Easy Question about Fluid Statics

[secret2]

I am a newbie to fluid mechanics, and I am confused about the "hydrostatic paradox". To begin with, consider three containers. All three have the same base area (all circular), but the angle between the base and the "wall" are all different:

Container 1: obtuse angle
Container 2: right angle
Container 3: acute angle

And the textbook says that the FORCES on the base of the three containers are identical. Why should container 3 have the same as the rest?

 

[ramollari]

Yes, the forces will be the same, provided the additional condition that the heights of the fluids are equal. Then, since pressures and areas are equal at the bottom, the same forces will apply.

 

[minger]

I did a quick search and found this website that can prolly explain it much better than I can:
http://www.scubageek.com/geek/articles/wwwparad.html

 

[Doc Al]

And the textbook says that the FORCES on the base of the three containers are identical. Why should container 3 have the same as the rest?

Unfortunately, the link that minger provided does not illustrate the 3rd case, which is the most interesting one. In cases 1 and 2, there is an unobstructed column of fluid directly above the base, so it's easy to imagine that the force exerted on the base equals the weight of that column. But in case 3, some of the column of water is obstructed by the sides of the container; it turns out that the sides of the container exert a downward force that just compensates for the truncated height of the column of fluid. The net result is that the force exerted on the base is equal in all three cases (as long as the height of the fluid is the same).

 

[secret2]

Thanks a lot Doc Al! That's exactly my concern! But why would a force be exerted by the wall anyway?


Just one more scenario. Imagine that we have the following device:

[FONT=System]
|     |              |  |           \                 /
|     |              /   \            \              /
|     |            /       \           \           /
|     |          /           \           \       /
|     |        /__       __\           \    /
|     |_______|      |________|  |_______
|__________________________________|
[/font]

Does the column in the middle have enough pressure (or force) to balance the others so that all three keep up the same level? If so, is it because for the column in the middle, once again, a force is exerted by the wall?

Thank you!

edited by enigma to add [ code ] and [ font ]tags for clarity

 

[secret2]

No good, the diagram doesn't come out right...I'll try describing it in words. The middle one has a circular base area, all of which is connected to the base. And the wall of the middle column, of course, makes an acute angle with the base.

 

[Doc Al]

But why would a force be exerted by the wall anyway?

Because the fluid pushes against the wall and the wall pushes back.

Does the column in the middle have enough pressure (or force) to balance the others so that all three keep up the same level?

The fluid reaches the same height in all three columns.

If so, is it because for the column in the middle, once again, a force is exerted by the wall?

The wall will exert a downward force on the fluid.

 

[pervect]

A couple of suggestions - if you wrap your ascii diagram in

[tex]\mbox{

}[/tex]

tags, it won't get reformatted.

The way I'd approach pressure is to look at a small volume element of fluid. The net force on the box will be the gradient of the pressure. But this approach may be a bit advanced, it's not the simplest possible approach.

http://astron.berkeley.edu/~jrg/ay202/node6.html

outlines this approach, having some nice diagrams.

The math requires vector calculus, but the link derives the fact that for any fluid in equilibrium in a conservative force field Phi (such that the force is the gradient of the potential - the Earth's gravitational field is an example of such a field, where the force is the gradient of -GM/R), surfaces of constant pressure, constant density, and constant potential Phi must all coincide.