Basics of Fluid Mechanics, 2014a

114 CHAPTER 4. FLUIDS STATICS
and element moment is
{ }}( { ( θ θ
dM = dF × l = dF 2 r sin cos
2)
2)
and the total moment is
l
or
M =
∫ θ0
0
M =
dM
∫ θ0
The solution of the last equation is
0
( ( θ θ
ρg sin θbr2 r sin cos dθ
2)
2)
The vertical force can be obtained by
or
M = grρ (2 θ 0 − sin (2 θ 0 ))
4
F v =
F v =
∫ θ0
0
F v = gr2 ρ
2
∫ θ0
0
P
PdA v
dA v
{ }} { { }} {
ρgr sin θ rdθ cos θ
(1 − cos (θ 0 ) 2)
Here, the traditional approach was presented first, and the direct approach second.
It is much simpler now to use the second method. In fact, there are many programs
or hand held devices that can carry numerical integration by inserting the function and
the boundaries.
End Solution
To demonstrate this point further, consider a more general case of a polynomial
function. The reason that a polynomial function was chosen is that almost all the
continuous functions can be represented by a Taylor series, and thus, this example
provides for practical purposes of the general solution for curved surfaces.
Example 4.18: