How can I find the distance and centroid of a hatch in fluid mechanics?

AI Thread Summary
The discussion focuses on calculating the distance from the fluid surface to the center of a hatch, labeled as H_c, using geometric principles. The area of the hatch was determined to be 0.160 m², and the specific weight was calculated as 8.83 kN/m³, leading to a resultant force of 1213 N. The user initially struggled with finding H_c but eventually solved it using trigonometric functions, resulting in H_c being 0.855 m. For finding the centroid of the hatch, the user combined the centroids of the individual shapes, confirming that this method was effective. The conversation highlights the importance of geometry and trigonometry in fluid mechanics calculations.
Jason03
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Im working on the problem below...

http://img90.imageshack.us/img90/4939/dfadfgs5.jpg

Im having trouble finding the distance I have labled in the problem...H_c...which is the distance from the fluid surface to the center of the hatch...It appears to be found just using geometry but I am not getting it...

I found the area of the hatch to be .160 m...by adding the areas of the rectangle and two semi circles...and I have the specific weight found by multiplying (.90 x 9.81 Kn/m^3) = 8.83 Kn/m^3...

so the resultant force formula is

F_r = specific weight x area x H_c...

and the correct answer for the resultant force is 1213 N...but I can't get H_c...

Also later in the problem I have to find the centroid of the hatch...so how would I go about doing that...would I take the centroid for each shape and average them?
 
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Jason03: Study some trigonometry. I.e., memorize the definition of sine, cosine, and tangent for a right triangle. Remember those? You must learn those.

The length of the slanted side would be (0.6 m)/cos(40 deg), right? See if you can figure out how to find H_c and F_r now.
 
yea actually i figured it out shortly after i posted...but i did it slightly differently..

Cos40 = y/.450

Y= .450cos40

Y= .344 m

So ….(.6 +.6) - .344 = .855

h_c = .855 m


and the centroids were easily enough found by just adding the sum of the composites of a circle and square...

Thanks!
 
Nice work. That is a good approach you used.
 

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