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Chapter Static (continue)
4.6 Buoyancy and Stability
Fig. 4.34 Schematic of Immersed Cylinder.
One of the oldest known scientific research on fluid mechanics
relates to buoyancy due to question of money was carried by
Archimedes.
Archimedes principle is related to question of density and
volume.
While Archimedes did not know much about integrals, he was
able to capture the essence.
Here, because this material is presented in a different era, more
advance mathematics will be used.
While the question of the stability was not scientifically
examined in the past, the floating vessels structure
(more than 150 years ago) show some understanding.
The total forces the liquid exacts on a body
are considered as a buoyancy issue.
To understand this issue, consider a cubical and a cylindrical
body that is immersed in liquid and center in a depth of, $h_0$
as shown in Figure 4.34.
The force to hold the cylinder at the place must be
made of integration of the pressure around the surface of the
square and cylinder bodies.
The forces on square geometry body are made only of vertical
forces because the two sides cancel each other.
However, on the vertical direction, the pressure on the two
surfaces are different.
On the upper surface the pressure is $\rho \, g \, (h_0a/2)$.
On the lower surface the pressure is $\rho \, g \, (h_0+a/2)$.
The force due to the liquid pressure per unit depth (into the
page) is
\begin{align}
F =
\rho \, g \, \left( (h_0a/2)  (h_0+a/2) \right)\,ll\, b =
 \rho\, g\, a\, b\, ll = \rho \, g \, V
\label{static:eq:qPforce}
\end{align}
In this case the $ll$ represents a depth (into the page).
Rearranging equation \eqref{static:eq:qPforce} to be
\begin{align}
\dfrac{F} {V} = \rho \, g
\label{static:eq:qPforcePerell}
\end{align}
The force on the immersed body is equal to the weight of the displaced liquid.
This analysis can be generalized by noticing two things.
All the horizontal forces are canceled.
Any body that has a projected area that has two sides, those will cancel each other.
Another way to look at this point is by approximation.
For any two rectangle bodies, the horizontal forces are canceling each other.
Thus even these bodies are in contact with each other, the
imaginary pressure make it so that they cancel each other.
On the other hand, any shape is made of many small rectangles.
The force on every rectangular shape is made of its weight of the volume.
Thus, the total force is made of the sum of all the small
rectangles which is the weight of the sum of all volume.
Fig. 4.35 The floating forces on Immersed Cylinder.
In illustration of this concept, consider the cylindrical shape
in Figure 4.34.
The force per area (see Figure 4.35) is
\begin{align}
dF = \overbrace{\rho\,g\, \left(h_0  r\, \sin \theta\right)}^{P}
\overbrace{\sin \theta\, r\, d\theta }^{dA_{vertical}}
\label{static:eq:cylinderElement}
\end{align}
The total force will be the integral of the equation
qref {static:eq:cylinderElement}
\begin{align}
F = \int_{0}^{2\pi} {\rho\,g\,\left(h_0  r\, \sin \theta\right)}
{\,r\, d\theta\, \sin \theta}
\label{static:eq:cylinderElementIa}
\end{align}
Rearranging equation \eqref{static:eq:cylinderElement} transforms
it to
\begin{align}
F = r\,g\,\rho\int_{0}^{2\pi} { \left(h_0  r\, \sin \theta\right)}
{\,\sin \theta\, d\theta }
\label{static:eq:cylinderElementI}
\end{align}
The solution of equation \eqref{static:eq:cylinderElementI} is
\begin{align}
F = \pi \,{r}^{2}\,\rho\,g
\label{static:eq:cylinderElementISolution}
\end{align}
The negative sign indicate that the force acting upwards.
While the horizontal force is
\begin{align}
F_v = \int_{0}^{2\,\pi} { \left(h_0  r\, \sin \theta\right)}
\cos\theta\,d\theta = 0
\label{static:eq:cylinderhorizontalForce}
\end{align}
To what depth will a long log with radius, $r$, a length, $ll$ and density, $\rho_w$
in liquid with density, $\rho_l$.
Assume that $\rho_l>\rho_w$.
You can provide that the angle or the depth.
Fig. 4.36 Schematic of a thin wall floating body.
Typical examples to explain the buoyancy are
of the vessel with thin walls put upside down into liquid.
The second example of the speed of the floating bodies.
Since there are no better examples, these examples are a must.
