Pressure, by definition, is force per unit area.

• The SI unit of pressure is the pascal (Pa) \((1 Pa = 1 N/m^2)\).
• Mass density (mass per unit volume) is denoted \(\rho\) (Greek rho).
• The factor \(\rho g\) is the density by weight, where \(g=9.8 m/s^2\) is the acceleration due to gravity.

Fluid Pressure

The pressure \(p\) at depth \(h\) in a fluid of mass density \(\rho\) is \[ \boxed{\bbox[#FAF8ED,5pt]{p = \rho gh}}\tag{1} \]

The pressure acts at each point on an object in the direction perpendicular to the object's surface at that point.

EXAMPLE 1

Calculate the fluid force on the top and bottom of a box of dimensions \(2\times 2\times 5\) m, submerged in a pool of water with its top 3 m below the water surface (Figure 8.12). The density of water is \(\rho = 10^3\) kg/m\(^3\).

Figure 8.12: Fluid pressure acts on each side in the perpendicular direction.

Solution The top of the box is located at depth \(h=3\) m, so, by Eq. (1) with \(g=9.8\), \[\textrm{Pressure on top} = \rho g h = 10^3(9.8)(3) = 29{,}400 \mathrm{Pa} \]

The top has area \(A=4 \mathrm{m}^2\) and the pressure is constant, so \[ \textrm{Downward force on top} = pA = 10^3(9.8)(3)\times 4 = 117{,}600 \textrm{N} \]

The bottom of the box is at depth \(h=8\) m, so the total force on the bottom is \[ \textrm{Upward force on bottom} = pA = 10^3(9.8)(8) \times 4 = 313{,}600 \mathrm{N} \]

EXAMPLE 2 Calculating Force Using Integration

Calculate the fluid force \(F\) on the side of the box in Example 1.

Solution Since the pressure varies with depth, we divide the side of the box into \(N\) thin horizontal strips (Figure 8.13). Let \(F_j\) be the force on the \(j\)th strip. The total force \(F\) is equal to the sum of the forces on the strips: \[ F = F_1 + F_2 + \cdots + F_N \]

Step 1. Approximate the force on a strip.

We'll use the variable \(y\) to denote depth, where \(y=0\) at the water level and \(y\) is positive in the downward direction. Thus, a larger value of \(y\) denotes greater depth. Each strip is a rectangle of height \(\Delta y = 5/N\) and length 2, so the area of a strip is \(2\Delta y\). The bottom edge of the \(j\)th strip has depth \(y_j=3+j\Delta y\).

If \(\Delta y\) is small, the pressure on the \(j\)th strip is nearly constant with value \(\rho gy_j\) (because all points on the strip lie at nearly the same depth \(y_j\)), so we can approximate the force on the \(j\)th strip: \[ F_j \approx \underbrace{\rho g y_j}_{\mathrm{Pressure}}\times \underbrace{(2\Delta y)}_{\mathrm{Area}} = (\rho g) 2 y_j \Delta y \]

Figure 8.13: Each strip has area \(2\Delta y\).

Step 2. Approximate total force as a Riemann sum. \[ F = F_1 + F_2 + \cdots + F_N \approx \rho g\sum_{j = 1}^N 2 y_j \Delta y \]

The sum on the right is a Riemann sum that converges to the integral \({\rho g \int_3^8 2y\, dy}\). The interval of integration is \([3,8]\) because the box extends from \(y=3\) to \(y=8\) (the Riemann sum has been set up with \(y_0=3\) and \(y_N =8\)).

Step 3. Evaluate total force as an integral.

As \(\Delta y\) tends to zero, the Riemann sum approaches the integral, and we obtain \[ F = \rho g\int_3^8 2y\, dy = (\rho g)y^2 \bigg|_3^8= (10^3)(9.8) (8^2- 3^2) = {539{,}000} \textrm{N} \]

THEOREM 1 Fluid Force on a Flat Surface Submerged Vertically

The fluid force \(F\) on a flat side of an object submerged vertically in a fluid is \[ F = \rho g \int_a^b y f(y)\, dy\tag{2} \]

where \(f(y)\) is the horizontal width of the side at depth \(y\), and the object extends from depth \(y=a\) to depth \(y=b\).

EXAMPLE 3

Calculate the fluid force \(F\) on one side of an equilateral triangular plate of side 2 m submerged vertically in a tank of oil of mass density \(\rho = 900 \text{kg/m}^3\) (Figure 8.15).

Figure 8.15: Triangular plate submerged in a tank of oil.

Solution To use Eq.(2), we need to find the horizontal width \(f(y)\) of the plate at depth \(y\). An equilateral triangle of side \(s = 2\) has height \(\sqrt 3 s/2 = \sqrt 3\). By similar triangles, \(y/f(y)=\sqrt 3/2\) and thus \(f(y) = 2y/\sqrt{3}\). By Eq.(2), \[ F = \rho g \int_0^{\sqrt 3} y\, f(y)\, dy = (900)(9.8) \int_0^{\sqrt 3} \frac{2}{\sqrt 3}y^2 \, dy = \left(\frac{17{,}640}{\sqrt 3}\right) \frac{y^3}{3}\bigg|_0^{\sqrt 3} = 17{,}640 \textrm{N} \]

EXAMPLE 4 Force on an Inclined Surface

The side of a dam is inclined at an angle of \(45^\circ\). The dam has height 700 ft and width 1500 ft as in Figure 8.16. Calculate the force \(F\) on the dam if the reservoir is filled to the top of the dam. Water has density \(w = 62.5\) lb/ft\(^3\).

Hoover Dam, with recently completed Colorado river bridge

Solution The vertical height of the dam is 700 ft, so we divide the vertical axis from 0 to 700 into \(N\) subintervals of length \(\Delta y=700/N\). This divides the face of the dam into \(N\) strips as in Figure 8.16. By trigonometry, each strip has width equal to \(\Delta y/\sin(45^\circ) = \sqrt 2 \Delta y\). Therefore, \[ \textrm{Area of each strip} = \textrm{length}\times\textrm{width} = 1500(\sqrt 2\, \Delta y) \]

As usual, we approximate the force \(F_j\) on the \(j\)th strip. The factor \(\rho g\) is equal to weight per unit volume, so we use \(w = 62.5\) lb/ft\(^3\) in place of \(\rho g\): \[ \begin {eqnarray} F_j &\approx &\overbrace{wy_j}^{\textrm{Pressure}}\times \overbrace{1500\sqrt 2 \Delta y}^{\textrm{Area of strip}} = wy_j\times 1500\sqrt 2\,\Delta y\,\,\,\textrm{lb}\notag\\ F = \sum_{j = 1}^N F_j &\approx &\sum_{j = 1}^N wy_j\bigl(1500\sqrt 2 \,\Delta y\bigr) = 1500\sqrt 2\, w \sum_{j = 1}^N y_j \Delta y \notag \end {eqnarray} \]

This is a Riemann sum for the integral \({1500\sqrt 2 w\int_0^{700} y\,dy}\). Therefore, \[ F = 1500\sqrt 2 w\int_0^{700} y\, dy = 1500 \sqrt{2}(62.5)\frac{700^2}{2} \approx 3.25\times 10^{10} \textrm{lb} \]