Calculate the water pressure force on a vertical

Calculate the water pressure force on a vertical trapezoidal plate whose upper base $a=70$ m lies at the water surface, whose lower base is $b=50$ m, and whose height is $h=20$ m.

Step 1: Width as a function of depth

\[w(y)=a-\frac{(a-b)y}{h}.\]

Check: $w(0)=a$, $w(h)=a-(a-b)=b$, as required.

Step 2: Force on a thin strip and total force

\[F=\int_0^h\rho gy\left(a-\frac{(a-b)y}{h}\right)dy=\rho g\left[\frac{ah^2}{2}-\frac{(a-b)h^2}{3}\right]=\rho gh^2\cdot\frac{3a-2(a-b)}{6}=\frac{\rho gh^2(a+2b)}{6}.\]

\[\boxed{F=\frac{\rho gh^2(a+2b)}{6}}\]

Step 3: Numerical value

\[F=\frac{1000\times9.8\times400\times(70+100)}{6}=\frac{1000\times9.8\times400\times170}{6}\approx111.07\times10^6\ \text{N}\approx111.07\ \text{MN}.\]

Step 4: Consistency checks against known special cases

If $a=b$ (rectangle): $F=\rho gh^2(3a)/6=\rho gah^2/2$, matching Problem 188's rectangle formula. If $b=0$ (triangle): $F=\rho gh^2a/6$, matching Problem 189's triangle formula. Both checks confirm the general trapezoid formula.

The lower base $b$ enters with twice the weight of the upper base $a$ in $(a+2b)$, reflecting that the lower edge sits at greater depth, where pressure is higher.