A cylindrical body, shown in Figure 4.36 ,is
floating in liquid with density, $\rho_{l}$.
The body was inserted into liquid in a such a way that the air
had remained in it.
Express the maximum wall thickness, $t$, as a function
of the density of the wall, $\rho_s$ liquid density, $\rho_{l}$
and the surroundings air temperature, $T_1$ for the body to float.
In the case where thickness is half the maximum, calculate the
pressure inside the container.
The container diameter is $w$.
Assume that the wall thickness is small compared with the other
dimensions ($t << w$ and $t<
Solution
The air mass in the container is
\[
m_{air} = \overbrace{\pi\,w^2\,h}^{V}
\overbrace{\dfrac{P_{atmos}}{R\,T}} ^{\rho_{air}}
\]
The mass of the container is
\[
m_{container} =
\left(\overbrace{\pi\,w^2 + 2\,\pi\,w\,h}^{A}\right)
\,t \,\rho_{s}
\]
The liquid amount enters into the cavity is such that the
air pressure in the cavity equals to the pressure
at the interface (in the cavity).
Note that for the maximum thickness, the height, $h_1$ has to
be zero.
Thus, the pressure at the interface can be written as
\[
P_{in} = \rho_{l}\,g\,h_{in}
\]
On the other hand, the pressure at the interface from the
air point of view (ideal gas model) should be
\[
P_{in} = \dfrac{m_{air}\,R\,T_1}
{\underbrace{ h_{in}\,\pi\,w^2}_{V}}
\]
Since the air mass didn't change and it is known, it can be
inserted into the above equation.
\[
\rho_{l}\,g\,h_{in}+ P_{atmos} = P_{in} =
\dfrac{{\left(\pi\,w^2\,h\right)
\overbrace{\dfrac{P_{atmos}}{R\,T_1}}^{\rho}
}\,R\,T_1}
{ h_{in}\,\pi\,w^2}
\]
The last equation can be simplified into
\[
\rho_{l}\,g\,h_{in} + P_{atmos} =
\dfrac{h \, {P_{atmos}} }
{h_{in}}
\]
And the solution for $h_{in}$ is
\[
h_{in}= 
\dfrac{P_{atmos}
+\sqrt{4\,g\,h\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}}
}{2\,g\,\rho_l}
\]
and
\[
h_{in} =
\dfrac
{\sqrt{4\,g\,h\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}}P_{atmos}}
{2\,g\,\rho_l}
\]
The solution must be positive, so that the last solution is the only
physical solution.
Advance Material
Calculate the minimum density an infinitely long equilateral triangle
(three equal sides) has to be so that the sharp end is in the water.
Solution
The solution demonstrates that when $h \longrightarrow 0$ then
$h_{in} \longrightarrow 0$.
When the gravity approaches zero (macro gravity) then
\begin{align*}
h_{in}= \dfrac{P_{atmos}}{\rho_l\,g}+h
\dfrac{{h}^{2}\,\rho_l\,g}{P_{atmos}}
+\dfrac{2\,{h}^{3}\,{\rho_l}^{2}\,{g}^{2}}{{P_{atmos}}^{2}}
\dfrac{5\,{h}^{4}\,{\rho_l}^{3}\,{g}^{3}}{{P_{atmos}}^{3}}
+\cdots
\end{align*}
This ``strange'' result shows that bodies don't float in the normal sense.
When the floating is under vacuum condition, the following
height can be expanded into
\begin{align*}
h_{in}=\sqrt{\dfrac{h\,P_{atmos}}
{{g\,\rho_l}}}
+\dfrac{P_{atmos}}{2\,g\,\rho_l}
+ \cdots
\end{align*}
which shows that the large quantity of liquid enters into the
container as it is expected.
Archimedes theorem states that the force balance is at displaced weight liquid
(of the same volume) should be the same as the container, the air.
