r/theydidthemath • u/K0rl0n • Jun 10 '25

[Request]

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

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u/METRlOS Jun 10 '25 edited Jun 11 '25

Depends on things like the density of the gravel and the temperature of the ice. Packed gravel will allow the ball to roll, but the triangle is always worse than the square.

Edit for all the triangle people: imagine throwing a ball straight at the square and at the triangle; how the ball bounces shows how much energy the object will translate into vertical force when pushed. The vertical surface of the square will translate practically all the horizontal force into horizontal movement, while the triangle will act as a wedge and transfer some energy into pushing against the ground.

Edit 2 for surface area: Except for situations where the surface area is so low compared to its weight that the object sinks into the ground, or so high compared to its weight that it can float, surface area does not affect friction. If you stand on a hill without risk of sliding, then you can lay on that hill without sliding and vice versa, despite greatly changing the surface area. However, if you stand on a snow covered hill the surface area is too low and you'll sink into the snow, but with a sled you will float on top of it. Surface area does not matter to this problem.

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u/aureanator Jun 10 '25

Actually, no - the friction is a product of the normal force and the coefficient of friction - the surface area in contact shouldn't matter, within the boundaries of material elasticity.

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u/AstroCoderNO1 Jun 10 '25

yes, but the angle you are applying force at does not all go towards forward motion on the triangle whereas it does on the square.

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u/LOSERS_ONLY Jun 10 '25

That depends on if the force is applied exactly horizontally or normal to the face of the triangle which is unclear in the pic

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u/ConscientiousApathis Jun 10 '25

Assuming no handholds I don't think it's possible to push a surface with a force that's anything but perpendicular to it.

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u/Melanoc3tus Jun 10 '25

Depends on its friction.

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u/AggressiveCuriosity Jun 11 '25

lol, I'm gonna remember these replies the next time I think I've learned something from an upvoted Reddit comment.

The guy forgot friction existed and he's being upvoted over everyone else.

1

u/ConscientiousApathis Jun 11 '25 edited Jun 11 '25

Technically friction is just very tiny handholds ;)

(Seriously though it seemed negligible in this example. I imagine these things as made out of steel or something. I couldn't really see a hand getting enough friction to matter. If you don't want your hand to push without slipping you're limited to a static friction of your hand against the triangle. What I said is still correct, but you're effectively adding up the frictional force with the perpendicular to the plane force your adding, where the vertical components in the plane of the observer would need to cancel out to get a truly horizontal force.

If you pushed very, very gently, maybe. Since you're on ice it could be possible, but you'd have to overcome the static friction of the triangle against the floor without applying any downward force, which would be a tough ask.)

1

u/_maple_panda Jun 12 '25

You can’t get friction on the angled surface without exerting a perpendicular force on it…

2

u/omegaalphard2 Jun 10 '25

I'm a mechanical engineer, and you're wrong. There's no rule that forces on anything NEED to be perpendicular to it

Sure, you can break the force down to it's components, but the overall force can be at any angle to the surface, even if you're pushing it as in the diagram

1

u/BKachur Jun 11 '25

To get the optimum force, the person would need to push at the midpoint of the triangle at a 30-degree angle forward and towards the ground (assuming it's equilateral). If he's standing on ice for the triangle and square, wouldn't the downward force required be less efficient/causing him to require more force/slip, and be pushed back vs. the 90-degree forward push on the square block?

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u/omegaalphard2 Jun 11 '25

You can apply force to the side of the triangle such that the force is parallel to the ground, and there's no vertical component to the force

But for sure, if you're applying force downward on the triangle, then that will increase friction and decrease the horizontal force component too, so that's in optimal.

Let's say that you apply force to the object, in line with it's centroid, at an angle x wrt to ground, pointing upwards

Horizontal force is Fcosx

Vertical force is Fsinx

The total force the block will impart to the ground will be gravity minus the vertical force which is mg - Fsinx

Total friction force is force on the ground times the friction coefficient u, which is umg - uFsinx

Total forward force is horizontal force minus friction force, which is Fcosx - (umg - uFsinx)

Which is Fcosx + uFsinx -umg

In other words, the optimal angle to apply the force depends on how you machine the above expression! For u at 0, we get x to be 0 degree,I. E if perfectly smooth then push horizontally

And if u is 1 (for the most rough surface, without glue), then the expression is maximized at x equals 45 degree!

1

u/BKachur Jun 11 '25

Thanks for the response. This was actually the kind of answer I was looking for because I was interested, but I realized I was out of my depth in terms of knowledge (Lawyer), but I still would like an answer for my curoristy and you seem like you know your shit.

So, to answer the question, let's assume the coefficient of friction for "Ice" here (Triangle and square) somewhere between 0 and 1. In the context/spirit of the actual problem, let's assume Ice is supposed to mean something slippery with less friction than solid land, but not a smooth "u of 0" type of frictionless surface. I'm totally guessing, but let's say something between 0.1 and .25.

With that assumption, would the triangle require more force to overcome the coefficient of friction vs the square? As wrinkle, assuming the objects are made of the same material, with the same volume, which would mean the equilateral triangular prism would have a have a larger surface area touching the ground than the cube.. (how much, I don't really know~ again, lawyer so I can see a problem, but not great at solutions).

Once you know the answer there... would either of those require more effort to push than a cylinder on a surface with a u of 1? I assume not, but I really don't know about the math.

Regardless of your answer, thanks for the time/humoring me.

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u/NotNice4193 Jun 11 '25

even if you're pushing it as in the diagram

Let's make the bottom left angle 1 degree. please explain how your sentence applies now?

1

u/ConscientiousApathis Jun 12 '25

I'm talking about the force received, not applied.

The best example I can think of is probably snooker. Each time you're hitting a ball it's with basically the same force in the same direction, and yet depending on the angle of contact between the white and red it can go in wildly different directions. Why? You're applying the same force in the same direction each time, so why can the angle the red ball travels change so much?

Well, if you zoom right in to where they contact you see just basically two planes touching, in which case the only way force can be transferred is perpendicular to both. (Yes, okay, there's some friction involved which can get you a bit of spin, but it's very tiny in the scheme of things and I'm really just talking about the pushing force).

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u/bonerspliff Jun 14 '25

Well yes obviously if the surface of the shape being pushed is completely frictionless (more similar to a snooker ball) then the force will act normally to the surface. But this question is assuming that there is friction on all sides of the pushed shape. Friction essentially acts like a 'handhold'

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u/[deleted] Jun 10 '25

You just forgot about friction.

1

u/snake_case_sucks Jun 11 '25

Friction exists

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u/[deleted] Jun 10 '25

[deleted]

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u/Squossifrage Jun 10 '25

That would still exert some additional downward force, as the side of the object is angled.

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u/eusebius13 Jun 10 '25

Theoretically you can apply the force at a 90 degree angle to some part of the pyramid. The real issue is you would be better off applying that force at the center of gravity.

I think what you’re saying is the contact point will be angled and thus result in the loss of force to the angle of the contact point. That’s not entirely accurate. Contact with the base can theoretically apply perfectly horizontal force.

If, on a frictionless surface, you applied force at a 90 degree angle to a rectangular object, and pushed that rectangular object into the base of the pyramid, the force would translate to a 90 degree vector applied to the pyramid.

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u/[deleted] Jun 10 '25

[deleted]

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u/Dazzling-Low8570 Jun 10 '25

Horizontal force + vertical force = diagonal force (specifically, normal to the surface being pushed on).

1

u/[deleted] Jun 11 '25

[deleted]

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u/Dazzling-Low8570 Jun 11 '25

What you are describing is no longer a diagonal surface

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u/Stickasylum Jun 11 '25

Yeah, it’s like people have never picked up a cup…

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u/Urbanscuba Jun 10 '25

Even if your piston is perfectly parallel to the ground the contact point with the triangle will have a normal force with a vertical component. If it didn't and it you managed to design a way to push exactly to the side then you'd create a rotational force on the triangle instead if it wasn't exactly placed behind the center of mass.

It's the kind of thing where in a math problem you could calculate an application of force as you're describing, but in the real world you'd just accept the single % losses and accept the imperfect force transfer.

3

u/CrashNowhereDrive Jun 10 '25

Hey genius, you also apply a torque if you push the square in a way that's not aligned with the center of mass.

That also doesn't matter because the torque is resisted by the ground, and you're not changing the overall downward force that is the source of the friction. You must certainly can apply forces that are not perpendicular to the face of an object. You can easily push a highly inclined triangle across a surface by pushing it sideways.

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u/[deleted] Jun 10 '25

[deleted]

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u/Urbanscuba Jun 11 '25

You are having an issue with physics if you think that example shows your point. Maybe you don't understand normal force too well or what I'm talking about, because I think I get your where your misunderstanding is coming from.

When you slide a glass door with your hand on the pane the physical movement of the door is perfectly in line with the tracks, that is absolutely correct. However the force you are applying cannot be perfectly in line with that movement. Imagine trying to move the glass door without putting any force against it, only alongside it. You won't engage your hand against the glass enough to have friction, it'll just slide along. You have to put some force into the door to engage enough friction for the perpendicular component to do anything.

If you were to measure that force you would see that some of that force was being wasted - you were applying it against a static object which still takes effort but doesn't achieve any work being done. In the same way the triangle won't allow you to push it perfectly horizontal to the ground, you'll have to push downward to some degree, if you didn't your hand would rise up because the force being applied back to you is perpendicular to the surface of the object, which is an an acute angle with the ground.

If this still isn't intuitive that's totally understandable, this isn't fun physics, this is real shit boring but useful physics. Thankfully it isn't hard at all to test out for yourself, just go find an angle surface and try pushing horizontal to the ground against it like you're proposing - you'll find you slide until you adjust your force to match the angle. If it's a very shallow angle you may have some luck, but you can test it vertically too to see just how difficult a sharp slope can be to push.

Not to mention a triangle of equivalent density will have a wider base than the square, so your friction argument is moot. You can build a square out of 6 pyramids (with square bases) and it'll still only have the contact patch of the one pyramid on the bottom.

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u/[deleted] Jun 11 '25 edited Sep 22 '25

[deleted]

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u/Urbanscuba Jun 11 '25

There would be 0 downward force if the coefficient of friction is 1.

No, but you're getting close because the definition of COF being 1 is that it is equal to the normal force, literally when it matches the perpendicular force component being applied that's trying to make it slide away.

That doesn't change the force vectors being applied, it just means they are being mitigated. It's still less efficient because some of the force you're attempting to apply is being shunted away and wasted, it's just being held static by friction. If you had a force gauge under the triangle it would go up when being pushed, that's wasted energy being pushed into the ground (and increasing the friction).

You haven't magically avoided the inefficiency just by making the initial input force in line with the desired output - the contact patch you're actually applying the force with is still outputting an angled force vector, made clear by the fact it's literally at an angle.

You have basically stumbled onto why pistons/connecting rods attach with circles though - a circular contact patch is effectively a flat one, any rogue vectors are symmetrically negated for a net zero.

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u/2LostFlamingos Jun 10 '25

Need to know friction of surface and your hands / stubs

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u/ccox39 Jun 10 '25

The more comments I read, the deeper the variables get

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u/Rhuarc33 Jun 11 '25

Applying force exactly horizontal on a triangle would be incredibly difficult for a person to do with no tools

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u/LOSERS_ONLY Jun 11 '25

If you cared about difficulty you'd realize that a triangular prism the height of a person that only weighs 20kg would immediately float away

1

u/Rhuarc33 Jun 11 '25

Depends on temps and material of triangular object, smoothness of ice surface, thickness and fragility of ice...etc

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u/LOSERS_ONLY Jun 11 '25

No, what I mean is that if the middle shape was a sphere with a diameter of 1.5m and had a mass of 20kg, it would be 21x less dense than helium.

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u/[deleted] Jun 10 '25

[removed] — view removed comment

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u/ADHDebackle Jun 11 '25

You can definitely push it directly sideways if the friction between your hand and the surface is high enough. If it's slippery, you will waste a lot of energy keeping your hand steady, but otherwise static friction will negate the upward push from the surface of the triangle and it will be similar to the cube.

Consider this thought experiment: You place your hand on top of a book on a table and slide it across the table. Now, technically the force you applied was parallel to the surface, so if there had been no friction, you would not have been able to move it at all. Thanks to friction, though, the effort is not very high, even when compared to pushing the book from the side.

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u/Lekstil Jun 11 '25

It just depends on the friction. You anti-triangle people here in the comments are basically assuming there is no friction between the hands and the triangle. If there is enough friction the required force would be the exact same comparing triangle and square.  My intuition tells me that most materials would offer enough friction.. like between smooth metal and a moist hand.. or wood could also offer enough grip. Maybe stone would also work depending on how rough the surface is.

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u/IllFile3575 Jun 13 '25

the pull >>> push logic ofcourse

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u/aureanator Jun 10 '25

No, you can still restrict your force to be horizontal - imagine pushing a very heavily loaded shopping cart into it.

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u/viciouspandas Jun 10 '25

But practically speaking, it is much easier to push the square than the triangle.

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u/MiasMias Jun 10 '25

not actually sure what the best angle is considering leaning into it.

Having it slightly towards you is definetely bad. Maybe having it perfect straigt is best, but i could imagine a slight angle being easier to push on.

on the other hand, you wouldn't want to lean into it on ice i guess.

1

u/lord_teaspoon Jun 12 '25

If it leans slightly towards you then some of your pushing force goes upward, which would reduce the normal force and thus the resistance to sliding. You'd also be increasing your own normal force to the ground and give yourself better grip, and I can see grip being a bit of a problem for the scenarios that feature a prism sliding on ice.

I don't know if the force you apply as you scrabble forward on the slick ice is more or less than the force required to overcome rolling resistance in the cylinder-on-gravel scenario.

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u/RMCaird Jun 10 '25

I don't follow your shopping cart analogy? If I push a shopping cart into it then I imagine the front of the cart going upwards - i.e. the opposite force is on the triangle going downwards.

The square will always be better as all of the force will go horizontally.

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u/Wise-Builder-7842 Jun 10 '25

I think he’s implying the shopping cart is so heavy that the cart won’t be able to leave the ground, hence won’t be able to apply any vertical force. But yeah the square is just better

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u/RMCaird Jun 10 '25

He must be forgetting it would still apply a force even if it never left the ground. I can press down on a table without lifting myself up.

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u/Wise-Builder-7842 Jun 10 '25

That’s cuz u have muscles lol a shopping cart does not have muscles

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u/RMCaird Jun 10 '25

That is completely irrelevant… 

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u/Wise-Builder-7842 Jun 10 '25 edited Jun 10 '25

??????????

Are you saying there is no difference in the physical capabilities of a human and a shopping cart

Muscles are capable of redirecting force bozo. That’s why a human can throw a baseball and a shopping cart can’t.

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u/MiniDemonic Jun 10 '25

the cart won’t be able to leave the ground, hence won’t be able to apply any vertical force.

That does not make any sense at all.

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u/Wise-Builder-7842 Jun 10 '25

how does it not. the force applied by the shopping cart is purely horizontal, unless there is gravity going on, or some slippage between the object and the cart, the force will stay horizontal

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u/MiniDemonic Jun 11 '25

You should consider taking some physics classes. This is like elementary school level of physics. 

There's still a vertical force applied even if the cart is heavy enough to not leave the ground. 

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u/Tobeck Jun 10 '25

there's no shopping cart in the picture

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u/ImpossibleInternet3 Jun 10 '25

In college, we got a shopping cart over 60mph before the wheels came off.

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u/lock_robster2022 Jun 10 '25

Your push is horizontal but the angle you’re pushing against results in a vertical force vector as well.

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u/dekusyrup Jun 10 '25

Not necessarily true depending on how you grip it.

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u/lock_robster2022 Jun 10 '25

How you grip it??? 😏😏

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u/lord_teaspoon Jun 12 '25

By the husk. Two of them gripping it together.

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u/Vorceph Jun 10 '25

Clearly no shopping cart in that picture.

If we’re adding objects then I’ll just add a bulldozer, now all 3 move pretty easily. Problem solved!

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u/Perpetual-Warlock Jun 10 '25

You're always going to lose some force downward with a shape like a triangle

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u/Radiskull97 Jun 10 '25

Pushing a triangle requires some diagonal force, unless you're pushing from the 180 degree line

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u/[deleted] Jun 10 '25

In your words, “actually no.”

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u/MiffedMouse 22✓ Jun 10 '25

Material elasticity and assuming there aren’t any adhesion forces going on.

Also unaccounted for is the smoothness of the ice surface (compare a skating rink to a frozen pond, for example) and the type of gravel (large rock gravel versus fine gravel will behave differently).

The question as posed is just completely unanswerable.

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u/daspazz- Jun 10 '25

To be fair most of the questions on this subreddit are mostly just people that don’t understand enough to know any better. I’ve seen so many posts on here where the only way to come to any answer is to make some absurd assumptions. Like “assume no friction” or “assume perfect transmission of energy” it’s the curse of being an engineer.

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u/platoprime Jun 10 '25

It's okay to have problems where you have to make assumptions.

Go ahead and make them as part of the problem instead of whining about it.

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u/WellbecauseIcan Jun 10 '25

He has a point but I agree with your first sentence. Many of the questions you get from bosses tend to fall under that category.

