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_Mil96 _Mil96
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11 years ago
I got two different answers and I dont understand how you do this. I know generally you integrate the fluid force of water per cubic foot (62.4) times the area of the hemisphere.
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wrote...
#1
11 years ago
Since the water is static the only fluid force is due to height differential. The force on a unit area in the hemisphere is equal to rho*g*h where rho is the density of water, g is the gravitational constant and h is the hight of the water directly above the unit area.

You can do a double integral to sum up all the forces across area of size dx by dy. Once you do that you'll find that the total force is equal to the weight of the water in the hemisphere which equals rho*volume of hemisphere.

hope that helps.

/m
wrote...
#2
11 years ago
This is a "weight-depth-area" concept....area  here is the area of a strip of the hemisphere at a "constant depth"....assume the hemisphere is found by rotating x^2 + y^2 = 1 about the y axis and taking the bottom half. Then depth is " -y " and area is  "2 pi radius thickness" or 2 pi x delta s or 2 pi x sqrt[ 1 + (x ')^2] dy, over [-1,0]....x ' = -y/x...sqrt[1 + {-y/x}^2] = 1 / x....so integral is 2 pi (-y) "weight" dy.....pi weight
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