Thus,
\begin{align*}
\overbrace{\pi\, w^2\, (hh_{in}) \,g}^{\text{net displayed water}}=
\overbrace{\left(\pi\,w^2 + 2\,\pi\,w\,h\right)\,t\,\rho_{s}\,g}
^{\text{container}}
+
\overbrace{{\pi\,w^2\,h}\,
\left(\dfrac{P_{atmos}}{R\,T_1} \right) \,g} ^{\text{air}}
\end{align*}
If air mass is neglected the maximum thickness is
\begin{align*}
t_{max} =
\dfrac{
2\,g\,h\,w\,\rho_l+P_{atmos}\,w
w\,\sqrt{4\,gh\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}}
}
{\left( 2\,g\,w+4\,g\,h\right) \,\rho_l\,\rho_s}
\end{align*}
The condition to have
physical value for the maximum thickness is
\begin{align*}
2\,g\,h\,\rho_l+P_{atmos} \ge
\sqrt{4\,gh\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}}
\end{align*}
The full solution is
\begin{align*}
\begin{array}{cc}
t_{max} = &
\dfrac{\left(
w\,R\,\sqrt{4\,gh\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}}
2\,g\,h\,w\,R\,\rho_lP_{atmos}\,w\,R\right)
\,T_1+2\,g\,h\,P_{atmos}\,w\,\rho_l}
{\left( 2\,g\,w+4\,g\,h\right) \,R\,\rho_l\,\rho_s\,T_1}
\end{array}
\end{align*}
In this analysis the air temperature in the container immediately
after insertion in the liquid has different value from the final temperature.
It is reasonable as the first approximation to assume that the process is
adiabatic and isentropic.
Thus, the temperature in the cavity immediately after the insertion is
\begin{align*}
\dfrac{T_i}{T_f} =
\left( \dfrac{P_i}{P_f} \right)
\end{align*}
The final temperature and pressure were calculated previously.
The equation of state is
\begin{align*}
P_i = \dfrac{m_{air}\,R\,T_i}{V_i}
\end{align*}
The new unknown must provide additional equation which is
\begin{align*}
V_i = \pi\,w^2\,h_{i}
\end{align*}
Thickness Below The Maximum
For the half thickness $t= \dfrac{t_{max}}{2}$ the general
solution for any given thickness below maximum is presented.
The thickness is known, but the liquid displacement is still unknown.
The pressure at the interface (after long time) is
\begin{align*}
\rho_l \,g \, h_{in} +P_{atmos} =
\dfrac{\pi\,w^2\,h \dfrac{P_{atmos}}{R\,T_1} R\, T_1}
{\left(h_{in}+h_1\right)\,\pi\,w^2}
\end{align*}
which can be simplified to
\begin{align*}
\rho_l \, g\,h_{in} + P_{atmos} =
\dfrac{h\,P_{atmos}}{h_{in}+h_1}
\end{align*}
The second equation is Archimedes' equation, which is
\begin{align*}
\pi\,w^2\left(hh_{in} h_1\right) =
\left( \pi\,w^2 +2\,\pi\,w\,h)\,t\,\rho_s\,g \right)
+\pi\,w^2\,h\,\left( \dfrac{P_{atmos}}{R\,T_1}\right)\,g
\end{align*}
End Advance Material
A body is pushed into the liquid to a distance, $h_0$ and left at rest.
Calculate acceleration and time for a body to reach the surface.
The body's density is $\alpha\, \rho_{l}$ , where $\alpha$ is
ratio between the body density to the liquid density
and ($0 < \alpha < 1$).
Is the body volume important?
Solution
The net force is
\begin{align*}
F = \overbrace{V\,g\,\rho_l}^{\text{liquid weight}} 
\overbrace{V\,g\,\alpha\,\rho_l}^{\text{body weight}} =
V\,g\, \rho_l \,( 1 \alpha)
\end{align*}
But on the other side the internal force is
\begin{align*}
F = m\,a = \overbrace{V\,\alpha \rho_l}^{m}\, a
\end{align*}
Thus, the acceleration is
\begin{align*}
a = g \left( \dfrac{1\alpha}{\alpha}\right)
\end{align*}
If the object is left at rest (no movement) thus time will
be ($h=1/2\,a\,t^2$)
\begin{align*}
t = \sqrt{\dfrac{2\,h \alpha}{g(1\alpha)}}
\end{align*}
If the object is very light ($\alpha \longrightarrow 0$) then
\begin{align*}
t_{min} =
\sqrt{\dfrac{2\,h\,\alpha}{g}}
+\dfrac{\sqrt{2\,g\,h}\;{\alpha}^{\dfrac{3}{2}}} {2\,g}
+\dfrac{3\,\sqrt{2\,g\,h}\,{\alpha}^{\dfrac{5}{2}}}{8\,g}
+\dfrac{5\,\sqrt{2\,g\,h}\,{\alpha}^{\dfrac{7}{2}}}{16\,g}
+\cdots
\end{align*}
From the above equation, it can be observed that only the density
ratio is important.
