Hydrostatic Pressure and Force — Question 2

Question 2

An inverted triangular gate has top width (6) ft at the surface and vertex (4) ft deep. Find the force $(\gamma=62.4\,\mathrm{lb/ft^3})$.

Question 2 – Solution

Step 1: Choose the depth variable. Let $y$ be depth below the water surface. The gate extends from $y=0$ to $y=4$ ft.

Step 2: Find the width of the gate at depth $y$. The width decreases linearly from $6$ ft at $y=0$ to $0$ ft at $y=4$. Its slope is $$\frac{0-6}{4-0}=-\frac32,$$ so $$w(y)=6-\frac32y.$$

Step 3: Write the force on a horizontal strip. At depth $y$, $$p(y)=62.4y\quad\mathrm{lb/ft^2}, \qquad dA=w(y)\,dy.$$ Thus $$dF=62.4y\left(6-\frac32y\right)dy.$$

Step 4: Integrate and evaluate. $$\begin{align*} F&=62.4\int_0^4y\left(6-\frac32y\right)dy\\ &=62.4\int_0^4\left(6y-\frac32y^2\right)dy\\ &=62.4\left[3y^2-\frac12y^3\right]_0^4\\ &=62.4(48-32)=62.4(16)=998.4\ \mathrm{lb}. \end{align*}$$ $$\boxed{F=998.4\ \mathrm{lb}}$$