# Fluid Mechanics and distance

1. Jan 27, 2009

### Jason03

Im working on the problem below...

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 im 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 cant 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?

Last edited by a moderator: May 3, 2017
2. Jan 28, 2009

### nvn

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.

3. Jan 28, 2009

### Jason03

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!

4. Jan 28, 2009

### nvn

Nice work. That is a good approach you used.