I know this has been asked before, but I don't fully understand the given answers. A common answer goes like this:
"Let us suppose we remove the object and fill the space left (in the fluid) with the same fluid. Assume this portion of fluid become solid without changing its volume or density. It will be in equilibrium with the fluid. Now suppose the buoyancy force on this solidified portion is off the centre of mass. This would imply non vanishing resultant force or torque and the solidified portion would not be in equilibrium." (Credit to Diracology)
What I don't understand with this argument is that zero net torque would be satisfied if the buoyancy force acted from any point along the line of action of the weight of this solid (provided that it is equal and opposite in direction). Therefore, this argument only justifies that the centre of buoyancy must be directly above or below the centre of mass of the displaced fluid.
So what is the reason that the centre of buoyancy must be at the centre of mass of the displaced fluid?
Best Answer
You are right only the line of action matters. Following is the proof.
So consider the object to be partially submerged in water. Let $\Sigma$ denote the surface of the part of the body submerged and let $\Gamma$ denote the submerged volume. Consider the $xy$ plane of the coordinate system to be aligned with the surface of the water and the $z$-axis to be pointing upward (the origin can be placed anywhere subject to these constraints). Let $\vec{dA}$ denote the surface area element of the body and by convention let it point outwards. Now that notations and conventions are defined let's proceed with the computation.
We want to find the action point of the force (which is the centre of buoyancy). For this the total torque due to infinitesimal pressure forces on each surface element must equal the torque due to the combined buoyant force acting at the centre of buoyancy i.e $$\int_{\Sigma} \vec r\times\vec{dF}=\vec r_{cb}\times\vec F_{tot} \tag{1}$$ where,
$\vec r_{cb}$ is the position of the centre of buoyancy
$\vec{dF}$ is the infinitesimal pressure force and
$\vec F_{tot}$ is the total buoyant force
Now in writing $\vec{dF}$ we need to be a bit careful with our signs. The pressure exerted by a fluid at depth $h$ is $h\rho g$. Here in this case the depth is $-z$ since below the surface $z$ is a negative quantity and depth is positive. Hence the pressure at coordinate $z$ is $-z\rho g$. The direction of the pressure force is inward i.e in the direction opposite to $\vec{dA}$. Hence we have $$\vec{dF}=(-z\rho g)(-\vec{dA})=z\rho g\vec{dA} \tag{2}$$ Using this we can also express $\vec F_{tot}$ as the following integral $$\vec F_{tot}=\int_{\Sigma}z\rho g\vec{dA} \tag{3}$$ Using eqs $(2)$ and $(3)$ and substituting into eq $(1)$ we get $$\int_{\Sigma} \vec r\times z\rho g\vec{dA}=\vec r_{cb}\times \int_{\Sigma}z\rho g\vec{dA}$$ The $\rho g$ factors cancel and it is sufficient to prove that $$\int_{\Sigma}z(\vec r-\vec r_{cb})\times \vec{dA}=0$$ Now note that $\Sigma$ does not enclose a volume. To prove the above integral vanishes we want to use a corollary of Gauss' Law. To get a surface that encloses a volume consider the surface obtained by intersection of the water surface and the object. Let this be denoted as $\Sigma_w$. We have $$\begin{aligned}\int_{\Sigma}z(\vec r-\vec r_{cb})\times \vec{dA}&=\int_{\Sigma}z(\vec r-\vec r_{cb})\times \vec{dA}+\int_{\Sigma_w}z(\vec r-\vec r_{cb})\times \vec{dA} \\&=\oint_{\Sigma\cup\Sigma_w}z(\vec r-\vec r_{cb})\times \vec{dA} \end{aligned}$$ The first line is justified since $z=0$ on $\Sigma_w$ and thus the second integral is $0$.
Now our integral is ripe for the application of the following corollary of Gauss' law$\int_V\nabla\times\vec v\ d\tau=-\oint_S\vec v\times\vec{dA}$. It can be derived by replacing the vector in Gauss' law with $\vec v\times\vec c$ where $\vec v$ is a vector field and $\vec c$ is a constant vector. Applying to our integral we get $$\begin{aligned}\int_{\Sigma}z(\vec r-\vec r_{cb})\times \vec{dA}&=-\int_{\Gamma}\nabla\times (z(\vec r-\vec r_{cb}))\ d\tau \\&=\int_{\Gamma}(x_{cb}-x)\ \hat y-(y_{cb}-y)\ \hat x\ d\tau \end{aligned}$$ This integral vanishes if and only if $x_{cb}$ and $y_{cb}$ coincide with the centre of mass of the displaced water. There is no condition on the $z$ coordinate. Thus only the line of action matters.