1. Jan 29, 2006

### physicsss

The two open tanks have the same bottom area, A, but different shapes.
When the depth, h, of a liquid in the two tanks is the same in accordance with P1 = pgh + P2 the pressure on the bottom of the two tanks will be the same. However, the weight of the liquid in each of the tanks is different. How do you account for this apparent paradox?

picture:
http://img296.imageshack.us/my.php?image=untitled7jl2.jpg

Is it because weight of the liquid is the specific weight times the volume of the liquid, which the tank on the right has more of?

Last edited: Jan 29, 2006
2. Jan 29, 2006

### durt

Consider that pressure (hence force) is exerted by the liquid on the walls of the containers.

3. Jan 29, 2006

### Cyrus

Not if the densities of the two liquids are different. The rho term will not be the same in both cases.

4. Jan 29, 2006

### Valhalla

I really don't think that there is any paradox right here if you think about what pressure is. The amount of matter on top of the bottom of the containers is the the same in both situations (it would help if I could draw a picture).

As per Resnick, Halliday, and Walker "Fundmental of Physics" 7th edition "The pressure at a point in a fluid in static equilibrium depends on the depth of that point but not on any horizontal dimension of the fluid or its container."

Basically, the matter that is causing the pressure at the bottom of the container is only the matter that is directly above that area. So it doesn't matter if the container kind of "V"s out like the picture shows. The way I like to think about it (I may be completely wrong in this) is the pressure at a point with area "A" is only caused by the amount of material above it with area "A".

As far as having different weights, well one has more volume hence more mass is in it (because they are filled to the same height being key here). More mass equals more weight. More mass does not mean more pressure. A higher density will cause a higher pressure though.