there are plenty of 100 ton TABLES that are required to be mobile, 10 inch wheels or not.
 
  • #8
The gross weight limit of a standard US rail freight car is about 286,000 pounds, supported on eight 33" diameter steel wheels (35,750 pounds each). The rolling resistance coefficient of steel wheels on a steel rail (without deformation) is about 0.1%.
 

Related to Calculating the Force to Move a 200,000 lb Table on 12 Wheels

1. How do you calculate the force required to move a 200,000 lb table on 12 wheels?

The force required to move a 200,000 lb table on 12 wheels can be calculated using the formula F = μ * N, where F is the force, μ is the coefficient of friction, and N is the normal force. The coefficient of friction is a constant value that depends on the type of surface the table is being moved on, while the normal force is the weight of the table divided by the number of wheels it is resting on.

2. What is the coefficient of friction for moving a 200,000 lb table on 12 wheels?

The coefficient of friction for moving a 200,000 lb table on 12 wheels can vary depending on the surface the table is being moved on. Generally, for a smooth, flat surface, the coefficient of friction can range from 0.01 to 0.05. However, for surfaces with more friction, such as carpet or grass, the coefficient of friction can be higher.

3. How do you determine the normal force for a 200,000 lb table on 12 wheels?

The normal force for a 200,000 lb table on 12 wheels can be determined by dividing the weight of the table (200,000 lbs) by the number of wheels it is resting on (12). This will give you the amount of force pushing down on each wheel, which is important in calculating the overall force required to move the table.

4. Is the force required to move a 200,000 lb table on 12 wheels the same on all surfaces?

No, the force required to move a 200,000 lb table on 12 wheels can vary depending on the surface it is being moved on. This is because the coefficient of friction changes based on the type of surface, as well as other factors such as the weight distribution of the table and the condition of the wheels.

5. Are there any other factors that can affect the force required to move a 200,000 lb table on 12 wheels?

Yes, there are several other factors that can affect the force required to move a 200,000 lb table on 12 wheels. These include the condition of the wheels (e.g. worn out or properly lubricated), the angle of the surface (e.g. if the table is being moved on an incline), and any additional forces acting on the table (e.g. wind or other external forces).

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