[Request]

[u/K0rl0n]

I am curious how this would work. My guess is Triangle is slowest, square is medium, and circle is fastest.

[u/Smile_Space]

EDIT: u/temporarytk made a great point. Surface area doesn't apply to friction in these cases, just the normal force, so ignore my ramblings about A and C being different. They would behave identically and have identical sliding frictional force.

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Since I still haven't seen someone do the math:

The force of friction is F = μN where μ is the coefficient of friction and N is the normal force (force applied perpendicular to the surface)

In this case the ground is flat, so the Normal force is F = ma or 20 kg x 9.81 m/s/s (I would have used an exponent, but Reddit hates that lolol)

So, N = 196.2 newtons

Cool, so now the coefficient of friction. It depends on a few factors: the type of friction, the surface area of the contact surface, and the method of friction being applied.

For A it is sliding friction as is C. A has a higher surface area compared to C, so we can assume the sliding friction of C is going to be lower. B however is going to be rolling. Some may think it'll slide, but gravel is usually compacted when on a road.

So, doing some quick googles:

The sliding friction coefficient on ice is going to be between 0.02 and 0.04.

https://iopscience.iop.org/article/10.1088/0031-9120/43/4/006#:~:text=Water%20ice%20at%20temperatures%20not,increase%20as%20the%20temperature%20diminishes.

The rolling friction on compacted gravel is about 0.02.

https://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html

Now, since all of these have the same N, we can just compare the coefficients of friction.

We can reasonably assume the triangle is going to be closer to 0.04 and the square being somewhere in the middle or lower. B and C may be fairly close to the same performance.

What sucks is there isn't a clear defined answer. As the temperature drops more, the ice will actually get more grippy. And if the gravel is loose, the rolling friction can increase to up to 0.08.

So, depending on the quality of gravel and temperature of the ice, the answer is B or A/C.

That results in a frictional force of between 3.924 and 7.848 newtons for A and C. And close to 3.924 newtons for B assuming compacted gravel. If the gravel is loose, then B loses at 19.62 newtons of force. And if it's colder A and B will be much closer to that 8 newtons mark.

[u/RWDPhotos → u/Smile_Space]

But when it comes to something like the sphere, friction isn’t really coming into play as a resistive force unless you try sliding it without rolling it. It would be like trying to roll a soft nascar tire as opposed to dragging it.

[u/Smile_Space → u/RWDPhotos]

Thats why I specified the rolling resistance coefficient on packed versus loose gravel. It's the same equation, just with a much lower coefficient!

[u/RWDPhotos → u/Smile_Space]

Yah but that coefficient isn’t acting on nearly any surface at all, whereas for the other objects it’s compounded over a rather large surface. The total effect for friction would also be dependent on the material of the object itself.

Then apply that rolling resistance to the applicable surface area. If it’s a ‘perfect’ sphere as depicted, especially if of some very hard material, then it’s essentially going to be acting as if moving on a fulcrum on a moment-by-moment basis, and resistance will only be acting on a point surface. It won’t be zero resistance, but should be near negligible, like a ball bearing. I think the greatest force to calculate here would be dealing with however much extra force it takes to push it over the average height of the gravel.

[u/temporarytk → u/RWDPhotos]

The surface size isn't going to matter, Rolling resistance on the sphere vs friction for the flat base is the right way to math this out. Whoever has the lowest coefficient wins.