This idea can lead to experiment in ``large gravity''
because the acceleration can be magnified and it is
much more than the reverse of free falling.
In some situations, it is desired to find equivalent of force
of a certain shape to be replaced by another force of
a ``standard'' shape.
Consider the force that acts on a half sphere.
Find equivalent cylinder that has the same diameter that has the same force.
Solution
The force act on the half sphere can be found by integrating the forces around
the sphere.
The element force is
\[
dF = (\rho_L  \rho_S) \, g\, \overbrace{r\, \cos\phi\, \cos\theta}^{h}
\overbrace{\cos\theta\,\cos\phi\,
\overbrace{r^2\,d\theta\,d\phi}^{dA} }^{dA_x}
\]
The total force is then
\[
F_x = \int_0^{\pi} \int_0^{\pi}
(\rho_L  \rho_S) \, g\, {\cos^2\phi \cos^2\theta} \,
{r^3\,d\theta\,d\phi}
\]
The result of the integration the force on sphere is
\[
F_s = \dfrac{{\pi}^{2}\, (\rho_L  \rho_S)\, r^3 }{4}
\]
The force on equivalent cylinder is
\[
F_c = \pi\,r^2 \, (\rho_L  \rho_S)\,h
\]
These forces have to be equivalent and thus
\[
\dfrac{{\pi}^{\cancel{2}}\, \cancel{(\rho_L\rho_S)}\,r^{\cancelto{1}{3}}}{4}
=
\cancel{\pi}\,\cancel{r^2} \, \cancel{(\rho_L  \rho_S)}\,h
\]
Thus, the height is
\[
\dfrac{h}{r} = \dfrac{\pi}{4}
\]
In the introduction to this section, it was assumed that above liquid
is a gas with inconsequential density. Suppose that the above layer is another liquid which
has a bit lighter density.
Body with density between the two liquids, $\rho_l < \rho_s < rho_h$ is floating
between the two liquids. Develop the relationship between the densities of liquids and solid
and the location of the solid cubical.
There are situations where density is a function of the depth.
What will be the location of solid body if the
Solution
In the discussion to this section, it was shown that net force is the body volume times the
the density of the liquid.
In the same vein, the body can be separated into two: one in first liquid
and one in the
second liquid.
In this case there are two different liquid densities.
The net force down is the weight of the body $\rho_c\, h\, A$.
Where $h$ is the height of the body and $A$ is its cross section.
This force is balance according to above explanation by the two liquid as
\begin{align*}
\rho_c\, \cancel{h\, A} = \cancel{A \,h}\,\left( \alpha\,\rho_l + (1\alpha) \rho_h \right)
\end{align*}
Where $\alpha$ is the fraction that is in low liquid.
After rearrangement it became
\begin{align*}
\alpha = \dfrac{ \rho_c  \rho_h}{\rho_l  \rho_h}
\end{align*}
the second part deals with the case where the density varied parabolically.
The density as a function of $x$ coordinate along $h$ starting at point $\rho_h$ is
\begin{align*}
\rho (x) = \rho_h  \left( \dfrac{x}{h} \right)^2 \left( \rho_h  \rho_l \right)
\end{align*}
Thus the equilibration will be achieved, $A$ is canceled on both sides, when
\begin{align*}
\rho_c\, h = \int_{x_1}^{x_1+h} \left[ \rho_h  \left( \dfrac{x}{h} \right)^2
\left(\rho_h\rho_l\right) \right]dx
\end{align*}
After the integration the equation transferred into
\begin{align*}
\rho_c\, h = \dfrac{\left( 3\,\rho_l3\,\rho_h\right) \,{x1}^{2}+
\left( 3\,h\,\rho_l3\,h\,\rho_h\right) \,x1+{h}^{2}\,\rho_l+2\,{h}^{2}\,\rho_h}
{3\,h}
\end{align*}
And the location where the lower point of the body (the physical), $x_1$, will be at
\begin{align*}
X_1 = \dfrac{\sqrt{3}\,\sqrt{3\,h^2\,{\rho_l}^{2}+\left( 4\,\rho_c6\,{h}^{2}\,\rho_h\right)
\,\rho_l+3\,{h}^{2}\,{\rho_h}^{2}12\,\rho_c\,\rho_h}+3\,h\,\rho_l3\,h\,\rho_h}
{6\,\rho_h2\,\rho_l}
\end{align*}
For linear relationship the following results can be obtained.
\begin{align*}
x_1=\dfrac{h\,\rho_l+h\,\rho_h6\,\rho_c}{2\,\rho_l2\,\rho_h}
\end{align*}
In many cases in reality the variations occur in small zone compare to the size of the body.
