I have asked this question earlier and got some conflicting answers in the beginning but then got the answer that the friction force is what accelerating the center of mass.

Here's the question

Take the example of a man (say of mass 50kg) and a tank (of negligible mass) filled with water kept at some distance (whose total weight is also about 50 kg). Now suppose there is a pipe of uniform cross section (whose mass can be ignored) connecting the mouth of the man to an opening at the bottom of the tank. Also the surface on which they are standing is sufficiently rough (so that they don't move).

enter image description here

At $t=0$ the system is released from rest. So the center of mass of the system is exactly in the midway at $t=0$ (represented by the dashed line). But as water comes out, the center of mass of the system shifts towards the man . This means that the center of mass accelerated from its initial position both horizontally and vertically.

But I am confused on this one again and I'll try to clearly put out my confusion.

@lineage mentioned that the thrust force on the tank and the friction force can be calculated as "$\rho v_{exhaust}^2 A$".

Now this was the force on the tank only and I think the force on the man in the answer was ignored.

But if we analyse the motion of the water once it reaches towards the man we can similarly say that the force on the man by the water is equal to the force on water by the man in magnitude which is in turn equal to its change in momentum per second (which is again ($ \rho A v_{exhaust}^2)$ given the fact that the tube is of uniform cross section.

This means that the water will push the man in the left direction and thus the man experiences a friction force in the rightward direction.

$\Rightarrow $ So we got two external forces from here acting on our system (man+ tank + water) :

1. Friction on the tank in left direction

2. Friction on the man in the right direction

But notice the coincidence that the magnitudes of both the forces are equal and thus we get zero external force on our system.

Now this is something very confusing and unclear to me because if there is no external force in the horizontal direction then how come the center of mass accelerate horizontally?

So once again can someone explain the stuff happening here ?

Note : Notice the fact I mentioned in the question linked above that the ground on which they are placed are sufficiently rough such that neither the tank nor the man accelerates anyhow.

  • $\begingroup$ Your analysis seems completely accurate to me and (to be honest) unsurprising. Can you elaborate on why you wouldn't expect the magnitudes of both forces to be equal? Or is your main question something else? $\endgroup$ Jul 1, 2021 at 18:06
  • $\begingroup$ @Michael Seifert hello. Finally someone responded. Btw I do believe the forces to be equal in magnitudes.. but if that happens there would be no horizontal external force on the system to make the center of mass accelerate horizontally.. is it clear now ?? $\endgroup$
    – Ankit
    Jul 1, 2021 at 18:08
  • $\begingroup$ That seems clearer. I don't have a good answer, but at least now I know what your question is. $\endgroup$ Jul 1, 2021 at 18:24
  • $\begingroup$ Can you include the relevant information in this post so it's all self-contained? $\endgroup$ Jul 2, 2021 at 1:01
  • 1
    $\begingroup$ @Ankit There is no maximum length a post can be. And I don't see why a longer post would be boring. I mainly suggested it because if something happens to the other question/link (for whatever reason) then this question becomes harder to understand. Also, I personally don't like digging through multiple links, questions, answers, etc. to understand a question, but that's a secondary point. $\endgroup$ Jul 2, 2021 at 11:50

1 Answer 1


Sometimes it is easier to imagine what would happen in friction were absent.

Let's put the system (the tank and the man) on a tray with zero friction against the ground.

As the water comes out of the tank, there is an unbalanced force on the tank. So the entire tray accelerates to the right.

When the water finally strikes the man, the forces are balanced, but that just stops the acceleration, not the speed. So the tray is moving slowly to the right.

When the water leaves the tank, it no longer has any horizontal force from the water, but some is still striking the man. So there is now a net force to the left which slows the tray. After all the water is gone, the tray has slowed to zero. There was a net acceleration for short periods of time, but not at steady state.

If we put friction back onto the tray, then the frictional forces are not in balance at the beginning and at the end of the flow. The forces on the tank and the forces on the man do not happen simultaneously. This non-simultaneity is responsible for the acceleration of the center of mass.

  • $\begingroup$ Good approach and good analysis. +1. I would recommend tying the tray scenario back to the original scenario at the end. Including the friction and the acceleration of the center of mass of the original system. $\endgroup$
    – Dale
    Jul 1, 2021 at 22:21
  • $\begingroup$ Yes, on re reading it the last paragraph is sufficient. Good answer! $\endgroup$
    – Dale
    Jul 1, 2021 at 23:42
  • 1
    $\begingroup$ @BowlOfRed so if I got your point correctly , the center of mass accelerates till the time water just hits the man and once it hits the man the net force becomes 0 but because of its initially gained speed it continues to go towards the man and once the tank is emptied till the mouth of pipe , there is now only one force which stops the center of mass.. Right ?? $\endgroup$
    – Ankit
    Jul 2, 2021 at 7:17
  • $\begingroup$ Yep. And that acceleration is driven by the unbalanced frictional forces.. $\endgroup$
    – BowlOfRed
    Jul 2, 2021 at 7:37
  • 2
    $\begingroup$ Feel free to ask a different question if you need. This answer is correct at a very simplified model. If you start caring more about the details, you'll need to add to it. Yes the water starting right by the man has very little acceleration (and moves very little). But the water further away will have a greater difference between the two accelerations. $\endgroup$
    – BowlOfRed
    Jul 2, 2021 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.