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If I have object that is heavy, I can pull it with rope but cannot push it. Why? What breaks the symmetry of the system? I can push or pull anything if I choose, so why is the push possibility not possible?

Qmechanic
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Iriene
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    Is buckling due lack of compression strength and stiffness the answer you are looking for? Because if your rope was as wide as the object then you could push it with the rope as long as it wasn't too long. Conversely, you can use loose sand to push something but as soon as you pull it disintegrates. In that case there is compressive strength but no tensile strength. This is a strength of materials question. Your background would help here. – DKNguyen Dec 16 '20 at 15:22
  • @Iriene does this answer your question https://physics.stackexchange.com/a/583332/271783 – Ankit Dec 16 '20 at 15:26
  • But then why does it have buckling then? why do other things not do the same? If the atoms are completely symmetrical and there is no parity in nature apparently, then why is one direction "privileged" or am I not understanding? – Iriene Dec 16 '20 at 15:29
  • I think it might be useful to consider an even simpler system, if you are interested in the issue at the most fundamental level. Consider a mass connected via a freely rotating rod to a fixed point, subject to gravity. There are two equilibrium positions, one stable and one unstable. What is creating the difference? Buckling in compression is an instability, analogous to the condition of an upwards pointing pendulum. – Smerdjakov Dec 16 '20 at 15:32
  • @Ankit so it is because of the atoms repelling, that is all? Is it related to entropy can I also think of it? That there are more states with positions of the rope to the sides than there are along the straight direction? So that is why there is a "breaking of symmetry" of actions? – Iriene Dec 16 '20 at 15:37
  • @Smerdjakov but here I can see that gravity which pushes down prefers the down direction, yes? There is no symmetry in this case. – Iriene Dec 16 '20 at 15:39
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    @Iriene, yes you pointed out an unfortunate accidental difference. However difficult, you could look at the stability aspect only: if you pull a rope, any changes in its shape will cost energy: to stretch the rope, if deformable, and to push you in the direction opposite the one you are entering force. If you push it, and it gives way, the stretching energy will vanish, and you will perform work (move along the direction you are pushing towards). The situation is not symmetric at all. – Smerdjakov Dec 16 '20 at 15:41
  • @Smerdjakov I don't understand, but maybe you are saying something similar to the others? because the energy of the rope when pushing is deflected to the sides? – Iriene Dec 16 '20 at 15:54
  • @DKNguyen ah I see, OK. If the rope was wider then this would be possible. So this also relates to what Ankit has said too. I think I understand now. Thank you. My background is no formal training in physics, if that is what you mean? – Iriene Dec 16 '20 at 15:56
  • Yes. Because someone might slam you with a giant physics hammer in an answer. – DKNguyen Dec 16 '20 at 15:59
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    @Iriene, yes that is the case. When you pull, little changes in the rope shape cost energy: the rope is stable. When you push, little changes lower the energy: it is hence unstable. – Smerdjakov Dec 16 '20 at 15:59
  • @Iriene The bonds between atoms in molecules can be more or less rigid and allow flexibility, movement, and distortion. Might be better to look at the molecule or arrangement of atoms rather than just two atoms next to each other. Molecules generally aren't symmetrical. – DKNguyen Dec 16 '20 at 16:04
  • The linked answer in the close vote is quite useful. Let me add one bit of commentary.

    Suppose you have a stone rod. Clearly you can push or pull, because the structure is rigid, and more important, reasonably stable. Now make that stone rod, still a strong crystalline structure, extremely thin in cross-section. Extension force is still stable, whereas shear and bending forces have a tiny region (angular) of stability.

     – Carl Witthoft Dec 16 '20 at 16:45
  • PART2 When you pull something, at worst you are causing internal transverse forces (or the transverse component of the forces) to become more aligned with the longitudinal force you're applying. If you deliberately pulled and applied a bending force, you might break the rod.
    Now push the rod. This increases the chance that compression causes internal forces to go transverse, and break the rod. Further, it's hard to push directly along the axis, so any transverse force applied will break the rod (or collapse the rope)     – Carl Witthoft Dec 16 '20 at 16:45
  • @DKNguyen yes, I did not consider that molecules would break symmetry anyway, but would not even the exactly symmetrical atoms also still not push because of the above reasons explained by the others? – Iriene Dec 16 '20 at 16:56
  • @CarlWitthoft yes, thank you. I think this explanation also is related to what the others have said. – Iriene Dec 16 '20 at 16:56
  • @DKNguyen I think the answer to question I asked you is yes because of what you and others have said. Correct me if I am wrong. But thank you to you and everybody else, I think I understand the answer to my question now. – Iriene Dec 16 '20 at 17:04
  • @Iriene This is getting beyond me knowledge but the way atoms bond together is...not straightforward and of varying strengths and not necessarily rigid, if I am not mistaken. It's like a cloud of stuff with some give to it I think. It's not a stick stuck into a hole onto a side of a ball. – DKNguyen Dec 16 '20 at 17:10

1 Answers1

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I think this is the most complex answer possible to OP's question; I don't know if I should be proud or ashamed of myself.

I can pull it with rope but cannot push it. Why? What breaks the symmetry of the system?

The system has symmetries, but not the one you are thinking about, you are confusing yourself. The true symmetries arise from the following facts:

  1. The robe has fixed lenght $l$
  2. The rope is bendable but not stretchable

Think about it: you have an object and a rope attached to it, this divides 3D space in two regions: the first one is a sphere of radius $l$ and center in the position of the object; if you have the other end of the rope, the one not attached onto the object, in this region then everything is fine! In fact the rope is bendable!

But you cannot have the other end of the rope outside the sphere, because this would mean that the rope has broken since it has a fixed lenght $l$! This implies that if you try to get the other end of the rope outside the sphere then the center of the sphere must move to prevent you from getting into that impossible configuration! (Impossible without breakage or deformation)

Noumeno
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  • This sounds like a mathematician using logic to explain physics so he can sidestep the physics...because y'know...mathematicians don't care about the real world. – DKNguyen Dec 16 '20 at 16:11
  • I think I understand what you are saying with the regions and the possible motions in each region, but is not what you have said is that the rope is bendable because it is bendable? but I think what you have said is similar to the other answers because it is all about the possible motions in each region (straight and to the sides) and that there is an asymmetry here. – Iriene Dec 16 '20 at 17:01