A gate in an irrigation canal is constructed in the form of a trapezoid 2 m wide at the bottom, 32 m wide at the top, and 2 m high. It is placed vertically in the canal so that the water just covers the gate. Find the hydrostatic force on one side of the gate.
I know that you have to find an expression for the definite integral from 0 to 2 with the gravitational constant, the constant 1000, and the expression for the change in area. I'm not sure how to find an expression for the change in area.
2026-04-06 10:57:13.1775473033
Hydrostatic force on one side of a gate of a trapezoid irrigation canal?
5.3k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
The expression for the change in area
Impose a coordinate system consisting of an $h$-axis that is orthogonal to the top of the trapezoidal gate. The top of the gate is at $h = 0$ and the bottom is at $h=32$. Assume the trapezoid is an isosceles trapezoid (both of the non-parallel sides are equal in length), and that the $h$-axis coincides with its axis of symmetry. You will eventually integrate along the $h$-axis to find total force, so we are interested in finding an expression for the area of a thin strip of the trapezoid cut orthogonal to the $h$-axis. The width of this strip is $dh$, and the height is the distance between the two non-parallel sides of the trapezoid (which is a linear function $h$). Let's now find that linear function. The width is 32 at the top, and 2 at the bottom, and the height is 2, so the slope of this linear function is $\frac{2-32}{2} = -15$. At a height of 0, the width is 32, so the constant term of this linear function is 32. So, for a height $h$, the area of the thin strip of the trapezoid is given by
$$ (-15h+32) dh$$
This term multiplied by the expression for pressure at a depth $h$ is the complete integrand for your definite integral:
$$ \int_0^2 \rho g h \left( -15h+32 \right) dh$$