physics problem, help me solve please.

Can anyone help me solve this?

Pretend I had a tube of water 10 feet tall 6" dia (14.67 gallons). The bottom has a slightly smaller diameter than the tube lets say 1/4" dia smaller. inside the tube are hollow cylinders (floats) connected together to a plug in the bottom. The floats and the plug are the same dia. The floats and plug displace 11.67 gallons leaving all but all but 3 gallons (25 lbs) of water. The plug seals water but has no friction to pull it up when no water is in the tube. Will the floats being in the tube connected to the plug lift the plug up in the bore or will the pressure against the plug from only 3 gallons of water at 10 foot of column be enough to hold it down. assume the plug weighs nothing by itself.

is there any arrangement (for instance 20 foot tube or only 1 gallon of water or any other change) that the plug would lift up out of the bore? A good answer to this question will save me at least a hundred dollars and lots of time spent testing this concept to apply it to an alternative energy application.
Here is a cross-section.

It's been too long since college I would think that the floats will easily lift the plug but I can't mathematically prove it. I know it would be very hard to physically hold a 12 gallon empty drum underwater. Not even sure if you could do it if you tried. Now imagine pulling a plug out of a hole in the bottom of a 12' tall pipe. I'd think you could easily do that. So it seems like the force up on the floats would greatly exceed the force holding the plug down. Maybe I'm missing an interaction so I'd love to hear a smarter answer as well.

One way to test it more easily would be to temporarily glue a string to the stopper on your bathroom sink. Attach an empty milk jug to the string and start adding water. The jug would probably lift the stopper well before the jug is underwater. That would tell me that less than a jug's worth of air can lift the stopper against however much water is in the sink (likely more than a gallon).

I found a website that walks us through the calculations of buoyancy: http://www.wikihow.com/Calculate-Buoyancy

You can modify the volume of the floats to find the exact upward force required that will lift your plug.

Hope this helps!

Thank you for fast replies, I think I could muddle through the buoyancy part but the force on the plug has me stumped. I think I need to estimate force pushing down on the plug if the buoys were held in place but not attached to the plug. how does the volume/weight/height of the water affect the plug if nothing is connected but the buoys are fixed inside the tube for instance bolted in place. is it simply 25 lbs/area of the plug ~6" dia? if I can figure out what it would take mathematically I can size parts and build one for real to test (instead of guess and check which gets expensive).

This is like one of those two trains are heading toward each other problems:)

in practice I have noticed 30 gallon drum of air will support about 200 lbs before sinking. so I would think it would be in the ballpark of 80 lbs pulling up.

great!

everything following is physics... short and hopefully precise.

Area of the plug: A_p = pi * (5.75" / 2)² = 167.5 cm²
Density of water: p_w ≈ 1 g / cm³
gravitation constant: g ≈ 10 Newton / kilogram
Water pressure at the bottom: P = water height * p_w * g ≈ 300cm * 1 g/cm³ * 10N/kg = 3 N/cm²
Mass(Weight) of buoyant parts: M_p
Volume of the buoyant parts: V_p =11.67 gallons = 44,176 cm³

Force on the plug downwards by water: F_w = A_p * P ≈ 502N
Force on the plug upwards (lift): F_l = V_p * p_w * g ≈ 440N
Force on the plugs by gravitation: F_g = M_p * g > 0 N

So even when the plugs have no weight at all, they will stay at the bottom.


Edit: second part of the question:
To get them to lift up, you need F_w + F_g < F_l
meaning a lower water pressure and thus lower water level. However the plugs have to remain below the water surface.

Another edit: A floating object at the water surface or an overflow pipe would do the same job… I am not sure if this is what you wanted.

That is exactly what I was looking for, I am going to suggest an apple for that post. I am now wishing I was raised on the metric system! Going to try to digest this and see how I can use it to calculate the minimum it would take to get the thing to lift and then try to test it. If it works out I will show you why I needed to do this calc! Thank you!

Hi,

As long as the weight of the buoy/floats and plug and the remaining water in the tube is less that the weight of the water displaced by the floats then the plug will seal.

If you were to add more water and go over the weight of the water displaced then the plug would open until the two weights were equal.

