# Fluid Statics Problem

1. May 10, 2007

### Dafe

1. The problem statement, all variables and given/known data
The gate shown is 3m wide and for analysis can be considered massless. For what depth of water will this rectangular gate be in equilibrium as shown?

2. Relevant equations

These are the ones that I'm trying to use, my guide does it differently and gets the right answer. I'm trying to figure out where my mistake is..

$$F_Ry'-F*L=0$$
$$F_Ry'=\int{pydA}$$

3. The attempt at a solution
NB! The variable y in my equation is the êta in my drawing. Please ignore the y and h symbols in the drawing.

$$d=ysin{\theta}$$

$$p=\rho gysin{\theta}$$

$$F_Ry'=sin{\theta}\rho gw \int_0^{sin{\theta}y}y^2dy$$
=
$$\frac{3}{16}\rho gwy^3$$

I put this into the momentum equation and solve for d giving me 2,44m.
The answer in the book is 2,66m.
Does anyone see some major errors in my thinking, I'm sure there are some :)

I hope my formulation is good enough..

Last edited: May 10, 2007
2. May 10, 2007

### variation

Hello,

The pressure in the depth $$h$$ is $$\rho gh$$.
In your formulation, the depth in your pressure $$p$$ should be $$d-y\sin\theta$$.
The torque by the water is
$$\int_{0}^{\frac{d}{\sin\theta}}\rho g(d-y\sin\theta)y(w\cdot dy)=\frac{1}{6}\frac{\rho gwd^3}{\sin^2\theta}$$