# Followup to easy question

1. Oct 7, 2007

### A(s)

ok, so lets say i have a buoyant force of 100N confined in a really long straw.. with 1kg of water on top. the water would accelerate out of the tube at 100m/s squared?

the thing that has me confused is that the origin of the force is acceleration due to gravity, but the application of the force should technically exceed 9.8 m/s^2

is this possible? or even rational?

Last edited: Oct 8, 2007
2. Oct 8, 2007

### stewartcs

No...Not unless you're on another planet that has gravitational acceleration of 100 m/s/s.

You're answering your own question. On Earth, gravitational acceleration cannot exceed (depending on altitude) approixmately 9.8 m/s/s. So, in order to have a 100N force with an object that has a mass of 1kg, the acceleration must be greater than that of Earth's.

3. Oct 8, 2007

### A(s)

k if there is any confusion,

you have a vertical straw, submerged in water.(sealed at point X) The length of the straw is such that when submerged, (with air in the tube) the buoyant force contained within the tube below point X is 100N (i.e. 10.2 kg of water is displaced)

10.2 kg of water *9.8 acceleration = 100 newtons of buoyant force

Above point x is a column of water within the straw such that the mass is 1kg.

If the seal at point X is removed, the 100N force should start acting on the 1kg of water. this should accelerate the water above the bubble at a rate of 100m/s second correct? and if so, how is it that a force originating with the acceleration due to gravity can create an acceleration greater than that of gravity

Last edited: Oct 8, 2007
4. Oct 8, 2007

### stewartcs

Sorry, missed the "buoyant" in your original post...the question makes more sense now!

The buoyant force generated by gravity that the fluid is exerting on the 1 kg of water in the straw once the seal is removed creates the higher acceleration. The buoyant force is a result of the potential energy of fluid at that depth.

I suppose you can think of it as the fluid pressure at X depth being similar to pressure stored in an accumulator, and then suddenly released. The accumulator would accelerate (depending on the pressure or depth in your case) the fluid to a value greater than that of gravitational acceleration.

5. Oct 8, 2007

### A(s)

thank you, i just wanted to make sure that i wasnt making a false assumption it was hard to rationalize it in my head, so thanks

6. Oct 8, 2007

### A(s)

there is a followup to this question too 1 second while i post it

7. Oct 8, 2007

### A(s)

now here is the kicker... heh

i was trying to find the average acceleration of a bubble as it rises in a straw. and i have the equation

v = volume of the bubble in m^3
1.205 = kg/m^3 of air
V(1.205) = mass of air in kg
pi (r^2) h (1000) = mass of a cylindrical column of water

Buoyant force = 9.8*V*1000
Resistant mass = pi (r^2) h (1000) + V(1.205)

for the starting acceleration equation with a 100m straw i have

9.8V*1000
pi (r^2) (100m)(1000)+ V(1.205)

but for the ending acceleration where the height of the column of water approaches 0 i have

9.8V*1000
pi (r^2) (0m)(1000)+ V(1.205)

or
9.8V*1000
v(1.205)

or
8132.7

therefore regardless of the volume it seems that the acceleration right at the surface, where the only mass being acted on is its own, the acceleration is this ridiculous number and the average acceleration is at least half of that

by the way, please disregard drag for now

Last edited: Oct 8, 2007
8. Oct 8, 2007

### stewartcs

First thing I see wrong is that you have forgotten g in this equation. The weight of a column of water is equal to pVg, where p is the density, g is gravitational acceleration, and V is the volume. All you have in your equation is the volume times the density of water.

Also, the density of air at STP is about 1.2 kg/m^3, not 1.02. The volume of air (V) is also decreasing (approaching zero) as the "bubble" moves up the tube. So from your equations you would actually have division by zero which mean an infinite acceleration. This should alert you to a problem.

The whole approach to the problem is wrong. You need to draw a free body diagram of the problem and look at the NET force.

F = ma really means Fnet = ma.

Last edited: Oct 8, 2007
9. Oct 8, 2007

### A(s)

i suppose i should edit that post.. i was in a rush to get to class, and made some mistakes... the 1.02 was a typo, and i meant to say that the mass of water was equal to
pi (r^2) h (1000)

in that way i could say that force divided by the mass (being acted on) should equal the acceleration

the volume of air is actually increasing as the bubble moves up the tube due to the decrease in pressure, but i spared you that part of the equation, by simply saying V to simplify things and look at the root of whatever mistake I may be making

The free body diagram should help, i suppose i will try that now

Last edited: Oct 8, 2007
10. Oct 8, 2007

### stewartcs

I was assuming that V(1.02) was the atmospheric head pressure. I was also assuming once the seal was removed the water would shoot up the tube and push the air out of the tube. If so, the head pressure due to the atmosphere would be decreasing.

