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

[A.T.]

...but the bucket would be moving upward because of losing weight.

What?

 

[Dale]

There has got to be some huge communications gap going on here. By your explanation, it would be self-contradictory to have a U-shaped tube mounted to the bottom. The jet thrust would be downward, but the bucket would be moving upward because of losing weight.

If you had a U-shaped tube mounted on the bottom such that the jet of water was squirting upwards then the bucket would be measured to weigh more than an otherwise identical bucket without the jet.

This is all just Newton's laws.

The drop of water just leaving the tube is accelerating in the direction of the tube and therefore has a net force acting on it. By Newton's 3rd law there is an equal and opposite force acting on the bucket, if the jet is vectored up then the force on the bucket is down, if the jet is vectored down then the force on the bucket is up, if the jet is vectored left then the force on the bucket is right, ...

So the three forces acting on the bucket are the force from the scale, the weight, and the force from the jet. These three forces sum to zero since the bucket is not moving. So, the force on the scale is equal to the vector sum of the weight and the force from the jet. If the force from the jet is down (jet pointed up) then the measured weight will be greater than the weight without the jet. If the force from the jet is up (jet pointed down) then the measured weight will be less than the weight without the jet.

 

[Danger]

The answer to that depends what is considered to be "the bucket". Just the vessel, or the vessel + content?

That should be pretty obvious to anyone who speaks English. (KFC doesn't count; in their case, a bucket is obviously meant as the bucket plus the chicken in it but that's a commercial term.)

in both cases there is thrust, as defined in the Variable-mass system equation, that acts either on the content or the vessel + content system.

You just contradicted yourself in the course of 2 sentences. Where does that equation mention just the vessel without content?

What?

That was in reference to Dale specifying that the bucket is on a scale, which obviously would be imparting an upward variable force. Without a scale, with the bucket simply sitting on the floor as the OP stated, the closest thing to an upward force would the the increase of the bucket's buoyancy in air and the decrease of both gravity attracting it and the floor pushing up without movement. I admit to both, but don't see how it can be considered a "jet reaction".

edit: Dale, I just saw your new post. In the case of a vectoring tube, that provides a structural connection between the tube and the bucket to transfer the force. In the case of a simple hole in the bottom, there isn't one. The reaction would be between Earth and the water, not between the water and the bucket. The same thing would happen to the water if it didn't have a bucket around it.

 

[Dale]

If in the bottom of the bucket with water we create an outlet, then water will flow down from it. Is the thrust of the water on the bucket? Explain the falsity of the positive answer to this question.

The positive answer is not false: http://en.wikipedia.org/wiki/Water_rocket

Think about suspending the bucket from a string and vectoring the jet to the side. The bucket will be pushed off the vertical axis due to the thrust. That same thrust exists if the jet is pointing down also.

 

[Dale]

In the case of a vectoring tube, that provides a structural connection between the tube and the bucket to transfer the force. In the case of a simple hole in the bottom, there isn't one.

What is the difference between a hole and a short straight vectoring tube?

 

[A.T.]

You just contradicted yourself in the course of 2 sentences

I merely gave two answers for both possible interpretations of the question.

 

[Danger]

What is the difference between a hole and a short straight vectoring tube?

Simple... a short straight down tube is not a vectoring device. Its existence is totally irrelevant because it causes no different effect than a hole without it.
Do you not understand the irony of linking to the example of a water rocket? That is a pressurized system where the water reaction mass is forceably ejected by an action against the interior or the bottle.

 

[Danger]

I merely gave two answers for both possible interpretations of the question.

No, you didn't. You made one sweeping statement that turned around and bit itself. I merely quoted it in separate sections.

 

[sophiecentaur]

If the dribble out of the bottom of the bucket produces an upward force then the same thing could be said to occur if I am carrying a 1kg mass and then drop it. It would be true that the scales I stand on would read 10N less, once I have dropped the mass. Would that count for you? Is a decrease in downward force the same as a force upwards? If the experiment were carried out on a weigh bridge, there would be a temporary drop of 10N in weight measured until the mass hits the floor, when the original value will be restored.

