# Making sense of the Buoyancy force [duplicate]

I was reading about how to find the buoyancy force when I came across this explanation:

Imagine a box submerged in a tank of liquid.

The distance from the surface of the liquid to the bottom surface of the box is greater than the distance from the surface of the liquid to the top surface of the box. In short, $$h2 > h1$$.

Because $$p = \rho g h$$, the pressure on the bottom surface due to the liquid must be greater than the pressure on the top surface. This results in there being more force on the bottom surface of the box than there is on the top surface.

The difference between these two forces causes there to be a net force upwards called the buoyancy force.

(source: bbc.co.uk)

I understand how the force acting downwards onto the top surface of the box is found. This is simply due to the column of water above the box. But how is the force acting upward onto the bottom surface found?

I specifically want to to understand where the force acting onto the bottom surface of the box, upwards, derives from.

## marked as duplicate by DanielSank, ACuriousMind♦, Danu, user10851, Ryan UngerSep 8 '15 at 23:38

• This has already been answered in the first paragraph of the accepted answer on another question. – DanielSank Sep 8 '15 at 18:29

This is a matter of putting a few bits together.

Pressure is a scalar quantity, so it has no direction. You can find the pressure at any given depth by dividing the weight ($mg$) of water above some surface area by that same surface area. Because the mass is proportional to the surface area, the area will cancel out (so it doesn't matter what area you choose). You should find the nice simple expression

$$P = \rho gh$$

where $\rho$ is the density of the liquid, $g$ is the gravitational acceleration and $h$ is the depth.

The pressure applies a force when there is a boundary, in this case the edges of the box. The force is always perpendicular to the boundary surface. It should be easy to see that the force on the left side of the box will cancel out that on the right (equal and opposite), and similarly for the front and back. That leaves the top and bottom. The magnitude of the force will be $F = PA$ where $A$ is the surface area. The direction on the top surface is down, on the bottom surface is up. Since $P_{\rm bottom} > P_{\rm top}$, there will be a net force upwards. This is the buoyant force.

More intuitively, you might think of this as the water column "around" the box pushing down on the water around and below the box, which "pushes" on the water under the box, which "pushes" upwards on the box. It may seem counterintuitive that the weight of the water can push up, but this is pressure, which is caused by many collisions between microscopic particles. Because the collisions are randomly oriented, the net effect is to in some sense spread out the direction of the force so that it can push in any direction (when confronted with a surface to push against).

• Thanks I'm sort of getting it now, my only issue now is how to find P(bottom)? Thanks. – J.Gudal Sep 8 '15 at 18:40
• You can just use $P = \rho gh$, with $h$ set to the depth at the bottom of the block. – Kyle Oman Sep 8 '15 at 19:56