With those assumptions, once the seal is removed the pressure acting on control volume of water (1kg mass in this case) would start to decrease proportionally with the height of the fluid column due to the increasing weight of the fluid column (water is entering the tube from below since the pressure is higher than atmospheric). That is to say that the pressure differential at point X (where the seal was) between the fluid column in the tube and the hydrostatic force of the tank of water at point X (or whatever the tube is submerged in) is decreasing (and will eventually equal zero).

If the pressure (P = F/A) at the bottom of the tube is great enough initially, the acceleration will exceed (at least temporarily) g. However, it will be constantly decreasing as the fluid is pushed up the tube. It is slowing down due to gravity and fluid drag. Eventually the fluid will stop moving up the tube and be in static equilibrium with a "U-Tube" effect.

11. Oct 8, 2007

### A(s)

i am going to class now, if you have any other input, about anything else that could be wrong with the acceleration at the surface, please let me know and i will check it when i get back

12. Oct 8, 2007

### stewartcs

Also, the fluid may actually shoot out of the top of the tube, in which case the control fluid would still have some acceleration remaining (or deceleration in this case).

13. Oct 8, 2007

### A(s)

those assumptions probably arose from my fault in not being specific enough...

the system's initial starting point could be at any depth say 200m, and with a larger bubble, bigger straw/tube radius, ect.

so to put it in perspective for you, you could have a 6m^3 initial bubble in a tube starting at a depth of 250 meters, with a radius of say .5meters.

as the bubble rises in the tube, it grows in size and pushes water out the top ect, so essentially it is an increasing force, acting on a decreasing weight. which is why i am looking for average acceleration. it is an acceleration that is increasing, not decreasing.

the point where the bubble is nearly at the surface (i.e. 1 kg of water on top or even no water on top) is the example i gave for simplicity based on the fact that that is where my calculation ran into trouble

14. Oct 8, 2007

### A(s)

i have in my calculations, allowances for increasing bubble size included in V
this equation above allows for decrease in resistant mass of the water on top of the bubble

also, you are correct, water does follow the bubble out of the tube with a decreasing acceleration, but i want to worry about what happens before that first

essentially i am looking at the point before the bubble breaks the surface and starts losing volume

Last edited: Oct 8, 2007
15. Oct 8, 2007

### stewartcs

Ohhh, I see what you are after now. This is a very interesting question, and of course, lots of studies have been done on it. If memory serves me correctly, R. M. Davies and Sir Geoffrey Taylor did some experiments and showed that the velocity of a rising bubble in a tube is approximately equal to (2/3) * sqrt(g*R), where R was the upper radius of the surface of the bubble.

Check out Stokes' Equations on the terminal velocity of falling spheres. You have the same situation here except they are rising instead of falling. Once the buoyant force is equal to the drag force, the bubble reaches its terminal velocity. Therefore the acceleration is zero at that point.

16. Oct 8, 2007

### A(s)

so do you think that nearly at the surface, the acceleration to that terminal velocity is almost instant or is my equation flawed

17. Oct 9, 2007

### stewartcs

I'm not quite sure what you question is. I will say that the acceleration is zero when the bubble has reached its terminal velocity. Is that what you are asking?

If you are asking at what point the bubble will reach its terminal velocity instead, then I would say it would reach it rather quickly, and, depending on the length of the tube, be nearer the bottom.

18. Oct 9, 2007

### Michele_Verona

This is my first post over here.

I liked this forum entry the most though, the way you said it was just amazing!
See you Later ;)

Regards.
Michele

19. Oct 10, 2007

### A(s)

If you are referring to a statement made by me, I am honored and thank you.

20. Oct 10, 2007

### A(s)

I was looking online and i finally found where everything goes wrong.. With buoyancy, you cannot use simply F=ma to calculate acceleration of a bubble. It is not that simple. it ends up giving nearly unlimited acceleration in many circumstances. so i guess my "easy question" was not so easy. ill get back about it when i have time