Hydrostatic Force

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lil_hawk

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I've already worked this problem out but somehow I've got the double the answer. The question is:

A swimming pool is 20 ft wide and 40 ft long and its bottom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9 ft. If the pool is full of water, find the hydrostatic force on (a) the shallow end, (b) the deep end, (c) one of the sides, and (d) the bottom of the pool.

I've already got the answers for (a) and (b) but the problem is with (c) and maybe (d). I found the equation of the line at the bottom of the pool which is
y = (3/20)x + 3. I then factored out 1/20 and multiplied that by (3x + 60). So what I've got the pressure as 62.5y and the area of a cross section as dx times y except in terms of x. So I've got (1/400)(62.5) times the integral from 0 to 40 of (3x + 60)^2 dx equal to 97500 lbs. I know there's other ways to handle this problem but please help me figure out what happened with this method. I don't know if the problem is with the equation for the line I got because it looks right to me. I think there's another method where the slope of that line is (20/3) but with the way I structured it and having the y direction going downward till you hit the point (0, 3) and then the x directon going across over to the point (40, 9) with the origin at the corner on the shallow end near the surface, it makes sense that my slope is fine. Then for part (d) I've got 300,000 lbs using basically the same method as above with just one integral instead of 2 and the area of the cross section as 20 times dx but my book has 3.03 times 10^5 lbs. from using trig. Is 300,000 lbs ok for an answer? Thanks for any help.

Peace
 
Part d:

The other dimension of the pool is:

\(\displaystyle \L\\\sqrt{40^{2}+3^{2}}=\sqrt{1609}\)

\(\displaystyle \L\\\frac{h(x)-3}{6}=\frac{x}{\sqrt{1609}}\)

\(\displaystyle \L\\h(x)=\frac{6x}{\sqrt{1609}}+3\)

\(\displaystyle \L\\F=\int_{0}^{\sqrt{1609}}62.4(\frac{6x}{\sqrt{1609}}+3)20dx\)

\(\displaystyle \L\\=1248\int_{0}^{\sqrt{1609}}(\frac{6x}{\sqrt{1609}}+3)dx\)

\(\displaystyle \L\\=7488\sqrt{1609}\approx{300361.22}\)

Is this close to what you got?. I hope I didn't get my proportion mixed up.
 
hydrostatic force

I got 300,000 lbs exactly but the main problem is still with part (c). Using the method I described above I get 97,500 but the actual answer is 48,750 lbs. and I can't figure out where I went wrong. Again I understand that there's other methods and I know how they work but I need your help figuring what happened with the way I did it, please. Thanks for your help.

Peace
 
The Skeeter is better at these than me. I am not arriving at the correct answer either. Maybe find the centroid of your region(the side).
 
looking at the side of the pool, I placed the origin at the top left of the deep end.

the line that defines the pool's bottom would then be y = (3/20)x - 9 ... so
x = (20/3)(y + 9)

water depth for a representative cross-section is just (-y)

\(\displaystyle \L F = 62.5 \int_{-9}^{-3} -y \cdot \frac{20}{3}(y+9) dy + 62.5 \int_{-3}^0 -y \cdot 40 dy = 48750\)
 
I was setting up a little different than Skeeter.

I used the top of the shallow end as the origin. This gave me a line equation of

\(\displaystyle \L\\y=\frac{-3x}{20}-3\)

\(\displaystyle \L\\x=\frac{-20}{3}(y+3)\)

I wouldn't think it would make much difference, but it did.

\(\displaystyle \L\\62.5\int_{-9}^{-3}-y\cdot\frac{-20}{3}(y+3)dy=52500\)
 
if done with the origin there, then your horizontal strip has length 40-x for -9 < y < -3

x = (-20/3)(y+3)

40-x = 40 + (20/3)(y+3)

\(\displaystyle \L 62.5 \int_{-9}^{-3} -y \left[40 + \frac{20}{3}(y+3)\right]dy = 37500\)

add the force of the last 3 ft of depth ...

\(\displaystyle \L 62.5 \int_{-3}^0 -40y dy = 11250\)

and you get 48750 lbs.
 
Yes, I see. Thanks Skeeter. It's so obvious once you point it out.
 
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