Whilst you are considering what happens to the water that is exiting the hole in the bottom, perhaps you should also consider what happens to the upper surface of the water in the bucket. It is also moving downwards at a rate where the ratio of the speeds is inversely proportional to the cross sectional areas of the hole and the top of the bucket. What about the dynamic pressures involved and the relevance of the Bernoulli effect? ;)

 

[A.T.]

You made one sweeping statement that turned around and bit itself.

No idea what you mean.

 

[Dale]

Simple, a short straight down tube is not a vectoring device. Its existence is totally irrelevant because it causes no different effect than a hole without it.

I think you are making a wrong conclusion. It causes no different effect than a hole without it, therefore there is no difference between a hole and a vectoring tube. Consider, if you put a hole on the side near the bottom then you clearly get thrust. Therefore a hole on the side is the same as a short straight vectoring tube on the side, both cause thrust to the side. Placing it on the bottom doesn't change any of that.

Do you not understand the irony of linking to the example of a water rocket? That is a pressurized system where the water reaction mass is forceable eject by an action against the interior or the bottle.

The water on the bottom of a bucket is also pressurized and the water reaction mass is also forceably ejected. The water rocket simply increases the pressure to make the effect obvious, but the water at the bottom of a bucket is already inherently pressurized by the weight of the water above it.

 

[A.T.]

If the dribble out of the bottom of the bucket...

If it's just a dribble, then u is 0, and so is the thrust. But the water can be shooting out with high speed due to the pressure, and then u > 0.

 

[Danger]

No idea what you mean.

I'm not surprised...

there is no difference between a hole and a vectoring tube. Consider, if you put a hole on the side near the bottom then you clearly get thrust. Therefore a hole on the side is the same as a short straight vectoring tube on the side, both cause thrust to the side. Placing it on the bottom doesn't change any of that.

The water on the bottom of a bucket is also pressurized and the water reaction mass is also forceably ejected. The water rocket simply increases the pressure to make the effect obvious, but the water at the bottom of a bucket is already inherently pressurized by the weight of the water above it.

Good grief, this is maddening!
Of course, there's a difference between a hole or straight-down tube (non-vectoring) and a vectoring one (vectoring, obviously). There's no point in even mentioning a hole in the side, since the OP has clarified that the hole is in the bottom.
The water in the bucket is not pressurized to above atmospheric, and certainly has no connection to the non-existent top of the bucket.

 

[Danger]

If it's just a dribble, then u is 0, and so is the thrust. But the water can be shooting out with high speed due to the pressure, and then u > 0.

Again, there is no pressure that is associated with the bucket other than that supplied by the ambient air. You can't pressurize an open container.

 

[A.T.]

Google translate says the bucket is

For the bucket alone you don't need the variable-mass equation.

 

[sergiokapone]

There are many opinions, but truth is one.
Only math can help.

 

[A.T.]

The water in the bucket is not pressurized to above atmospheric,

The water has a higher pressure than atmospheric at the bottom of the bucket:
http://en.wikipedia.org/wiki/Fluid_statics#Hydrostatic_pressure

 

[sergiokapone]

For the bucket alone you don't need the variable-mass equation

No, bucket is with the water of course.
I thought that might be the wrong word I used.

 

[Dale]

The water in the bucket is not pressurized to above atmospheric

It most certainly is. Ask any scuba diver if water has greater than atmospheric pressure.

For a bucket of about 10 m depth the pressure at the bottom will be twice the atmospheric pressure at sea level.

 

[A.T.]

No, bucket is with the water of course.

If "bucket" = bucket + water in bucket, then there is thrust on the "bucket". If your book says there is no thrust on the "bucket", they apparently define "bucket" as bucket without water.

