7.5: Hydrostatics

If you dive deeper into a body of water, then you feel more pressure on your body due to the increased weight of the water above you. Hydrostatics is the study of water's behavior on a stationary, submerged object. It enables us to test whether a submarine will be crushed in the deep water, or whether a shark can survive at a certain depth.

Suppose that a thin, stationary object with area \(A\) is submerged in a body of fluid of density \(\rho\) at a distance of \(h\) below the surface (Figure 1). The hydrostatic force that acts on this body equals the weight of the water above it. The volume of the water above the object is \(Ah,\) and its mass equals its density multiplied by its volume—namely, \(\rho Ah.\) Because weight is the product of mass and the acceleration due to gravity, \(g = 9.8 \undiv{m}{sec}^2\) \(= 32 \undiv{ft}{sec}^2,\) the hydrostatic force acting on the object is \begin{equation} F = \rho A g h \pd \label{eq:h-force} \end{equation} Pressure is defined by the force exerted on an object divided by its area. (A sharp knife exerts high pressure because its edge is so thin.) So the hydrostatic pressure that acts upon the submerged object is \(P = F/A,\) or \begin{equation} P = \rho g h \pd \label{eq:h-pressure} \end{equation} We use \(\eqref{eq:h-force}\) and \(\eqref{eq:h-pressure}\) for a thin object whose depth is uniform. But what if we submerge a bigger body whose depth can't be taken as uniform?

Let's generalize how to calculate the hydrostatic force of a vertical body, as in Figure 2. Suppose that the body is bounded between the lines \(y = a\) and \(y = b.\) We cut the body into \(n\) equal-height horizontal rectangles of endpoints \(a = y_0,\) \(y_1, \dots, y_{n - 1},\) \(y_n = b\) and equal height \(\Delta y.\) In the general subinterval \(\parbr{y_{i - 1}, y_i},\) we pick \(y_i^*\) to be a sample point. If \(w(y)\) is the approximating rectangle's width at \(y,\) then its area is \[A_i = w(y_i^*) \Delta y \pd\] Let \(h(y)\) be the rectangle's depth beneath the water's surface at some \(y.\) If each of the \(n\) rectangles is thin, then the hydrostatic pressure acting on an approximating rectangle is, by \(\eqref{eq:h-pressure},\) roughly \[P_i \approx \rho g h(y_i^*) \pd\] (We assume that each rectangle's depth is uniform.) Then by \(\eqref{eq:h-force},\) the hydrostatic force acting on the rectangle is roughly \[F_i = P_i A_i \approx \rho g h(y_i^*) w(y_i^*) \Delta y \pd\] Summing the hydrostatic forces of all \(n\) rectangles gives a total hydrostatic force of approximately \[F \approx \sum_{i = 1}^n F_i = \sum_{i = 1}^n \rho g h(y_i^*) w(y_i^*) \Delta y \pd\] If we make \(n\) larger, then the rectangles become thinner and so better model the vertical body. Thus, by increasing \(n\) the calculation of the total hydrostatic force increases in accuracy. In other words, \(F\) is the limiting value of the sum as \(n \to \infty \col\) \[F = \lim_{n \to \infty} \sum_{i = 1}^n \rho g h(y_i^*) w(y_i^*) \Delta y \pd\] This is a Riemann sum for the function \(\rho g h(y) w(y)\) from \(y = a\) to \(y = b,\) so we attain \begin{equation} F = \rho g \int_a^b h(y) w(y) \di y \pd \label{eq:force-h-body-constants} \end{equation}

Units The International System of Units (SI) is the international standard for measurement. The SI unit for force is the newton, whose symbol is \(\un N.\) One newton is equivalent to a kilogram-meter per second squared; that is, \(1 \un N\) \(= 1 \undiv{kg m}{sec}^2.\) Additionally, water has a density of \(\rho =\) \(1000 \undiv{kg}{m}^3.\) To simplify \(\eqref{eq:force-h-body-constants},\) let's write \(\rho g\) as one number \(\gamma\) (gamma), called the specific weight. In the SI system, \(\gamma = 9.8 \times 1000\) \(= 9800 \undiv{N}{m}^3.\) In the imperial system, \(\gamma = 62.5 \undiv{lb}{ft}^3.\) Hence, we present only a single formula for the hydrostatic force: \begin{equation} F = \gamma \int_a^b h(y) w(y) \di y \pd \label{eq:force-h-body} \end{equation}

A reservoir of water has a cross section in the shape of an isosceles triangle whose slanted sides are \(5\) feet and horizontal length is \(8\) feet (Figure 3). Calculate the hydrostatic force on a cross section of the reservoir when the water level is \(2\) feet.