Thus, the calculations can be carried out under the assumption of sharp change.
However, if the body is smaller compare to the zone of variation, they have to accounted for.
A hollow sphere is made of steel ($\rho_s/\rho_w \cong 7.8$) with a $t$ wall thickness.
What is the thickness if the sphere is neutrally buoyant?
Assume that the radius of the sphere is $R$.
For the thickness below this critical value, develop an equation for the
depth of the sphere.
Solution
The weight of displaced water has to be equal to the weight of the sphere
\begin{align}
\label{sphere:gov}
\rho_s\,\cancel{g} \, \dfrac{4\,\pi\, R^3}{3} =
\rho_w \,\cancel{g} \, \left( \dfrac{4\,\pi\, R^3}{3} 
\dfrac{4\,\pi\, \left(Rt\right)^3}{3} \right)
\end{align}
after simplification equation \eqref{sphere:gov} becomes
\begin{align}
\label{sphere:govR}
\dfrac{\rho_s\,R^3 }{\rho_w} =
3\,t\,{R}^{2}3\,{t}^{2}\,R+{t}^{3}
\end{align}
Equation \eqref{sphere:govR} is third order polynomial equation which it's solution
(see the appendix) is
\begin{align*}
\label{sphere:completSol}
t_1&=&\left( \dfrac{\sqrt{3}\,i}{2}\dfrac{1}{2}\right) \,{\left( {\dfrac{\rho_s}{\rho_w}R}^{3}
{R}^{3}\right) }^{\dfrac{1}{3}}+R \\
t_2&=&\left( \dfrac{\sqrt{3}\,i}{2}\dfrac{1}{2}\right) \,{\left( {\dfrac{\rho_s}{\rho_w} R}^{3}
{R}^{3}\right) }^{\dfrac{1}{3}}+R\\
t_3&=& R\,\left( \sqrt[3]{ \dfrac{\rho_s}{\rho_w}  1 } + 1 \right)
\end{align*}
The first two solutions are imaginary thus not valid for the physical world.
The last solution is the solution that was needed.
The depth that sphere will be located depends on the ratio of $t/R$ which similar analysis to the above.
For a given ratio of $t/R$, the weight displaced by the sphere has to be same as the sphere weight.
The volume of a sphere cap (segment) is given by
\begin{align}
\label{sphere:capV}
V_{cap} = \dfrac{\pi\,h^2\,(3Rh)}{3}
\end{align}
Where $h$ is the sphere height above the water.
The volume in the water is
\begin{align}
\label{sphere:waterV}
V_{water} = \dfrac{4\,\pi\, R^3}{3}  \dfrac{\pi\,h^2\,(3Rh)}{3}
= \dfrac{4\,\pi\,\left( R^3 3\,R\,h^2 + h^3 \right) }{3}
\end{align}
When $ V_{water}$ denotes the volume of the sphere in the water.
Thus the Archimedes law is
\begin{align}
\label{sphere:archimedes1}
\dfrac{\rho_w\,4\,\pi\,\left( R^3 3\,R\,h^2 + h^3 \right) }{3} =
\dfrac{\rho_s\,4\,\pi\,\left( 3\,t\,{R}^{2}3\,{t}^{2}\,R+{t}^{3} \right)}{3}
\end{align}
or
\begin{align}
\label{sphere:archimedes}
\left( R^3 3\,R\,h^2 + h^3 \right) = \dfrac{\rho_w}{\rho_s}
\left( 3\,t\,{R}^{2}3\,{t}^{2}\,R+{t}^{3} \right)
\end{align}
The solution of \eqref{sphere:archimedes} is
\begin{multline}
\label{sphere:solArc}
h = \left( \dfrac{\sqrt{fR\,\left( 4\,{R}^{3}fR\right) }}{2}\dfrac{fR2\,{R}^{3}}{2}\right)^{\dfrac{1}{3}}
\\ + \dfrac{{R}^{2}}
{{\left( \dfrac{\sqrt{fR\,\left( 4\,{R}^{3}fR\right) }}{2}\dfrac{fR2\,{R}^{3}}{2}\right)}^{\dfrac{1}{3}}}
\end{multline}
Where $fR = R^3 \dfrac{\rho_w}{\rho_s}\,(3\,t\,R^23\,t^2\,R+t^3)$
There are two more solutions which contains the imaginary component.
