A dam is inclined at an angle of 30° from the vertical and has the shape of an isosceles trapezoid 200 ft wide at the top and 100 ft wide at the bottom and with a slant height of 160 ft. Find the hydro static force on the dam when it is full of water. ( Recall that the weight density of water is 62.5 lb/ft3.) The answer for this question is 83,354,954 lb but I get something else.Any help is appreciated.
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This looks like a simple integration. What did you do? – almagest Mar 21 '16 at 10:22
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$$62.5\int_{0}^{160}\frac{\sqrt{3}}{2}\left ( 100+\frac{5x}{8} \right )\left ( 160-x \right )dx$$ @almagest – mike Mar 21 '16 at 17:08
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We fought with this one.
To get the quoted numerical answer it is necessary to take 160 to be the length of the long (diagonal) edge of the inclined trapezoid. (Ambiguous use of slant height, perhaps). What I normally think of as the slant height is then $A=\sqrt{160^2-50^2}$, and the vertical height is $H=A cos(30)$.
Let's model good problem solving by drawing it. (Drawings are tools).
https://i.stack.imgur.com/QklQv.jpg
Then mike's integral becomes:
$$F = 62.5\int_{0}^{A}\frac{\sqrt{3}}{2}\left ( 100+\frac{100}{A}x\right )\left ( A-x \right )dx,$$ where $$A=\sqrt{160^2-50^2}.$$ Or alternatively as a vertical integral: $$F = 62.5\int_{0}^{H}y\left ( 200-\frac{100}{H} y\right ) \frac{2}{\sqrt{3}}dy$$ where $$ H=\sqrt{160^2-50^2} cos(30).$$
Dr. Z
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