Taking the downward direction to be the positive \(y\)-axis, let the water surface be \(y = 0.\) The reservoir's bottom is therefore \(y = 2.\) At some level \(y,\) a thin horizontal strip of water has width \(w(y)\) and is a depth of \(h(y) = y\) beneath the water's surface. (See Figure 4.) The reservoir's height is \(\sqrt{5^2 - (8/2)^2}\) \(= 3 \un{ft}.\) By similar triangles, we see \[\frac{2 - y}{3} = \frac{w(y)}{8} \pd \] Solving for \(w(y)\) yields \[ w(y) = \tfrac{8}{3} \par{2 - y} = \tfrac{16}{3} - \tfrac{8}{3} y \pd\] We therefore find the hydrostatic force on the reservoir to be, by \(\eqref{eq:force-h-body},\) \[ \ba F &= 62.5 \int_0^2 y \par{\tfrac{16}{3} - \tfrac{8}{3} y} \di y \nl &= 62.5 \par{\tfrac{8}{3} y^2 - \tfrac{8}{9} y^3} \intEval_0^2 \nl &= \boxed{\tfrac{2000}{9}} \approx 222.222 \un{lb} \pd \ea \]

A dam is constructed in the shape of a trapezoid that is \(10\) meters tall, \(20\) meters long on the bottom, and \(30\) meters long on the top, as shown by Figure 5. Calculate the force on the dam due to the water when the water height is \(8\) meters.

If we take the downward direction to be the positive \(y\)-axis, then it's most convenient to let the water surface level be \(y = 0\) and the dam's bottom be \(y = 8.\) (See Figure 6.) A thin horizontal strip of water has width \(w(y) = 2(10 + a)\) \(= 20 + 2a.\) Its depth below the surface is \(h(y) = y.\) By similar triangles, we see \[\frac{a}{5} = \frac{8 - y}{10} \pd \] Then solving for \(a\) gives \[a = \tfrac{5}{10} \par{8 - y} = 4 - \tfrac{1}{2} y \pd\] Accordingly, \[w(y) = 20 + 2 \par{4 - \tfrac{1}{2}y} = 28 - y \pd\] Hence, the force on the dam is, following \(\eqref{eq:force-h-body}\) with \(\gamma = 9800,\) \[ \ba F &= 9800 \int_0^8 y (28 - y) \di y \nl &= 9800 \par{14y^2 - \tfrac{1}{3}y^3} \intEval_0^8 \nl &\approx \boxed{7.108 \times 10^6 \un N} \ea \]

A right triangle has a horizontal length of \(2\) feet and a vertical length of \(3\) feet. Calculate the hydrostatic force on the triangle if its top vertex is \(1\) foot beneath the water's surface (Figure 7).

At any \(y,\) a thin horizontal strip of the triangle has width \(w(y)\) and is a depth of \(h(y) = y\) below the water's surface. (See Figure 8.) The triangle is between \(y = 1\) and \(y = 4.\) By similar triangles, we see \[\frac{y - 1}{3} = \frac{w(y)}{2} \pd \] Then solving produces \[w(y) = \tfrac{2}{3} (y - 1) \pd\] We use \(\eqref{eq:force-h-body}\) with \(\gamma = 62.5\) to find the hydrostatic force to be \[ \ba F &= 62.5 \int_1^4 y \parbr{\tfrac{2}{3} (y - 1)} \di y \nl &= \tfrac{2}{3} (62.5) \par{\tfrac{1}{3} y^3 - \tfrac{1}{2} y^2 } \intEval_1^4 \nl &= \boxed{562.5 \un{lb}} \ea \]

A circular disk of radius \(4\) meters is submerged in a body of water such that the disk's top is located \(5\) meters below the surface (Figure 9). Calculate the hydrostatic force on the disk.

The water's surface is the line \(y = 9.\) Consider a thin horizontal strip of the disk. Since the strip is a distance of \(y\) above the \(x\)-axis, its depth is \(h(y) = 9 - y\) below the water's surface (Figure 10). An equation for the circle is \(x^2 + y^2 = 16,\) so \(x = \sqrt{16 - y^2}.\) Accordingly, the rectangle's width is \[w(y) = 2x = 2 \sqrt{16 - y^2} \pd\] Because the body is bounded between \(y = -4\) and \(y = 4,\) using \(\eqref{eq:force-h-body}\) with \(\gamma = 9800\) gives the hydrostatic force to be \[ \ba F &= 9800 \int_{-4}^4 \par{9 - y} \par{2 \sqrt{16 - y^2}} \di y \nl &= 9800 \cdot 18 \int_{-4}^4 \sqrt{16 - y^2} \di y - 2 \cdot 9800 \int_{-4}^4 y \sqrt{16 - y^2} \di y \pd \ea \] Since \(y \sqrt{16 - y^2}\) is an odd function, the second integral is \(0.\) The first integral represents the area of a semicircle of radius \(4.\) (We could also use a trigonometric substitution, but possessing geometric intuition is far easier.) We therefore have \[ \ba F &= 9800 \cdot 18 \par{\frac{\pi}{2} \cdot 4^2} + 0 \nl &= \boxed{1\cmaNum411\cmaNum200 \pi} \approx 4.433 \times 10^6 \un{N} \pd \ea \]