Please read the following. A cube submerged in water exerts a force on the water. The cube exerts a force equal to its weight on the water below it, and by Newton's third law it should feel a reaction force to this weight.
This means that there are four forces on it: one due to the pressure and weight of the water on top, one due to the pressure of the water below, one due to gravity and one due to the reaction force to its weight.
But the pressure from below is due to the weight of the water above it pushing downwards, but there is no water because it was displaced, so the pressure from below can only be due to random collisions. The pressure from above is due to random collisions and the weight of all the water. The displaced water also moves up, so its weight adds to the pressure on the upper surface of the cube.
The weight of the object is balanced by the reaction force of the water below, like when you place something on a table and it does not fall. So the sum of all the forces on the object is simply the weight of the water above, plus the weight of the water displaced. Therefore no object can ever sink because the net force is downwards.
What is wrong with my line of thought? I can see that it is obviously wrong but I have no idea why.
Best Answer
This is incorrect, and leads to many other errors in your thinking.
The force on a submerged object in a gravitational field is easily calculated by considering the pressure in the fluid at each vertical position. That being the case, all the horizontal forces add to zero leaving only the forces on the top and the bottom which are different.
Now lets consider a submerged cube oriented with vertical and horizontal faces. The downward force on top ($\rho_w g d_{\mathrm{top}}A$) is less than the upward force on the bottom ($\rho_w g d_{\mathrm{bottom}}A$), where $A$ is the area of the face of the cube, $\rho_w$ is the density of water, and the $d$s are the depths. There is also the weight force on the cube which is totally independent of the liquid, so there are 3 important forces.
Now, the sinking or floating of the cube depends on the density of the cube. Summing the forces, taking down as positive, we have $$F_{\mathrm{down}}=\rho_w g d_{\mathrm{top}}A-\rho_w g d_{\mathrm{bottom}}A +(\rho_{\mathrm{cube}}Vg).$$
Now the length of one side is $A^{1/2}$, so the volume is $A^{3/2}$ and $d_{\mathrm{bottom}}-d_{\mathrm{top}}=A^{1/2}.$ Substituting gives us $$F_{\mathrm{down}}=\rho_{\mathrm{cube}}gA^{3/2}-\rho_wgA^{3/2}=gA^{3/2}(\rho_{\mathrm{cube}}-\rho_w).$$
So, if the density of the cube is greater than the density of water, there is a net downward force. If it's less, there is a net upward force.