# I need more Brain Juice here

The water level would be the same in both vertical tubes in the above
static fluid model regardless of the diameter of each tube.
F = density x Area x height x g ..... (Area x height = volume)
Pressure = F/Area = density x height x g = Fluid's Mass x g / Area
What I don't get is:
If one of the 2 tubes of water was a large ***CONE*** shape at its top....then how come the water level would still be the same level in both tubes..... Since the Pressure should be higher there?
Pressure = Force/Area (= Fuid's Mass inside the CONE/ Narrower Area at the Cone's bottom)..... Therefore how come the pressure at the base of the CONE isn't much higher than that of a simple vertical tube (since if that was the case then the water wouldn't be the same level in both tubes)????
\ / | | \ / | | | | | | | ------------------- | |________________|
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The short answer is that the sides of the cone support some of the weight of the water in the cone.
Best of luck - Mike
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Mike Yarwood wrote:

Up to a certain point, a heavy block on a slope will still exert a downward force. (Not 100% force but fy = force x sin (slope angle))???
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Yep but your water is a static fluid so we are neglecting any friction forces (and surface tension) - draw some arrows but make sure they are all normal to surfaces. Best of Luck - Mike
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Mike Yarwood wrote:

For a block on a slope the arrows forces are not null, and therefore there would be a force excerted at the base of the slope as well as the walls of the slope...
If there was a heavy block on a slope it would excert a force on the slope (cone's walls), as well as a vertical and horizontal force:
Fy = block's weight x g
F along slope = Fy / sin (slope's ange)
Fx = Fy / cos (slope's angle)
So if you has 3 springs:
1. Fy= One at the same angle as the cones wall would be compressed (representing the force excerted by the block on the wall itself) 2. F = One just below the block on the slope would also be compressed
3. Fx = and One just to the left of the block would also be compressed
????
Unless perhaps Fx of adjacent fluid molecules and Fy of cone's wall are sufficient to nullify any F going down along the cones walls???
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Look at the outcome and use it to guide your 'interpretation' of the physical model
Yes there is a force transmitted to the material of the tube, and the fact that it is "***cone***" shaped means that the "***support***" for the material of the tube has to be stronger than if it was vertical. However, the hydrostatic pressure in the liquid column is not affectted by this effect. - why? (1)
(1) exercise for the reader.
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No - the F goes down the cone walls all right ( and so does the weight of the cone itself ) but it's all redistributed in the structure at the base of the cone.
Try this one :
|~~~~~~~~~| |~| | | | | | | | | |___ ____| | | | |___________| | |_____________|
sorry for my lousy ascii art : down at the bottom of the big fat cylinder there's a whole lot of force acting to distort the bottom of the cylinder but it all goes down the walls of the little tube which is supporting all that weight (except for a little thin cylinders worth of water directly above above the little thin tube) none of it somehow magically gets back into the water .
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Mike Yarwood wrote:

All right it seems more than plausible, so i guess the force is 100% absorbed by the walls of the cone.
Thanks to all !
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Draw straight lines up the cones from thinnest point, call cone tube.
Recall (static) fluid pressure acts equally in all directions therefore we have horizontal equillibrium up effective heights of cone. hence fluid in cone in cone, supported by the coine vertically. Only active vertical forces in the cylinder are those in the 'tube' which is not conical.
kind of.
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The mechanics of fluids have little in common besides the physical properties. Fluid friction (viscosity) is measured in terms of the velocity it travels as a body proportional to other parts of it away from the wall. du/dy and it's resolution is pretty complicated.
Fluid statics does not involve viscosity.
Pressure is constant at any given depth regardless of the shape of the vessels.
c.f. water levels
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Mistake, this should have read something like:
The mechanics of fluids have little in common with solid state mechanics besides the physical properties (of the materials in the systems).
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wrote:

Fluids and solids are, by definition, different physical things. The definition of a fluid is something which will shear under it's own weight, solids will not.
The equillibrium of static fluids results in the forces exerted by a fluid element (or similarly, by virtue of the fact of static equilibrium, the forces exerted upon the fluid element) are equal in all directions.
\ | / -- o -- / | \
for a wee diagram.
Hence the vertical force on an element is equal to a horizontal force on the element (or exerted by the element). Hence static fluid has horizontal and vertical (and indeed any planar) equilibrium. It does not want to 'slide' under friction.
I explained above how the fluid is in horizontal equilibrium along the plane of a vertical line up the bottom of the cone's incline. The only vertical forces acting on the fluid in the none conical sections are those in line vertically with it.
A hole of 1 foot diameter at 27 foot depth displays the same pressures in the water which fills it as does a hole of 1 square mile at 27 feet of depth of water, whether either has inwardly, outwardly or non sloping sides.
Quirky stuff is fluid mechanics, wierd and wonderful. Go study it, the stuff is 'cool'.
--
Billy H

The spirit is not the letter, 2 corinthians 3,6
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the pressures are equal.
the heads should be at the same level.
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snipped-for-privacy@hotmail.com wrote:

The pressure just depends on depth, on the weight of water directly above. Water in the slanted part of the cone is not pressing on the water in the vertical tube. It just causes pressure on the cone- shaped parts of the walls.
- Randy
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snipped-for-privacy@hotmail.com wrote:

(Pressure = Force/Area )
Note thar area is below the fraction bar.
The expresson of pressure inherrenty includes the area that the weight of the fluid will be distributed over.
1 kg/ cm^2 = 2kg / 2(cm^2) = 4 kg / 4(cm^2)
These are all the same pressure but may represent different cross sections of a cone.
Sue...
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|_ _| |_ _| | |
Pitiful to watch. Androcles
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I have thought about it, and here is the simplest answer that I can come up with. The reason that you don't have a difference in the water level is simple, much more simple than what everybody else is trying to propose.
The water remains the same level because it is at it's lowest potential energy state. Since water is allowed to flow between the two sides the water will flow towards it's lowest potential energy state. If there was a difference in height of the water, then it would have to be flowing somewhere.
For example, if you were able to put one end of a pipe at the bottom of the ocean and the other end above the water level, you wouldn't create a water spout. It is the same thing in this situation. Since the air pressure is the same above the cone as over the cylinder, the water level remains the same. Unless there is another force involved the water will remain the same.
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The other answer, that's basically as simple, is that pressure always acts normal to the solid surface.
People often ignore that because so many fluids problems use vertical walls where the net force at any level integrates to zero. This is *not* true for sloped walls.
Tom.
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wrote

It's about horizontal and vertical equilibrium in statics.
If you want an explanation I'll write one and email it, it's hard to draw figures on usenet.
--
Billy H

The spirit is not the letter, 2 corinthians 3,6
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water pressure increases with depth, not surface or planar area, the only effective variable is the depth. Hence the fluid is subject to same pressures at equal depths.
If you had two tube of different diameters you can use the system to amplify force and increase pressure at heads of cylinders but static presures at equal depths remain the same.