Expert: Daniel Mazur - 6/29/2009

Question
Hi daniel,

Archemedes principle says that the a bouyant force equal to the weight of the volume of the fluid displaced by the body immersed acts on the body.

My question is that why only the weight of the displaced fluid?? why not the entire volume of the fluid?? whys isn't the force of the weight of the entire fluid transmitted to the body immersed??

Also suppose we take a piece of wood to the bottom of the sea. According to archemedes principle, there will be a an upward force on the piece due to the weight of the displaced fluid. But at the same time there is a very high column of water above this piece of wood which is at the bottom of the sea. The entire column of water above the piece of wood till the surface of the sea exerts aa tremendous amount of force on this piece.
This force is surely greater than the bouyant force due to the displaced fluid. But still the piece of wood will eventually come to the surface. why ????

Thanks,
Shikhin

Answer
Dear Shikhin,

I think that the most straightforward way to illuminate this is via the energy bilancing. It is a curious thing that in fluids the pressure acts as "volume density of potential energy", so when we speak of one, we speak of the other. Also one must accept the observation that fluids/liquids flow and so if there is any force that will push your log up from the sea's bottom, the water will yield.

Allow me to use a model of our log to have a simple shape: let it be a cuboid of horizontal cross-section area A and height h. Naturaly then its volume will be V=A*h and mass m_log=A*h*rho_log. In the same spirit we can prepare the mass of displaced water as m_w=A*h*rho_w for later use.

We learn about liquids in a container that pressure is the same in the whole volume: you press a pedal of a hydraulic brake system on one end and the car brakes engage on the other end of a pipeline, simple. That is true for liquids outside gravitational field. If we assume our liquid in a container in a homogeneous grav. field, the water pressure will be zero on the upper surface and maximal by the bottom and it goes as P=rho_w*g*z, where g is grav. acceleration and z is the vertical coordinate measured from the top surface of the water body. This "pressure" pushes in all directions at once independently of how much water there is in any particular direction: pressure is a measure of potential energy of the liquid, so its magnitude depends only on depth (coordinate z).

When we immerse the log in the water - completely for start, but not necessarily to the bottom - we will observe a different pressure at the log's (cuboid's) top and bottom surface. If log's top is at the depth z0 then log's bottom will be at depth z0+h. Pressure at the top will be p_top=rho_w*g*z0, at the bottom p_bot=rho_w*g*(z0+h). Observe that while both pressures depend on depth, the pressure DIFFERENCE does not: p_bot-p_top = rho_w*g*h ! Next, we recall that pressure exerts force depending on the surface area it acts on. The p_top will act on the log pushing it down with force F_top=A*p_top and the p_bot will act on the log pushing it up F_bot=A*p_bot. I have intentionally chosen a simple cuboid shape for the log, so an osimple calculus suffices. What I am showing here can, of course, be proven for any shape of log, but it would require integrations and thus become less transparent.

Well then, you see that there are two forces acting on the log (apart from gravitation), F_top and F_bot and that those forces have opposite direction and NOT the same magnitude. Hence they mutually subtract and we will observe an effect of the result force. F_bot is always larger than F_top, so the resulting force will always be buoyant:
F_buoy=F_bot-F_top=A*p_bot-A*p_top=A*rho_w*g*(z0+h)-A*rho_w*g*z0=A*rho_w*g*h
And behold: the last part of the equation is equal to grav. force of "the mass of water displaced", F_buoy=A*h*rho_w*g=m_w*g.

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. You can also look at it in such a way that the CONTAINER of the water is also part of the system. As watter exerts pressure on the container to break through, the container exerts pressure on the water to hold it in = Newton's 3rd Law. I think, though, that the best way to consider this problem is that of pressure as potential energy and if you have an energy difference somewhere (I've shown you the pressure difference here) there must be a force resulting from it.

I believe that the consideration of the case of rho_log<rho_w and the concept of floating is trivial, if you've got all the way to this point.

Take care!
Daniel