Problem 3.51
[Difficulty: 2]
F
R
dy
a
= 1.25 ft
SG
= 2.5
y
b
= 1 ft
y’
w
Given:
Geometry of access port
Find:
Resultant force and location
Solution:
Basic equation
F
R
A
p
⌠
⎮
⎮
⌡
d
=
dp
dy
ρ
g
⋅
=
Σ
M
s
y' F
R
⋅
=
F
R
y
⌠
⎮
⎮
⌡
d
=
A
y p
⋅
⌠
⎮
⎮
⌡
d
=
or, use computing equations
F
R
p
c
A
⋅
=
y'
y
c
I
xx
A y
c
⋅
+
=
We will show both methods
Assumptions:
Static fluid;
ρ
= constant; p
atm
on other side
F
R
A
p
⌠
⎮
⎮
⌡
d
=
A
SG
ρ
⋅
g
⋅
y
⋅
⌠
⎮
⎮
⌡
d
=
but
dA
w dy
⋅
=
and
w
b
y
a
=
w
b
a
y
⋅
=
Hence
F
R
0
a
y
SG
ρ
⋅
g
⋅
y
⋅
b
a
⋅
y
⋅
⌠
⎮
⎮
⌡
d
=
0
a
y
SG
ρ
⋅
g
⋅
b
a
⋅
y
2
⋅
⌠
⎮
⎮
⌡
d
=
SG
ρ
⋅
g
⋅
b
⋅
a
2
⋅
3
=
Alternatively
F
R
p
c
A
⋅
=
and
p
c
SG
ρ
⋅
g
⋅
y
c
⋅
=
SG
ρ
⋅
g
⋅
2
3
⋅
a
⋅
=
with
A
1
2
a
⋅
b
⋅
=
Hence
F
R
SG
ρ
⋅
g
⋅
b
⋅
a
2
⋅
3
=
For y'
y' F
R
⋅
A
y p
⋅
⌠
⎮
⎮
⌡
d
=
0
a
y
SG
ρ
⋅
g
⋅
b
a
⋅
y
3
⋅
⌠
⎮
⎮
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 Fall '07
 Lear
 Fluid Mechanics

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