Examples

Fluid force on a triangular plate

Find the total force on the submerged vertical plate with the following shape: Equilateral triangle, sides $2\mathrm{m}$, top vertex at the surface, liquid is oil with density $\rho =900\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$.

Solution

(1) Write the width function:

Establish coordinate system: $y=0$ at water line (also the vertex), and $y$ increases going down.

Method 1: Geometry of similar triangles

Top triangle with base at $w(y)$ is similar to total triangle with base $\sqrt{3}$.

Therefore, corresponding parts have the same ratios.

Therefore:

$$\begin{array}{c}\frac{w(y)}{y}=\frac{2}{\sqrt{3}}\\ \\ \gg \gg w(y)=\frac{2}{\sqrt{3}}y\end{array}$$

Method 2: Quick linear interpolation function

$$\begin{array}{c}w(y)=0+\frac{2-0}{\sqrt{3}}\cdot y\\ \\ \gg \gg w(y)=\frac{2}{\sqrt{3}}y\end{array}$$

Generalized “quick linear interpolation function”

Generalization:

$$w(y)=e_1+\frac{e_2-e_1}{\text{Height}}(\pm y-a)$$

where:

• $e_1$ is edge 1 and $e_2$ is edge 2 (increasing order of $y$)
• $\pm y$ is $+y$ when $e_1$ comes earlier (smaller $y$), and $-y$ if it comes later
• $a$ is created to force the quantity $(\pm y-a)$ to equal $0$ for the given $y$ value at $e_1$

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(2) Compute integral using width function:

Bounds:

• $y=0$ (shallow)
• $y=\sqrt{3}$ (deep)

Integral formula:

\begin{gather*} F \;=\; \rho g\int_0^{\sqrt{3}}y\, w(y)\,dy\\\\ \gg\gg \quad 900\cdot9.8\int_0^{\sqrt{3}}y \left(\frac{2y}{\sqrt{3}}\right)\,dy \quad \gg\gg \quad 10184.5\int_0^\sqrt{3} y^2\,dy\\\\ \gg\gg \quad 10184.5\frac{y^3}{3}\Big|_0^{\sqrt{3}} \quad \gg\gg \quad \colorbox{cyan}{$17640$} \end{gather*} ParseError: Got function '\sqrt' with no arguments as superscript at position 178: … 10184.5\int_0^\̲s̲q̲r̲t̲{3} y^2\,dy\\\\…

Triangular plate deeply submerged

Set up an integral to calculate the fluid force of water on the following plate, using $\rho$ and $g$ as appropriate:

Solution

The coordinate system has been established as $x=0$ at the vertex, increasing downwards.

Set ${\mathrm{Edge}}_1$ at the vertex and ${\mathrm{Edge}}_2$ at the base of this triangle.

Width function using quick linear interpolation:

$$w(x)=0+\frac{4-0}{3}(x-a)$$

Now find $a$ such that $(x-a)=0$ at ${\mathrm{Edge}}_1$: at that edge, $x=0$, so we must have $a=0$ too. Therefore $w(x)=\frac{4}{3}x$.

Water depth is $h(x)=x+2$ since we must have $h(x)=0$ at the water surface where $x=-2$. So the solution is:

$$F={\int }_0^3\rho g(x+2)\left(\frac{4}{3}x\right)dx$$

Fluid force on circular disk plate

Set up an integral to find the force on the following circular plate suspended deep under water:

Solution

By setting $x=0$ at the center of the circle, we have the equation of the circle:

$$x^2+{\left(\frac{w}{2}\right)}^2=1^2$$

Solve this to obtain:

$$w=2\sqrt{1-x^2}$$

So the integral is:

F \;=\; \colorbox{cyan}{$\int_{-1}^{+1} (x+3)\Big(2\sqrt{1-x^2}\Big)\,dx$} ParseError: Can't use function '$' in math mode at position 73: …1-x^2}\Big)\,dx$̲}

Fluid force on a trapezoid

Set up an integral to find the force on the following trapezoidal plate partially suspended in water:

Solution

Use quick linear interpolation to find the width function with ${\mathrm{Edge}}_1$ at the top:

$$\begin{array}{c}w(x)=100+\frac{40-100}{30}(x-a)\\ \\ \gg \gg 100-2(x-a)\end{array}$$

Since we need $(x-a)=0$ at $x=0$, we have $a=0$.

Now the depth is $h(x)=x-10$. Then $h(10)=0$ at the water line. Therefore:

$$F={\int }_{10}^{30}\rho g(x-10)(100-2x)dx$$

Weight of water on angled dam

Find the total hydrostatic force on an angled dam with the following geometric description: Tilted trapezoid. Base $=\mathrm{2,000}\mathrm{m}$, Top $=\mathrm{3,000}\mathrm{m}$, and vertical height $185\mathrm{m}$. The base is tilted at an angle of $\theta ={55}^{\circ }$.

Solution

(1) Write the width function:

Establish coordinate system: $y=0$ at water line (also the top edge), and increases going down.

“Quick linear interpolation function”:

$$\begin{array}{c}w(y)=3000+\frac{2000-3000}{185}y\\ \\ \gg \gg 3000-\frac{1000}{185}y\end{array}$$

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(2) Incorporate angle of incline in strip thickness:

$$dz=\csc {55}^{\circ }dy$$

So the area of a strip is:

$$\begin{array}{c}dA=w(y)dz\\ \\ \gg \gg (3000-\frac{1000}{185}y)\csc {55}^{\circ }dy\end{array}$$

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(3) Compute total force using integral formula.

Plug data into formula:

$$\begin{array}{c}F=\rho g{\int }_a^{b}h(y)w(y)dz\\ \\ \gg \gg \rho g\csc {55}^{\circ }{\int }_0^{185}y(\mathrm{3,000}-\frac{\mathrm{1,000}}{185}y)dy\\ \\ \gg \gg 4.777\times {10}^{11}\end{array}$$