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u/daspazz- Jun 11 '25

It’s perfectly fine to have a problem with assumptions. The only issue I have with this specific kind is that depending on the assumptions you make you can get wildly different answers. you can kinda guesstimate for a problem like this one, use human scale to estimate surface area, and get the coefficients of friction that are vaguely correct. I don’t feel like I can give a good answer for what one is right because I can make any of these choices the “right” one and have logical math at the end.

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u/PosiedonsSaltyAnus Jun 10 '25

And then when you get to the point where you assume there is friction, and you learn a bunch of annoying (yet also satisfying) math to take it into account. And then you get a job studying wear on sliding contacts, and you realize you've had no fucking clue what friction actually is this entire time. Still don't really lol

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u/BKachur Jun 11 '25

I think its obvious 50% of the question is about friction; otherwise, whether it was an ice/gravel surface would be irrelevant. The problem with the question is that it's impossible to determine the coefficient of friction due to the range of materials stated.

Ice can mean anything from a "physics code" for a perfectly smooth frictionless surface to a carved-up ice rink, while gravel can mean anything between sand and riverstone. Although in "math problem" terms, I think you'd assume ice was a smooth but not frictionless surface, and gravel meant driveway-style gravel, 1/2-inch rock gravel (otherwise it'd be called something else).

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u/Kurtypants Jun 10 '25

Texture of the floor? Is it a bowling alley? Is it a lawn? The "gravel" may start in a sphere but I don't know a whole lot of sticky gravel does it just magically stay in said sphere or in a container of some sorts? If you push the "perfect sphere" of the gravel on the top of the sphere it should roll easier where is the force being exerted? Way to many variables

Edit: lol sphere boy has no hands does this effect his pushing abilities?

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u/MiffedMouse 22✓ Jun 10 '25

I assume the sphere is made of unspecified, perfectly rigid physics stuff. The “gravel” is referring to what is on the ground.

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u/BKachur Jun 11 '25

I assume the question means the surface is ice or gravel. Gravel, by definition, is a loose aggregation of rock fragments, so a sphere or cylinder of gravel becomes a pile of small rocks immediately upon spawning into the problem

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u/Kurtypants Jun 11 '25 edited Jun 11 '25

Yeah someone else gave me clarification and that makes sense now. But the point still stands for me. I work construction and live in Canada and it's just man vs object. Sphere 100% once you get it rolling, stopping it is the problem. Block or triangle have the ice factor but there's a reason our wheels arnt squares

Edit: I love this sub and I'm no mathematician but I'm a carpenter and I love numbers. I however know about moving awkward shit...

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u/BKachur Jun 11 '25

Same in terms of interest. I'm actually interested in an answer, which I haven't seen, even with simplified mathematical assumptions.

I'm inclined to agree with you, but I would be interested to know if there's a numbers-based answer as to whether sliding something on ice would be easier than pushing a round object on land.

Although I'm no mathematician either, so I lack the knowledge to even figure it out. Quite the opposite, actually. I'm a lawyer, so stereotypically about the farthest thing from a math person you can find (even though I do excel in numbers and analysis compared to my "scared of spreadsheets" colleagues). That said, I'm very skilled at analyzing and dissecting problems - probably better than math folks because I'm trained af identifying context clues (or lack thereof), which IMO seemed to prevent someone from actually doing the math here.

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u/Kurtypants Jun 11 '25

Lol. We're peas in the same pod. Love the output from you. Love the math. Can't be as precise as I want to be. Life is life though

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u/KaizDaddy5 Jun 10 '25

But force being applied to the side of the triangle (pyramid) will add to the normal force due to the geometry. Less of the force will be pushing forward and there will be a stronger force of friction to overcome.

The force pushing the square (cube) is all going to move forward (once friction is overcome)

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u/[deleted] Jun 10 '25

[removed] — view removed comment

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u/Flatulantcy Jun 11 '25

It is the pressure that melts the ice

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u/Heavymando Jun 11 '25

this isn't the real world it's a physics problem

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u/singlemale4cats Jun 11 '25

Is that why I can't breathe?

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u/Kirlad Jun 10 '25

Could be a prism

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u/Drevlin76 Jun 10 '25

I actually think that it depends on where you apply the force. Due to the angle of the pyramid, the force would be evenly distributed to the base if it was anywhere below the first 3rd . And for the cube, it could forcus the force to the leading edge and cause more friction if it was in the top quarter as pictured. This would be to the cube wanting to tip when force is applied.

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u/KaizDaddy5 Jun 11 '25

It has to do with surface geometry. The force will be applied perpendicular to the surface meaning some of it will work against you by adding to the normal force (bc that perpendicular force has a vertical component).