 

[Danger]

A 10 m deep bucket is getting pretty far afield, but regardless... the water is not pressurized relative to the top of the bucket (because there isn't one). The pressure is unevenly (but laterally symmetrically) distributed to the walls and bottom. If all of those forces are balanced, as they must be, there is no thrust. That would occur if there were an imbalance between the top and bottom, and again there is no top. Would a rocket work if you cut the top off?

 

[Danger]

they apparently define "bucket" as bucket without water.

That's pretty obvious from his enhanced translation, if not the first couple of questions.

 

[A.T.]

The pressure is unevenly (but laterally symmetrically) to the walls and bottom. If all of those forces are balanced, as they must be, there is no thrust.

The forces on the water aren't balanced if there is a hole in the bottom, so there is thrust.

 

[Danger]

so there is thrust.

Against what...?

 

[A.T.]

Against what...?

Explained many times. See for example post #23.

 

[Danger]

Explained many times. See for example post #23.

And as you said your very own self in that very same post, the thrust is upon the water, not the bucket, which is exactly what I've been saying all along. You just made my argument for me.

 

[A.T.]

...which is exactly what I've been saying all along.

Unfortunately along with some completely wrong stuff.

 

[Dale]

It doesn't matter if the thrust of the jet is directly against the water or the bucket. Either way, there is a different net force on the bucket with the jet than without.

 

[sophiecentaur]

Against what...?

I introduced Bernouli on the last page but no one took it up., What does Bernoulli say about the pressure change when a fluid flows into a narrow aperture? The pressure in the narrow bit is less than the pressure in the wide bit. So there is a higher pressure just above the hole than if there were no flow. So the rate of flow will be less than if the diameter of the bucket were the same as the diameter of the hole. (Naturally the water level would soon drop with such a narrow tube but just consider the initial situation) Is that not what's responsible for the 'mysterious' reaction force?

 

[Dale]

A 10 m deep bucket is getting pretty far afield, but regardless...

Nothing far afield whatsoever. It is just a big bucket to make the point clear. There is without a doubt a pressure greater than 1 atmosphere at the bottom of a 10 m bucket of water.

the water is not pressurized relative to the top of the bucket (because there isn't one).

Danger, this is simply factually wrong. If the top of the bucket is at 1 atm then the bottom of the bucket will be at 1 atm + the hydrostatic pressure of the water column. If you have spent any time diving or swimming then this should be clear to you.

Whether it is a hole or a nozzle, there is atmospheric pressure on the outlet and above-atmospheric pressure inside the bucket. This pressure difference causes the water to accelerate. That is what a jet is, regardless of whether the nozzle is a complicated tube or a simple hole.

Newton's 3rd law does the rest. The force that accelerates the water has a 3rd law partner which is equal and opposite. This is the thrust.

 

[Danger]

Sophie, perhaps I'm misunderstanding Bernouli; I admit that I've never designed a carburator, although I've rebuilt a few. I can't quite see how a hole in a piece of sheet metal is long enough to be considered a venturi, but I'll take your word for it. Still, my understanding is that a venturi doesn't increase the pressure ahead of it; it just converts some of it to speed. In either case, though, the water instantly reverts to the same atmospheric pressure as is experienced at the surface of the reservoir. How can thrust be imparted to the bucket in that instance? If Dale cuts the top off of his rocket bottle before launch, will it still fly?

 

[Dale]

If Dale cuts the top off of his rocket bottle before launch, will it still fly?

I never claimed that the thrust from the bucket's jet is greater than the weight, just that it is non zero.

The point of the 10 m bucket and the water rocket is to emphasize physical features that you are ignoring, but that are already present in a standard bucket with a hole.

 

[sophiecentaur]

I never claimed that the thrust from the bucket's jet is greater than the weight, just that it is non zero.

The point of the 10 m bucket and the water rocket is to emphasize physical features that you are ignoring, but that are already present in a standard bucket with a hole.

