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If I push on a table with some force 'f' then by newtons third law of motion the table also applies the same magnitude of force on me in the opposite direction.

I get that there may be various forces that would be acting on me but then those forces (like friction) would also be acting on the table then why is it that the table moves but I don't?

To summarize I am asking that if I push a table then why is it that i don't move but the table does?

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If I push on a table with some force 'f' then by newtons third law of motion the table also applies the same magnitude of force on me in the opposite direction.

This is a true statement.

I get that there may be various forces that would be acting on me but then those forces (like friction) would also be acting on the table

This is not a true statement. Newton's third law doesn't say "If I push on an object, then all forces acting on me must act on that object as well." If you push on the table, then the only force acting on the table related to you is that push (ignoring negligable gravitational attraction). If there is friction involved, then the friction force on you is not related to the friction force on the table.

why is it that I don't move but the table does?

This is due to friction. If the friction force between you and the floor is strong enough to stop you from sliding, then you will not move. If friction is not strong enough to stop you from sliding, you will move. The same is true for the table.

In fact, if you pushed on the table with both you and the table on a frictionless surface, both you and the table would move.

Essentially any of the four scenarios are possible (you move/don't move and table moves/doesn't move) depending on the applied force and the friction that is present between the various surfaces in question.

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You could of course be pushing on the table with nothing moving (yourself or the table).

Suppose you're standing on the ground pushing a relatively light table. There is friction between your feet and the floor, but also between the table legs and the floor. A simple model of friction tells us that its magnitude is related to the reaction force of the object on the surface. In other words, you have some friction on you given by $F_\text{you}$ (limited by $\mu R_\text{you}$), and the table also has some friction given by $F_\text{table}$ (limited by $\mu R_\text{table}$).

Again keeping things simple, if you position yourself such that you are pushing down on the ground more than the table (e.g. because you're heavier) then eventually after applying enough force yourself, you can overcome the table-floor friction and move it.

Now imagine you're sitting on a wheelie chair whilst trying to move the table. Now the friction between you and the floor has changed, due to the wheels on the chair (the details I'll leave for another question). Now if you try to push on the table, then (assuming it's not too light) you will move instead.

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  • $\begingroup$ Thank you for the answer. So you are suggesting that when I am pushing on the table, I am also applying some force on the ground which is in turn increasing the force of friction on me and is hence preventing me from moving.. right? $\endgroup$ – Karan Mehta Nov 16 '18 at 14:08
  • $\begingroup$ @KaranMehta Yes. Another thought experiment you could do is to picture yourself and the table in space. If you push the table, then to conserve momentum, you must also be pushed in the opposite direction. This doesn't happen on Earth because the ground (via friction or reaction forces) stops you (and in fact you then change the momentum of the Earth) $\endgroup$ – Garf Nov 16 '18 at 14:11
  • $\begingroup$ @AaronStevens Yes, it's just a very simple treatment of the problem. The fact is there is still friction present (regardless of how you want to model it) that will stop you from sliding. $\endgroup$ – Garf Nov 16 '18 at 14:12
  • $\begingroup$ Sorry yes I see what you mean now. My mistake. $F=\mu N$ in that simple model is indeed the "limiting friction". $\endgroup$ – Garf Nov 16 '18 at 14:17
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    $\begingroup$ I've edited my answer to correct that $\endgroup$ – Garf Nov 16 '18 at 14:19

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