Basics of Fluid Mechanics, 2014a

4.5. FLUID FORCES ON SURFACES 101
The liquid total moment on the gate is
M =
∫ b
0
gρ(l + ξ) sin βadξ(l + ξ)
The integral can be simplified as
∫ b
M = gaρ sin β (l + ξ) 2 dξ (4.106)
0
The solution of the above integral is
( 3 bl 2 +3b 2 l + b 3 )
M = gρa sin β
3
This value provides the moment that F 1 and F 2 should extract. Additional equation is
needed. It is the total force, which is
F total =
The total force integration provides
F total = gρa sin β
∫ b
0
∫ b
0
gρ(l + ξ) sin βadξ
( ) 2 bl+ b
2
(l + ξ)dξ = gρa sin β
2
The forces on the gate have to provide
( ) 2 bl+ b
2
F 1 + F 2 = gρa sin β
2
Additionally, the moment of forces around point “O” is
( 3 bl 2 +3b 2 l + b 3 )
F 1 l + F 2 (l + b) =gρa sin β
3
The solution of these equations is
F 1 =
(3 l + b) abgρ sin β
6
F 2 =
(3 l +2b) abgρ sin β
6
The above calculations are time consuming
and engineers always try to make
life simpler. Looking at the above calculations,
it can be observed that there is
a moment of area in equation (4.106) and
End Solution
β
ξ
"O"
ξ
ξ
l0
l1
dξ
Fig. -4.21. Schematic of submerged area to
explain the center forces and moments.