Problem *3.121
[Difficulty: 3]
Given:
Rectangular container of base dimensions 0.4 m x 0.2 m and a height of 0.4 m is filled with water to a depth of d =
0.2 m. Mass of empty container is M
c
= 10 kg. The container slides down an incline of
θ
= 30 deg with respect to
the horizontal. The coefficient of sliding friction is 0.30.
θ
y
x
α
Find:
The angle of the water surface relative to the horizontal.
Solution:
We will apply the hydrostatics equations to this system.
a
g
p
G
G
ρ
=
+
∇
−
Governing Equations:
(Hydrostatic equation)
a
M
F
G
G
=
(Newton's Second Law)
Assumptions:
(1) Incompressible fluid
(2) Rigid body motion
x
a
x
p
=
∂
∂
−
x
a
x
p
−
=
∂
∂
y
a
g
y
p
=
−
∂
∂
−
()
y
a
g
y
p
+
−
=
∂
∂
Writing the component relations:
dy
y
p
dx
x
p
dp
∂
∂
+
∂
∂
=
We write the total differential of pressure as:
Now along the free surface of the water dp = 0. Thus:
y
x
a
g
a
y
p
x
p
dx
dy
+
−
=
∂
∂
∂
∂
−
=
and
α
atan
dy
dx
−
⎛
⎝
⎞
⎠
=
To determine the acceleration components we analyze a freebody diagram:
MM
c
M
w
+
=
M
c
ρ
w
V
w
⋅
+
=
M1
0
k
g
⋅
999
kg
m
3
⋅
0.4
×
m
⋅
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 Fall '07
 Lear
 Fluid Mechanics

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