Suppose there is a spherical container with high-pressure air inside, and there is a small hole on its left that sprays air outward. We all know that this will generate a left thrust F1.

enter image description here

I believe that the reaction force F2 of F1 is ultimately applied to the entire right wall of the container, but some people do not believe so. They believe that F2 is applied locally to the right wall corresponding to the small hole. Am I wrong?

enter image description here

The forces at each point on the container point towards the outlet, and these forces are not of the same magnitude. Their combined force is F1, so the reaction force F2 of F1 is applied to each point on the container.Am I wrong?

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    $\begingroup$ "We all know that this will generate a left thrust F1." On what? This is a crucial first step. It's even more useful to draw a boundary around that recipient of the force you're describing. $\endgroup$ Commented May 17 at 23:15
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    $\begingroup$ They believe that F2 is applied locally to the right wall corresponding to the small hole. Do you feel the whole weight of an airplane as it passes directly over your head? $\endgroup$
    – RC_23
    Commented May 18 at 0:04
  • $\begingroup$ @Chemomechanics I didn't understand what you meant, could you please describe it again? $\endgroup$
    – enbin
    Commented May 18 at 0:11
  • $\begingroup$ @RC_23 So that means what I think is right? $\endgroup$
    – enbin
    Commented May 18 at 0:13
  • $\begingroup$ The reaction force for F! is the force that the expelled gas has on the gas within the opening that is about to be expelled; it is at the left side, and is directed to the right. $\endgroup$ Commented May 18 at 10:43

1 Answer 1


Whenever in doubt, think fundamentally at a basic level. Clearly define the system that you are applying forces on.

Case 1: enter image description here

Here the force on the escaping gas is applied by System 2. So, if system 2 applies F force on system 1, so, system 1 will also apply F(equal in magnitude) on system 2. This force is what you are saying will "ultimately" push the balloon, which is a bit wrong as the force on system 2 is actually at the left side (open side) and the rest of the forces are just internal. This is why people are saying that the reaction force acts at the left end.

But your point is also not false . Infact your point is more precise because you took the balloon without the air as the system. In Case 1, the balloon was pushed by internal forces. But if you take balloon as the system, you will observe that the gas inside it pushes the balloon in all directions, except the left end which is open. The gas cannot push the balloon through the left end, therefore it simply escapes. Therefore all the forces inside the balloon will be cancelled except the force pushing it directly to the right.

Newton's 3rd law does not specifically say anything about where the force is being applied, for that we will have to actually see how the force is being applied. The point of application of force on the balloon is actually at all points inside it, creating a net force forward. You cannot say that the net force acts on "this" one point. The small components of force act all over the surface. Both the arguments in the questions are slightly wrong from a conceptual point of view.

  • $\begingroup$ Is it because the force applied to the inner surface of the left hemisphere is smaller than that on the inner surface of the right hemisphere that a pure force to the left is generated? $\endgroup$
    – enbin
    Commented May 19 at 0:53
  • $\begingroup$ it would be better if you think in vacuum first. The air will push the surface in all directions equally. So, if a force acts on any point, it will cancel out the force which is acting diametrically opposite point of the balloon. The left end has a hole, so the air leaks out instead of exerting a force on the balloon to the left. So there is no force to cancel the force on the right side of the balloon. $\endgroup$ Commented May 19 at 6:27
  • $\begingroup$ I added another image and explanation, was I wrong? $\endgroup$
    – enbin
    Commented May 19 at 8:25
  • $\begingroup$ Now its better, but still a bit wrong. The forces should be perpendicular to the balloon surface. $\endgroup$ Commented May 19 at 12:05
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    $\begingroup$ "The forces at each point on the container point towards the outlet". I do not get what force are you talking about here. Also pressure is a scalar quantity, its acts in all directions. $\endgroup$ Commented May 21 at 7:34

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