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I've seen some variations of this question on the forum, but none has satisfactorily cleared up my confusion, so I made this post with information about what I take issue with specifically. Thank you!


For simplicity, let's neglect atmospheric pressure. Consider a cylinder object submerged underwater (picture below). The forces applied by water on the side of the object cancel out, so only the forces applied vertically (by water) are of interest. Call the net force $F$, which is equal to the difference between the force applied at the top and the force at the bottom:

$$F = F_\text{bottom} - F_\text{top}$$

$F_\text{top} = W_\text{1}$ (the weight of the column of water directly above. The volume of this column of water is $V_\text{1}$ in the picture.)

$F_\text{bottom}$ (the force exerted by the water from below; the actual value is to be discussed.)

When formulating $F_\text{bottom}$, people often make the argument that $F_\text{bottom} = P_\text{bottom}*A$, and argue that $P_\text{bottom} = g*\rho _\text{water}*H$ with $H$ being the depth at the bottom (also indicated in the picture).

The problem is, when I look at how people drive the formula for water pressure at a certain depth, the argument is always that the downward force at that point is caused by the weight of the column of water directly above. Divided both sides by area $A$ to get pressures, the equation becomes $P = g*\rho _\text{water}*H$. By Newton's 3rd law, the upward pressure is of equal measure.

However, this argument is contingent on the premise that the volume above is ONLY occupied by water. It's not clear how the same result could be obtained for $F_\text{bottom}$ if the volume above is partly occupied by an object of a different density.

I've read the thought experiment that suggests replacing the object with water. The water would be in equilibrium so the force applied upward must equal the weight of that body of water. However, I don't think it should necessarily be the case for the object, because with an object of a different density, the force exerted by the object on the water below could hypothetically be different, suggesting a potentially different return force.

enter image description here

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  • $\begingroup$ If the density of the liquid is not constant, then we have to integrate the pressure of the column, which boils down to an integral over the density. The only "problem" that is non-trivial in this case is to find the density distribution itself. I suspect you are intuitively understanding that there is a non-trivial problem here for so called "compressible fluids" (like air), but you have not seen how it can be solved by means of a differential equation, yet? $\endgroup$ Mar 30, 2023 at 18:24

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However, this argument is contingent on the premise that the volume above is ONLY occupied by water. It's not clear how the same result could be obtained if the volume above is partly occupied by an object of a different density.

Just move horizontally until there’s water directly above. This doesn’t change the pressure (fluids can’t sustain a shear load, so if the pressure varied horizontally, the fluid would rearrange to eliminate the variation). Then move up until reaching the free surface (zero gauge pressure). The process can be repeated if you run into something other than the free surface.

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The sum of the forces on the object in the picture is $F=-mg -\rho A h g +\rho A H g$, and is not zero in general. If the object is denser than water it will sink, if it is less dense it will rise, and if it has the same density as water it will stay put. The first term of $F$ is the downward gravitational force on the object. The second term is the weight of the column of water above the object. The last term is the upward force at the bottom of the object. To see this, imagine at a particular moment of time (while the object is sinking, rising, or staying put) a cylinder of water of identical shape at the same depth as the object. The upward force at the base of this imaginary cylinder is equal in magnitude to the weight of the column of water above it, which is $\rho A H g$. The pressure at this height in the liquid is therefore $\rho H g$. Since the base of the object is at this height, the force at bottom surface is $\rho A H g$. The sum of the last two terms in $F$ is $\rho A (H-h)g$, which is equal to the weight of the water "displaced" by the object, and is the bouyant force.

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  • $\begingroup$ I'm not sure if I wasn't clear in the post, but you're essentially using the thought experiment I take issue with (clarified in my post). Also, without an imaginary approach, how could we arrive at the same result? $\endgroup$ Mar 30, 2023 at 19:40
  • $\begingroup$ The argument is not "contingent on the premise that the volume above is only occupied by water". It assumes that the water is stationary, which is true if the pressure doesn't vary horizontally at the depth of the object's base. However, this is an approximation. In reality, a sinking or rising object disturbs the water around it in a way that depends on its shape. That is a more complicated problem in hydrodynamics. However, if the density of the object is close to that of water, it will sink or rise at about the same rate as if we had neglected the extra flow. $\endgroup$
    – tneulinger
    Mar 31, 2023 at 5:08

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