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Imagine a rectangular box placed at the tip of an iceberg; the box contains a man who can freely walk/jump/rotate inside the box. The man walks up to one of the corners of the box and starts jumping. Intuitively, the box must topple over thus bringing some momentum in the box+man system. The situation doesn't seem to violate any of the laws; how is it then possible for the man, an internal force, to bring any change in momentum (the net force at any time seems zero)?

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We shall consider the man to be at the middle of the box initially. If the man starts walking left, the box shifts towards right (assuming no friction) such that the centre of mass of the man box combined system remains on the tip. If the man reaches the end of the box and starts jumping, the box begins to oscillate about the centre of mass and lf cases are ideal he doesn't topple.

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The short answer to your question: Yes, they can. But in the particular example you are considering, they don't. As mentioned by Jahan, it is the gravity that gives the system (man+box) a net non-zero momentum. A rather interesting point to worry about in this scenario is 'who gives the system a net non-zero angular momentum?' Since gravity acts through the center of mass of the system, it can't provide the system any net angular momentum! The answer is 'the normal reaction force from the tip of the iceberg'. If the man stands in the middle then this force passes through the center of the mass of the system, but when the man moves the center of mass also moves. But the point of application of the normal reaction force can't change accordingly - it has to remain the tip of the iceberg. So it doesn't pass through the center of mass anymore and provides the system a net angular momentum.

There exist some systems in which only internal forces provide the system a net momentum change. Forces in such systems don't follow the third law of Newton.

Edit: I have assumed some friction to be present which will cause the horizontal position of the center of mass of the man+box system to displace from the tip of the iceberg. Since in the OP it was mentioned that the box topples, I considered it to be an implicit assumption that the friction is present.

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    $\begingroup$ What are those systems? $\endgroup$
    – Jahan Claes
    Commented Jun 7, 2016 at 4:26
  • $\begingroup$ Indeed, what are those systems? $\endgroup$
    – Swami
    Commented Jun 7, 2016 at 4:27
  • $\begingroup$ @JahanClaes Many electrodynamic systems have this characteristic. For the simplest case, consider two charges. Take the velocity of one of them towards the other and that of the second perpendicular to the line joining two charges. $\endgroup$
    – user87745
    Commented Jun 7, 2016 at 4:29
  • $\begingroup$ @Dvij but that's discounting the momentum of the EM field, yes? $\endgroup$
    – Jahan Claes
    Commented Jun 7, 2016 at 4:31
  • $\begingroup$ @JahanClaes No. I am not saying that the conservation of momentum is violated. I am just saying that the momentum of the system of charges (which are simply particles) increases on their own because the mutual forces are not equal and opposite. Of course, the reason this happens is that the fields take away the corresponding momentum and the net momentum of the world is conserved (even locally). $\endgroup$
    – user87745
    Commented Jun 7, 2016 at 4:34
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The net force isn't zero. There is an external force, namely gravity. If you did this experiment in outer space, where there truly was no external force, you wouldn't see the box topple over.

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  • $\begingroup$ Isn't it countered by the normal reaction of the iceberg on the box? Thus making net force along the vertical direction zero? $\endgroup$
    – Swami
    Commented Jun 7, 2016 at 4:09
  • $\begingroup$ @Swami No, if the guy is off center, the normal force is less than the gravitational force. $\endgroup$
    – Jahan Claes
    Commented Jun 7, 2016 at 4:22

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