# Why the fluid pressure does not affected by the object submerged?

1. Apr 10, 2012

### tsw99

As we know $P=P_{0}+\rho gh$
If now I put an object of height h into the water, just completely submerged,
why the pressure below the object not $P=P_{0}+\rho_{obj}gh$ instead?
Is it because it may not be in equilibrium or is it due to some other properties of fluid?
Thanks!

2. Apr 10, 2012

### nasu

Why do you think it is not?

3. Apr 10, 2012

### HallsofIvy

The water pressure acting on an object is, by definition, the force (per unit area) the water is applying to the object, which is just the weight of the water above the object, not the other way around. The density of the object has nothing to do with the weight of the the water.

4. Apr 10, 2012

### tsw99

But if we see the derivation of the first equation, we use the relation of the weight of the column of water (not other objects) equals to the net force arises from the water above and below.
If the density of the object is greater than that of water, it will be sinking obviously, so the system is not in equilibrium. Why can we still use the relation $P=P_{0}+\rho gh$ which comes from the equilibrium condition of water?

5. Apr 10, 2012

### Whovian

In my opinion, here's the most intuitive way to think about "buoyancy." First, take a bit of water with the same shape as the object being put in, and put it in instead of the object. We'll assume the system will be in equilibrium. The weight of this bit of water will be pulling down on it, so an equal force must be pushing it up for the system to be in equilibrium. We call this force buoyancy. Now replace the water with the object again. The same force should be being applied. This should help you understand what's going on.

Anyways, no fluid will ever be in equilibrium; these formulae are more of an approximation anyways.

6. Apr 10, 2012

### nasu

The pressure at the bottom of the (barely) submerged object is
$P_{bottom}=P_{0}+\rho gh$
The pressure on the top surgace of the object is
$P_{top}=P_{0}$
The difference is $\rho gh$
The buoyant force is due to the difference so it does not matter if you include atmospheric pressure or not, at least in the first approximation. And the difference is the same at any depth, again in the first approximation.
So including the atmospheric pressure does not change the buoyancy of the object.

7. Apr 13, 2012

### tsw99

Yes, I understand what you say, but as I said when we derive the Pressure-depth relation, we consider the weight of the water to find out the pressure. But if we replace the water by an object with same mass, my question is, why we can say that the pressure exerted by the water is still the same?

8. Apr 13, 2012

### rcgldr

Any increase in pressure at some specific point in the container will correspond to the increase in the height of the surface of the water due to the object being placed onto or in the water, which is an indirect effect of the object being placed in the water. If a floating object is forced to be submerged, it only increases the height of the water by a small amount.

9. Apr 15, 2012

### ForemanS87

Thanks a lot guys..
this helped me a lot.. I Googled this and didn't expect finding great answer like this one.

And I registered specifically to thank you guys