A gate in an irrigation canal is constructed in the form of a trapezoid $4 \mathrm { ft }$ wide at the bottom, $8 \mathrm { ft }$ wide at the top, and $2 \mathrm { ft }$ high. It is placed vertically in the canal, with the water extending to its top. Find the hydrostatic force on one side of the gate.
A) $166 \mathrm { lb }$
B) $21 \mathrm { lb }$
C) $671 \mathrm {~b}$
D) $666 \mathrm { lb }$
E) None of these

Question

Quizplus · Accepted Answer

First, we need to find the centroid of the trapezoid. Using the formula for the centroid of a trapezoid, we get:

$x = \frac{b_1 + b_2}{2} = \frac{4 + 8}{2} = 6 	ext{ ft}$

$y = \frac{h}{3} = \frac{2}{3} 	ext{ ft}$

So, the centroid is located 6 ft from the left edge of the trapezoid and 2/3 ft from the bottom edge.

To find the hydrostatic force on one side of the gate, we need to find the pressure at each point along the height of the gate and then integrate to get the total force.

At the bottom of the gate, the pressure is:

$p_1 = \gamma h = 62.4 	ext{ lb/ft}^3 \cdot 2 	ext{ ft} = 124.8 	ext{ lb/ft}^2$

At the top of the gate, the pressure is:

$p_2 = \gamma h = 62.4 	ext{ lb/ft}^3 \cdot 4 	ext{ ft} = 249.6 	ext{ lb/ft}^2$

We can assume that the pressure varies linearly between these two points, so we can use the average pressure to calculate the force. The average pressure is:

$p_{avg} = \frac{p_1 + p_2}{2} = \frac{124.8 + 249.6}{2} 	ext{ lb/ft}^2 = 187.2 	ext{ lb/ft}^2$

Now we need to find the height of each small vertical element along the gate. We can divide the height of the gate into small increments, $\Delta y$, and find the height of each element by adding $\Delta y$ to the height at the bottom of the element. The force on each element is then:

$dF = p_{avg} \cdot w \cdot \Delta y$

where $w$ is the width of the element.

We can use the location of the centroid to find the width of each element as a function of the height. Let $y$ be the vertical distance from the bottom of the gate to the centroid, and $b(y)$ be the width of the gate at that height. Then,

$b(y) = \frac{b_1 + (b_2 - b_1) \frac{y}{h}}{2} = \frac{4 + (8 - 4) \frac{y}{2}}{2} = 2 + 2y$

The total force is then:

$F = \int_0^2 p_{avg} \cdot b(y) \cdot dy = p_{avg} \int_0^2 (2 + 2y) dy = p_{avg} \cdot \left[2y + y^2 ight]_0^2 = 666 	ext{ lb}$.

Therefore, the hydrostatic force on one side of the gate is 666 lb, which is answer choice D.

A Gate in an Irrigation Canal Is Constructed in the Form