Basics of Fluid Mechanics, 2014a

4.5. FLUID FORCES ON SURFACES 109
The second equation is (4.136) to be written for the moment around the point “O” as
F 1 a 1 + F 2 b 1 = P atmos
′
The solution for the above equation is
x c A { }}
total
{
(b 2 + a 2 )
l(b 2 − a 2 )+g sin β
2
3∑
ρ i+1 I x ′ x ′ i
i=1
F 1=
2 b 1 g sin β P 3
i=1 ρ i+1 x ci A i −2 g sin β P 3
i=1 ρ i+1 I x
′ x
′ i
2 b 1 −2 a 1
−
(b 2 2 −2 b 1 b 2 +2 a 2 b 1 −a 22 )lP atmos
2 b 1−2 a 1
F 2=
2 g sin β P 3
i=1 ρ i+1 I x
′ x
′ i
−2 a 1 g sin β P 3
i=1 ρ i+1 x ci A i
2 b 1−2 a 1
+
(b 2 2 +2 a 1 b 2+a 2 2 −2 a 1 a 2)lP atmos
2 b 1−2 a 1
The solution provided isn’t in the complete long form since it will makes things messy.
It is simpler to compute the terms separately. A mini source code for the calculations is
provided in the the text source. The intermediate results in SI units ([m], [m 2 ], [m 4 ])
are:
x c1 =2.2892 x c2 =3.5355 x c3 =4.9497
A 1 =2.696 A 2 =3.535 A 3 =3.535
I x ′ x ′ 1 =14.215 I x ′ x ′ 2 =44.292 I x ′ x ′ 3 =86.718
The final answer is
and
F 1 = 304809.79[N]
F 2 = 958923.92[N]
End Solution
4.5.2 Forces on Curved Surfaces
The pressure is acting on surfaces
perpendicular to the direction of
the surface (no shear forces assumption).
At this stage, the
pressure is treated as a scalar
function. The element force is
z
dA y
dA x
dA
d F = −P ˆn dA (4.138)
Here, the conventional notation is
used which is to denote the area,
dA z
y
x
Fig. -4.26. The forces on curved area.