Basics of Fluid Mechanics, 2014a

4.5. FLUID FORCES ON SURFACES 115
For the liquid shown in Figure 4.32 ,calculate
the moment around point “O” and
the force created by the liquid per unit
depth. The function of the dam shape is
y = ∑ n
i=1 a i x i and it is a monotonous
function (this restriction can be relaxed
somewhat). Also calculate the horizontal
and vertical forces.
Solution
b
y = n ∑
i=1 a ix i o
y
x
dA
dy
dx
Fig. -4.32. Polynomial shape dam
description for the moment around
point “O” and force calculations.
The calculations are done per unit depth (into the page) and do not require the actual
depth of the dam.
The element force (see Figure 4.32) in this case is
P
{ }} {
h
dA
{ }} { √{ }} {
dF = (b − y) gρ dx2 + dy 2
The size of the differential area is the square root of the dx 2 and dy 2 (see Figure 4.32).
It can be noticed that the differential area that is used here should be multiplied by the
depth. From mathematics, it can be shown that
√
√
dx2 + dy 2 = dx 1+
The right side can be evaluated for any given function.
For example, in this case describing the dam function
is
√
1+
( ) 2 dy
=
dx
( n
) 2
∑
√ 1+ ia(i) x (i) i−1
i=1
( ) 2 dy
dx
O
y
y
dy
dx
θ
l
dF
b
The value of x b is where y = b and can be obtained by
finding the first and positive root of the equation of
x
x
n∑
0= a i x i − b
i=1
Fig. -4.33. The difference between
the slop and the direction
angle.
To evaluate the moment, expression of the distance and
angle to point “O” are needed (see Figure 4.33). The
distance between the point on the dam at x to the point “O” is
l(x) = √ (b − y) 2 +(x b − x) 2