Hydrostatic Pressure and Force — Question 7

Question 7

A rectangular dam face is $b$ meters wide and extends from the surface to depth $h$. Derive $F(h)$. If $h$ doubles, by what factor does $F$ change?

Question 7 – Solution

Step 1: Choose the depth variable. Let $y$ measure depth below the surface. The dam face extends from $y=0$ to $y=h$.

Step 2: Write the force on a horizontal strip. The pressure is $p(y)=\gamma y$, and a strip of width $b$ and thickness $dy$ has area $dA=b\,dy$. Therefore, $$dF=p(y)\,dA=\gamma by\,dy.$$

Step 3: Integrate from the surface to depth $h$. $$\begin{align*} F(h)&=\gamma b\int_0^h y\,dy\\ &=\gamma b\left[\frac{y^2}{2}\right]_0^h =\frac12\gamma bh^2. \end{align*}$$ $$\boxed{F(h)=\frac12\gamma bh^2}$$

Step 4: Determine the effect of doubling the depth. $$F(2h)=\frac12\gamma b(2h)^2 =4\left(\frac12\gamma bh^2\right)=4F(h).$$ $$\boxed{\text{Doubling }h\text{ multiplies the force by }4.}$$