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Water displacement.

For the barge to float and neither rise nor fall, there must be equilibrium of forces.

For it to float, the resistance of the water must be greater than the force applied by the barge, hence it will settle in the water, until it reaches that point where the resistance is equal and opposite to the pressure over the surface area of the barge in contact with the water.

So, take a solid ball of metal, say cricket ball sized and it will sink like a stone. Take the same weight of metal, make it in the shape of a barge over a larger surface area and it will float.

The displaced water must go somewhere, as it does not compress. So, the volume outside the footprint of the barge will rise; the amount that this rises depends on whether or not the water is confined (think of getting in the bath, as opposed to getting in a swimming pool). As the water has risen beyond the barge, then at its base, the pressure will be slightly greater.

There will be less water under the barge, so the water pressure alone will be less at its base; however, there is the pressure of the barge at the top of this water, so this must also be added to the base pressure, for equilibrium to exist. But, as the variation in height of the water is equal to the displacement due to the weight of the barge, there will be no increase in pressure at the base of the water, relative to the base pressure beyond the barge, although the pressure as a whole on the viaduct will have increased proportionally to the mass of the barge divided by the area of the viaduct.

Theoretically.

In reality, as the water is not confined, the water will actually have a slight slope in the water level as it runs away from the barge - to remain dead level, all the connected water in the world would have to rise instantaneously by the requisite amount - so the pressure will be slightly greater beneath and within the immediate vicinity of the barge, tailing off with an increase in distance from the barge. Thus, there will be some slight variation in pressure as the barge moves along.
 
i seem to recall a "jug" that had a spout pointing down wards (not up like a tea pot) you fill it with water to the same level as the spout.

you then place an object in the jug, the water that is displaced runs out the spout.

suppose you could fill a cup to the brim, stand it on a saucer, and put a small object in it.

that way you can find out for yourself

i also think some famous bloke said something like

a body of water displaces it volume, me, i think it should say when a body is imersed in water , the phone rings.

i really think its volume not weight.

say a cube of polystyrene 10x10x10 cm

That must if you push it under displace 10 cubic centimetres of water, but equaly so would the same size of a cube of lead
 
if archemedes had a phone in the bath why didnt he put it to answer machine :?
 
Yes, but that is not applying the same weight to the water, but the same volume. Suppose that you arranged it so that the top of both objects was to be level with the surface of the water. You would have to push the polystyrene into the water to displace the water equivalent to the volume of the polystyrene; conversely, you would have to support the lead to stop it sinking. Thus the pressure at the underside of the two vastly different materials would be the same for them to be at the same level.

Simple maths for a given volume of water:

flat bottom barge 5m x 2m x 1m, weighs 200kN. Weight of water = about 10kN/m^3
Therefore volume of water displaced = 200/10 = 20m^3 and thus height of water displaced = 20m^3/(5x2)m^2 = 2m

barge size increased to 10m x 2m x 1m, same weight; volume of water displaced is still 200/10 = 20m^3, but height of water displaced is now 20/(10x2) = 1m.

Volume of first barge is 10m^3, second barge volume is 20m^3, but displaces same volume of water.
 
i said way back bouyancy and mass bear, in this instance no relationship.
regardless mass is mass.
put an object on a coffee table
its an additional mass
cut it in half and stick half above and half below (thats the water element) the coffee table is plus the additional mass
 
Ah yes, but as the table is not a fluid, if you put that closer to one pair of legs than the other, then the load would be greater to the nearest legs than the furthest. If you move it along, it would vary those loads, as you're not displacing anything with that weight.

However, with a table on four legs that was full of water, with the thing bobbing along on the top of it, the load to all four legs would remain equal, regardless of where it was positioned, as it's displacing a volume of water.

If you think that goes against my earlier comment, the water on the viaduct is not confined, so there is a slight variation in pressure and hence load to the structure as the barge moves along. If it had watertight gates at either end, the pressure would remain constant, but increased by the weight of the barge over the surface area of the water on the viaduct.
 
ok put in a more simple format.

tank of water on table with scales underneath.
put any object in the tank and the scales will register its increased weight
 
ok put in a more simple format.

tank of water on table with scales underneath.
put any object in the tank and the scales will register its increased weight
Yep, agree (providing that the water doesn't flow over the sides, of course). What was the original question?!
 
that's not measuring the weight on the viaduct

It's measuring the weight on the total area of canal.
 
Not so: I gave two answers, one for water confined to the viaduct (equal increase in pressure), the other with no confinement (varying increase in pressure).
 
i meant this nonsense
ok put in a more simple format.

tank of water on table with scales underneath.
put any object in the tank and the scales will register its increased weight
 
Oh. Well he's right if it's got watertight gates either end, though. Seemed quite a good analogy to me for that circumstance :)
 
but completely wrong for an ordinary viaduct on a canal with a barge passing through it
 
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