Basics of Fluid Mechanics, 2014a

130 CHAPTER 4. FLUIDS STATICS
Substituting equation (4.XXVIII.e) into (4.XXVIII.d) yield the solution when GM =0
0= ρ l d tan 2 θ
ρ s 192
⎛
− ⎜
⎝
d 3 √
ρl
ρ c
4
⎞
− d ⎟
4 ⎠ =⇒ ρ l tan 2 θ
ρ s 48
= 3 √
ρl
ρ c
− 1
(4.XXVIII.f)
Since ρ l >ρ c this never happened.
End Solution
h
a
h 1
L
Fig. -4.40. Cubic body dimensions for stability analysis.
To understand these principles consider the following examples.
Example 4.29:
A solid block of wood of uniform density, ρ s = αρ l where ( 0 ≤ α ≤ 1 ) is floating in a
liquid. Construct a graph that shows the relationship of the GM as a function of ratio
height to width. Show that the block’s length, L, is insignificant for this analysis.
Solution
Equation (4.161) requires that several quantities should be expressed. The moment of
inertia for a block is given in Table 3.1 and is I xx = La3
12
. Where L is the length into the
page. The distance BG is obtained from Archimedes’ theorem and can be expressed as
immersed
V
{ }} {
volume
{ }} {
W = ρ s ahL= ρ l ah 1 L =⇒ h 1 = ρ s
h
ρ l