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Problem *3.123
[Difficulty: 3]
Given:
Cubical box with constant acceleration
Find:
Slope of free surface; pressure along bottom of box
Solution:
Basic equation
In components
x
p
∂
∂
−
ρ
g
x
⋅
+
ρ
a
x
⋅
=
y
p
∂
∂
−
ρ
g
y
⋅
+
ρ
a
y
⋅
=
z
p
∂
∂
−
ρ
g
z
⋅
+
ρ
a
z
⋅
=
We have
a
x
a
x
=
g
x
0
=
a
y
0
=
g
y
g
−
=
a
z
0
=
g
z
0
=
Hence
x
p
∂
∂
SG
−
ρ
⋅
a
x
⋅
=
(1)
y
p
∂
∂
SG
−
ρ
⋅
g
⋅
=
(2)
z
p
∂
∂
0
=
(3)
From Eq. 3 we can simplify from
pp
x
y
,
z
,
()
=
to
x
y
,
=
Hence a change in pressure is given by
dp
x
p
∂
∂
dx
⋅
y
p
∂
∂
dy
⋅
+
=
(4)
At the free surface p = const., so
dp
0
=
x
p
∂
∂
dx
⋅
y
p
∂
∂
dy
⋅
+
=
or
dy
dx
x
p
∂
∂
y
p
∂
∂
−
=
a
x
g
−
=
0.25 g
⋅
g
−
=
Hence at the free surface
dy
dx
0.25
−
=
The equation of the free surface is then
y
x
4
−
C
+
=
and through volume conservation the fluid rise in the rear
balances the fluid fall in the front, so at the midpoint the
free surface has not moved from the rest position
For size
L8
0
c
m
⋅
=
at the midpoint
x
L
2
=
y
L
2
=
(box is half filled)
L
2
1
4
−
L
2
⋅
C
+
=
C
5
8
L
⋅
=
y
5
8
L
⋅
x
4
−
=
Combining Eqs 1, 2, and 4
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 Fall '07
 Lear
 Fluid Mechanics

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