Hydrostatic Pressure and Force — Question 4

Question 4 – Solution

Step 1: Choose the depth variable. Let $y$ be depth below the oil surface. The plate begins at $y=1$ and ends at $y=3$ m.

Step 2: Find the width at depth $y$. The width grows linearly from $1$ m at $y=1$ to $3$ m at $y=3$. The slope is $$\frac{3-1}{3-1}=1,$$ so the linear width function is $$w(y)=y,\qquad 1\le y\le3.$$

Step 3: Write the force on a horizontal strip. $$p(y)=8000y,\qquad dA=w(y)\,dy=y\,dy.$$ Hence $$dF=p(y)\,dA=8000y^2\,dy.$$

Step 4: Integrate and evaluate. $$\begin{align*} F&=8000\int_1^3y^2\,dy =8000\left[\frac{y^3}{3}\right]_1^3\\ &=\frac{8000}{3}(27-1) =\frac{208000}{3}\ \mathrm N. \end{align*}$$ $$\boxed{F=\frac{208000}{3}\ \mathrm N\approx69{,}333.3\ \mathrm N}$$