# Hydrostatic Force

1. Jul 9, 2009

### Shaneo58

You have a rectangle box filled with a liquid to height H. How will the force on the sidewall of the box change if the height (H) is increased without adding anymore fluid? (i.e. a smaller solid box is added to displace the liquid). Assuming the equation F=specific weight*depth of center point*Area of the sidewall, holds true then the force would increase based on the height increase. I have a hard time understanding this because the weight of the water in not increased. How would this same situation affect the force on the bottom of the box?

Thanks for the input!

2. Jul 9, 2009

### Naty1

It does take some thought!!! Often an "extreme" thought experiment yields clues:

Think of this situation: what is the "sidewall" pressure containing the water when you open the box and all the water spills out in a huge, maybe infinite, flat puddle....any "pressure" on the non existent "sidewalls"...none, right, maybe a smidgeon of capilary something or other at the edge...so somehow depth DOES seem to make a difference!!!!

Try reading the introduction at Wikipedia: http://en.wikipedia.org/wiki/Hydrostatic_pressure#Hydrostatic_pressure

it has your answers....and the basic assumptions (such a unchanging gravity with height and an incompressible liquid, uniform temperature,etc ); you want to think in terms of hydrostatic pressure (and force, if you like). (P = pgh)

That pressure is dependent in your example only on the height of the column of water...the pressure, for example, ten feet down in the ocean is the same as in your container if it is ten feet high and filled with water, right??

Will the sidewall pressure change if it's a cylinder or triangle shaped container with the same water depth??

3. Jul 9, 2009

### Shaneo58

Ok, so if you are trying to find the force at the bottom of the box, which would be the weight of the liquid (weight=volume*density*g), and had a situation where there was a pocket of air right in the middle of the liquid.

Would you take the total volume, including that pocket of air, in your volume calculations even though its weight is almost nothing compared to the liquid?

Or

Would you calculate the total volume and subtract the volume of air, saying that the weight at the bottom of the box is due almost entirely to the liquid?

Because the the air has very little mass I dont see how it can effect the force on the bottom of the box.