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Friction coefficient of an object on wheels 2

Dylanuk22

Mechanical
Aug 13, 2022
4
Hello everyone,

I was wondering if you could point me in the right direction.

I have an object on 4 x 100mm hard rubber castor wheels on a flat concrete. The object weighs roughly 1000kg.

What force would be required to pull the object?

I have been looking at this website as an example https://www.omnicalculator.com/physics/friction

Looking more specifically at calculator 2 and 3 on the website, I get that one uses a dimensioned friction coefficient and the second uses dimensionless, but why do the values vary so much? Even when comparing that the rolling resistance value is 0.35 https://www.bulldogcastors.co.uk/blog/castor-wheels-roll-resistance/ and for rubber even being a value of 1 (Example of friction values https://web.mae.ufl.edu/designlab/Class Projects/Background Information/Friction coefficients.htm)

Surely it would be easier to pull an object with 4 rubber wheels vs the surface being just rubber on concrete?

Thankyou
 
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Your gut feel is right. Using the static friction coefficient is equivalent to locking all 4 wheels. You need the rolling resistance. The good news is that for large sensibly designed wheels it is perhaps of the order of 2% on smooth hard surfaces. Note that the wheel bearings will have some impact on this, as will the exact properties of your tires. I'm a bit surprised by that 0.35 figure for castors, that must be a worst case scenario.

Small diameter castors often have problems with stones and gravel.
 
Last edited:
Have a look at the attached -
 

Attachments

  • White Paper Rolling Resistance.pdf
    286.4 KB · Views: 10
I think gut feeling, sanity checks, or a simple test will get you answers quicker that a coefficient of friction calculation.

What you describe there sounds awfully similar to a pallet jack. Moving a 1 tonne load on smooth concrete isn't too hard. Moving a 2500kg load requires a bit of muscle maybe 50kg of force. So 50kg/2500kg = 2%. Which coincidentally exactly aligns with GregLocock's figure.

You could readily double that or halve that depending on the quality of the casters and the surface it is being moved on.
 

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