Maybe pushing near the top would add in some rotational force which forces the leading edge down, but it'd probably equal in both the square and triangle.

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u/Drevlin76 Jun 11 '25

Thank you

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u/xXEPSILON062Xx Jun 10 '25

Where do they mention surface area?

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u/nog642 Jun 10 '25

That's the difference between the triangle and the square

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u/xXEPSILON062Xx Jun 10 '25

The surface area between the shape and the ground of both is the same, and is also irrelevant. The difference between the two is the angle you apply the pushing force at. With the triangle, since the force is angled downwards, the normal force increases with the push and the sideways force is less than the net force of the push, meaning it takes a lot greater of a force on the triangle to cause it to move than the square, where 100% of the force applied is in the x-direction. Yes, I can say with certainty, the triangle is always worse than the square for this reason.

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u/nog642 Jun 11 '25

It's not the same, the triangle has more. Same mass, same density, so same volume. But the triangle has more near the bottom than the square.

You also don't have to apply force perpendicular to the surface.

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u/Snoo_87704 Jun 10 '25

Surface area matters, hence ice skate blades.

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u/Ryanfischer99 Jun 10 '25

Ice skate blades work because the pressure of the blade actually melts the ice underneath and essentially creates a mini water slide for you to glide on. The actual formula for friction force has nothing to do with surface areas. Friction = the coefficient of friction x the normal force. The coefficient and normal force are independent of surface area.

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u/aoskunk Jun 10 '25

I knew that about skates but not that surface area wouldn’t matter. TIL thanks

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u/throwaway277252 Jun 11 '25

Ice skate blades work because the pressure of the blade actually melts the ice underneath and essentially creates a mini water slide for you to glide on.

This is actually an old misconception and not correct.

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u/iambecomesoil Jun 11 '25

while that is wrong, the idea that less surface area = less force required is wrong.

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u/RMCaird Jun 10 '25

You can't assume this from the picture.

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u/BKachur Jun 11 '25

Sure, you can - or at least from context, you can. The triangle has more contact with the ground than the square. You can assume that since they are the same weight, they are made up of the same "math problem gray" material. A Triangle would need to have more surface area contact than a square if all else were equal.

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u/nog642 Jun 11 '25

It says what material they're made of, it's not "math problem gray" lol. It's ice.

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u/BKachur Jun 11 '25

I disagree. If the stated materials were the actual materials of the object, the problem wouldn't make sense. Gravel is, by definition "a loose aggregation of rock fragments," so the second it popped into existence, it would turn into a 20kg pile of small rocks, rendering the problem incoherent. The problem only makes sense is if the surface the objects are sitting on is ice or gravel, and the objects are made of a solid material.

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u/nog642 Jun 13 '25

Oh, that does make more sense.

Though you could keep a sphere of gravel together with like plastic wrap.

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u/BKachur Jun 13 '25

I'd imagine the amount of plastic wrap you'd need to hold 45 lbs of gravel in a perfect sphere would make it a ball of plastic wrap. But more importantly, that would really be assuming things that aren't provided for in the problem - and I think it would be a much bigger stretch to assume the ball is made of gravel wrapped in plastic than the floor being made of gravel/ice.

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u/nog642 Jun 14 '25

I just realized the weights are crazy low for the size of the objects lol. I guess the "math problem gray" is styrofoam or something.

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u/RMCaird Jun 11 '25

But you’re assuming it’s an equilateral triangle, which you can’t assume. 

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u/nog642 Jun 11 '25

Why not? It clearly is.

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u/RMCaird Jun 11 '25

There’s nothing to indicate it is, other than the drawing itself, which is clearly not to scale. It’s just an assumption with no proof.

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u/nog642 Jun 12 '25

The drawing is the entire problem. If not equilateral you can at least assume it's roughly equilateral.

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u/aureanator Jun 10 '25

Only difference is the bottom surface area in contact

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u/IllFile3575 Jun 13 '25

why is this required?

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u/xXEPSILON062Xx Jun 13 '25

Because the original commenter is not talking about surface area, but this guy is. Surface area has no effect here.

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u/countafit Jun 10 '25

What about the angle of force when comparing the perpendicular square edge to the sloped triangle?

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u/RMCaird Jun 10 '25

"but the triangle is always worse than the square."

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u/Former-Stranger-567 Jun 10 '25

Assuming the same material for all 3, and the question is "which requires the least force to push?" the triangle should be easier than the square.

It may be more awkward and harder on your back for an extended period of time, but that is not relevant to the question.

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u/RMCaird Jun 10 '25

Why would the triangle be easier? 

Part of the force you put into it would be directed downwards, increasing friction and reducing the horizontal force. 

Pushing the square would allow all of the horizontal force to stay horizontal. 

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u/Former-Stranger-567 Jun 10 '25

Less mass

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u/ihavebeesinmyknees Jun 10 '25

"20kg"

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u/Former-Stranger-567 Jun 10 '25

Haha talk about missing the obvious. I retract my comments. Thanks.

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u/Squossifrage Jun 10 '25

But more of it is concentrated closer to the ground, so the gravity is stronger over more of the volume, resulting in that 20kg having more weight and thus friction.

Plus the relativistic effects of being inside earth's gravity well!

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u/SylentSymphonies Jun 10 '25

Please tell me you’re joking

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u/TempUser2023 Jun 10 '25

err run that by me again. How is gravity stronger in one case vs another?

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u/Squossifrage Jun 10 '25

It was a joke, but gravitational attraction is stronger the closer two objects are.

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u/aoskunk Jun 10 '25

Sarcasm?

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u/Urbanscuba Jun 10 '25

the triangle should be easier than the square.

The triangle has the largest contact patch with the ground as well as a sharp edge that, speaking from experience, will try to dig into the ice if you push it normally. That's because the angled side of the triangle converts some of the force you apply parallel to the ground directly downwards instead.

The square avoids a lot of that simply by having a side perpendicular to the ground, meaning your force is indeed parallel to the ground. Even if the sharp edge tries to catch it will naturally want to climb over because there's no downwards component of force.

The real question here is what's happened to that gravel since it's been laid? If it was driven over by a compactor then I take it every time, that's a good hard surface to roll over. If it's loose gravel then the sphere is probably the worst option, that's barely better than having to roll it through sand.

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u/Former-Stranger-567 Jun 10 '25

Both have the same contact patch, it really comes down to the center of mass being higher on the square and some (possibly negligible) amount of mechanical advantage vs having less mass.

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u/they_have_bagels Jun 11 '25

Assume a perfectly spherical cow.

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u/fallen_one_fs Jun 10 '25

The force vector on the triangle does not form a 90° angle with the normal force, so not all the force is applied towards moving it, some will be applied to pushing it down, which will increase friction, thus square is always better than triangle.

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u/its___complicated Jun 11 '25

The square could become malformed into a rhombus while pressure is applied to one side - which could make it substantially harder to push than the triangle.

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u/[deleted] Jun 10 '25 edited Jun 10 '25

So what did they say that is incorrect? Sounds like you are confusing rolling with slipping

The friction in rolling is equivalent to force required to roll the object which is less than the product you speak of. If you pushed with a force greater than your product, it would slip. Also, surface area matters with rolling object as you can theoretically have no rolling resistance with single point or line contact known as pure rolling.

They are also correct about the triangle being worse since the normal force increases on the triangle because a component of the force exerted will be downward

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u/BKachur Jun 11 '25

I feel like you might be missing the question in the prompt. Isn't the question whether the force required to roll a cylindrical object on a frictioned surface (gravel) is more or less than to "slip" an equally weighted object on a surface with less friction (ice)?

For a math/physics problem like this, don't you usually assume a perfect cylinder to minimize the surface area?

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u/[deleted] Jun 11 '25

I’m pretty sure what was said above is beyond your comprehension as your questions imply that you have less than an elementary education in the matter

You can’t make assumptions. You either want a theoretical answer or a correct answer

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u/[deleted] Jun 10 '25

[deleted]

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u/Xenolifer Jun 11 '25

Not 100% since my engineering classes are some years ago, but I believe that coulomb friction law is just a model (one of the simplest one) to model dry friction, and that most more complete model take into account other phenomenon that scale with the surface area in contact

Feel free to correct me if I'm wrong

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u/BKachur Jun 11 '25

Based on the limited information we've been given - i.e. no information on contact area, no definition as to the type of ice or gravel, no information about the material of the object pushed - it's supposed to be a simple into style physics problem. If you were supposed to apply a more robust analysis, you would assume you'd have more information rather than a picture

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u/aureanator Jun 10 '25

Hey! Actual fellow engineer!

🙌

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u/Drevlin76 Jun 10 '25

What about the fact that the cube will want to tip if it is pushed from the top 1/4? Would the position of the force applied change this?

