Chapter 6.6, Problem 41E

To determine

To Calculate: The hydrostatic force of water on irregularly shaped plate using Simpson’s Rule.

Answer to Problem 41E

The force on the irregular object estimated from Simpson’s Rule is $298782.4N$

Explanation of Solution

Given Information: The width of the plate at various depths are

  

Method:

  

Calculation: The hydrostatic pressure at a depth $x$ inside the water is

  $P_x=\text{ρ}gx$

The hydrostatic force on a thin infinitesimal horizontal strip of the plate of thickness $dx$ and width $w_x$ at depth $x$ is

  $dF=P\cdot A=\text{ρ}gx\times w_{x}dx$

Therefore, the total force on the plate is the integral of $dF$ from $x=0$ to $x_{max}$ . So

  $F=\underset{0}{\overset{x_{max}}{{\displaystyle \int }}}dF=\underset{0}{\overset{x_{max}}{{\displaystyle \int }}}\text{ρ}g{w}_{x}xdx$

From the above integral, assume $f\left(x\right)=\text{ρ}g{w}_{x}x$ as the function of Simpson’s Rule and calculate the values of $f\left(x\right)$ for the given pairs of $\left(x,w_x\right)$ as shown below

  $\begin{array}{ccc}x& w_x& f(x)\\ 2.0& 0& 0\\ 2.5& 0.8& 2\\ 3.0& 1.7& 5.1\\ 3.5& 2.4& 8.4\\ 4.0& 2.9& 11.6\\ 4.5& 3.3& 14.85\\ 5.0& 3.6& 18\end{array}$

Also

  $\Delta x=\frac{3.6-0}{6}=0.6$

So,

  $F=\rho g\times \frac{0.6}{3}\left[1\times 0+4\times 2+2\times 5.1+4\times 8.4+2\times 11.6+4\times 14.86+1\times 18\right]$

  $=\rho g\times 0.2\times 152.44=98000\times 30.488$

  $=298782.4$

Therefore, the force on the irregular object estimated from Simpson’s Rule is $298782.4N$