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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)

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


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

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

$\displaystyle {\bf F} =- \rho\,g\,{\bf e}_z$ (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:

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


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


$\displaystyle W_x$ $\displaystyle = W_y = 0,$ (2.10)
$\displaystyle W_z$ $\displaystyle =-\int_V \rho\,g\,dV.$ (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

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

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

next up previous
Next: Equilibrium of Floating Bodies Up: Hydrostatics Previous: Hydrostatic Pressure
Richard Fitzpatrick 2016-03-31