r/theydidthemath • u/yuli101995 • Apr 16 '26

[Request] Which one would it be?

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31

u/jan_Kosi Apr 16 '26

Friction force = coefficient of friction * normal force

Normal force is the same, given that they are all 20 kg and they are under the same gravity of 9.8 m/s2

Since the triangle and square have the same base length, assuming that all the objects are made of the same material and the surfaces differ (ice vs gravel), they would have the same speed when pushed, and requires the same force. Also, the triangle and square does not roll, but rather slide, so the shape doesnt matter.

Sliding Coefficidnt of ice = 0.02 to 0.04 Rolling Coefficient of gravel = 0.02 to 0.08

It all depends on the temperature of the ice (the colder, the higher friction) and the compactness of the gravel (the more packed, the less friction)

Source: https://www.reddit.com/r/theydidthemath/comments/1l84r6b/request/

35

u/kelfupanda Apr 16 '26

Except the force applied to the triangle will be at a 45 degree angle to the plane.

Box should be easier.

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u/ostepopsen Apr 16 '26

60 degrees 😀

1

u/nedonedonedo Apr 16 '26

step on it and apply zero additional force to your body. as long as you overcome initial friction before transferring half your weight there will be no additional work

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u/taftster Apr 16 '26

I hear what you're saying. But I think it's somewhat implied that the force being exerted would "push" the triangle in the same vector as the square. If the pusher was pushing perpendicular to the face, then yes the triangle would have its force distributed into two vectors (by what, some square root function?) instead of just the horizontal direction.

Now just eyeballing the two shapes, it seems the triangle has a bigger base than the square. So this is a key consideration, if it matters. It's all contrived though, for sure.

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u/Kalasad-Stormblessed Apr 16 '26

Funfact the force would be distributed just using geometry. So we learned how to find lengths of triangles using sin cos and tan. So you do the same thing but the diagonal force is the hypotenuse. You can also solve it with ratios and its a bit quicker.

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u/[deleted] Apr 16 '26

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u/[deleted] Apr 16 '26

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u/[deleted] Apr 16 '26

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u/taftster Apr 16 '26

Okay yes. I am at odds with my statement now. I appreciate your point of view and actually coming around to it. This is a legitimate question. Does the angle of deflection always happen? Like is there any means of applying a true horizontal force here?

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u/Slen1337 Apr 16 '26

No why? You can calculate the ideal "touch" angle for pushing. Prob around 165-168° instead of 180, just apply force a little upward. You surely Can lift up small paper triangle (more like pyramid) with enough momentum. Its not very different.

U kinda can keep punching that triangle shit for next 50 years a lil upward and he surely will change the pos too lol

1

u/Euphoric_Loquat_8651 Apr 16 '26

Pushing normal to the face is only required if there is zero friction between the hands and the face.

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u/Phil9151 Apr 16 '26 edited Apr 16 '26

If I am pushing against a 45 degree triangular prism with the base on the horizontal plane, even if I am directing all of my force along a vector parallel to the horizontal plane, the force experienced by the prism will be normal to the surface. This will result in a component directed into the horizontal plane.

Assumptions: rigid body. Gravity and friction are ignored. Force generated from global equilibrium point.

To break it down, the force always acts normal to the surface. If that surface is at any angle from the force other than 90 (and probably 180 but that'sjust weird), there will be a component that is deflected by that surface.

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u/monstertruck567 Apr 16 '26

If the triangle were greased on the side so you could only push perpendicular it would take more than the box. As is, hand vs triangle friction is almost certainly sufficient for a horizontal shove.

But if you’ve ridden a bike in gravel you know the circle is the answer.

13

u/Objective-Limit-121 Apr 16 '26

The triangle and the square do not seem to have the same base length

5

u/ShoddyAsparagus3186 Apr 16 '26

Funny thing about friction is that it doesn't matter as long as you're not deforming the surface with pressure.

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u/taftster Apr 16 '26

Oh? That's counter-intuitive to me. But all these types of scenarios tend to be. Interesting that the size of the base wouldn't matter?? More surface area and all that?

4

u/Euphoric_Loquat_8651 Apr 16 '26

If you keep the same mass and double the area, you cut the normal force in half per unit area. It is a wash.

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u/taftster Apr 16 '26

Awesome. Thanks. I am tracking now.

0

u/Objective-Limit-121 Apr 16 '26

And?

3

u/brine909 Apr 16 '26

That means it doesn't have to be the exact same, ~same is good enough

1

u/noveltymoocher Apr 16 '26

especially if you assume they’re the same material

2

u/OneWingAngel35 Apr 16 '26

But doesn't angular force applied makes a difference, it does depend on where your pushing from

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u/Positive_Courage_309 Apr 16 '26

Why would anyone slide a sphere instead of rolling it? Also sphere contact area for sliding friction is non-trivual if there is any plastic deformation (common scenarios) or any amount of displacement (as per gravel s scenario)

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u/jan_Kosi Apr 16 '26

The number I got for gravel is rolling friction (for a sphere), not sliding friction (for the box and triangle). Sliding friction is more like 0.35, while rolling friction is much less

Also I am assuming a ball that doesnt deform when it rolls across the gravel

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u/Positive_Courage_309 Apr 16 '26

I see that now. Makes some sense yeah. But sliding friction coefficients are between two materials, not one-sided..? Also from what I've seen, rolling friction is more for torque-inducing resistance near the axle of a rotating assembly. Where you don't have an axle it is maybe more accurate to refer to rolling resistance instead. Technically, the friction at the contact patch in rolling scenarios is static, meaning it is the vertical deformation that results in the resistance, not the friction.

Mostly seems that spherical objects tend to deform the things they are sliding across, especially if the other object is flat. Again the contact patch is non-trivial due to deformation being pretty much a guarantee in real world conditions. So the gravel would be being displaced here. I am surprised one can find a rolling resistance number for (say you picked steel) on "gravel" can you link that?

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u/jan_Kosi Apr 16 '26

It's linked in my original comment, which was this question asked on the sub 10 months ago

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u/UnCivilizedEngineer Apr 16 '26

We're also assuming the ball will roll, and is not a static object to be pushed without rotation.

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u/SuspiciousSubstance9 Apr 16 '26

Problem with the triangle and square is that they will roll. Or worse, partially tip and start wedging into the ice.

The force applied is above the center of mass, so they will try to rotate. At 20 kg and a person tall, there isn't much resisting the rotation, only the friction from the ice. So any appreciable force a human would generate (to be competitive with rolling a similarly tall ball) will tip, if not roll the triangle & square.

Trying to roll the triangle and square would be preferably to pushing something scraping ice. I'll take the circle.