Let's consider the the bucket with water, which has a small hole at the bottom. Let the bucket lift up with a constant force $\vec F$. Water in the bucket, of course, flows out of it.

Questions are as follows:

How can we write the equation of motion of the bucket?

Like (a) $$m(t)\frac{d\vec v}{dt} = \vec F +m(t)\vec g$$

or like


$$ m(t)\frac{d\vec v}{dt} = \vec F +m(t)\vec g + \vec u \frac{dm}{dt}$$

e.g. with the thrust $\vec u \frac{dm}{dt}$

In other words, as mentioned in the title, does the outflowing water create a thrust on the bucket? If not (I hope so), What reasons why we should not consider a term $\vec u \frac{dm}{dt}$?

  • $\begingroup$ I'd say the extra term is CM preservation. If the cointainer (bucket, gas chamber) struggles to preserve the geometry of the CM position there is always a thrust that you could call in some sense fake. $\endgroup$
    – arivero
    Aug 1 '15 at 17:26

There is the force of gravity on the water, the force of lifting the bucket, and the force of the pressure of water on the bottom of the bucket.

For a "normal" bucket, the pressure times area on the bottom equals the weight; but in this case, the area is reduced so the force on the bottom of the bucket is less than the weight of the water. This is the force you need to hold the bucket in the air. But the additional force of gravity on the water is what propels the water out of the hole.

If force $F$ is not exactly equal to the force needed to hold the bucket still, then the system will accelerate - but we could easily transform into a frame of reference (accelerating) in which the bucket is stationary, and the apparent acceleration is not $g$ but $g'$.

For liquid height $h$ and bucket area $A$ with hole $a$, we can write

$$P = \rho g' h$$

for the pressure, and

$$F = (A - a) P$$

for the force.

Then the net force on the mass of water in the bucket is $$mg - F = \rho g' h A - (A - a) \rho g' h\\ = a \rho g' h$$

which is exactly the weight of the column of water above the hole. That column of water is therefore "free falling" out of the hole, and giving rise to the flow that is observed. The mass flow rate at the hole is $\rho a v = \rho a \sqrt{\frac{2g'}{h}}$

Do you need a diagram or is this clear?

  • $\begingroup$ The formula involving both g and g' implies that if the area of the hole is cero, then F=mg??? Ah, this F is not the original F or the question. Not very clear. $\endgroup$
    – arivero
    Aug 1 '15 at 17:36
  • $\begingroup$ is it $mg - F$ or $mg' -F$ ? Is F the same external force, or just coincident notation? $\endgroup$
    – arivero
    Aug 1 '15 at 17:44

you are right, it is your formula a), as there is no thrust of the flowing water against the bucket, although the official NASA formula for the rocket chamber thrust implies that there is.

Imagine that you instantly open the bucket floor, would the bucket jump to equalize he momentum of the flushed water?

If I'm not mistaken, this problem is related to the rockets thrust in vacuum. It is a taboo unfortunately.

Edited: The rocket equation, ie the thrust equation, (https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation) is good if an astronaut would be throwing weights out into the space, one after another. However, the heated gas in the rocket chamber escapes through the throttle (ie adiabatic free expansion) into the vacuum without any work done against anything (including the wall of the chamber). Also the center of mass of the gas remains near to the chamber, as the gas molecules quickly go in every direction including in front of the ship, as its volume goes to infinity and pressure to zero.

So, the hot gas simply escapes the chamber (it is not pushed out by the ship) not pushing anything and not doing any work, similarly to the horizontal outflow of water doesn't push bucket in the opposite direction, or in the original problem of this tread, the leaking water doesn't push the bucket upwards.

To be precise, in the buckets examples the flowing out water does perform a work by pushing the air around the hole, but this is negligible for this problem.

Conclusion: So the question was: "does the outflowing water create a thrust on the bucket?", and I think the answer is "No". The answer accepted above doesn't answer the question, it rather explains the diminishing weight of the bucket due to the leak, but that was not the question.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Jul 30 '15 at 9:32

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