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what am i supposed to do? If you please explain to me step by step. I'm lost after long time trying and my answer is always being different.

Q: A vertical dam has the form of an isosceles trapezoid with horizontal sides parallel. The dam is 30 m high, 20 m in its lower part and 40 m in its upper part. Finally, the dam retains 25 m of water, as indicated in the figure below.

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a) Let $y$ denote the height in meters measured from the base of the dam. The hydrostatic force exerted by the water on the portion of the door comprised between $y$ m and $y+\Delta y$ m is approximately $p(y)\Delta y$ N.

What is $p(y)$? Note that the density of water is $\rho=1000$ Kg/m$^3$ and the acceleration due to gravity on the earth’s surface is $g=9.8$ m/s$^2$. Express your answer as a formula.

b) In Newtons, what is the total hydrostatic force exerted on the dam? Give the answer correct to 3 significant digits.

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    sorry for that. is it unclear? – 2017-01-31
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    now? is it good? – 2017-01-31
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    Not saying that's a big deal, but not giving any feedback about an answer seems kind of ungrateful – 2017-01-31

1 Answers 1

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The hydrostatic force is the product of water pressure and surface, i.e. $$F=P\times S$$ Now between $y$ and $y+\Delta y$ the portion of surface is: $$\Delta S=w(y)\Delta y$$ where $w(y)$ is the width of dam at $y$: $$\frac{w-20}{w_{max}-20}=\frac y{y_{max}}\implies w(y)=20+20\frac y{30}=20+\frac 23 y$$ and $$P=\rho gh=9800(30-y)$$ So just write: $$dF=P\,dS=p(y)dy=9800(30-y)(20+2y/3)dy$$ Now with $p(y)=9800(30-y)(20+2y/3)$ you can do the rest: $$F=\int dF=\int_0^{30} p(y)dy=(5880000 y- \frac{19600}9 y^3)|_0^{30}=1.176\times 10^8$$