According to Archimedes principle, the buoyant force is equal to weight of the fluid displaced.

Buoyant force = Weight of water displaced

             =  (Mass of water displaced) x g

             =  (Density of water x volume of water displaced) x g

             =   volume x density x g

We can express this relation in the equation:$$F_B = v *\rho* g$$ where $F_B$ = Force of buoyancy , $v$= volume of water displaced, $\rho$ is density of water and g is acceleration due to gravity

So, buoyant force is independent (not dependent) of the height of water above it according to the formula. $$\text {Pressure} = \text {force} / \text {area}$$ Pressure by formula is equal to,$$ p = h* \rho*g$$ where h is the height of the water column above. So, pressure is dependent on height of water column. Area doesn't change in both cases.

Summary: Buoyant force is independent of height of fluid column above the object , whereas the pressure on the same object which is simply (force/area) is dependent on height, But Area in both cases is same ,How is this possible?

[NOTE: I've interchangeably used water sometimes to mean fluid]

  • $\begingroup$ Sentences in all caps can be potentially perceived as shouting. $\endgroup$
    – Chemist
    Mar 3, 2020 at 13:04

3 Answers 3


Short and direct answer:

Buoyant force is not equal to simply pressure×area. It is, in a simplistic manner, (difference in pressure on upper and lower surface)×area.

Since pressure varies linearly with depth, difference in pressure would be the same for two surfaces at separation of 5 m anywhere in the liquid, be it just below the surface or at a depth of 100m.

Since the difference in pressure doesn't change with depth, the buoyant force, too, wouldn't change.

  • $\begingroup$ So, pressure difference always remains constant throughout the liquid and so does the buoyant force. I agree. $\endgroup$ Mar 3, 2020 at 16:21

Think about a cuboid immersed so that its top and bottom faces (each of area $A$) are horizontal. Let the cuboid's height (separation of top and bottom faces) be $H$, and let its top face be at depth $h$ below the water surface.

Since Force = pressure $\times$ area and the pressure at depth $h$ is $h \rho g$, the force of the fluid on the top face of the block is

$$F_\text{top}=-Ah \rho g$$

The minus sign signifies that the force is downwards,

But the fluid also exerts an upward force on the bottom face of the cuboid, at depth $(h+H)$, where the pressure is $(h+H) \rho g$...

$$F_\text{bottom}=A(h+H)\rho g$$

So the resultant force due to the fluid on the cuboid is

$$F_B=A(h+H)\rho g-Ah\rho g = AH\rho g=v \rho g$$

Here, $v$ (your notation) is the volume of the cuboid, so we have just derived the Archimedes principle from the basic laws of fluid pressure, but only for the special case of a nicely-orientated cuboid. It's not that difficult to do it for any shape of immersed body.

As you will have seen, greater depth of immersion does increase the pressure around the cuboid, and therefore the forces on its faces, but the upward and downward forces are increased equally, so the resultant force isn't affected!


Remember that $V=A*h$ so that $p=\frac{F_B}{A}=\frac{V*\rho*g}{A}=\frac{A*h*\rho*g}{A}=h*\rho*g$. Does this answer your question?

  • $\begingroup$ 'h' in Ahpg is simply the height of the object whereas according to my textbook 'h' in hd*g (for calculating pressure) is the height of the liquid above the object(or the depth ). This does not answer my question. $\endgroup$ Mar 3, 2020 at 16:19

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