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Problem *3.90
[Difficulty: 3]
Given:
Data on sphere and weight
T
F
B
W
Find:
SG of sphere; equilibrium position when freely floating
Solution:
Basic equation
F
B
ρ
g
⋅
V
⋅
=
and
Σ
F
z
0
=
Σ
F
z
0
=
TF
B
+
W
−
=
where
TM
g
⋅
=
M1
0
k
g
⋅
=
F
B
ρ
g
⋅
V
2
⋅
=
WS
G
ρ
⋅
g
⋅
V
⋅
=
Hence
Mg
⋅
ρ
g
⋅
V
2
⋅
+
SG
ρ
⋅
g
⋅
V
⋅
−
0
=
SG
M
ρ
V
⋅
1
2
+
=
SG
10 kg
⋅
m
3
1000 kg
⋅
×
1
0.025 m
3
⋅
×
1
2
+
=
SG
0.9
=
The specific weight is
γ
Weight
Volume
=
SG
ρ
⋅
g
⋅
V
⋅
V
=
SG
ρ
⋅
g
⋅
=
γ
0.9
1000
×
kg
m
3
⋅
9.81
×
m
s
2
⋅
Ns
2
⋅
kg m
⋅
×
=
γ
8829
N
m
3
⋅
=
For the equilibriul position when floating, we repeat
the force balance with T = 0
F
B
W
−
0
=
WF
B
=
with
F
B
ρ
g
⋅
V
submerged
⋅
=
From references (trying Googling "partial sphere volume")
V
submerged
π
h
2
⋅
3
3R
⋅
h
−
()
⋅
=
where h is submerged depth and R is the sphere radius
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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

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