Question

Determine the hydrostatic force on one side of the gate.

Short answer

The resultant answer is \(458.33\;lb\).

Step-by-step solution

Given data

The trapezoid shape of the canal has the bottom width of\(3{\rm{ft}}\), the top width of\(5{\rm{ft}}\), and the height of\(2{\rm{ft}}\).

Concept of Hydrostatic force

Hydrostatic forces are the resultant force caused by the pressure loading of a liquid acting on submerged surfaces.

Simplify the expression

Set the coordinate system so that \(x = 0\) is at the bottom of the gate and up increases \(x(x\) is height from the bottom). If the water surface is \(2{\rm{ft}}\) from the bottom, then the depth is: \(d = 2 - x\).

The width of one triangular part at height \(x\) from the bottom is found by similar triangles

\(\frac{w}{1} = \frac{x}{2} \to w = \frac{1}{2}x\)

Since there are two triangular parts, we multiply that by 2.

Total width of two triangle\( = x\). Add that to the constant width of the rectangular part, the total width at height \(x\) from the bottom is\(f(x) = x + 3\).

Determine the hydrostatic force

The hydrostatic force is given by \(F = \int_a^b \rho gdf(x)dx\)

For non-metric, we use \(\rho g = \delta = 62.5{\rm{lb}}/{\rm{f}}{{\rm{t}}^3}\) for water

\(\begin{aligned}F &= \int_a^b \delta df(x)dx\\F &= \int_0^2 6 2.5(2 - x)(x + 3)dx\\F &= 62.5\int_0^2 {\left( {2x + 6 - {x^2} - 3x} \right)} dx\\F &= 62.5\int_0^2 {\left( {6 - x - {x^2}} \right)} dx\end{aligned}\)

Similarly, calculate further;

\(\begin{aligned}F &= 62.5\left( {6x - \frac{1}{2}{x^2} - \frac{1}{3}{x^3}} \right)_0^2\\F &= 62.5\left( {6(2) - \frac{1}{2}{{(2)}^2} - \frac{1}{3}{{(2)}^3} - (0 - 0 - 0)} \right)\\F &= 62.5\left( {12 - 2 - \frac{8}{3}} \right)\\F &\approx 458.33{\rm{lb}}\end{aligned}\)

The hydrostatic force is \(F \approx 458.33{\rm{lb}}\).