Does the Bernoulli Effect Apply to the Jet Reaction Force in a Leaking Bucket?

[sophiecentaur]

I would guess you guys do know you are talking about the same thing, no, from different perspectives.
One is the weight of bucket as a whole which changes due to dm/dt.
The other is the force from an ejected dm/dt at the jet.

But there is a significant difference between just letting a ball drop down off the bottom (just a dm/dt) and ejecting water through a hole (with a dx/dt, as well as a dm/dt). The ball leaves the bottom at zero velocity (which was my reason for doing the ball thing) but the jet of water has velocity. If you want to get a sideways force from a dropping ball, you would need to run it down a curved track, inside, to give it a velocity - but that's not my experiment.

 

[256bits]

The ball is not just dropped off the bottom, but released from a height equal to the surface of the liquid.
Molecules are just really tiny little balls after all.

The dx/dt comes from the pressure which is a function of the height of liquid.
The exit velocity of the stream of water is √2gh, where h is the depth of water.

Similarily, for a ball dropped from a height h, the velocity of the ball will be √2gh.

 

[Wucko]

Have you tried using Bernoulli equation? Here is mine solution of the problem. The difference is I haven't used any derivations in my equation. Hope it will help and excuse me if my english is bad. And yes, there is no jet force. Hope you will get the picture

 

[sophiecentaur]

Have you tried using Bernoulli equation? Here is mine solution of the problem. The difference is I haven't used any derivations in my equation. Hope it will help and excuse me if my english is bad. And yes, there is no jet force. Hope you will get the picture

That's fine and it involves Energy conservation (when there is laminar flow and no energy loss in the fluid). If the exit velocity is (ideally) √2gh, that must involve a step change in velocity as the molecules speed up on exit. So there must be a force on each one, to accelerate it. This involves no energy transfer to the bucket as there is no vertical movement - so no Force times Distance. There is no violation of any conservation law, even though a reaction force appears. People are looking for a paradox when there is none.

If the water were to leak out of the bottom, at the same rate, via a sponge bung in a larger hole, there would be no upward force as the water would be starting from zero speed on leaving the sponge.

 

[sgphysics]

Would one not need to consider also the change ov momentum of the water insise the bucket? The change d(mv)=mdv+vdm. For a straight bucket the first term is zero, but the second is not. It would counteract the momentum at the outlet, so if the bucket was hanging in a rope, the weight woluld not be lessened by the jet, only by the loss of water.

 

[sophiecentaur]

This has all been discussed earlier but the thread is long enough for you to have missed it. There is a change of momentum of the water as it passes through the constriction. Bernouli described the effect, which accounts for an increase in velocity. This change in momentum means that an impulse (downwards) is imparted to every drop of water that is ejected. There must be an equal and opposite reaction against the water in the bucket - making it ( ever so slightly) lighter.

Is there anyone who can argue that the Bernoulli effect does not happen in this circumstance?