1. Hydrostatic Forces on Curved, Submerged Surfacesx
Pressure is always acting perpendicular to the solid surface since there is no shear motion in static condition. Pz P q Px Z q q
dAx=dAcos(q)
dAz=dAsin(q)

2. Projected Forces x h
dAz Z
Integrated over
all elements

3. Buoyancy Force acting down FD= rgV1 from Buoyancy = FU-FD
=rg(V2-V1)=rgV
V: volume occupied by the object
Force acting up FU = rgV2
from

4. Horizontal Forces x Z h Projected area Ax dAz
Finding Fx is to determine the force acting on a plane submerged surface oriented perpendicular to the surface. Ax is the projection of the curved surface on the yz plane. Similar conclusion can be made to the force in the y direction Fy. Z
Integrated over
all elements h
dAx
Equivalent system:
A plane surface perpendicular to the free surface

5. Examples
Determine the magnitude of the resultant force acting on the hemispherical surface. x
2 m
R=0.5 m z
equals
minus

6. Line of Action zC z’ Projection in x-direction
Horizontal direction: line of action goes through z’ zC z’
Projection
in x-direction
Vertical direction: the line of action is 3R/8
away from the center of the hemisphere z
The resultant moment of both forces with respect to the center of the hemisphere should be zero:
Fx( )-Fz(0.1875)
=15386( )-2564(0.1875)=0
location of the centroid for a hemisphere
is 3R/8=0.1875(m) away from the equator plane