Basics of Fluid Mechanics, 2014a

4.5. FLUID FORCES ON SURFACES 103
x c A
{ }} {
∫ ξ1
I x
′ x
′
{ }} {
M y = P atmos ξdA+gρ sin β ξ 2 dA (4.112)
ξ 0
ξ 0
The moment of inertia, I x ′ ′
x
, is about the axis through point “O” into the page.
Equation (4.112) can be written in more compact form as
∫ ξ1
Total Moment in Inclined Surface
M y = P atmos x c A + gρ sin βI x ′ x ′
(4.113)
Example 4.14 can be generalized to solve any two forces needed to balance the area/gate.
Consider the general symmetrical body shown in figure 4.22 which has two forces that
balance the body. Equations (4.109) and (4.113) can be combined the moment and
force acting on the general area. If the “atmospheric pressure” can be zero or include
additional layer of liquid. The forces balance reads
and moments balance reads
F 1 + F 2 = A [P atmos + ρg sin β (l 0 + x c )] (4.114)
F 1 a + F 2 b = P atmos x c A + gρ sin βI x ′ x ′ (4.115)
The solution of these equations is
[(
)
]
ρ sin β − P atmos
gb
x c + l 0 ρ sin β + P atmos
g
bA−,I x ′ x ′ ρ sin β
F 1=
(4.116)
g (b − a)
and
)
]
I x ′ x
[(ρ ′ ρ sin β − sin β − P atmos
ga
x c + l 0 ρ sin β + P atmos
g
aA
F 2=
(4.117)
g (b − a)
In the solution, the forces can be negative or positive, and the distance a or b can
be positive or negative. Additionally, the atmospheric pressure can contain either an
additional liquid layer above the “touching” area or even atmospheric pressure simply
can be set up to zero. In symmetrical area only two forces are required since the
moment is one dimensional. However, in non–symmetrical area there are two different
moments and therefor three forces are required. Thus, additional equation is required.
This equation is for the additional moment around the x axis (see for explanation in
Figure 4.23). The moment around the y axis is given by equation (4.113) and the total
force is given by (4.109). The moment around the x axis (which was arbitrary chosen)
should be
∫
M x = yPdA (4.118)
A