Yes. That's the whole point. The existence of a force doesn't imply it is a great force.
There is an intuitive feel about this that is misleading people. It can 'feel right' that a hole in the side of the bucket would push it sideways with a considerable force but there is no reason to think it actually would be so. So we are, in fact, only looking for a small force. (Bernoulli based) How would this force reveal itself when in the vertical direction? It would be that the measured weight (bucket hung on a sensitive balance) would be a bit less than expected (say against a volume scale on the side). The difference would be greater, the higher the water level in the bucket and would approach zero as the bucket empties. A control experiment, run beside the leaking bucket, could consist of small masses (ball bearings?) released by a subtle system of magnets of shutters from the bottom of the bucket. Without the fluid flow, the 'weight' of the control bucket would always be a bit heavier if the rate of mass release were kept the same as the jet of water.

 

[A.T.]

A control experiment, run beside the leaking bucket, could consist of small masses (ball bearings?) released by a subtle system of magnets of shutters from the bottom of the bucket. Without the fluid flow, the 'weight' of the control bucket would always be a bit heavier if the rate of mass release were kept the same as the jet of water.

That's complicated. Just fill the bucket to different marked levels, and note down the scale readings. Then open the hole and
compare the scale readings at the same levels.

 

[sophiecentaur]

That's complicated. Just fill the bucket to different marked levels, and note down the scale readings. Then open the hole and
compare the scale readings at the same levels.

I guess that could do it; small hole and bigger hole and bigger hole etc. I think the bucket with the whole bottom taken away would be subject to 'experimental problems'. The ball bearing method is a bit lumpy, I have to admit but I just wanted to eliminate the effect of the fluid dynamics.

 

[Danger]

That's complicated. Just fill the bucket to different marked levels, and note down the scale readings. Then open the hole and
compare the scale readings at the same levels.

It's nice to see that you've been paying so much attention that you attributed to me a quote that was written by Sophie. How can you even do that with a "Reply" function button?

Anyhow, now that you're fighting amongst yourselves, I'm going to get the hell out of this thread and leave you to it. I still say, however, that if this thing was in microgravity it wouldn't go anywhere. (And there is nothing in the question, if you'll notice, that says gravity exists in the experimental location...)

 

[Dale]

The whole OP wouldn't work in microgravity.

 

[256bits]

I guess that could do it; small hole and bigger hole and bigger hole etc. I think the bucket with the whole bottom taken away would be subject to 'experimental problems'. The ball bearing method is a bit lumpy, I have to admit but I just wanted to eliminate the effect of the fluid dynamics.

Your ball bearing analogy is valid, as a liquid can be considered of being composed of tiny molecular balls of mass dm. Pressure is just the energy gained by a dropping a mass from a height h, so we get mgh, and it doesn't matter if we initially drop the mass dm from the height h, or squirt the mass dm out by the pressure - the velocity at exit is the same.

The vectoring is a side-shoot. Any change in momentum, or a change in direction of velocity, will of course have a associated forces. While a stream of water coming out of a bucket and directed to the side will produce a sideways force from the velocity ( from mgh ) and direction ( just to be clear even though velocity includes direction ) of the stream being altered, both on the bucket and velocity stream, it is not so clear on intutive grounds whether an L-shaped tube with a downward exit, would would supply a or thrust on the system as a whole, any different than that from a hole exiting directly out th botom. If that were so, though, we should be able to configure a system with many twists, turns and lengths of pipe, where internal forces can defy physical laws to produce an anti-gravity machine, which no one would consider to be likely. Certainly a torque is accomplished, but that just shifts the centre of moment. An added complication is whether the pipe will droop or rise upward - ie the weight of the fluid in the pipe verus the exit momentum change.

 

[256bits]

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.

 

[A.T.]

And there is nothing in the question, if you'll notice, that says gravity exists in the experimental location...

Wrong again. The question clearly suggests that gravity exists in the experimental location:

If in the bottom of the bucket with water we made a hole, then water flows out of it.

 

[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?