These solutions are rejected.
One of the common questions in buoyancy is the weight with variable cross section and fix load.
For example, a wood wedge of wood with a fix weight/load.
The general question is at what the depth of the object (i.e. wedge) will be located.
For simplicity, assume that the body is of a solid material.
Solution
It is assumed that the volume can be written as a function of the depth.
As it was shown in the previous example, the relationship between the depth and
the displaced liquid volume of the sphere.
Here it is assumed that this relationship can be written as
\begin{align}
\label{FixVariableW:dV}
V_w = f(d,\mbox{other geometrical parameters})
\end{align}
The Archimedes balance on the body is
\begin{align}
\label{FixVariableW:archimedes1}
\rho_{ll} V_{a}= \rho_{w} V_{w}
\end{align}
\begin{align}
\label{FixVariableW:archimedes}
d = f^{1} \dfrac{\rho_{ll} V_{a}}{ \rho_{w}}
\end{align}
In example 4.26 a general solution was provided.
Find the reverse function, $f^{1}$ for cone with $30^{\circ}$ when the tip
is in the bottom.
Solution
First the function has to built for $d$ (depth).
\begin{align}
\label{woodenCone:gov}
V_{w} = \dfrac{\pi\,d\,\left(\dfrac{d}{\sqrt{3}} \right)^2}{3} = \dfrac{\pi\,d^3}{9}
\end{align}
Thus, the depth is
\begin{align}
\label{woodenCone:d}
d = \sqrt[3]{\dfrac{9\,\pi\, \rho_w}{\rho_{ll}\,V_a} }
\end{align}
Fig. 4.37 Schematic of floating bodies.
4.6.1 Stability
Figure 4.37 shows a body made of hollow balloon
and a heavy sphere connected by a thin and light rod.
This arrangement has mass centroid close to the middle of the sphere.
The buoyant center is below the middle of the balloon.
If this arrangement is inserted into liquid and will be
floating, the balloon will be on the top and sphere on the bottom.
Tilting the body with a small angle from its resting position
creates a shift in the forces direction
(examine Figure 4.37b).
These forces create a moment which wants to return the body
to the resting (original) position.
When the body is at the position shown in Figure
4.37c ,the body is unstable and any tilt from
the original position creates moment that will further
continue to move the body from its original position.
This analysis doesn't violate the second law of thermodynamics.
Moving bodies from an unstable position is in essence like a potential.
Fig. 4.38 Schematic of floating cubic.
A wooden cubic (made of pine, for example) is inserted into water.
Part of the block floats above water line.
The cubic mass (gravity) centroid is in the middle of the cubic.
However the buoyant center is the middle of the volume
under the water (see Figure 4.38).
This situation is similar to Figure 4.37c.
However, any experiment of this cubic wood shows that it
is stable locally.
Small amount of tilting of the cubic results in returning to the
original position.
When tilting a larger amount than $\pi/4$ , it results in a flipping
into the next stable position.
The cubic is stable in six positions (every cubic has six faces).
In fact, in any of these six positions, the body is in situation
like in 4.37c.
The reason for this local stability of the cubic is that other
positions are less stable.
If one draws the stability (later about this criterion) as a function
of the rotation angle will show a sinusoidal function with
four picks in a whole rotation.
Fig. 4.39 Stability analysis of floating body.
So, the body stability must be based on the difference between the
body's local positions rather than the ``absolute'' stability.
That is, the body is ``stable'' in some points more than others
in their vicinity.
These points are raised from the buoyant force analysis.
When the body is tilted at a small angle, $\beta$,
the immersed part of the body center changes to a new location,
B' as shown in Figure .
The center of the mass (gravity) is still in the old location
since the body did not change.
The stability of the body is divided into three categories.
If the new immerse volume creates a new center in such way
that couple forces (gravity and buoyancy) try to return
the body, the original state is referred as the stable body
and vice versa.
