I need to find the hydrostatic force on one side wall of the pool when it's full of water.
I'm not sure how to start this problem off because of the different depths. Any ideas?
I'll try to do this without a diagram...
Pressure is a function of the depth only, so
$\displaystyle P = \rho g d$, where d is the depth and $\displaystyle \rho$ is the water density.
We also know that $\displaystyle F = \int \int P(d) dA$ where we are integrating over the surface we want the force on.
Since the pressure is constant for a constant depth, we can break the surface into thin horizontal strips. So let x be the horizontal direction and y the vertical. P(d) doesn't depend on x, so
$\displaystyle F = \int \int P(d) dx dy = \int_0^{16} P(d) \cdot 50 dy = 50 \rho g \int_0^{16} y dy$
$\displaystyle F = 50 \rho g \frac{1}{2}(16)^2 = 6400 \rho g$
(Remember we need $\displaystyle \rho$ and g in English units! Since I don't have $\displaystyle \rho$ for water in those units handy, I'll leave the rest to you.)
-Dan
Hello, topsquark,
in general I can understand your solution. But I have one point which I don't understand completely:
The pressure in a (calm, non moving) fluid works always perpendicular to the surface. So the force to the surface depends on the depth of the fluid and the size of the surface.
In the given problem the surface in question is a trapezoid. If I understand your calculatuions right you considered the sidewall to be a rectangle. (But maybe I've misread something)
EB
First pressure is a scalar, not a vector, quantity so it can't point perpendicular to anything. The force exerted, on the other hand, is a vector and by symmetry must be pointing perpendicular to the surface.
To answer your second question: I misread the problem! I worked out the pressure on one END of the pool, not one of the side walls.
The method I used can be rearranged to work this out. Now, however, the thin rectangular strips will have a length depenent on the slope of the bottom surface:
$\displaystyle F = \rho g \int_3^{16} \int_0^{-\frac{80(y-16)}{13}}y dxdy + \rho g \int_0^3 \int_0^{80} y dx dy$
$\displaystyle F = -\rho g \int_3^{16} \frac{80(y-16)}{13}y dy + \rho g \int_0^3 80 y dy$
$\displaystyle F = -\frac{80}{13} \rho g \int_3^{16} (y^2-16y) dy + 80 \rho g \int_0^3 y dy$
$\displaystyle F = -\frac{80}{13} \rho g \left ( \frac{1}{3}(16)^3-8(16)^2 \right ) + $ $\displaystyle \frac{80}{13}\rho g \left ( \frac{1}{3}(3)^3-8(3)^2 \right ) + 80 \rho g \frac{1}{2}(16)^2 $
$\displaystyle F = 14053.33 \rho g$
-Dan