Buoyancy

Consider the air/water system described in the previous section. Let $V$ be some volume, bounded by a closed surface $S$ , that straddles the plane $z=0$ , and is thus partially occupied by water, and partially by air. The $i$ -component of the net force acting on the fluid (i.e., either water or air) contained within $V$ is written (see Section 1.3)

$$f_i = \oint_S \sigma_{ij}\,dS_j + \int_V F_i\,dV,$$ (2.5)

where

$$\sigma_{ij} = - p\,\delta_{ij}$$ (2.6)

is the stress tensor for a static fluid (see Section 1.5), and

$${\bf F} =- \rho\,g\,{\bf e}_z$$ (2.7)

the gravitational force density. (Recall that the indices $1$ , $2$ , and $3$ refer to the $x$ -, $y$ -, and $z$ -axes, respectively. Thus, $f_3\equiv f_z$ , et cetera.) The first term on the right-hand side of Equation (2.5) represents the net surface force acting across $S$ , whereas the second term represents the net volume force distributed throughout $V$ . Making use of the tensor divergence theorem (see Section B.4), Equations (2.5)-(2.7) yield the following expression for the net force:

$${\bf f}= {\bf B} + {\bf W},$$ (2.8)

where

$$B_i =- \int_V\frac{\partial p}{\partial x_i}\,dV,$$ (2.9)

and

$$W_x = W_y = 0,$$ (2.10)

$$W_z =-\int_V \rho\,g\,dV.$$ (2.11)

Here, ${\bf B}$ is the net surface force, and ${\bf W}$ the net volume force.

It follows from Equations (2.4) and (2.9) that

$${\bf B} = M_0\,g\,{\bf e}_z,$$ (2.12)

where $M_0 = \rho_0\,V_0$ . Here, $V_0$ is the volume of that part of $V$ which lies below the waterline, and $M_0$ the total mass of water contained within $V$ . Moreover, from Equations (2.2), (2.10), and (2.11),

$${\bf W} = -M_0\,g\,{\bf e}_z.$$ (2.13)

It can be seen that the net surface force, ${\bf B}$ , is directed vertically upward, and exactly balances the net volume force, ${\bf W}$ , which is directed vertically downward. Of course, ${\bf W}$ is the weight of the water contained within $V$ . On the other hand, ${\bf B}$ , which is generally known as the buoyancy force, is the resultant pressure of the water immediately surrounding $V$ . We conclude that, in equilibrium, the net buoyancy force acting across $S$ exactly balances the weight of the water inside $V$ , so that the total force acting on the contents of $V$ is zero, as must be the case for a system in mechanical equilibrium. We can also deduce that the line of action of ${\bf B}$ (which is vertical) passes through the center of gravity of the water inside $V$ . Otherwise, a net torque would act on the contents of $V$ , which would contradict our assumption that the system is in mechanical equilibrium.