Hydrostatic Pressure and Force — Question 6

Question 6 – Solution

Step 1: Set up the hydrostatic force. Let $y$ be depth below the surface. The gate has constant width $2$ m and extends from $y=0$ to $y=2$. Thus $$dF=(9800y)(2\,dy).$$ Therefore, $$\begin{align*} F&=9800\int_0^2 2y\,dy =19600\left[\frac{y^2}{2}\right]_0^2\\ &=39{,}200\ \mathrm N. \end{align*}$$

Step 2: Find the moment of the pressure force about the surface. The moment arm of the strip force is $y$, so $$\begin{align*} M&=\int_0^2y\,dF =9800\int_0^2 2y^2\,dy\\ &=19600\left[\frac{y^3}{3}\right]_0^2 =\frac{156{,}800}{3}\ \mathrm{N\,m}. \end{align*}$$

Step 3: Divide moment by total force. $$d=\frac{M}{F} =\frac{\frac{156800}{3}}{39200} =\frac43\ \mathrm m.$$ $$\boxed{d=\frac43\text{ m below the surface}}$$ The center of pressure lies below the geometric center at depth $1$ m because pressure increases with depth.