OK, Here is the hemisphere.
we have a hemisphere with base 16 feet below ground and bottom 21 feet below ground.
We need its volume and the distance of the cg below ground.
The volume is easy, half a sphere
(1/2) (4/3) pi r^3 = (2/3) pi 125 = 250 pi/3
the centroid of a sphere is 3/8 r from the base as derived here:
Therefore the center of mass of the hemisphere is
therefore we must lift a weight of water of
rho g (250 pi/3) a distance of (21+15/8) meters
that is in Joules
use rho = 10^3 kg/m^3 and g = 10 m/s^2
rho g (250 pi/3) a distance of (16+15/8) meters
Now do the cone the same way
base is at 6 meters
volume = (1/3) pi r^2(4)
cg is at [6 - (1/4)4] meters below ground
Can you do the rest now?
Why would you do the distance of hemisphere from 0 to 16, when we are doing just the hemipsher alone then adding it to the rest after.. wouldnt distance by 5-dy
You are lifting the water from the cg of the hemisphere all the way to the surface.
that is 16 meters to the top of the hemisphere plus another 15/8 to the cg
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