Findings need confirmation about change of weight

[Crowxe]

Admin please don't delete, this is not homework nor part of any curriculum

A water tank put on scale measuring 50 Kg , inside the tank a fully submerged balloon tied with a thread to the tank bottom.

if the thread was cut, will there be a different reading on the scale (momentarily until the balloon reach the surface) ?

i believe that the forces downward are not only the mass but the water pressure at the tank bottom multiplied by the area. A balloon being submerged will cause displacement and raise water level thus total pressure of tank bottom, that can be observed during the process of submerging the balloon and until you tie it to the tank bottom, only then the buoyancy force will cancel the added pressure until that thread is cut .

 

[mfb]

this is not homework nor part of any curriculum

It is still homework-like, therefore I moved it to our homework section.

i believe that the forces downward are not only the mass but the water pressure at the tank bottom multiplied by the area.

Is that either-or, or is it a different way to say the same thing?

A balloon being submerged will cause displacement and raise water level thus total pressure of tank bottom

Right, but the balloon also adds a different force that is relevant. The balloon is attached somewhere initially.

 

[sophiecentaur]

During the period that the balloon is rising, some water is also falling. The two masses are not the same even though the volumes are. Some accelerations are involved during the process and acceleration means force. Does this help?

 

[Crowxe]

During the period that the balloon is rising, some water is also falling. The two masses are not the same even though the volumes are. Some accelerations are involved during the process and acceleration means force. Does this help?

i looked into the water acceleration in that model. the water above the balloon will accelerate downward and then comes to a halt after the balloon passes, so it will be like astronaut in ISS pushing a brick forward but then not letting it go, he won't travel in space by doing that.

 

[haruspex]

the water above the balloon will accelerate downward and then comes to a halt after the balloon passes

Yes, but so long as the balloon is accelerating up there will be more water moving down even faster.
Think about the mass centre of the water+balloon system. Is that accelerating? If so, which way?

 

[CWatters]

i looked into the water acceleration in that model. the water above the balloon will accelerate downward and then comes to a halt after the balloon passes, so it will be like astronaut in ISS pushing a brick forward but then not letting it go, he won't travel in space by doing that.

Correct. That would be due to Conservation of Momentum. However the problem statement said..

(momentarily until the balloon reach the surface)

So it's asking what happens while the water is moving.

 

[Crowxe]

Water movement due to balloon movement is not my main research , anyway, I think the key player is water level , here's my analysis which i admit sounds odd :
1- understanding the forces when object submerged in the tank
If you force a 1 L volume balloon in 49 L water tank, the whole system would weigh 50 Kg , how? don't simply say it's because we are using force to force the balloon down, the balloon has no contact to the scale. but it's the displacement caused by the balloon made the water level rise and simulates an extra 1 L of water , at this point it won't matter if it's a balloon or 1000 cm³ of iron (around 8 Kg) . The extra reading will still be 1 Kg

2- Releasing the object
When the balloon is released , water level won't change so the scale should maintain the new reading (50 Kg) , and this is the main stage that I'm investigating, it's odd part yet it is consistent with physics laws, it's the displacement effect if i may call it so.

3- object travel
as the balloon accelerate upward, positive pressure forms on top of it and negative pressure under it increase until it maintain fixed speed, at that point , the displacement effect will be reduced to minimum or even get canceled by the balloon kinda pushing water upward with it.

please give feed back or if you can perform the experiment , would be great. sorry if I'm not using proper physics language, and take it easy with that if you going to explain something. thanks

 

[mfb]

The scale will change its reading briefly when you release the balloon. It will go back to its previous value once the balloon reaches its terminal velocity.

Water height alone does not tell you everything, especially in a dynamic system.

 

[haruspex]

If you force a 1 L volume balloon in 49 L water tank, the whole system would weigh 50 Kg ,

That is only true if the 1L object is being held in position by an external force (whether that is holding down an object that would otherwise float or holding up an object that would otherwise sink). But in this case the balloon is tethered to the bottom of the tank, so is exerting an upward pull on the tank. As a result, the scale reading will correspond to the weight of the 49L of water, regardless of the immersed object.

When the balloon is released , water level won't change so the scale should maintain the new reading

You cannot ignore the accelerations occurring inside the tank. As I suggested in post #5, consider the mass centre of the water plus balloon system. Will that accelerate when the thread is cut? If it does, there must be a net force acting on it. Gravity hasn't changed, so what has?

 

[Crowxe]

The scale will change its reading briefly when you release the balloon. It will go back to its previous value once the balloon reaches its terminal velocity.

