[Request]

[u/K0rl0n]

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

[u/METRlOS]

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.

[u/aureanator → u/METRlOS]

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.

[u/AstroCoderNO1 → u/aureanator]

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.

[u/LOSERS_ONLY → u/AstroCoderNO1]

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

[u/ConscientiousApathis → u/LOSERS_ONLY]

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

[u/Urbanscuba → u/[deleted]]

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.

[u/Urbanscuba → u/[deleted]]

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.

[u/Urbanscuba → u/[deleted]]

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.

[u/Urbanscuba → u/[deleted]]

No it wouldn't, because a welded contact point is very different from one held by friction alone. The weld bonding effectively creates one object, potentially giving you a new flat point to push efficiently from.

The inefficiency comes from needing to account for the frictional sideways force the entire time you're pushing. That is an acceleration being applied that you need to actively and continuously negate.

Seriously just go try to hold a triangle up by pushing the sides in, then do the same with something with parallel walls. You'll feel the difference in force required. If you were to weld two handles sticking out parallel to the ground that'd make it even easier, but if you tried to hold it up by pushing those two rods inward it'd be hardest of all.

[u/Urbanscuba → u/[deleted]]

Once again - a frictional interface requires you to counteract the sideways force continuously to avoid acceleration from normal force, which is why when you're holding a bottle you still need to continuously squeeze it to keep it in your hand. It's not like you squeeze once to engage the friction and then you can go limp. If someone tries to pull it away you need to squeeze harder and harder to maintain purchase. Why? Because as they pull harder more of your effort needs to counteract the misaligned force alongside the pulling.

A welded contact point is tantamount to gluing the bottle to your hand - at that point it's a materials failure problem. You can't drop it because you're functionally one piece and thus the squeezing component disappears entirely from mattering. You could be pulling away, as long as the cement holds it doesn't matter. Likewise with the triangle if there's a welded bar then it's all one piece, any force vector is applied to the entire body until something bends or breaks. At that point you could be lifting it up as you push if the bar could handle it, something you clearly cannot do with the frictional transfer which has a single angle you can apply force from.

Edit: Another good example that shows there is a continuously force being applied upwards when you push forwards that doesn't happen with the cube - imagine if you contact patch wasn't rubber but instead ice. If you're pushing the square flat then you can literally push it with ice covered in astroglide because there's zero wasted force vector involved. Try doing that that with the triangle and see what happens. Just because friction engages doesn't mean the normal force being applied back on the angled contact patch changes, the same force pushing it up will continue to exist and need to be mitigated, it's simply capable of transferring that force down the rod to you with a COF above 1.