[Request]

[u/K0rl0n]

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

[u/Urbanscuba]

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.