Basics of Fluid Mechanics, 2014a

116 CHAPTER 4. FLUIDS STATICS
The angle between the force and the distance to point “O” is
( ) ( )
dy
b − y
θ(x) = tan −1 − tan −1
dx
x b − x
The element moment in this case is
dF
{ √}} {
( ) 2 dy
dM = l(x) (b − y) gρ 1+ cos θ(x) dx
dx
To make this example less abstract, consider the specific case of y =2x 6 . In this case,
only one term is provided and x b can be calculated as following
x b = 6 √
b
2
√
Notice that 6 b
2
is measured in meters. The number “2” is a dimensional number with
units of [1/m 5 ]. The derivative at x is
dy
dx =12x5
and the derivative is dimensionless (a dimensionless number). The distance is
( √ ) 2
l =
√ (b − 2 x6 ) 2 6 b
+
2 − x
The angle can be expressed as
The total moment is
M =
⎛ ⎞
θ = tan −1 ( 12 x 5) − tan −1 ⎝ √
b − 2 x6 ⎠
6 b
2 − x
∫ 6 √ b
0
l(x) cos θ(x) ( b − 2 x 6) gρ √ 1+12x 5 dx
This integral doesn’t have a analytical solution. However, for a given value b this integral
can be evaluate. The horizontal force is
F h = bρg b 2 = ρgb2
2
The vertical force per unit depth is the volume above the dam as
F v =
∫ 6 √ b
0
(
b − 2 x
6 ) ρgdx= ρg 5 b 7 6
7