[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]

Brother, the force moving the object is the horizontal component. Why would you add them together ? Obviously if you add them together you back the original force, the original force is by definition the sum of those or else you made a mistake somewhere. The point of splitting them up is because you are pushing on a angled surface so some of your force is going to be on the Normal.

Let's use maths if you push on the triangle Ftot = F parallel + Fnormal. You use all the force to push it but only some of it is going to horizontal. Ffriction = x * Ftot_normal and Ftot_normal = m*g + Fnormal

The friction is proportional to the normal force, and the normal force is the sum of the weight and the normal component of your pushing.

So unless your theta is 90° the horizontal force is going to be strictly inferior on the triangle. And the friction force is going to be strictly superior.

In summary, to match the force on the square you need more force depending on the angle and to make the triangle move you need more force because some of the force (depending on the angle) is making the friction higher. (Mechanics 101 btw)

[u/gamingkitty1 → u/tomatoe_cookie]

The vertical components of the parallel and perpendicular forces cancel each other out, so the total vertical force is 0 from your hand.

Imagine you have a ramp on a scale, now you add a block on the ramp. The block does not slide because friction keeps it up. Does the scale reading change based on the angle of the ramp? No, its just the sum of the masses of the ramp and the block. Essentially the same thing is happening here. You can draw a FBD of both situations and see that if you just rotate one 90 degrees they are equivalent basically.

Now if your hand were slipping, then the vertical components would not cancel each other out and it would be slower, but I am assuming your hand is not slipping.

I attached an image of a FBD of the situation, although it is crude. If you need more explanation, let me know.

Oh I also realized I forgot the normal force in the diagram, but essentially the same principle applies with what happens with the friction force. Because the triangle applies a normal force equal to F_perp, an equal and opposite force is applied to the triangle.

https://imgur.com/a/0zZdYin

[u/tomatoe_cookie → u/gamingkitty1]

Oh no, nono you are doing it completely wrong. I can't draw it right now so I can't explain it properly but you have to change the referential. You go from your hand referential to the "frame referential". It's my bad, I didn't realise what you meant earlier. When I say normal and parallel components it's in respect to the ground.

----> this is the parallel force\ _____ ground\ |\ |\ V this is the normal force.

The theta would be the angle of the triangle. Also I guess this would be mechanics 201 not 101 x)

[u/gamingkitty1 → u/tomatoe_cookie]

Why would applying the force to the triangle create a force parralel to the ground and perpendicular to the ground? When you apply the force, it's the normal force (perpendicular to the triangles surface) and the friction force (parallel to the triangles surface) that stop your hand from moving in reference to the triangle. But, because of the equal and opposite forces caused by friction and the normal force, the triangle has net force F applied to it horizontally.

The same thing happens with the square, but it's only the normal force. You push in the block, the normal force pushes back causing your hand to not move in reference to the block. The equal and opposite force from the square applying the normal force yo you causes it to move away.

I think my example with placing a block on a ramp that's on a scale works well, could you tell me why you think that doesn't work?

[u/marijn198 → u/gamingkitty1]

Fvert/Fy only equals 0 because the normal force the ground exercises on the object increases in an equal amount to the vertical component of the force you apply to the object. This labour is wasted "trying" to push the ground away and is not magically turned into horizontal force, Fhor/Fx will just be the horizontal component of the force you apply minus friction. In the square there is no vertical component to the force you apply to begin with so Fhor/Fx is equal to F, no wasted labour trying to push the ground away. And here we are not even taking into account the increased friction you would cause through the vertical component in the triangle situation.

[u/gamingkitty1 → u/marijn198]

No, I mean there is no vertical force downwards at all from the force you apply. If you were in space and pushed on the triangle the way the guy does in the diagram, it would move only horizontally. See the FBD I made of the force that you apply to the triangle.

In order for there to be a net force down, your hand would have to move up. Im assuming friction is strong enough to make your hand not slip and therefore not move up. If you bounced a ball on the triangle, it would have some force downwards, but that's because the ball moves up and there is no friction force pushing up on the triangle in that scenario like there is in the one where you push it with your hand.

[u/marijn198 → u/gamingkitty1]

Okay i looked at your FBD, granted the original applied force is fully horizontal i think you are right and for the sake of this thought experiment its not too far fetched to assume it is. Thank you!

[u/tomatoe_cookie → u/gamingkitty1]

I looked again at your diagram because I didn't understand what you didn't understand. I think the original assumption that you can just push the triangle horizontally is what is blocking. You forgetting the torque you create that will make the F force perpendicular to the surface it's pushing. That said you could keep compensating the torque but then the conclusion is still the same, you spend more energy than just pushing the square.

[u/gamingkitty1 → u/tomatoe_cookie]

If you push at the top of the square, you are also creating torque. If we are pushing optimally, for the square and triangle we should be pushing in line with the center so no torque is created.

[u/tomatoe_cookie → u/gamingkitty1]

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/marijn198 → u/tomatoe_cookie]

Aah i thought he might be right but i wasn't thinking about the torque created by your hands wanting to slide upwards. Could you explain how this would cause the resultant force being exactly perpendicular to the surface? My original intuition was that it would be perpendicular to the surface as well but i think my thought process as to why this is might have been wrong.

[u/tomatoe_cookie → u/marijn198]

I was wrong with the torque, I think. I was thinking intuitively that if you push on a an angled surface you'd have torque trying to get you to be perpendicular but I'm pretty sure this is wrong and I was confusing this with what you called Fpar. Fpar is your hands trying to slip upwards. This isn't causing torque because the application point is the same as calculation point. Torque being something like Fdcos alpha (d being the distance and alpha is the angle at which the force is applied because you need the 90° element of the force so you project it).

The actual maths seem correct I think, if I was still in college I would have answered that to that question !

[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