How High Can Water Rise Before Tipping a Dam?

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CGI
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Homework Statement


Find height (h) of the water level which can result in failure of dam by tipping it over due to water pressure at U/S of dam. Also draw the free body diagram.

Density of water to be = 1000 kg/m^3 and its specific weight=62.4 lb/ft^3
Hints: The specific weight of concrete is = 150 lb/ft^3.
Width of dam = 1ft.

IMG_1005.JPG


Homework Equations



Fr = (Specific Weight)(Height to Centroid)(Area)

The Attempt at a Solution



I've thought about rearranging the Fr equation so that I get
h = Fr/(Specific Weight * Area)
But the part where I'm confused is where to get the Fr from.

I thought that I could get Fr for the concrete by using
150 lb/ft^3 for the specific weight
(12ft * 1 ft*) As the Area
If I were to go about it this way(which I'm not even sure if it is right) how would I go about finding the height
to the centroid?

I tried to draw the free body diagram and I know that the pressure due from the water is in a prism-like shape as shown, but this is all I have. The resultant force from the water is also shown on there.
IMG_1006.JPG


I hope I used the template right! If anyone could help me on this, I would really appreciate it!
 
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Okay, so once I find the centroid, I use that to calculate Fr? And then from there I can use that to find the height to the center of pressure?
 
CGI said:
Okay, so once I find the centroid, I use that to calculate Fr? And then from there I can use that to find the height to the center of pressure?
I would integrate the moment of the water pressure from top to bottom about the tipping axis A and equate that to the moment generated about the same axis by gravity. That is safer than assuming some sort of "center of pressure" since the moment arm of the water force is a function of depth, not only in length but also in direction.