Solutions - 0100-04

(a) Left:

Set $y=0$ at the water line, increasing downwards.

$$\begin{array}{c}w(y)=4+\frac{5-4}{3}(y-1)\gg \gg \frac{11}{3}+\frac{1}{3}y\\ \\ \gg \gg {\int }_1^4\rho gy(\frac{11}{3}+\frac{1}{3}y)dy\end{array}$$

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Alternative: Set $y=0$ at the top of the trapezoid. Obtain:

$${\int }_0^3\rho g(y+1)(4+\frac{5-4}{3}y)dy$$

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(b) Center:

Set $y=0$ at the water line, increasing downwards.

$$\begin{array}{c}w(y)=5+\frac{4-5}{3}(y+1)\gg \gg \frac{14}{3}-\frac{1}{3}y\\ \\ \gg \gg {\int }_0^2\rho gy(\frac{14}{3}-\frac{1}{3}y)dy\end{array}$$

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Alternative: Set $y=0$ at the top of the trapezoid. Obtain:

$${\int }_1^3\rho g(y-1)(5+\frac{4-5}{3}y)dy$$

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(c) Right:

Set $y=0$ at the water line, increasing downwards.

$$\begin{array}{cccc}& w_S(y)& =2& \text{(square)}\\ & w_T(y)& =2+\frac{0-2}{3}(y-2)& \text{(triangle)}\end{array}$$ $$\gg \gg {\int }_0^2\rho gy2dy+{\int }_2^5\rho gy(2-\frac{2}{3}(y-2))dy$$