# Regarding Liquid Pressure

1. Jul 31, 2013

### B4ssHunter

i have been studying pressure and hydrostatics
and i stumbled upon this
now imagine a container filled with water , and there is a cube in the water
now i know that the weight of the displaced water = the upthrust force
but what bothers me is , how is the water pressure at a certain level the same even after the Cube is inserted , for instance water pressure at POINT B is the same even when the bottom of the cube is exactly at point B * 3 meters from the top of water for instance* , shouldn't it increase ? and if it increases shouldn't it cause more upthrust force and therefore suspend the cube in water ? according to newtons law every action has a reaction , now i suppose water reaction should be equal to the cubes weight and therefore suspend it ?
if the pressure doesn't change then this explains why the cube sinks , but again how is it constant ?
if i am wrong after all , can you tell me what causes pressure at the bottom of the cube , if it is atoms being compressed , then i should be right because the cube will compress them more causing them to suspend the cube

Last edited: Jul 31, 2013
2. Jul 31, 2013

### B4ssHunter

okay since most of you guys didn't reply because you probably did not understand my question
basically , if water pressure increases when you go deeper because there is more mass of water
why doesn't water pressure increase right under a sinking object since there is also great mass that is the water pressure + the pressure caused by the sinking object ? and if it does increase why doesnt it lift the object then ?

3. Jul 31, 2013

### sophiecentaur

There are different forces on each part of the object, depending upon the depth. The vertical components of all the forces will be equal to the weight. If the object is not a sphere, there will be a position (or several) where the potential is at a minimum and that will be where it settles down. The obvious one is when a cylindrical stick comes to rest with its axis horizontal. Th rotate it away from that position will involved doing work on it - hence the potential minimum when horizontal.

btw, if the cube sinks, it isn't in equilibrium. It will keep sinking unless the cube compresses less than the water at great depth, when the upthrust could rise to equal the weight as the water density increases but that of the cube doesn't increase as much.

4. Jul 31, 2013

### B4ssHunter

just another question , if there is a body in the water , with a relativity high mass , will it increase the pressure in water ? and in which area ? is the area directly under it ?

5. Jul 31, 2013

### sophiecentaur

If it did, wouldn't the water flow away from it, leaving a gap? OR, as the pressure is exerted by the water, there will be pressure pushing down (on the top) and pressure pushing up (on the bottom). If the object is heavy enough then there will be excess weight and it will sink, if it is not heavy enough, it will rise until some of it is above the surface and it will float.

6. Jul 31, 2013

### B4ssHunter

okay this is a really great answer , but what if the cube is big enough , like a piston , where there is no way for water to escape , THEN it wont sink and water will generate more upward pressure right ?

7. Jul 31, 2013

### sophiecentaur

Yes. You then have a piston and the hydrostatic pressure is no longer just gh rho.

8. Jul 31, 2013

### cjl

The pressure below an actively sinking object will be higher (and similarly, the pressure behind it will be lower) than you would expect based on a pure hydrostatic equilibrium. This manifests itself as a drag force on the object, and this is why the object achieves a terminal velocity while sinking. This increased pressure serves to basically push the water out of the way in front of the object.

9. Jul 31, 2013

### sophiecentaur

It's more complicated than that. You need to consider the water that will be flowing past the object, the viscosity and turbulence, to account for the pressure difference. If it weren't for that, the pressure difference would just be the same as the static pressure difference, I think. Take a streamlined object and the drag can be as low as you care to make it, yet for a given overall cross sectional area, the vertical up and down components would be the same for any shape..

10. Jul 31, 2013

### cjl

I don't think I really understand what you're trying to say here. It's true that it's fairly complicated (and I completely neglected to account for viscous/skin friction drag), but at least at a really rudimentary level, the pressure ahead of a sinking object will be higher than predicted by static equilibrium, and the pressure behind will be lower simply due to the flow around the object. Any statement more detailed than that starts to get into some fairly complex fluid dynamics though...

(Also it might just be that my brain is really fried at the moment due to lack of sleep)

11. Jul 31, 2013

### sophiecentaur

What I mean is that the force imbalance you are implying can only be due to energy imparted to the fluid (i.e. accelerating it and causing it to flow from the bottom facing part, over to the top facing part. I can't think of any mechanism that will actually produce the effect you are suggesting if the fluid can move without loss of energy (i.e. work being done). Granted, there will be some displacement of fluid from an incremental space at the bottom to a space at the top but can't the amount of this can be arbitrarily small by choosing the right shape? If you ignore energy losses due to friction forces then what work is done?