Problem 4.68
[Difficulty: 2]
CS
c
R
x
d
V
V
θ
y
x
Given:
Water flowing past cylinder
Find:
Horizontal force on cylinder
Solution:
Basic equation: Momentum flux in x direction
Assumptions:
1) Steady flow
2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
Hence
R
x
u
1
ρ
⋅
u
1
−
A
1
⋅
(
)
⋅
u
2
ρ
⋅
u
2
A
2
⋅
(
)
⋅
+
=
0
ρ
V
−
sin
θ
( )
⋅
(
)
⋅
V a
⋅
b
⋅
(
)
⋅
+
=
R
x
ρ
−
V
2
⋅
a
⋅
b
⋅
sin
θ
( )
⋅
=
For given data
R
x
1000
−
kg
m
3
⋅
3
m
s
⋅
⎛
⎜
⎝
⎞
⎠
2
×
0.0125
×
m
⋅
0.0025
×
m
⋅
sin 20 deg
⋅
(
)
×
N s
2
⋅
kg m
⋅
×
=
R
x
0.0962
−
N
=
This is the force on the fluid (it is to the left).
Hence the force on the cylinder is
R
x
R
x
−
=
R
x
0.0962 N
=

Problem 4.75
[
Difficulty:
2]

Problem 4.83
[
Difficulty:
2]

Problem 4.88
[
Difficulty:
3]