Hydrostatic Pressure and Force — Question 1

Question 1 – Solution

Step 1: Choose the depth variable. Let $y$ be the depth below the water surface, measured in meters. The window extends from $$y=0.5\quad\text{to}\quad y=0.5+1.5=2.$$

Step 2: Write the pressure at depth $y$. Freshwater has weight density $\gamma=9800\,\mathrm{N/m^3}$, so $$p(y)=\gamma y=9800y\quad\mathrm{N/m^2}.$$

Step 3: Find the area of a horizontal strip. The window is $2$ m wide. A strip of thickness $dy$ has area $$dA=2\,dy.$$ Therefore, $$dF=p(y)\,dA=(9800y)(2\,dy).$$

Step 4: Integrate over the window. $$\begin{align*} F&=9800\int_{0.5}^{2}2y\,dy =9800\left[y^2\right]_{0.5}^{2}\\ &=9800(4-0.25) =9800(3.75) =36{,}750\ \mathrm N. \end{align*}$$ $$\boxed{F=36{,}750\ \mathrm N}$$