Problem *3.101
[Difficulty: 3]
(
L + c
)/2
L
c
F
BB
W
B
F
BR
W
R
L
/2
a
θ
F
hinge,y
F
hinge,x
Given:
Geometry of block and rod
Find:
Angle for equilibrium
Assumptions:
Water is static and incompressible
Solution:
Basic
equations
Σ
M
Hinge
0
=
F
B
ρ
g
⋅
V
⋅
=
(Buoyancy)
The free body diagram is as shown.
F
BB
and F
BR
are the buoyancy of the
block and rod, respectively; c is the (unknown) exposed length of the rod
Taking moments about the hinge
W
B
F
BB
−
()
L
⋅
cos
θ
⋅
F
BR
Lc
+
2
⋅
cos
θ
⋅
−
W
R
L
2
⋅
cos
θ
⋅
+
0
=
with
W
B
M
B
g
⋅
=
F
BB
ρ
g
⋅
V
B
⋅
=
F
BR
ρ
g
⋅
−
⋅
A
⋅
=
W
R
M
R
g
⋅
=
Combining equations
M
B
ρ
V
B
⋅
−
L
⋅
ρ
A
⋅
−
⋅
+
2
⋅
−
M
R
L
2
⋅
+
0
=
We can solve for M
B
ρ
A
⋅
L
2
c
2
−
⋅
2M
B
ρ
V
B
⋅
−
1
2
M
R
⋅
+
⎛
⎝
⎞
⎠
⋅
L
⋅
=
M
B
ρ
A
⋅
2L
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 Fall '07
 Lear
 Fluid Mechanics

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