One other issue i see is that at very low water levels in the tube then the wight of the floats and plug have to be overcome before the plug will seal but will not seal unless there is water in the tube, So the seal would have to be manually done until there is sufficient water in the tube to support the weight of the plug and floats.

Sam

Two things came into my mind:

- John, what is going on at the lower end of the tube?
My calculation assumes that there is air at the same pressure as at the top of the tube.
If this is not the case the F_w is incorrect.

The other thing is…

I wrote:A floating object at the water surface or an overflow pipe would do the same job…

is wrong. An overflow pipe overflows then the water is over the threshold.

Your arrangement opens the plug when the water is under the threshold.

Yes, Air on the bottom of the plug at the same pressure.  I used your calculations to prove out that a roughly 300cm buoy of about 5cm dia will lift a 5 cm plug out of its own 15cm tube of water.  Just got some more supplies today to do some more testing, thank you again so much for your help, your math was spot on as far as I can tell.

Is the force to lift a plug at the bottom of each of these containers the same regardless of the shape and volume of the container if the head is the same?  air on top and bottom...

The force on the plug arises form the pressure of the water above it and the area of the plug itself.
The water pressure determined by the height of the water above the point you are interested in.

pressure =force / area
force = pressure · area

When both water levels are at the same height above their respective plug and both plugs have the same area, the resulting force by the water on the plug will be equal.

You could think of the small container first. Then you add another bigger container around it (leaving the plug open). This has not changed anything. Now you fill the outer container to the same height with water. Still no change.
Now you add a closed valve between them. No change yet. Then you look at the valve: Both sides have water and the water has the same height and pressure. When you open the valve nothing changes. Now you take the ineer container away and it is still the same situation to the plug.
Or the other way around: You add the inner container with an open valve. It doesn't matter whether the valve is open or closed.

got it, I didn't know if the pipe above the plug was smaller than the plug if the pressure and the force would be less than if it were bigger.  going to do another test today but it looks like if the math is correct the idea I had might not work out for me.  assume 300 cm tube, 300 cm float 5cm dia and 5cm dia plug.

force up on buoy
areaxlengthxdensityxgravity
pi(2.5cm 2)* 300cm*1g/cm3*10N/kg

force down on plug
areaxheadxdensityxgravity
pi(2.5cm 2)* 300cm*1g/cm3*10N/kg

at best the forces appear to be the same and only if I can make the buoy larger in dia than the plug will I get it to pull out.

Why can't you make the plug smaller?

the reason for doing the math was to see if it could be the same or larger then the buoys.  Going to try it out with 600 cm tube and 600 cm buoy and see how it does.  right now I have 300 cm tube and 300 cm buoy and it feels like its close to lifting it.  If it works out, eventually I will post photos of why I would go through all the trouble to solve such a silly looking math problem.  If not I will throw it into my scrap pile of inventions that didn't function as planned:)  Maybe I could eventually open a museum!

Small changes in the diameter of the plug will have a large impact on the force needed to pull it up, since the force on it is proportional to the square of it's diameter.  For example, if you were to reduce the size of the plug to 1/2 of it's original diameter, the force needed to lift it would be 1/4 of the original...
If you have any latitude to change the bottom orifice diameter, then you can make the system work with nearly any arrangement of floats and water level.  The force on the plug is directly related to the height of the water column above it.

The following is a true story.

During WW2 a damaged airplane landed In a large pond and sank. Of course the USA wanted the plane back: it could either be repaired or used for parts. But, they were in RURAL China and there was no heavy equipment available The villagers offered to get it out, but nobody listened to them.

So, after the Americans had tried everything they could think of, the villagers again offered to get the airplane. This time people just shrugged, snickered  bit, and said "go ahead".

So then the villagers got the plane up. They had no problem at all. Men dove in with small bundles of bamboo and tied each bundle under the wing of the airplane. It took a huge amount of bamboo, but the next day the plane rose to the surface and was then towed to shore.

The moral of the story is, small mounts of lift adds up. If your first attempt does not work, keep adding things that will float. If you do not run out of room first that thing WILL* float!