Hydrostatic force problem with trapezoids?

Question

If I have a trapezoid with a $4$m and $8$m base that is partly submerged vertically in water so that the top is $2$m above the surface and the bottom is $2$m below the surface; how do I express the hydrostatic force against one side of the plate as an integral?

Answer 1

First you need to find an area equation for the triangle section of the trapezoid where the width is increasing from 4m to 8m, keeping in mind that we're only interested in the area that is submerged in water (2m).

Set up the equation so that you divide the total height (4m) by the maximum width of this section (2m since (8m - 4m)/2, there's 2m on each side). This is equal to the line from the base to the surface of the water (2 - x) divided by a, which is the width at 2-x.

$$ \frac{2-x}{a} = \frac{4}{2} $$

Isolate for a: $$ a = \frac{2-x}{2} $$

Now we can find the area of the trapezoid, which is the base added to double the width of the triangle (a), since the triangle is on both sides of the trapezoid. To find the area, we will multiply all of that by the change in height, which is Δx.

$$ A = (4 + 2\cdot\frac{2-x}{2})\cdot\Delta x $$

The force is calculated by taking the definite integral on the interval submerged in water of the pressure multiplied by gravity, the change in depth (2-x) and the area equation.

$$ F = \int_{0}^{2} pg(2-x)\text{d}A $$

Assuming p = 1000 and g = 9.8, we calculate:

$$ F = 1000\cdot9.8\int_{0}^{2} (2-x)\frac{10-x}{2} $$ $$ F = 9800 [28/3] $$ $$ F = 91466.6667 N $$