Water height alone does not tell you everything, especially in a dynamic system.

thanks a lot , i needed that confirmation before studying next step for possible application, however , the feeds i got is still 50% : 50% as Mr. Haruspex doesn't confirm it

 

[Crowxe]

That is only true if the 1L object is being held in position by an external force (whether that is holding down an object that would otherwise float or holding up an object that would otherwise sink). But in this case the balloon is tethered to the bottom of the tank, so is exerting an upward pull on the tank. As a result, the scale reading will correspond to the weight of the 49L of water, regardless of the immersed object.

You cannot ignore the accelerations occurring inside the tank. As I suggested in post #5, consider the mass centre of the water plus balloon system. Will that accelerate when the thread is cut? If it does, there must be a net force acting on it. Gravity hasn't changed, so what has?

water level changed , total water pressure on the tank floor changed to +50 while the balloon is affecting the tank floor by -1 , but when we cut the thread , that later is opted out and remains the +50 for certain limited period of time until the balloon reach its terminal velocity

 

[Crowxe]

@mfb and I are in agreement. The question as stated is:

Terminal velocity is never quite reached, and it certainly is not reached instantly when the thread is cut.
The scale reading will change immediately, and gradually return towards the original reading as the balloon's acceleration drops.
If you will just take my hint and consider the mass centre of the system you will see this easily.

now this really made my day. I'm glad i get another confirmation. and as for looking at the mass center. well, i see the empty half of the glass and you see the filled half, it's same result :) . physics is very tricky and i fooled my self many times unintentionally by dropping an element or factor, so i didnt want to start from scratch into a new approach. any way, thank you for your support and time

Next investigation
if we pumped air through a hole in the tank bottom , will we get higher reading ? as each air bubble is considered like the balloon. will the tank be heavier than it should be?

 

[haruspex]

i'm glad i get another confirmation

I am not confirming your view, and neither, it seems to me, is anyone else on this thread.
The question you asked was:

if the thread was cut, will there be a different reading on the scale (momentarily until the balloon reach the surface) ?

The answer to that question is that after cutting the thread the mass centre will accelerate downwards. Since gravity has not changed that means the normal force from the scale has reduced, so the reading will reduce.
As the balloon ascends its acceleration will drop, so the reading will gradually increase. Since terminal velocity is never quite reached, it will not quite return to the original reading until the balloon reaches the surface.

(What happens then is a bit complicated. At first, the buoyant force will not have gone down much, but the drag will drop quickly, so it will start to accelerate again. But now its upward acceleration is not equal, volumetrically, to the downward acceleration of the water.)

 

[haruspex]

I've thought of an analogy you might find easier to think about.
Consider a box on the scale - air-filled, no water this time - with a weight hanging from the lid inside. If the thread is cut, what happens to the scale reading between then and when the weight hits the bottom?

 

[Crowxe]

I've thought of an analogy you might find easier to think about.
Consider a box on the scale - air-filled, no water this time - with a weight hanging from the lid inside. If the thread is cut, what happens to the scale reading between then and when the weight hits the bottom?

the reading will drop for specific period of time then it will increase much more for a much less period of time . but this case include the weight hanged is directly affecting the scale through the box walls and the existence of the weigh inside the box has no secondary effect such as that of raising the water level. i think i'll have to study the center of mass thoroughly

 

[haruspex]

the existence of the weigh inside the box has no secondary effect such as that of raising the water level

It is completely analogous. In the original, you have air moving one way and water the other. In the box version you have air one way and the mass the other.

 

[Crowxe]

It is completely analogous. In the original, you have air moving one way and water the other. In the box version you have air one way and the mass the other.

true, but in the box, if you let the mass in, hanging by your hand, the reading won't change, unlike the original

 

[scottdave]

You can see a sort-of similar experiment here. And an explanation here:

 

[haruspex]

if you let the mass in, hanging by your hand

But that would be applying an external force to the mass+box system, so would be different from the tethered balloon in the tank.

 

[scottdave]

The thing to do is: identify all forces acting on the system. Then the vector sum of forces equals mass x acceleration. So when the balloon is tied to the bottom, it obviously is not accelerating. But when the string is cut, the center of mass of the system is accelerating downward. Think about a different situation. What happens when you pick up a heavy box . After you pick up the box, and the box's weight is added to the total: as you lift the box, your center of mass is accelerating upward. What does this do with the forces between your feet and the ground.
Also, think about when you are in an elevator: when it first starts moving, and when it slows to a stop. You know the feeling. If you were standing on a scale, how would the reading change?

