Archimedes

Pye_End

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No - it is not about pressure - it is about relative density.

You could discuss and resolve the forces if you wish - but I think there is a danger when talking about pressure that you think it is water pressure, which increases with depth. Depth is irrelevant and has no part in this thread.

The force exerted upward by the water on the floating body is g. Anything more and it will be accelerating upward, and anything less it will be sinking.

Buoyancy has its place, but is really a factor of relative density - buoyancy is the force exerted for example by a submerged float - a buoyancy aid of 100 newtons will exert a force of 100 newtons upwards when fully submerged. This is due to the relative density of the float compared with the fluid. It will continue to accelerate upward until no further force is experienced (ignoring friction) - ie when enough of it is out of the water so there is neutral buoyancy - ie it is floating.

The volume of water that it displaces or the water pressure have no bearing on whether it will float or sink.
 
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angelsson

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An excellent and accurate if technical explanation.

I remeber the principle being explained in this way,

Imagine a large tank say 20ft x 20 ft x 10ft deep

It is say 3/4 full of water, in which a boat is floating, depth not important so long as enough water under the keel.

In the boat is a cube of concrete block, weighing 1 ton.

The block is lifted out of the boat, and completely submerged into the water.

Q. Does the water level in the tank rise, stay the same, or fall?

It actually falls, because the block when in the boat displaces its weight of water, whereas when in the water displaces its volume, which is much less.

Apply this to the lock problem is why the barge floats, it has diplaced its weight of water whatever the gap around it and is therefore floating.
 

Avocet

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Richard, would it help if you thought not so much of the small weight of water IN your lock but rather thought of the huge weight of water MISSING from your lock when the barge is in? Also, I don't know if it would help but the walls of the lock are doing a lot of work too. IF you made them out of plywood they'd burst open.
 

achwilan

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O worthy forumites... Putting out to sea with so ragged arguments to keep your estimable yachts afloat is wonderful recklessness /forums/images/graemlins/grin.gif /forums/images/graemlins/grin.gif /forums/images/graemlins/grin.gif
 

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Pye_end,

It is all about pressure. Density is a property of the object, and cannot in itself generate a force. Buoyancy is simply the sum of all the pressure forces acting on the body. If these equal the mass of the object before it floods, it will float.
 

cindersailor

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Correct! I was offered the job, but turned it down. A good decision as it turned out as it was with the National Coal Board a year or so before it was abolished!
 

Stoaty

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Throw your teacup of water into your dock and the boat floats. Freeze the water and whip your boat out. Peer into the dock and you will see the frozen water with a big hole in the middle the shape of your boat. If that hole was filled with water it would equal the weight of your boat. Its the DISPLACED water that does the floating trick. The water that wasn't there in the first place.
 

jimbaerselman

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[ QUOTE ]
It is all about pressure.

[/ QUOTE ] Exactly.

Convert your boat into a light weight cube (I know,it won't sail very fast now). Put it into a box 1mm larger in each dimension. Allow some imperfections at the bottom so water can seep under the boat. Pour in water.

Water will spread along the bottom. Then it will rise up the sides. Water pressure at the bottom of the cube will depend on the 'head' of water - the height of the upper water surface relative to the cube bottom.

The net force due to water pressure at the sides cancels out. But as the water level rises, pressure acting upwards on the cube bottom increases. So far, only a small amount of water is being poured in, so little work is being done (cf earlier post on V x A, = kilowatts = work).

At a certain point, the pressure on the cube bottom multipied by its area will create an upward force equal to the downward force exerted by gravity on the mass of the boat.

The boat will rise - 'float'. From now on, vast quantities of water will be required to lift the vessel further. That's work (kilowatts). And over time, that adds up to energy (kw hours) which is the height through which you lift the boat.

The fact that boats have shape, and locks are a different shape, doesn't alter the physics of the matter. All side forces cancel (otherwise the vessel would rush to one side of the dock). And the horizontal area over which (various, depending on depth) upward pressures are acting create the net upward force of flotation.

Whether we're in a big sea, or a small container, those sums will be the same.

It's the vertical component of pressures multiplied by horizontal areas that make things float. Thank heavens. And chucking that brick overboard lifted the boat, but dropped the tide level. Now, does that mean my keel can still just scrape over the reef?
 

