Question

Irrigation Canal Gate The vertical cross section of an irrigation canal is modeled by \(f(x)=\frac{5 x^{2}}{x^{2}+4}\) where \(x\) is measured in feet and \(x=0\) corresponds to the center of the canal. Use the integration capabilities of a graphing utility to approximate the fluid force against a vertical gate used to stop the flow of water if the water is 3 feet deep.

Short answer

To find the Fluid Force exerted on the vertical gate, compute the integral \( F = 62.4 \int_{-1.5}^{1.5} 3h(x) dx \) using a numerical method or a graphing utility where \( h(x) = \frac{5x^2}{x^2+4} \). The result will be the Force in pounds. For the exact value, perform the calculation described in the steps.

Step-by-step solution

Understand the Fluid Force

The challenge is to find the fluid force against a gate. Fluid force is given by the equation \( F = w \times \int_{a}^{b} d \times h(x) dx \), where: - \( F \) is the fluid force (which we need to find) - \( w \) is the weight density of the fluid - \( d \) is depth of the fluid - \( h(x) \) is shape function of the gate - \( a \) and \( b \) are the limits of integration. In this case, the weight density of water (w) is 62.4 lb/ft³. We know the depth is 3 feet. And, the function \( h(x) \) representing the vertical cross section of the canal is given as \( h(x) = \frac{5x^2}{x^2+4} \).

Setup the Integral

We need to integrate from \( x = -1.5 \) to \( x = 1.5 \) feet (since the depth is 3 feet and \( x = 0 \) represents the center). This gives us the limits of integration. Therefore the integral to find the fluid force is given as \( F = 62.4 \int_{-1.5}^{1.5} 3h(x) dx \).

Compute the Force

Now, replace \( h(x) \) in the integral with the function from step 1. Then use a graphing utility or numerical integration (like Simpson's rule or Trapezoidal rule) to find the approximate value of the integral. This will give the fluid force exerted on the gate.