Archimedes

[richardabeattie]

I started this and am now losing the will to live.

Cindersasilor seems to be getting closest to an intelligible answer.

Let me try one last time.

Lock empty
Barge an almost perfect fit with the walls and jacked up of the floor just a tiny amount.

Barge weighs 100 tonnes. Air space all round the barge sides and bottom is only sufficient to contain one tonne of water.

Pour in the said water which brings the water level up to the barge's waterline and creates all this pressure (but no real tonnage of water)

Barge now floating? I doubt it.

Now, with the water inside the lock up to the barge waterline and the tide outside up to the same level, open the gate. Now it must be floating.

My theory is that for an object to float it must displace its own weight of water and that displaced water must be able to continue to exert force on the object. The lock gate cuts off that displaced water. So for the barge to float inside a closed lock the amount of water left in the lock when the barge enters must exceed the weight of the barge.

Come on, let's be havng you!

 

[mel80]

[ QUOTE ]


Barge an almost perfect fit with the walls and jacked up of the floor just a tiny amount.

Barge weighs 100 tonnes. Air space all round the barge sides and bottom is only sufficient to contain one tonne of water.

Pour in the said water which brings the water level up to the barge's waterline and creates all this pressure (but no real tonnage of water)

Barge now floating? I doubt it.


[/ QUOTE ]

It would float even if there were only 1kg of water. Try it yourself with a couple of tuppaware boxes with weights in them.

It's a reasonable theory that you have and it would probably be most peoples intuative starting point. However it is pretty easy to disprove empirically. You need to abandon the idea that it is the weight of displaced water that causes an object to float; it is the action of pressure (or more accurately inequalities of pressure) that causes the upward force.

 

[richardabeattie]

So we can float a battleship in a cup of water provided the dock is a close fit?

 

[gandy]

Yes.

I can see that this worries you, but it is correct. In your example, imagine you've poured in the water until the battle ship is on the very point of floating but is not quite afloat. Pour in another litre of water. You've just lifted thousands of tons of ship by pouring in one kilo of water. That sounds wrong, however your one litre of water has only lifted the ship through a tiny distance.

 

[Cruiser2B]

Equilibrium!

Perhaps if you visualize replacing the displaced volume of the barge with a block of water of the exact same dimensions - it has the exact same weight as the barge - do you think it will submerge in the 1" thick water box. No, of course not. Same reason the barge won't.

 

[johnalison]

If you haven't passed away yet; just imagine the ship etc with the lock walls taken away and the water around the ship extended infinitely. The water adjacent to the ship is the same and the ship is floating!

 

[Bergman]

If you start with the lock gate open and the barge floating you are happy.

But why then would the barge sink when you close the gate?

It wouldn't

Think perhaps its terminology

It is not kept afloat BY the water it has displaced

Its kept afloat because it takes the place of water that COULD BE in that place.

Perhaps consider that the barge in the tight fitting lock was loaded - what would happen if you started to unload it with the gate closed.

 

[Bajansailor]

Another similar situation happens every day with the Panama Canal locks - many ships are designed to be 'Panamax', ie to conform exactly to the length and breadth of the locks. The breadth limit for ships is 32.2 m. - that is why you will see many big container ships and cruise liners with overall beam = 32.2 m.

These big ships literally have only inches between them and the lock side, and not much room at each end either. The canal authority love these big ships because not only can they charge a higher fee (loosely based on gross tonnage) for transit, but also the passage of a Panamax ship through a lock uses a relatively small quantity of water (which has to come from the lakes, ie it is not infinite). because most of the lock is filled with ship, in the same manner as the rectangular barge in the examples illustrated above.

I know it has been said that 'the barge is not displacing its own weight of water', if you drop it into an almost empty lock with only a minimal amount of water of water in the bottom of the lock.

But it effectively is.

Lets say the barge is perfectly rectangular, 10m long, 1m wide, and 1 m deep. It has a total volume of 10 cubic metres (10 x 1 x 1).

And lets say that the lock is 10 cm (approx 4") bigger all round. So the size of the lock is 10.2m long, 1.2m wide, and 1.1m deep. There has been a lot of rain in the hills, and the river is running high; the water is just lapping against the edge of the lock, and is about to overflow.

The total volume of water in the lock is :
1.1 x 1.2 x 10.2 = 13.46 cubic metres.

Or 13.46 tonnes if it is fresh water.

Now let us assume for simplicity that the barge weighs 9.6 tonnes. In fresh water it will have a draft of 0.96 m, and a freeboard of 0.04m, assuming it has level trim.

If we lower the barge carefully into the lock, 9.6 tonnes of water will spill out over the edge.
Hence we have 13.46 - 9.6 = 3.86 tonnes of water left in the lock, 'supporting' 9.6 tonnes of barge.

There is still 1.1 - 0.96 = 0.14m (5.5") of water under the keel of the barge.

If we lift the barge out of the lock, that 3.86 tonnes of water will remain. The water depth in the lock will now be only 0.32m (12.5").

If we lower the barge gently back in to the lock, the water depth will increase again to 1.1m, from 0.32m.

Has anybody ever tried snorkelling? Even with a long snorkel, the deepest you can breathe in when underwater is probably about 12" or 30 cm. This is because of the pressure of the water - it is almost 1,000 times heavier than air.
(Yes, air has density! Just over 1 kg per cubic metre at atmospheric pressure).

So imagine the pressure at 1.1m. water depth. Even with a long hose going to the surface there is no way you could breathe, because of the water pressure on your body (and hence lungs).

