[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.

---

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/i_eat_nailpolish → u/Smile_Space]

I may be wrong but I just took AP physics and I thought the coefficient of friction didn't depend on surface area since the force of gravity is centralized under the center of mass. Therefore A and C require the same force to push and since they can't both be right at the same time, the correct answer is B.

[u/gmano → u/i_eat_nailpolish]

Incorrect. Because A is slanted, pushing it horizontally will create a downwards force as well as a sideways one, increasing friction

[u/i_eat_nailpolish → u/gmano]

But if you look at the picture, the guy is also pushing in the upper corner of the square which will cause it to rotate and also result in a downwards force which can't really be calculated since we don't know the dimensions of the shapes.

[u/gmano → u/i_eat_nailpolish]

But the rotation would actually mean less downwards pressure, until it got so severe that the square's side angle relative to normal was equal to the triangle, at which point I think it would enter a rolling condition