Here is a man in a box with wheels on a frictionless surface.

enter image description here

Case 1 - The person constantly pushes the box while standing near its wall. Why doesn’t it move? (I know it is because of the 3rd law, but I am not able to understand the “why” here, please explain.)

Case 2 - If he runs inside the box will it move backwards? (Due to the force of his feet?)

Case 3 - If he runs from the right side to the left one, and just before colliding with the left side of the box, he jumps and while he is in the air, he collides with the box. Will it move?

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    $\begingroup$ If the surface is frictionless, why does the box need wheels? $\endgroup$ Commented Oct 19, 2015 at 12:03
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    $\begingroup$ If the surface is not frictionless, it's possible to obtain a non-zero net velocity using method 3. In fact, an even better method would likely be for the man to repeatedly walk from one end of the box to the other, and run back. Or just stand in place and shift his weight back and forth, slowly in one direction and quickly in the other. $\endgroup$ Commented Oct 19, 2015 at 13:39

5 Answers 5


Case 1 - The box will not move. Although the man is pushing on the wall, the floor is pushing on the man though his feet. These cancel each other out, meaning there's no net force on the whole system.

Case 2 - The box will move as the man walks. As he steps forward, the box moves backwards. How much will depend on the relative masses of the box and the man. If they weigh the same, the box will move as much as the man.

Case 3 - The box will come to rest. As the man starts running, both he and the box start to move, with their momentum equal and opposite. When the man runs into the side, the momentum cancels out again. It doesn't matter if he jumps or is still on the floor.

  • $\begingroup$ Case 2 ought to explicitly say that the box will only move while the man is walking. As soon as he stops (e.g., when he reaches the wall at the other end of the box) the box stops too. $\endgroup$ Commented Oct 19, 2015 at 13:28
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    $\begingroup$ Case 1 is simpler than that - the man is pushing forwards with his hands, using the floor of the same object (pushing backwards) as a counterweight. These cancel out. He's literally pushing and pulling the same thing the same amount. You don't need to go into the whole "the floor is pushing back" thing, which I find rather unintuitive and I'm sure I'm not the only one! $\endgroup$ Commented Oct 19, 2015 at 17:26
  • $\begingroup$ Also be clear that in Case 1 the assumption is that the person is completely motionless - just pushing. In a box on a frictionless surface, even just leaning forward or backward would cause the box to slide to restore its center of mass. $\endgroup$
    – J...
    Commented Oct 20, 2015 at 0:48

Wonderful answer by Matt. I would like to add a few points just for the OP to know.

Case 1: This follows from Newton's First Law that an external agent or force is required to move a body or, rather the center of mass of a system. Internal forces cannot do so and in this case, the man belongs to the man-bus system. So his interactions with the bus are internal forces.

Case 2: This follows from the conservation of linear momentum. If the man has mass $m$ and velocity $v$ and bus has mass $M$ and velocity $V$, then $mv=MV$.If $m<<M$, then $V$ will be very small.

Case 3: This also follows from the conservation of linear momentum.As long as the man is running, the bus also moves in parity with the aforementioned equation and just when the man stops, $v=0$ and hence $V=0$ as well.

Case 2 and Case 3 can also be explained from Newton's third Law that as the man is running forward (right to left, that is), he is exerting a backward force on the floor of the bus. Thus the bus moves backward. And by Third Law, the bus gives a forward reaction to the man and hence he moves forward. And the rest of the explanation follows.


The center of mass of the the man-box system, which I'll abbreviate as $C$, is stationary relative to us. The man-box system isn't exchanging momentum with the outside world—in other words, there's no outside force acting on it. That means $C$ will remain stationary relative to us. This is enough to understand what happens in all three cases.

  1. When the man pushes on the wall, he doesn't move relative to the box. That means $C$ is stationary relative to the box. Since $C$ is also stationary relative to us, the box is stationary relative to us.
  2. When the man runs in the box, he moves relative to the box. That means $C$ moves relative to the box. Since $C$ is stationary relative to us, the box moves relative to us.
  3. When the man is running, jumping, and flying through the air, he's moving relative to the box, so the box is moving relative to us, just as in Case 2. When the man pacakes against the wall, he stops moving relative to the box, so the box stops moving relative to us, as in Case 1.

Case 1. If he is just pushing, he has nothing to brace against, therefore not having any power behind his push. It is unrelated to Newton's 3rd law. It's like in a theme park and you suddenly decelerate, you can't brace against anything.

Case 2. Yes just a little, to move a lot he could try walk back lightly but this method would take a lot of time to move anywhere.

Case 3. The problem is the box moves backwards as he is running and forwards when he collides with the box, so I am not sure which way the box will go. But it will definitely move.

  • $\begingroup$ can you explain case 1 please? $\endgroup$ Commented Oct 19, 2015 at 7:22
  • $\begingroup$ Case 1: Of course he has something to brace against. If he had nothing to brace against, he would not be able to push. The man in case 1 exerts equal and opposite forces on the wall and on the floor that he's standing on. Any time you exert a static force on something, you MUST exert an equal and opposite force on something else. In case 1, since the 'something else' is just a different part of the same box, the net force on the box is zero, and it doesn't move. $\endgroup$ Commented Oct 19, 2015 at 13:31
  • $\begingroup$ Case 3 and case 2 are the same. While the man moves relative to the box, the box will move relative to the floor upon which it rests. When the man stops (no matter how or why he stops), then the box stops too. $\endgroup$ Commented Oct 19, 2015 at 13:33

Case 1:

You can understand the fist point by thinking that you can not pull yourself up by your bootstraps. Maybe a better way to picture this point is to think at the result if you put your feet of the wall and your hand on the floor. This way you can understand that while you were pushing your hand against the wall your feet where pushing the floor in the opposite direction. Can you imagine any reason why there should be any difference between the two pushes? In that case, do you think that pushing your hand on the floor and your feet on the wall can change to motion?

The reason is obviously no, but I hope that this may help you to understand the third law.

Case 2 and 3:

An easy way to understand what happends in this case is to understand that the centre of mass of any system will not move unless there are some external forces action on it. Then, no matter how the man/woman/cat/hat/you/light/particle/energy is going to move, no matter if it moves using fancy gadgets of mysterious forces, as long as everything happens inside the box, the centre of mass of the box will not move.

Then let's try to explore this concept for your cases. Using our statement the only thing that you have to care is where is the man at the beginning and where he's at the end. Then in your last problem you can see that the box will be at rest, the only difference will be a shift in position in order to keep the centre of mass in the same position.

In the case where he's moving, at each moment the box will shift, keeping the centre of mass in the same position, he will eventually reach one of the wall and basically end like the third case.

Extra: What if the man start to run back and forth in a dumbbell shaped box?

enter image description here

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    $\begingroup$ That's a hell of a gif. $\endgroup$ Commented Oct 19, 2015 at 16:10

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