Water pressure is a function of depth and density. At a draught of 0.96m. the overall force exerted by the water supporting the barge it is equal to the downwards force exerted by the barge (due to its weight) on the water, and the barge is thus in equilibrium.

Totally confused now?

I hope not! /forums/images/graemlins/laugh.gif

I hope that somebody else has in the meantime come along and explained it in a simpler fashion than I have.

(Edited re final draught of 0.96m).

 

[Pye_End]

Yes.

The whole premis of your question is not right - it is all about density not displacement.

When the level is up to 'the waterline' then the object will float because it will be less dense (for the volume of water displaced) than the surrounding water.

See here - How to float

However, as I tried to say earlier, Archimedes said that an equal volume of water should be displaced, but in the case of 1" around the battleship I cannot see there being enough. I suppose that when he talked about displaced water then you should think not in terms of putting a boat into the water and where the water goes (ie a dynamic situation), but displaced water as would have filled the hole if the boat wasn't there.

 

[boomerangben]

A question to you Richard.

Two tanks of different sizes are connect via a tube and you fill the little tank with water, then let the levels settle. Where are the water levels in each tank?
a) Water level in Little tank higher than that in the Big tank
b) Water level in Little tank lower than that in the Big tank
c) Water levels are the same.

I hope your answer is c), but what your intuition is telling you about the barge in the lock is answer a)

Remember that the pressure at the bottom end of the tiny tube of a mercury baromter is the same as the pressure exerted by the whole atmosphere.

 

[MoodySabre]

Some of you seem to make things complicated when they need not be.

If it takes a thousand tons of water to be displaced in order that a ship can float then it will not float in less than a thousand tons of water. The water needs to be deep enough to prevent the ship from touching the bottom. Your tiny dock would have to be very very tall.

 

[Pye_End]

No. You do not need 1000t of water to float 1000t ship.

As said earlier you can float a battleship in a teacup of water if the dock was the right shape.

The Archimedes bit is that if the ship was not there, then the hole it leaves will take 1000t of water.

 

[gandy]

That's simply not the case. When the water in the dock reaches the ship's normal waterline then addition of further water will float the ship off the bottom. The alternative would mean the ship stays resting on the bottom while the water level rises above the topsides and over the superstructure until this thousand tons have been added.

 

[Sinbad2222]

You are incorrect. Archimedes Principle should be stated then all the above conundrums are clearly explained.
WHEN A BODY IS TOTALY OR PARTIALY IMMERSED IN A LIQUID IT EXPERIENCES AN UPTHRUST EQUAL TO THE WEIGHT OF LIQUID DISPLACED
Very simple really.

 

[mel80]

I think I said as much above.

What archimedes principle doesn't say (and this is the bit that seems to be causing the confusion) is that the upwards force is caused by the displaced water. In other words, you could gather up all the displaced water and carry it off miles away from the ship (send it into space if you like) without causing the ship to sink. That is where the apparant paradox above falls down.

 

[richardabeattie]

So we are all agreed. The cup of water does float the battleship in the tight dock because the pressure of the water depends only on depth and pressure is all that matters.

Apart from me. I think the battleship would sit on the bottom and squish the cup of water up a bit though the Panamax thing has me worried!

An analogy: high voltages (pressure) doesn't get much work done if there is no significant amperage. But I am clearly out of my depth and will have to go and get out the tupperware and experiment. Mere pragmatism! Goodnight all!

 

[boomerangben]

Sinbad,

Which is exactly illustrated by the point made by Pye_end above. Archimedes principle is all to do with the "hole" in which a vessel sits. It makes no mention of what has to happen to the displaced water or indeed how much water surrounds the ship.

Mr Beatties confusion (and I understand his confusion) lies within the principles of hydraulics (of which Achimedes' Principle is a special case).

Archimedes is perhaps best illustrated mathematically.

Imagine a regular rectangular box 100m long x 10m beam, drawing 1m.

It displaces 1000m^3 which assuming fresh water would indicate a mass of box of 1000tonnes = 1000 tonnes of displaced water.

This bottom of the box has an area of 10m x 100m = 1000m^2.

The upward pressure exerted by the water pressure (1m deep) = 1000 (denisty of water) x 1 (water column height) x g (acceleration due to gravity) say 10 (nearly 9.81!) = 10,000 Newtons per m^2.

Multiply this by the area of the bottom of box to give the force acting upwards (buoyancy force) = 10,000 x 1000 = 10,000,000 Newtons.


10,000,000 Newtons = 1,000,000kgs = 1000 tonnes which hey presto is the displacement of the ship and the mass of water displaced.

It works for other numbers too.

The difficult bit to get the head around is that such a small amount of water (in the case of a barge in a well fitting lock) can produce enough pressure (hydraulics) to make Archimedes works. Which it does. If it didn't, Archimedes Principle would be dead in the water.

 

[donm]

Isn't the trick to weigh less than the weight of water displaced if flotation is to be achieved?

One thing I'm sure of - I'm keeping away from them tricksy locks from now on! /forums/images/graemlins/confused.gif

 

[RichV]

"Another question in the same vein which I was once asked at a job interview. You are in a smal boat on a reservior and have a brick which you throw overboard. What happens to the water level in the reservoir?"

Assuming that the water is fully incompressible and that the reservoir is sufficiently shallow that the brick will rest on the bed of the lake.

The water level in the reservoir will fall.

 

[floatything]

Well.er ... but..... doesn't it depend on the relative densities of the brick and the water?