Consider the problem wherein we are told to find out the force of reaction of water flowing out of a vessel through an orifice. I attempt to compare it with a shotgun and a bullet as a conservation of momentum of the system. Another approach is Newton's third law that is inextricably related to momentum conservation. However, I can't picture what forces are interacting: how or in what directions. If there were no friction on the floor, would the vessel be pushed in the opposite direction of the water?

There's a second problem about figuring out the reaction moment of water flowing out of a tube, I think the tube's wall interacts with water; what other interactions would there be? Aren't those forces applied in all directions? Do the forces cancel each other or not? Here's a picture of what I think. I learnt a way to calculate the resultant force in this case using linear momentum, but I would like to know the direction of the net force, not only its modulus.

enter image description here

Here's the problem picture

enter image description here

Finally, there's a problem about finding out the horizontal force that tends to pull the tube out of the tank. I think the forces interacting should be the same as above with the difference that forces are acting on the tube this time, but it's kind of difficult to picture a resultant force since they appear to cancel each other. Ignoring what I just wrote, since the orifice is small we can assume pressure is the same in all the points of it (that's my idea), so:

$$ dF = ds P $$

I still can not locate the force. Here's the picture of the problem:

enter image description here


1 Answer 1


Consider the folowing image:


It depects the cross section of a pipe, where one should imagine the water tank on the left. Let's consider a control volume, the dashed lines. Due to the water pressure, the walls exert a force on the water, and so does the water outside the control volume. But let's look at the bottom of our control volume. There are no forces there, since there's nothing underneath it! This means that there is a resultant downwards force on our control volume, and it will flow out as expected

A keen reader will notice that I neglected air pressure. However, since the air pressure works both on the top of our water tank and the bottom of our tap, it will cancel out. This of course only applies to an open water tank.

  • $\begingroup$ I really appreciate your help. That's exactly what I was looking for. :) $\endgroup$
    – Omar
    Sep 16, 2015 at 1:37
  • $\begingroup$ Just a question. Are those forces you mentioned the ones that triggers a moment of torsion? Furthermore, in the case where there were no this downwards orifice (Only the horizontal tube), what would be the resultant force? :D $\endgroup$
    – Omar
    Sep 16, 2015 at 15:18
  • $\begingroup$ @OmarMedinaBautista A moment of torsion is defined around a certain point. Any force triggers a moment of torsion given any point that's not colinear with the force. For the horizontal case, one could just rotate the control volume by 90°. Remember: You're always free to choose your control volume. There's no harm in choosing a rectangular control volume inside a trapezoidal tube. $\endgroup$
    – Sanchises
    Sep 16, 2015 at 22:20
  • $\begingroup$ Thank you! I'm taking fluids mechanics but as general physics, so the explanations I've been give are very general and I have hundres of doubts especially the physical analysis of these phenomena like: How does water interact with the tube or the vessel. Why does water flow out of the vessel If there's an orifice or how does gravity play a role in this. I know they are basic, but I hope I will find the answers, and much better more questions. Thank you for your great help. :) $\endgroup$
    – Omar
    Sep 17, 2015 at 6:00

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