Pye_End

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[ QUOTE ]

At a certain point, the pressure on the cube bottom multipied by its area will create an upward force equal to the downward force exerted by gravity on the mass of the boat.



[/ QUOTE ]

Yes - but you don't define the point at which it happens.

You have resolved the forces, but the point I am making is that the point at which the object floats is when the relative densities are the same.

Pressure is simply force divided by area, and you have then mutiply it back up again to get back to force.

Buoyancy is the force that you are talking about, and it is dependant upon relative density.
 

mel80

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[ QUOTE ]


Buoyancy is the force that you are talking about, and it is dependant upon relative density.

[/ QUOTE ]

That isn't technicaly true. Buoyancy is independent of density; i.e. a 1m cube of lead has the same buoyancy as a 1m cube of polystyrene. The reason that one floats and not the other is because the weight of the lead is greater than its buoyancy whilst the reverse is true for the polystyrene. The point of transition is, of course, when the density of the object is equal to that of the water.
 

Pye_End

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buoyancy (boi'unsē, bOO'yun–) [key], upward force exerted by a fluid on any body immersed in it.

I would hate to have a buoyancy aid made of lead.

If this definitition of buoyancy is true then it must be dependant upon relative densisity.
 

mel80

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Yes, that definition is true; and in each case the upward force exerted by the water is the same. What varies is the downward force exerted by gravity upon the mass of the object. That is why the direction of the net force is different in each case, and one floats and one sinks.
 

Pye_End

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I think I understand your point but can't quite get my head fully around it.

How would you therefore measure buoyancy (or calculate it)?

It still does change my post that to measure or calculate when an object floats you have to look at the relative densities.

The implications of the previous post is that a deeper vessel (with the same mass and volume) will float before a shallower vessel due to increased water pressure as you go down, which is not the case. The head of water is irrelevant.
 

bbg

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Try this experiment. Try it in the sink unless you want a very wet countertop.

Take two bowls that stack together with a very close fit.

Fill them both almost to the top (say 3/4 full, maybe a bit more) with water.

Carefully place one bowl into the other.

Watch the water overflow from the bottom bowl.

Thank me for having told you to do this little experiment in the sink.

Notice that the upper bowl is floating in the lower bowl.

Take the bowls apart - carefully! Don't spill any water.

Notice that the lower bowl has much less water in it now.

Realise that the upper bowl, which weighed much more, was floating in water than weighed much less.

Conclude that the walls of a lock, dock etc. act in exactly the same way as water - they exert a sideways or upwards pressure on the water next to the boat preventing it from rushing away.

Curse when your wife sees you playing with bowls in the sink and assumes you will now be washing the dishes tonight.

Hope that helps.
 

mel80

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[ QUOTE ]

How would you therefore measure buoyancy (or calculate it)?


[/ QUOTE ]

Well, you could calculate the pressure acting at each point on the object and sum the forces, however that's a bit of a hassle. Fortunately this guy called Archimedes realised that the physics worked in such a way that the buoyancy was equal to the weight of water that the object displaces (see what I mean about it being a useful tool). Another way would be to measure the difference in the apparent weight when the object was out of the water compared to when it was in it (allowing for the fact that the object will still have some buoyancy in air).

[ QUOTE ]

The implications of the previous post is that a deeper vessel (with the same mass and volume) will float before a shallower vessel due to increased water pressure as you go down, which is not the case. The head of water is irrelevant.

[/ QUOTE ]

No, because a shallower vessel of the same volume must, by definition, have a greater surface area at it's minimum depth. Thus, although the pressure is lower, it is exactly compensated for by the increase in surface area over which the pressure can act.
 

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[ QUOTE ]
How would you therefore measure buoyancy (or calculate it)?

[/ QUOTE ] Attach a spring balance to the object with a line. Lift, and note the reading (actual weight plus trivial weight of line)).

Immerse the object. Note the reduced reading. Difference equals bouyancy.

Note that if the second reading is zero, the item floats, irrespective of the first reading. Whether the item floats or not does depend on its density. If its density is less than that of water, its waterline will depend on its density.
 

Pye_End

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So buoyancy is a force, but you can't calculate it, but is not dependant on relative densities!

Don't accept this.

