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Suppose I stand on a friction less floor, and another object of finite nonzero mass stand in front of me. Can I push the mass, so that it has a nonzero acceleration? Also, where does this force arise, since I cannot push the ground, since that would make me slip on the floor.

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Absolutely you can apply a force on some other object and make it accelerate. What will happen is that you too will accelerate - in the opposite direction. You push the thing, the thing pushes you, it's classic Newton's Third Law. This is akin to rocket propulsion.

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    $\begingroup$ An ice skating rink is sufficiently low friction to try the experiment is the object is sufficiently massive (e.g. a similar-sized friend). Skating rinks should all be opened up for science occasionally $\endgroup$ – Chris H Mar 30 at 15:48
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    $\begingroup$ In other words, objects will move away from each other but the center of mass will remain where it as since it is not acted upon by an external force. $\endgroup$ – wander95 Mar 30 at 19:43
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    $\begingroup$ I've skated on ice quite a bit @ChrisH - If you push something like a net or a friend, you will both be pushed back (Newton's Third Law ofc). However, Newton's First Law also plays a big part. Objects at rest want to stay at rest and the object that has more inertia that needs to be overcome will tend to move less. $\endgroup$ – VirxEC Mar 31 at 11:32
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    $\begingroup$ @VirxEC yes, I gave up trying to skate years ago, but recall trying it. With an outdoor rink you could use GPS to measure the speeds you and a friend (masses known) get and match theory pretty well if the force and all skate blades are parallel; you could also work out the friction. $\endgroup$ – Chris H Mar 31 at 11:36
  • $\begingroup$ Not in terms of the Question, you can't… What meets another object can of course apply a force against it but that has nothing to do with the Question. Feynstein clearly Asked about pushing something while standing on a frictionless floor. If you try that, neither you nor the pushed object will move. Where any force arises is in the pushing, not the result. You slip on the floor because you can push the object $\endgroup$ – Robbie Goodwin Apr 3 at 22:52
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You can apply a force to the object, but since you are on a frictionless surface, you will slide backwards.

In your example of standing on a frictionless surface and pushing another object on that same surface, you exert a force $F$ on the object and the object exerts a force $F$ of equal magnitude and opposite direction on you, according to Newton's third law. See the figure below.

Since you and the object are on a frictionless surface, there is no force opposing the force $F$ that you and the object exert on each other. Therefore there is a net force on each of you of $F$ and per Newton's second law you and the object experience an acceleration of $F$ divided by your respective masses.

So if your mass is $M$ and the object's mas is $m$, you experience an acceleration of

$$a_{M}=\frac{F}{M}$$

and the object experiences an acceleration of

$$a_{m}=\frac{F}{m}$$

Think of you and the object being skaters on an ice rink which is very low friction. What happens if you push against each other?

Hope this helps.

enter image description here

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  • $\begingroup$ This is a good answer, except for the phrase “equal and opposite force”. Vectors in opposite directions are not equal! $\endgroup$ – Brian Drake Mar 30 at 9:57
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    $\begingroup$ @Brian Drake how about “of equal magnitude and opposite direction”. See edit $\endgroup$ – Bob D Mar 30 at 10:26
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    $\begingroup$ @Flater This assumes that there are no other forces (or that any other forces sum to zero). I would refer to myradio’s answer, which covers this in detail. $\endgroup$ – Brian Drake Mar 30 at 11:45
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    $\begingroup$ Personally, I prefer the original "equal and opposite" phrasing. This wording is commonly used phrase in the sciences and it is understood that equal refers to magnitude, either due to context (equal and opposite is nonsensical if equal includes direction) or inexperience (i.e., not realizing that forces always have a direction). There's some value in forcefully specifying that direction is relevant when talking about forces, but calling it out explicitly adds extra cognitive load to an answer that was understandable without it. $\endgroup$ – Brian Mar 30 at 14:57
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    $\begingroup$ @BrianDrake If "æquales et in partes contrarias" was good enough for Newton... $\endgroup$ – user3067860 Mar 30 at 17:41
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To add to the other answers:

The reason you ask this question might be that you intuitively and from other physics teaching know that you can't move anywhere on a frictionless surface. That you will never be able to set off and move away from your starting point.

This is true.

Still, though, while you can't move as a whole, you can move your arms and legs about. Because they set off not from the floor but from your torso (when you stretch your arm, a force pushes the arm out while another but equal force simultaneously pushes the body in the opposite direction). Together your limbs of your body constitute a system - it is the system that as a whole can't move (there is nothing to set off from), while the individual parts can.

And a system is represented by its centre-of-mass. It is the centre-of-mass that remains stationary and can't move anywhere. Points of your body can move about by pushing off from each other, as long as an equal "portion of mass", so to say, simultaneously moves in the opposite direction.

In your present scenario you are able to push the box forwards by simultaneously pushing yourself backwards (the box sets off from you, and you set off from the box), so both you and box are able to move away from your starting point. That is because you and box are two parts of a system. Your combined centre-of-mass remains stationary.

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As others said, indeed you can. This is a typical exercise in Newtonian mechanics.

As @BobD said, it can be seen from Newton's third law aka action-reaction law. If your mass is $m_y$ and the mass of the object is $m_o$, both objects will experience a force of the same magnitude but in the opposite direction! The acceleration will magnitude will not be the same but it will be scaled with the mass,

$$a_{y}=\frac{F}{m_y}$$

$$a_{o}=-\frac{F}{m_o}$$

So far is what other including @BobD said (with the detail that the accelerations have opposite sign).

Now, this exercise typically is included in the unit covering conservation laws. If, as you said, there is no friction, we can assume there are no external forces to the system, hence, the momentum of the system consisting in you + the object 'o' is conserved at any time.

If you push the object, the forces that will act on both you and the object (and hence, the accelerations) will exist for a very short time (naturally, only while the contact exists). If we assume that both you and the object were standing still before you applied the force, looking only at the horizontal direction, the momentum before the force is applied is,

$p_o+p_y=m_o v_o+m_y v_y = 0$

which is to be conserved, so, if you apply the force to the object and then measure the constant speed you have after it, you can derive the speed that the object has (provided you know both $m_y$ and $m_o$).

This expression is true before and after the you apply the force,

$v_o = -\frac{m_y v_y}{m_o}$

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  • $\begingroup$ +1 for emphasising that the forces in Newton’s third law are in opposite directions! $\endgroup$ – Brian Drake Mar 30 at 10:00
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Just ignore the floor! If you push an object in front of you, by extending your arm, surely you and the object will have to move apart! So one or both of you will have to start moving! It makes no sense if only you move unless the other object is somehow fixed in place, since there is no critical difference between the object and you. (To see why, observe that the object could be another person, and you can replace your 'arm' by a mechanically extending stick in-between both of you.)

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Look up the impulse equation Force x time = change in momentum

then imagine what would happen to you when you push the object and think how conservation of momentum can still hold.

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    $\begingroup$ I think this answer needs more detail. I am not sure what it is trying to say, especially in the context of a question that asks whether it is possible to exert a force to begin with. $\endgroup$ – Brian Drake Mar 30 at 10:01
  • $\begingroup$ It looked like a homework question, the OP should think about it for themselves $\endgroup$ – John Hunter Mar 30 at 11:39
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    $\begingroup$ I see no comments on the question to that effect. And it is a bit late for that argument anyway, unless the question is deleted. $\endgroup$ – Brian Drake Mar 30 at 11:46

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