The third state is when the couple forces do have zero moment,
it is referred to as the neutral stable.
The body, shown in Figure 4.39, when given
a tilted position, move to a new buoyant center, B'.
This deviation of the buoyant center from the old buoyant
center location, B , should to be calculated.
This analysis is based on the difference of the displaced liquid.
The right green area (volume) in Figure 4.39 is
displaced by the same area (really the volume) on left since
the weight of the body didn't change
so the total immersed section is constant.
For small angle, $\beta$, the moment is calculated as the
integration of the small force shown in the Figure
4.39 as $\Delta F$.
The displacement of the buoyant center can be calculated by
examining the moment these forces creates.
The body weight creates opposite moment
to balance the moment of the displaced liquid volume.
\begin{align}
\overline{BB'}\, W = \mathbf{M}
\label{static:eq:momentBouyant}
\end{align}
Where $\mathbf{M}$ is the moment created by the displaced areas
(volumes), $\overline{BB'}$ is the distance between points
B and point B' , and,
$W$ referred to the weight of the body.
It can be noticed that the distance $\overline{BB'}$ is an
approximation for small angles (neglecting the vertical component.).
So the perpendicular distance, $\overline{BB'}$, should be
\begin{align}
\overline{BB'} = \dfrac{ \mathbf{M}}{W}
\label{static:eq:momentBouyantD}
\end{align}
The moment $\mathbf{M}$ can be calculated as
\begin{align}
\mathbf{M} = \int_{A}
\overbrace{g\,\rho_l\,\underbrace{x\,\beta\,dA}_{dV}}^{\delta F}\,x
= g \,\rho_l\, \beta \int_{A} x^2 dA
\label{static:eq:staticMoment}
\end{align}
The integral in the right side of equation
qref{static:eq:staticMoment}
is referred to as the area moment of inertia and
was discussed in Chapter 3.
The distance, $\overline{BB'}$ can be written from equation
qref{static:eq:staticMoment} as
\begin{align}
\overline{BB'} = \dfrac{g\,\rho_l\, I_{xx} }
{\rho_{s} V_{body} }
\label{static:eq:tiltdeX}
\end{align}
The point where the gravity force direction is intersecting with
the center line of the cross section is referred as metacentric
point, M.
The location of the metacentric point can be obtained from the geometry as
\begin{align}
\overline{BM} = \dfrac{\overline{BB'}} {\sin \beta}
\label{static:eq:metacentricP}
\end{align}
And combining equations \eqref{static:eq:tiltdeX} with
qref{static:eq:metacentricP} yields
\begin{align}
\overline{BM} = \dfrac{\cancel{g}\, \rho_l \beta I_{xx}}
{\cancel{g}\,\rho_{s}\,\sin\beta\,V_{body}}
=
\dfrac{ \rho_l\,I_{xx}}{\rho_{s}\,V_{body}}
\label{static:eq:GMIntermidiateA}
\end{align}
For small angle ($\beta \sim 0$)
\begin{align}
\lim_{\beta \rightarrow 0}
\dfrac{\sin \beta}{ \beta} \sim 1
\label{static:eq:lhopitalRule}
\end{align}
It is remarkable that the results is independent of the angle.
Looking at Figure 4.39, the geometrical
quantities can be related as
\begin{align}
\overline{GM} =
\overbrace{\dfrac{\rho_l\,I_{xx}}{\rho_{s} V_{body}}}
^{\overline{BM}}
 \overline{BG}
\label{static:eq:GMIntermidiate}
\end{align}
A solid cone floats in a heavier liquid (that is $\rho_l/\rho_c> 1$).
The ratio of the cone density to liquid density is $\alpha$.
For a very light cone $\rho_{c}/\rho_{l} \sim 0$,
the cone has zero depth.
At this condition, the cone is unstable.
For middle range, $1 > \rho_{c}/\rho_{l} > 0$ there could be a range
where the cone is stable.
The angle of the cone is $\theta$.
Analyze this situation.
Solution
The floating cone volume is $\dfrac{\pi\,d\,r^2}{3}$ and the center of
gravity is D/4.
The distance $\overline{BG}$ depend on $d$ as
\begin{align}
\label{coneStability:BG}
\overline{BG} = D/4  d/4
\end{align}
Where $D$ is the total height and $d$ is the height of the submerged cone.
The moment of inertia of the cone is circle shown in Table .
