Basics of Fluid Mechanics, 2014a

102 CHAPTER 4. FLUIDS STATICS
also a center of area. These concepts have
been introduced in Chapter 3. Several represented
areas for which moment of inertia
and center of area have been tabulated in
Chapter 3. These tabulated values can be used to solve this kind of problems.
Symmetrical Shapes
Consider the two–dimensional symmetrical area that are under pressure as shown
in Figure 4.21. The symmetry is around any axes parallel to axis x. The total force and
moment that the liquid extracting on the area need to be calculated. First, the force is
∫ ∫
F = PdA =
A
(P atmos + ρgh)dA = AP atmos + ρg
∫ l1
l 0
h(ξ)
{ }} {
(ξ + l 0 )sinβdA
(4.107)
In this case, the atmospheric pressure can include any additional liquid layer above
layer “touching” area. The “atmospheric” pressure can be set to zero.
The boundaries of the integral of equation (4.107) refer to starting point and
ending points not to the start area and end area. The integral in equation (4.107) can
be further developed as
⎛
⎞
x c A
{ }} {
∫ l1
F total = AP atmos + ρg sin β ⎜
⎝ l 0 A + ξdA⎟
⎠
l 0
(4.108)
In a final form as
Total Force in Inclined Surface
F total = A [P atmos + ρg sin β (l 0 + x c )]
(4.109)
The moment of the liquid on the area around
point “O” is
y
ξ 0
β
"O"
a
M y =
∫ ξ1
ξ 0
P (ξ)ξdA (4.110)
ξ 1
b
F 1
M y =
∫ ξ1
ξ 0
(P atmos + gρ
Or separating the parts as
ξ sin β
{}}{
h(ξ) )ξdA (4.111)
F 2
Fig. -4.22. The general forces acting
on submerged area.