# What is the physical origin of buoyancy?

Imagine a horizontal cross-section $$\Delta S$$ of a cylinder containing a liquid of density $$\rho$$. The downward pressure on $$\Delta S$$ at the depth $$d$$ from the surface of the liquid is $$P_{down}=\rho d g$$. This is fine and intuitive. But there is also an upward pressure of equal magnitude $$P_{up}=\rho d g$$.

Clearly there is a fluid of weight $$\rho d g$$ to exert downward pressure on $$\Delta S$$. What exerts the upward pressure of same magnitude from below? Clearly this is an upthrust. It can be shown (not explained) based on Archimedes principle and buoyancy (or by considering static equilibrium) that this upthrust will also be of magnitude $$\rho d g$$. But is this not counterintuitive? Why should the liquid exert an upward force when the weight $$\rho dg \Delta S$$ sits above $$\Delta S$$? I am sorry if the question is not clear.

It is impossible for 'fluid above' to be stationary unless it is not accelerating. When it IS stationary, and not accelerating, Newtonian mechanics requires zero total force, and that (in the presence of gravity) only happens if an upward force on the fluid is being exerted.

In the case of my tea, that force is exerted on the bottom by the tea cup, and in the middle by the tea underneath (which is a slightly compressed fluid). Removing the tea cup, or letting the tea leak out, would result in the fluid level accelerating, downward.

The liquid under pressure is compressed, and that compression, like the compression of spring, makes it generate the force we call 'fluid pressure', against fluid above, cup bottom, and cup sidewalls.

Pressure does not have a direction it is the force as a result of pressure which has a direction.

Imagine that your cylinder of fluid of horizontal cross-section area $$\Delta S$$ and height $$h$$ was removed and you were asked to find the forces which needed to be applied to the rest of the fluid so that none of the remaining fluid came into the void.
There would be horizontal forces on the sides of the void and vertical forces on the end caps.

The upward force on the fluid at the top would have to be $$d\rho g\Delta S$$ and the downward force on the fluid at the bottom would be $$(d+h)\rho g\Delta S$$.

Now the fluid is exerting equal in magnitude and opposite in direction forces on the "void" - Newton's third law.

So the net upward vertical force on the "void" is $$(d+h)\rho g\Delta S- d\rho g\Delta S = h\rho g\Delta S$$ which is the magnitude of the weight (downward force due to the gravitational attraction of the Earth) of the fluid which originally filled the void.

So when there is fluid in the void the net force on that fluid is zero ie it is in static equilibrium.

In the way you have formulated your question, now consider what happens as the height of the cylinder of fluid $$h$$ becomes smaller and smaller.
Not a lot - there is still a static equilibrium state with the weight of the fluid in the cylinder equal to the upward force due to the fluid surrounding it.

Fill a big bathtub with plastic balls and stones of equal sizes. What happens if you shake it a bit?

All the lighter plastic balls will fly around more vividly, while the stones move around less due to their higher mass. So the stones end up taking the bottom-most positions, and the plastic balls eventually go to the top. Think of the molecules of a liquid in that same way:

• If they are lighter, then they will eventually end on top. This causes sinking of heavy objects, since the water particles below will give away and squeeze away from the bottom of the object.
• If they are heavier, they will eventually fall to the bottom, pushing anything lighter upwards. This is buoyancy, since lighter objects give away as heavier water particles squeeze in from the sides towards the bottom of the object.