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

[Dale]

Vectored thrust does indeed provide lift if it is vectored downwards. This is the principle of VSTOL aircraft

 

[A.T.]

To make it more more intuitive think about a hole to the side, near the bottom. It clearly produces horizontal "thrust". Now attach a pipe to the hole that bends 90° downwards, and you have the same "thrust", but upwards, just like with the hole in the bucket bottom.

It imparts torque to the bucket, along with the lateral force, but not lift.

It does impart a vertical force.

If you were to add two more bends to that pipe to have it eventually exit straight down under the centre of the bucket,

None of that is need for the vertical force.

 

[Danger]

Vectored thrust does indeed provide lift if it is vectored downwards. This is the principle of VSTOL aircraft

VSTOL's, as I specified in my post, utilize balanced thrust. The Raptor has a balancing lift fan far ahead of the main engine, the Harrier has 4 properly located ducts, etc.. In the case of the simple bent pipe, the upward component of the thrust would merely be upon the part of the pipe where it bends. That would impart a torque factor to the bucket, with its far-side bottom edge as the fulcrum.

 

[A.T.]

VSTOL's, as I specified in my post, utilize balanced thrust.

That is completely irrelevant. The vertical force is there, regardless whether its torque is balanced by some other force.

That would impart a torque factor to the bucket

It would impart a vertical force, which creates torques around some points.

 

[Danger]

This has gotten so far off-track as to be ridiculous. The original question asked whether or not a hole in the bottom of the bucket would act like a jet (which implies a lift force); it would not. Introducing things like bent tubing just obfuscates the issue.

 

[A.T.]

The original question asked whether or not a hole in the bottom of the bucket would act like a jet ...

The original question was far more precise and specific than your, as usual, vague and ambiguous phrasing. It asked if a specific term still appears (is non-zero), if the variable mass system equation is applied to the bucket.

And the answer is: Yes, the term still appears (is non-zero) in that case.

 

[Danger]

The original question was far more precise and specific than your, as usual, vague and ambiguous phrasing.

That is unnecessarily rude, even for you. The original question, if you think back, was so incomprehensible that neither Dale nor I could understand it. The OP has since edited it. His first explanation, after we requested one, homed in on a hole in the bottom of the bucket and asked for the application of the movement equation due to jet effect, of which there is none. He also specified that the bucket is on the floor, which means that there would be no movement at all, even torque, unless it was a sealed and pressurized bucket.

 

[A.T.]

The OP has since edited it.

And that is the version I'm responding to.

jet effect, of which there is none.

The OP doesn’t ask about a "jet effect", whatever that might mean. It asks about a specific term:

$u\frac{dm}{dt}$ -- is jet reaction force

And that term is not zero or "none".

 

[Danger]

And that is the version I'm responding to.

Which I didn't even see until I checked back to see what you were going on about.

OP doesn’t ask about a "jet effect", whatever that might mean.

Oh, really...? Have you read the title of the thread? "About jet reaction force"

edit: I see that you have now also edited, to add that last quote since I wrote this. It doesn't alter the fact that he originally asked about it in a manner that suggest a "lift" factor and specified its relation to the bucket, not to the water in the bucket.

 

[A.T.]

Have you read the title of the thread? "About jet reaction force"

Yes, that how the OP calls the specific term it asks about. If your "jet effect" refers to the same term as "jet reaction force", then you are simply wrong, it's not zero.

 

[Dale]

VSTOL's, as I specified in my post, utilize balanced thrust.

Sure, it is balanced, but it is still vectored. As long as you vector thrust downwards you will have a lift force upwards, even if it is not balanced (in which case you will have a torque in addition to the lift). Having the torque is a bad idea for a vehicle design, but it doesn't mean that there is no lift from a downward vectored thrust.

Assuming that the jet of water is not hitting the scale, a bucket with a downward jet will be measured to weigh less than an identical one without a jet. It does in fact provide thrust, just not much.

 

[Danger]

Assuming that the jet of water is not hitting the scale, a bucket with a downward jet will be measured to weigh less than an identical one without a jet. It does in fact provide thrust, just not much.

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.

 

[sergiokapone]

About prehistory of my first post. The quastion was asked in book "http://libgen.org/book/index.php?md5=F3AC2AA73CE9CE76B9627FA6AB1B2E4C " by Matveev (book in russian). It sounds like (I try to translate more correctly):

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.
http://www.babla.ru/%D0%B0%D0%BD%D0%B3%D0%BB%D0%B8%D0%B9%D1%81%D0%BA%D0%B8%D0%B9-%D1%80%D1%83%D1%81%D1%81%D0%BA%D0%B8%D0%B9/outlet
Hope its quite correct.

 

[Danger]

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.

Thank you for posting that. It leaves my answer intact (although I suspect that other's might still argue about it). To whatever extent there is any "thrust" at all, it is upon the remaining water, not upon the bucket. (Unless you want to include viscous drag of the water against the sides of the bucket, which would be simply ridiculous.)

 

[A.T.]

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?

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

 

[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.