# Hydrostatic pressure on an equilateral triangle

The Triangle is point downward at water level, all sides are 2 feet.

I have no idea what to do, I know that the interval for my integration depends on the height of the triangle but I do not know how to find that and I am pretty certain this is some weird geometry trick that I do not know.

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You can use the Pythagorean theorem to solve for the height of an equilateral triangle in terms of the sides. Specifically, the height will obey $h^2+(s/2)^2=s^2$, hence $h=\frac{\sqrt{3}}{2}s$, where $s$ is the side length and $h$ the height. – anon Jun 17 '12 at 21:44
How does that follow the pythagorean thoerem? – user138246 Jun 17 '12 at 21:45
Draw a line to divide the triangle in half. It will hit the horizontal side of the triangle at a right angle and create one triangle on the left, the other on the right (they are reflections of each other across the dividing line). Each of these triangles has sides $h,s/2,s$... – anon Jun 17 '12 at 21:47
Strange, I am getting $4 = 1 + b^2$ Is that wrong? – user138246 Jun 17 '12 at 21:49
If you denote the height by $b$, then that is right. Since $s=2$ the equation for the height is $h^2+(2/1)^2=2^2$, which is what you have. – anon Jun 18 '12 at 1:27

Apply the Pithagorean theorem and find $c$ (see sketch). The width $w(h)$ of a rectangular strip drawn on the triangle at a distance $h$ from the top varies linearly with $h$ from $l$ to $0$ as $h$ goes from $0$ to $c$. Deduce the following formula $$w(h)=l\left( 1-\frac{h}{c}\right).\tag{1}$$ Since the hydrostatic pressure is given by

$$P(h)=\rho gh,\tag{2}$$ the hydrostatic force exerted on the triangle is the definite integral $$F=\int_{0}^{c}\rho ghw(h)\; dh=l\rho g\int_{0}^{c}h\left( 1-\frac{h}{c}\right) dh=\ldots =l\rho g\times \frac{1}{6}c^{2}\tag{3}$$

As for the meaning, numeric values and units of the symbols they are as follows:

• $P$ is the hydrostatic pressure at a generic point $H(\text{measured in }\textrm{Pa } \equiv$ Pascal above the atmospheric pressure)
• $\rho$ is the water density ($\approx 1000 \textrm{kg/m}^{3}$),
• $g$ is the gravitational acceleration ($9.81 \textrm{m/s}^{2}$),
• $h$ is the height of the fluid column above $H\; (\textrm{m}).$ $^1$
• $F$ is the hydrostatic force $(\text{ }\mathrm{N})$

$^1$ $1\text{ }\mathrm{ft\ }=0.3048\text{ }\mathrm{m}$