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If I place a cube in water, the force at the top of the cube, F1 will be Ahρwg.
Where,
A = cross-sectional area
h = height at the top
ρw = density of water
g = acceleration due to gravity
All this made sense, because this downward Force(F1) is actually the weight of the water above the cube.....but where did the upward force come from? is it the opposite force of the weight of the cube? or something else?

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  • $\begingroup$ Fluid exerts pressure in all directions. $\endgroup$ – Ankur Singh Apr 21 '18 at 17:26
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  • You have successfully described a downwards force on the cube, being the pressure at its top due to the weight of water above.

  • Another downwards force is the cube's own weight.

  • An upwards force comes from the water below the cube. Here the pressure is a bit larger than that at the top of the cube, since we are a bit deeper. This is what is called buoyancy.

If the downwards forces together equal the upwards force then the cube doesn't move. That would be the case if the cube's density equalled that of water. It would then weigh just as much as the same amount (volume) of water would.

If it's density is smaller (and thus it's weight smaller) than in this equilibrium scenario, we suddenly have less total downwards force. The upwards force due to the pressure at that depth does not change. So now the upwards force is larger, causing the cube to accelerate up.

You can think of this buoyancy effect as caused by water that really wants to move somewhere else due to the pressure. If something heavy is nearby, then that heavy object is more difficult to push away than neighboring water molecules. If something light is nearby (less dense than water itself), then it is easier for the water molecules to push this object away than to push other water molecules away. This effect is strongest where the pressure is strongest, which will be at the bottom. The net effect is thus an upwards motion.

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Lets assume you position a 10 cm cube one meter under the water.

The water pressure on the four sides cancel each other. So the cube is experiencing the difference between opposing pressures on the top and bottom surfaces.

The top pressure is $h g\times\rho_{water}.\times 100cm $

The bottom pressure is $(h+10cm)g\times\rho_{water}\times 100cm $

That excess pressure on the bottom surface is the buoyancy.

$$ Buoyancy = (h+10-h) g\times\rho\times 100cm = (10cm\times100cm)\times g\rho_{water} = cube.volume\times g\rho_{water} $$ If its less than the weight of cube the cube will sink to the bottom but will weigh les. If it is more than the weight of cube it will push the cube up above the surface of water until the volume of submerged part is offering enough buoyancy to keep the floating equilibrium.

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  • $\begingroup$ I think the top pressure will be g×ρ_water_×100cm.....if not, could you please tell me why? $\endgroup$ – Zarif Apr 22 '18 at 11:30
  • $\begingroup$ @Zarif, It was to be cancelled out, regardless. I edit my answer and you see the buoyancy is the same. $\endgroup$ – kamran Apr 22 '18 at 16:12

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