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

[sergiokapone]

If in the bottom of the bucket with water we made a hole, then water flows out of it. Bucket with water is the system with $m(t)$, thus we have to use equation:
$m(t)\frac{dv}{dt} = F + u\frac{dm}{dt}$.
Where $u\frac{dm}{dt}$ -- is jet reaction force.
But, I think the jet reaction force should not appear in this case. What is the solution of this paradox?

 

[Danger]

Hi, Sergio. If I understand the question correctly, there are two answers. A hole in the actually bottom would not provide thrust because the water would just be accelerating toward the ground as opposed to away from the bucket. That would lessen the effect of gravity upon the bucket because of weight loss, but that's not really a jet effect. If, on the other hand, the hole is in the side of the bucket near the bottom, there would be a tiny one because it would be pushing off from the water still in the container. I honestly think that I'm not correct in my interpretation of what you're asking though. Can you perhaps rephrase it?

 

[Dale]

I don't understand the question.

 

[sergiokapone]

A hole in the actually bottom would not provide thrust because the water would just be accelerating toward the ground as opposed to away from the bucket. That would lessen the effect of gravity upon the bucket because of weight loss, but that's not really a jet effect.

I do not completely understand this. How I can write the equation of motion in the case the bucket, which is on the floor?

 

[Danger]

How I can write the equation of motion in the case the bucket, which is on the floor?

No help here... I don't know any math. I can say that there will be no motion at all if the thing is on the floor. (Unless you count immeasurably small structural vibrations on the molecular scale.) In fact, the water will just stop flowing completely if the surfaces of the bucket rim and the floor are smooth enough to form a seal.

 

[sergiokapone]

It is strange to me, the in the case of the rocket the outgoing propellant create jet forse, but in the case of bucket - whater is not.

 

[Danger]

In the case of a rocket or air-breathing jet, or water jet or any other kind, the reaction mass is accelerated away from the container by either being expanded or being compressed and then released, or having the reaction chamber reduce in size to eject it (like in a syringe or balloon). In the bucket instance, gravity is simply pulling the water away from it. Earth is providing the force, not something in the bucket.

 

[A.T.]

It is strange to me, the in the case of the rocket the outgoing propellant create jet forse, but in the case of bucket - whater is not.

This is how a rocket produces thrust:

Where is the green force in the case of your bucket?

 

[sergiokapone]

Earth is providing the force, not something in the bucket.

Ok, I understand that.

But now the quastion in the math. Why should I use equation of the motion $m(t)\frac{dv}{dt}=F$ instead $m(t)\frac{dv}{dt}=F + u\frac{dm}{dt}$? Is the $u$ -equal zero in this case? But, obviously it is not.

 

[A.T.]

But now the quastion in the math.

Define your symbols. What is F, v, m, u ?

 

[Danger]

Ok, I understand that.

Is the $u$ -equal zero in this case? But, obviously it is not.

As mentioned, I don't know any math at all beyond simple geometry. What is "u"? If it's physical displacement on an x, y, or z axis, it is zero.

 

[sergiokapone]

$u$ - is the propellant relative velocity (relative to rocket or some other thing, bucket, for example e.t.c.)

 

[A.T.]

$u$ - is the relative velocity

Velocity of what and when? What are the other symbols representing?

 

[sergiokapone]

Define your symbols. What is F, v, m, u ?

$u$ - is the propellant relative velocity
$F$ - external force, gravity force, for example
$v$ - is the velocity of the rocket
$m$ - is the current mass of the rocket

 

[Danger]

$u$ - is the propellant relative velocity (relative to rocket or some other thing, bucket, for example e.t.c.)

Okay, but I rather suspect that the equation should apply only if there is some physical interaction between the two objects. In this case, there isn't. That would be like using it to determine the force between an aeroplane and canoe. What applies in your case is the gravity formula and how the bucket becomes lighter as water leaves it.

edit: You guys are posting too quickly for me to keep up with. I'm going to take a bathroom/TV break (it's the only room with a flatscreen) and I'll check in later.

 

[sergiokapone]

I mean this equation https://en.wikipedia.org/wiki/Variable-mass_system

 

[A.T.]

I mean this equation https://en.wikipedia.org/wiki/Variable-mass_system

Okay, that should work for you bucket too.

But, I think the jet reaction force should not appear in this case.

Why not? If you make the hole to the side, it will create thrust an propel the bucket horizontally.

 

[sergiokapone]

If you make the hole to the side, it will create thrust an propel the bucket horizontally.

