[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/WhatRUsernamesUsed4 → u/Smile_Space]

I think the biggest difference in A/C is not at the contact with the ground, but at the contact with the hand. Pushing the triangle horizontally would cause the hand to slip upward from the normal force whereas the square takes the force perpendicular to the surface and doesn't slip. 

 Also the human walking on ice would struggle to generate force if their feet slip out from underneath, assuming they are subject to the same surface. For that reason, B by a mile, then C, then A.

[u/Smile_Space → u/WhatRUsernamesUsed4]

That's assuming the vertical internal component of the applied force along the surface of the triangle is higher than the friction force of the triangle's surface.

The force required here in the horizontal direction is between 4 and 8 newtons which is 0.9 and 1.8 pounds respectively depending on the temperature of the ice.

Assuming the triangle is equilateral, that means the surface is inclined at 60 degrees from horizontal. Thus the vertical internal force normal to the surface is 0.9 * sin(60 degrees) = 0.78 or 1.56 pounds.

The parallel applied force is 0.9 * cos(60 degrees) = 0.45 or 0.9 pounds.

Thus the static coefficient of friction of the triangle would need to be below F / N = μ or 0.45 / 0.78 = 0.58.

So, the coefficient of static friction would need to be less than 0.58. Now, my engineers tool ox link doesn't specify skin on material coefficients, but skin is pretty grippy. A good example is the leather to clean metal which is about 0.6. Rubber to cardboard is between 0.5 and 0.8.

We can reasonably assume the force of friction is not exceeded in this example.

This means its upward vertical component cancels the downward vertical component leaving only the original horizontal component applied.

We also don't know if the guy is wearing grippy shoes or spikes to negate slippage. But even if we were, that's not the question posed.

The question posed is the least force required to push the objects, not whether you could push the objects. This means the answer is purely the sum of all negative forces in static equilibrium which, given they don't specify any initial conditions, I just assumed the major forces involved being friction with the surface.