Expert: Daniel Mazur - 1/21/2010

Question
QUESTION: Hi,

I'm currently studying hydrostatics and I have got a problem I just can't sort out myself. While surfing the web for an answer I stumbled upon one of your answers were you explain bouancy. In this explanation you say the following:

"And it doesn't matter if you place the log all the way at the bottom, so that there is no water left underneath, because as we said, pressure does not depend on the amount of water present in a given direction from an object, but only on depth."

I find it very hard to understand why this is true. My reasoning is that   if there isn't any water underneath it, there isn't any water who can push this object upwards. So, since water cannot pull the object up nor from the sides nor from the top it will only experience a downward force from the water above it so that it will stay at the bottom.

I would be very happy if you could tell me what's my error? 
I am 19 years old and i am currently in studying the second year of civil engineering.

Thanks!

Anthony   

ANSWER: Hello Anthony,

if you quoted me exactly, I should have written that "the pressure depends on the depth *of immersion* of the log". The reason is that once the entire log is under water, the pressure (buoyant force) does not change with further sinking of the log.

Now to your question. It puzzles almost everyone upon the first encounter. The big secret is in the walls of the container that holds the water body. It can be a barrel or an ocean bottom. Why does the water body hold in the shape we see it in? It is thanks to the container walls. Just as the water applies pressure on the walls, the walls apply the same pressure on the water. It is a form of Newton's 3rd Law. These pressures must be equal. If the water pushes stronger than the wall, then the wall yields - balls out and a rupture can happen.

When our log is pushed all the way to the bottom so that no water remains underneath the log, there is still the container (ocean) bottom. The bottom pushes up with the same pressure regardless if directly above it is water or the log. Only this way the water is kept in a container of a constant shape, see above. I know it is still a tough bite, but once you start considering all the forces that need to be kept in balance, you will eventually arrive at the conclusion that only the observed behavior of the log makes sense.

Please post a follow-up if needed.

All best!
Daniel


---------- FOLLOW-UP ----------

QUESTION: Hi,

Thanks a lot for your quick answer! It's indeed quite puzzeling to me, but i'll try to understand it.

I've got just one more question. When we would look at a situation where whe have got an object, let's say a cube which is positioned between a layer of water and a layer of oil (because of density oil < density cube < density water). Is it right to say that the oil will also push the part of the cube which is in the oil upwards according to the buoyant force? Or am I wrong again?

Thanks again,

Anthony  

Answer
Yes Anthony,

that is exactly right. A cube immersed in oil will be buoyed by it and the buoyant force has the same formula: F = V*Rho*g . In case it's only in oil, then Rho is the density of oil. If the cube is partially immersed in water and partially in a layer of oil, the buoyant force is F = g*(V_in_oil*Rho_oil + V_in_water*Rho_water).

So your instinct was correct.
Cheers!

Daniel