 

[Crowxe]

The thing to do is: identify all forces acting on the system. Then the vector sum of forces equals mass x acceleration. So when the balloon is tied to the bottom, it obviously is not accelerating. But when the string is cut, the center of mass of the system is accelerating downward. Think about a different situation. What happens when you pick up a heavy box . After you pick up the box, and the box's weight is added to the total: as you lift the box, your center of mass is accelerating upward. What does this do with the forces between your feet and the ground.
Also, think about when you are in an elevator: when it first starts moving, and when it slows to a stop. You know the feeling. If you were standing on a scale, how would the reading change?

thank you for the videos , they are so cool. i'll focus on the issue of center of mass shifting as mr. haruspex also suggested that.

 

[Crowxe]

i admit , the mass center change concept sounds compelling , I'm on it but i can't go further with it before sorting out couple of issues :

1- There's no component in the system that exert power on the mass center forcing it into action , so how can we expect a reaction (loss of reading on scale)
2- if we study the water particles around the balloon , they accelerate downward by gravity (external effect in regard of the system) then they decelerate when they collide with the standing still layer that was under the balloon (internal effect in regard of the system)

 

[scottdave]

... it will be like astronaut in ISS pushing a brick forward but then not letting it go, he won't travel in space by doing that.

This is a decent analogy, but there will be some movement of the astronaut (not much, as the center of mass of the man/brick system will stay fixed). This is also similar to the person walking on a canoe floating on a frictionless pond. A classic problem. The man can move himself relative to the pier, but is constrained.

 

[haruspex]

There's no component in the system that exert power on the mass center forcing it into action ,

There is. It is called gravity. If the balloon rises a little then the mass centre falls a little, so gravity is doing work on the system.

if we study the water particles around the balloon

It is not practical to figure out what will happen by drilling down to that level. There is too much going on. At all times after the start, some water particles will be accelerating while others are decelerating. How will you figure out the net result? This is why stepping back and looking at attributes of the system as a whole (such as mass centre) can be so powerful.
Archimedes' principle is similar in that way. You could figure out the buoyant force on an object by integrating the pressure vector over the surface, but so much simpler to use Archimedes.

 

[Crowxe]

There is. It is called gravity. If the balloon rises a little then the mass centre falls a little, so gravity is doing work on the system.

It is not practical to figure out what will happen by drilling down to that level. There is too much going on. At all times after the start, some water particles will be accelerating while others are decelerating. How will you figure out the net result? This is why stepping back and looking at attributes of the system as a whole (such as mass centre) can be so powerful.
Archimedes' principle is similar in that way. You could figure out the buoyant force on an object by integrating the pressure vector over the surface, but so much simpler to use Archimedes.

Gravity is not a part of the system , since the whole system is subjected to it. nothing in the system pushes the water down.

i guess the term "particle" was not accurate or misleading , what i meant is , for every amount of water that accelerate falling by gravity, there's equal amount that will decelerate by components of the system (water layers or tank floor)

i can't look to the bigger picture (mass center shift) and skip its analyses , at least to a considerable level of details. i can't go into vortex and tiny centrifugal forces whirls. but only looking into masses at smaller levels , what moves it and what stops it

 

[scottdave]

Gravity is not a part of the system , since the whole system is subjected to it. nothing in the system pushes the water down.

I guess the next question to investigate is "Where does the buoyant force come from?"

 

[haruspex]

Gravity is not a part of the system , since the whole system is subjected to it. nothing in the system pushes the water down.

It is part of the system and it does push the water down. Otherwise, why would anything move?
Consider a tilted seesaw with a weight placed on the upper end. According to your argument gravity will not cause the seesaw to tip.

 

[Crowxe]

It is part of the system and it does push the water down. Otherwise, why would anything move?
Consider a tilted seesaw with a weight placed on the upper end. According to your argument gravity will not cause the seesaw to tip.

i think I'm starting to get it , I'm thankful for all the support and time :)

 

[Crowxe]

I guess the next question to investigate is "Where does the buoyant force come from?"

as in general or that's related to my proposed experiment ?

 

[scottdave]

as in general or that's related to my proposed experiment ?

I was just trying to prompt you to think about things. So if it seems there is no external force, then think about what all is happening to see if that is true or not.

 

[Crowxe]

I was just trying to prompt you to think about things. So if it seems there is no external force, then think about what all is happening to see if that is true or not.

that what i did from the start, thinking about all components, actions and reactions across time , and that lead me to a result that was refuted by almost everyone through analysis that i can't quite comprehend. weird results don't stop me. nothing is weirder than space stretch, time dilation and delayed choice experiment , so i thought my conclusion could be right about having more force with same mass and acceleration, but may I'm wrong