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Problem 3.56
[Difficulty: 3]
R
y
R
x
F
R
F
n
Given:
Geometry of lock system
Find:
Force on gate; reactions at hinge
Solution:
Basic equation
F
R
A
p
⌠
⎮
⎮
⌡
d
=
dp
dh
ρ
g
⋅
=
or, use computing equation
F
R
p
c
A
⋅
=
Assumptions:
static fluid;
ρ
= constant; p
atm
on other side
The force on each gate is the same as that on a rectangle of size
hD
=
10 m
⋅
=
and
w
W
2 cos 15 deg
⋅
()
⋅
=
F
R
A
p
⌠
⎮
⎮
⌡
d
=
A
ρ
g
⋅
y
⋅
⌠
⎮
⎮
⌡
d
=
but
dA
w dy
⋅
=
Hence
F
R
0
h
y
ρ
g
⋅
y
⋅
w
⋅
⌠
⎮
⌡
d
=
ρ
g
⋅
w
⋅
h
2
⋅
2
=
Alternatively
F
R
p
c
A
⋅
=
and
F
R
p
c
A
⋅
=
ρ
g
⋅
y
c
⋅
A
⋅
=
ρ
g
⋅
h
2
⋅
h
⋅
w
⋅
=
ρ
g
⋅
w
⋅
h
2
⋅
2
=
Using given data
F
R
1
2
1000
⋅
kg
m
3
⋅
9.81
×
m
s
2
⋅
34 m
⋅
2 cos 15 deg
⋅
⋅
×
10 m
⋅
2
×
Ns
2
⋅
kg m
⋅
×
=
F
R
8.63 MN
⋅
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics, Gate

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