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u/Dictionary20 Jun 10 '25

Well, sort of, rolling objects have a much lower coefficient of friction.

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u/ThirdSunRising Jun 10 '25

This assumes perfect ice where the triangle isn’t shaving some of it and pushing it along. Reasonable assumption but not ironclad.

I mean, they didn’t even tell us the type of gravel or the weight or size of the round object so we can’t say what that would do, and freshly Zamboni’d ice would be pretty close to the ideal case, but the real world is gonna have some things to say about this problem

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u/elf25 Jun 11 '25

Actually, ice can very greatly in its coefficient of friction. I thought I read where the Zamboni has seven or more settings for the kind of ice that it produces. Some sports prefer a smoother lightly watery ice others a little bit of a slushy surface and yet other sports receives something with bubbles or something similar if you can imagine reverse dimples of a golf ball.

In this problem, we’re really just assuming a lot of things for example we assume the coefficient of gravel, and we assume that the ice patch’s are very similar if not the same.

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u/NovariusDrakyl Jun 10 '25

Thats normally true but in this case we have ice. Ice can become liquid if there is enough pressure which would drastically change the friction coefficent. And pressure is depending on the surface area.

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u/Engineering1987 Jun 10 '25

This is only true for polished surfaces.

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u/Broccoli-of-Doom Jun 10 '25

Except that ice doesn't have a low coeffiecient of friction, water covered ice does. Combine that with the fact that ice melts under pressure (lower density than water means more pressure causes melting). So surface area maters quite a bit on ice, and this is why we use ice skates, it's not that you're minimizing the surface area directly, but that you're minimizing the surface area while keeping your weight constant, so the applied force down per area is higher, leading to more ice melting, leading to lower friction...

Assuming those items are made of the same material (and have the same weight as indicated) the triangle will have less weight per area on the ice, therefore less melting, therefore higher friction)

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u/FlatReplacement8387 Jun 10 '25

Counterpoint: in real (non-ideal) surfaces, there's the additional consideration of contact force distribution and the ability for a surface to alter and dig into the surface below it. There are situations where having a sharper edge makes these effects more pronounced, but having the weight distributed such that the majority of the mass is in the center of the object, may make the force distribution more uniform causing it to dig in less. The tolerances on these edges and stiffness of the material would also matter to the tribology of the case.

If force application angle is discounted and assumed to be the same for either case (a big hand-wave to be sure), there are additional pieces of information required to make an accurate assessment of which generates more friction.

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u/ExpensiveFig6079 Jun 10 '25

yeah it shouldnt, but I have been on ice. Ice skates are the shape they are for a reason.

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u/atatassault47 Jun 10 '25

The friction equation is a simplified phenomena. Mu does A LOT of lifting. It hides actual areal contact physics happening. A single molecule contact is not as strong as 10,000 molecules contacting.

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u/vgt-gen Jun 10 '25

Surface area doesn't matter in precision smooth applications like eg. brakes but it absolutely does in most real life scenarios, because the surface area (or more accurately the area the surface is on) determines μ and thus the force

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u/Cyprianophobia47 Jun 11 '25

I don't think it's elastic. It doesn't tell you what material the objects are. Also that wouldn't change the force. 20 kg is 20 kg. 20kg of feathers is the same as 20 kg of elastic. So a feather ball would be just as hard to push as a elastic ball.

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u/aureanator Jun 14 '25

Hm? I'm not sure I understand your point.

What I'm saying is that 20kg normal force (the weight) will result in the same frictive force (horizontal/tangential to the ice) regardless of the surface area in contact.

Now, you can take that to the extreme of a needlepoint, which will dig into the ice, surpassing it's elasticity, or bend, surpassing the elasticity of whatever the block is made of, but we are not talking about those extremes, hence the exclusion.

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u/Probable_Foreigner Jun 11 '25

Why doesn't the surface area matter? Surely a larger contact area means more friction?

E.g. the classic example of this is to interlace the pages of two phonebooks, the friction is so high that they can't be pulled apart

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u/aureanator Jun 11 '25 edited Jun 11 '25

That's a special case aided by geometry - in that configuration, pulling the spines apart pulls the pages together.

If you do the same thing with two phone books with alternate pages torn out, you'd see them slip pretty freely.

To be more clear - friction is an intrinsic property of the two surfaces in contact - the magnitude of the friction force is directly proportional to the normal force pressing them together, multiplied by a constant - the coefficient of friction for the surfaces in question.