If you consider a tennis ball, and you put it under the sea, it bobs up. To look at the forces, mg is less than buoyancy (which is yet undefined).

Now it bobs into the air and it falls back down to the sea - ie it does not float in the air. So what has changed? mass is the same, g is the same, therefore buoyancy has been reduced in air.

So what is it about the change in medium that has done this? Volume is the same. It is the density of the medium. Therefore I can only conclude that the relative densities of the object and the fluid must be a factor in determining buoyancy.


Re. water pressure -

You miss the meaning of my post I think.

Consider a cube which is 2/3 the density of water.

It will float with 2/3 underwater (by volume). Now chop half the cube off and stick it together as a cuboid and put it back in the water. It will still float with 2/3 (by volume) underwater, but will be at a different depth. Yes I agree that the surface area has changed and therefore the pressure on that surface, but what I am trying to explain is that the 'water pressure' which increases with depth has no bearing on how far down or up it will float - it is entirely dependant on the relative densities of the object and the fluid. Tell me, using your water pressure theory how you would predict such where the water line mark would be on such an object?
 

Pye_End

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I sort of agree with your method - but if the object 'floats' then you can measure how much it wants to float using the spring balance 'upsidedown' and underwater. I completely concur but it is mel80 that does not.

An object will float when the mass of the object divided by the volume of water it diplaces equals the density of the fluid.

'Whether the item floats or not does depend on its density. If its density is less than that of water, its waterline will depend on its '

How can you say that it does not depend on density, but then say that it does. Confused.

Edit - changed commend on density calculation - can be done two different ways depending on what you are trying to achieve.
 

mel80

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[ QUOTE ]
So buoyancy is a force, but you can't calculate it

[/ QUOTE ]
More accurately it is the sum of many forces. Yes, you can calculate it directly from the forces exerted by the water, but this in inconvenient; that is why things like Archimedes principle are so useful.

[ QUOTE ]

So what is it about the change in medium that has done this? Volume is the same. It is the density of the medium. Therefore I can only conclude that the relative densities of the object and the fluid must be a factor in determining buoyancy.


[/ QUOTE ]

I can see where you're coming from with this, and it's certainly one way to look at it. However, the explanation most directly related to the forces involved is that the pressure changes associated with varrying depth are greater in water than they are in air. After all, a column of water just 30ft high exerts the same pressure as the entire atmosphere! That is why the pressure difference and thus the buoyancy is greater in water than it is in air.

[ QUOTE ]


Consider a cube which is 2/3 the density of water.

It will float with 2/3 underwater (by volume). Now chop half the cube off and stick it together as a cuboid and put it back in the water. It will still float with 2/3 (by volume) underwater, but will be at a different depth. Yes I agree that the surface area has changed and therefore the pressure on that surface, but what I am trying to explain is that the 'water pressure' which increases with depth has no bearing on how far down or up it will float - it is entirely dependant on the relative densities of the object and the fluid. Tell me, using your water pressure theory how you would predict such where the water line mark would be on such an object?

[/ QUOTE ]

Ok, first of all you can state that because the object is floating, its buoyancy is exactly equal to its weight. E.g. If the weight is one Newton, then the upwards force must also be one Newton.

In order to find the waterline, we must calculate the depth required to produce such a force on such a surface. To do this we divide the surface area (of the bottom of the cube) by the required force. That will give us the required pressure. It is a simple matter to calculate the depth of water that will give us that pressure.

So, for your example (lets make it a 1m cube and say that the fluid weighs 10 000N/m2 to make the maths easy). The weight of the cube is, therefore, 0.66 x 10 000=6600N. Thus the buoyancy must also be 6600N. Dividing that by 1 (the surface area of the bottom of the cube), we see that the pressure must be 6600N/m2. A column of water 1m tall exerts a pressure of 10 000N per m2. This means that we require a column of water 66cm high to create a pressure of 6600N/m2. The base of the cube must therefore be at that depth.

Turn the cube into a cuboid, it will still have the same weight 6600N, but its bottom surface will now be doubled in area, as such the required water pressure will be 6600/2=3300N/m2. A column of water just 33cm tall will provide this.

I can see where you’re coming from with the densities thing, and it is a pretty intuitive solution (it works as well). From the point of view of the physics it is a less satisfactory solution because densities cannot exert force; pressures can.
 
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