This is not interesting trivial case.
What if I will create the hole in the bottom?

 

[A.T.]

What if I will create the hole in the bottom?

Different direction of thrust.

 

[sergiokapone]

But as mentionet above, the thrust forcedoes not apear in this case.

 

[A.T.]

But as mentionet above, the thrust forcedoes not apear in this case.

Why not?

 

[Drakkith]

Hmm, a good question. My thoughts:

The reason that threre is no thrust is because there is no "top" of the bucket. If you poke a hole in the side, water still presses on the opposite side of the bucket and there is an imbalance of forces, which results in a net force on the bucket that makes it swing sideways.

However, since water isn't pressing on the top of the bucket, poking a hole on the bottom doesn't result in an imbalance of force from the water. Or, rather, there was already an imbalance of force since the water is pressing down but not up and we have simply lessened it.

 

[A.T.]

The reason that threre is no thrust is because there is no "top" of the bucket.

But there is water on top of the water that is being pressed out of the bucket. The "thrust" (as defined in those equations) is not on the bucket itself, but on the water inside it. It reduces the vertical support force, that the bucket must apply to the water. This can also be attributed to the fact, that the water column above the hole is not supported by the bucket anymore.

 

[Danger]

the water column above the hole is not supported by the bucket anymore.

Which is exactly what I meant when I said that you are simply reducing the effect of gravity...

I swear that this is like talking to a wall...

Drakkith, your last post makes it about as clear as it can be. It's what I've been trying to say, but you said it better.

 

[sergiokapone]

Drakkith,
But my quastion is how looks the equation of variable-mass motion in this case and why its looks like?

 

[sergiokapone]

I need hard mathematical provement.

 

[A.T.]

In the bucket instance, gravity is simply pulling the water away from it.

There is pressure in the bucket that presses the water out faster than just free fall from rest due to gravity pulling on it.

Earth is providing the force, not something in the bucket.

There is something in the bucket that provides a force: the water above the water that is being pressed out.

 

[Danger]

There is pressure in the bucket that presses the water out faster than just free fall from rest due to gravity pulling on it.

There is something in the bucket that provides a force: the water above the water that is being pressed out.

Both of which statements merely confirm what I've been saying; any impartation of force from the release is away from the water, not upon it and not upon the bucket. There is no physical transfer between the reservoir of water and the top of the bucket. The only force (other than the "negative" one) is upon the ground under the bucket.

 

[rcgldr]

For the hole on the bottom, the water flowing out of the hole experiences an acceleration due to the pressure at the bottom of the bucket (which is related to the density and height of the water in the bucket) and from the pull of gravity, so there's more than just gravity accelerating the water.

 

[Drakkith]

But there is water on top of the water that is being pressed out of the bucket. The "thrust" (as defined in those equations) is not on the bucket itself, but on the water inside it. It reduces the vertical support force, that the bucket must apply to the water. This can also be attributed to the fact, that the water column above the hole is not supported by the bucket anymore.

Hmm, I see. I wasn't aware that counted as "thrust".

 

[Danger]

Hmm, I see. I wasn't aware that counted as "thrust".

It doesn't, in any logical or physical sense of the term. (No wonder I hate math...)

edit: I wonder how sheepish NASA will feel when it's pointed out to them after all of these decades of spacecraft development that they could have just filled their ships with water and poked a hole in the bottom.

 

[Dale]

I need hard mathematical provement.

The Wikipedia link you cited earlier includes the proof.

 

[A.T.]

Hmm, I see. I wasn't aware that counted as "thrust".

It doesn't really matter how you call that term, the OP just asks whether it is zero in the bucket case. But the term "thrust" might be counter intuitive in the bottom hole bucket case, and the reason why the OP thinks it should be zero.

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.

 

[Drakkith]

Okay, that clears it up. Thanks, A.T.

 

[Danger]

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.

By the definitions that I'm used to, that is simply "thrust-vectoring" as is used in some aeroplane controls; the original thrust is still horizontal. It imparts torque to the bucket, along with the lateral force, but not lift.

edit: If you were to add two more bends to that pipe to have it eventually exit straight down under the centre of the bucket, it would impart a lift factor to the horizontal-to-vertical final bend of the pipe, but not to the bucket itself except for that due to the rigidity of the pipe. Think of the F-23 Raptor fighter jet. It swivels its main engine straight down for vertical take-off and landing, but that would simply cause it to somersault if it didn't have that honkin' huge lift fan in front of it to balance the thrust.

 

[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.