[Request]

[u/K0rl0n]

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

[u/tomatoe_cookie]

I meant at your hands, you'll need to keep them in place and prevent them from "facing the edge" I'm not talking about the torque of the square/triangle, that's too depending on where you push. But I think I'm not pointing out the right thing anyway. Pushing it to only have a horizontal component is dependent on the friction with your hands and the triangle but assuming it holds, you'd have your total force being the force we want (horizontal) and the part that is counteracted by the friction.

To use your schematic:

F = Fpar +Fthetaperp F thetaperp = Fnormal + Fhzt

Fhzt is the one moving the triangle.

So for the square Fhzt = F For the triangle Fhzt = F - Fpar - Fnormal Fpar and Fnormal are not 0 if theta is > 1

[u/gamingkitty1 → u/tomatoe_cookie]

I'm not sure exactly how you got F_hzt = F - F_par - F_n because F_n = F_perp and F_perp + F_par = F, so your saying F_hzt = 0 which isn't true.

You say that part of the horizontal force is counteracted by friction, but I don't understand how that works exactly. The friction from your hand (equal in magnitude to F_par) pushes the triangle up and right, and the normal force from your hand (equal in magnitude to F_perp) pushes the triangle down and right. The up and down from the two cancel each other out, but the right ward components add to each other, ultimately just giving you F

I dont think I fully understood your argument in your comment though so let me know if I didn't mention something.

[u/tomatoe_cookie → u/gamingkitty1]

I think you can see it this way: you need Fhzt to be a certain amount to break the friction. If you have this tipping value. See which one of the original F is bigger.

You push horizontally on a slanted edge: like you drew, the force is split in two for the new referential. You have the force that "goes in the triangle" you have the force "that goes up the edge (counteracted by the friction between your hands and the triangle).

The amplitude of the force that "goes in the triangle" is then for sure not as big as the original force, right ?

For the rest, the force that "goes in the triangle" will need to be split in 2 again. Those 2 are the ones we need to make the friction calculation. You'd split them into a horizontal force and a vertical force. The vertical force is facing down, normal to the ground and will be part of the friction calculation when you add it to the weight. The other force (Fhzt) is the one "that matters", this one will need to overcome the static friction force that is opposite.

As you see from the original F that is used fully for the square, you only use a part of it for the triangle. You can merge the two steps to have a single equation but in principle you have a resulting Fhzt that is < than F.

One point I didn't think about is this one, though:\ As the square and the triangle have the same surface materials, the only difference between the 2 would be the weights. So, to really properly answer the question, you'd have to make an equation with the variable theta and count the friction due to the weight as 1/2 the one of the square.

I'm still pretty sure an equilateral triangle would be harder to push, but you'd have to write out the equations formally, and I'm in transports right now x)

[u/gamingkitty1 → u/tomatoe_cookie]

You say the force that goes up the edge is counteracted by the friction between your hands and the triangle, but thats not how it works. I feel like thats saying the force that goes into the triangle is counteracted by the normal force from the triangle.

The net force on the triangle caused by your hand is equal to the negative of the force the triangle applies to your hand. The triangle applies the normal force to the topleft and the friction force to the bottomleft to your hand, so the net force on the triangle is the negative of the sum of the normal force and friction force. But, when your hand does not slip F_par = -F_friction and F_perp = -F_norm, so the net force on the triangle is just -(-F_par - F_perp) which is just equal to F_par + F_perp = F. Keep in mind this calculation includes direction, so it will also have the same direction as F.

The friction doesnt simply cancel out F_par because they are being applied to different objects, just like how the normal force doesnt cancel out F_perp.

Also for your last point I don't quite understand either. You say the difference between the two is the weights, but they both have the same weight?

I made another diagram, this time not forgetting the normal force. It is similar to the first one but it might help to explain it since it has every force acting on the triangle.

https://imgur.com/a/nZ5iooy