The relationship between the radius the depth is
\begin{align}
\label{coneStability:dr}
r = d\,\tan\theta
\end{align}
\begin{align}
\label{coneStability:GMini}
\overline{GM} =
\dfrac{\rho_l\,\overbrace{\dfrac{\pi\,\left( d\,\tan\theta\right)^4}{64}} ^{I_{xx}}}{\rho_{s}
\underbrace{\dfrac{\pi\,d\,\left( d\,\tan\theta\right)^2}{3} }_{V_{body}} }
 \overbrace{\left(\dfrac{D}{4}  \dfrac{d}{4} \right)}^{\overline{BG}}
\end{align}
Equation \eqref{coneStability:GMini} can be simplified as
\begin{align}
\label{coneStability:GM}
\overline{GM} =
\dfrac{\rho_l\,d\, \tan^2\theta }{\rho_{s}\,192}
 \left(\dfrac{D}{4}  \dfrac{d}{4} \right)
\end{align}
The relationship between $D$ and $d$ is determined by the density ratio ( as displaced volume
\begin{align}
\label{coneStability:dD}
\rho_l\,d^3 = \rho_c\, D^3 \Longrightarrow D = d \sqrt[3]{\dfrac{\rho_l}{\rho_c}}
\end{align}
Substituting equation \eqref{coneStability:dD} into qref{coneStability:GM} yield the solution when
$\overline{GM} = 0 $
\begin{align}
\label{coneStability:sol}
0 = \dfrac{\rho_l\,d\, \tan^2\theta }{\rho_{s}\,192}
 \left(\dfrac{d \sqrt[3]{\dfrac{\rho_l}{\rho_c}}}{4}  \dfrac{d}{4} \right) \Longrightarrow
\dfrac{\rho_l\, \tan^2\theta} {\rho_{s}\,48} = \sqrt[3]{\dfrac{\rho_l}{\rho_c}  1 }
\end{align}
Since $\rho_l > \rho_c$ this never happened.
Fig. 4.40 Cubic body dimensions for stability analysis.
To understand these principles consider the following examples.
A solid block of wood of uniform density,
$\rho_s = \alpha\,\rho_{l}$ where
( $0\le\alpha\le1$ ) is floating in a liquid.
Construct a graph that shows the relationship of the
$\overline{GM}$ as a function of ratio height to width.
Show that the block's length, $L$, is insignificant for this analysis.
Solution
Equation \eqref{static:eq:GMIntermidiate} requires that several quantities should be expressed.
The moment of inertia for a block is given in Table
and is $I_{xx}= \dfrac{La^3}{12}$.
Where $L$ is the length into the page.
The distance $\overline{BG}$ is obtained from Archimedes' theorem and can be expressed as
\begin{align*}
W = \rho_s \,\overbrace{a\,h\,L}^{V} =
\rho_l \,\overbrace{a\,h_1\,L}^{\text{immersed volume} }
\Longrightarrow h_1 = \dfrac{\rho_s}{\rho_l} h
\end{align*}
Fig. 4.41 Stability of cubic body infinity long.
Thus, the distance $\overline{BG}$ is (see Figure
4.38)
\begin{align*}
\overline{BG} = \dfrac{h}{2} 
\overbrace{\dfrac{\rho_s}{\rho_l}\, h}^{h_1}\,\dfrac{1}{2}
= \dfrac{h}{2} \left(1  \dfrac{\rho_s}{\rho_l} \right)
\label{static:eq:BGbar}
\end{align*}
\begin{align*}
GM = \dfrac{\cancel{g}\,\rho_l\,
\overbrace{\dfrac{\cancel{L}\,a^3}{12}}^{I_{xx}} }
{\cancel{g}\,\rho_s\,\underbrace{a\,h\,\cancel{L}}_V}
 \dfrac{h}{2} \left(1  \dfrac{\rho_s}{\rho_l} \right)
\end{align*}
Simplifying the above equation provides
\begin{align*}
\dfrac{\overline{GM}}{h} = \dfrac{1}{12\,\alpha}
\left(\dfrac{a}{h}\right)^2
 \dfrac{1}{2} \left( 1  \alpha \right)
\end{align*}
Notice that $\overline{GM}/{h}$ isn't a function of the
depth, $L$.
This equation leads to the condition where the maximum height
above which the body is not stable anymore as
\begin{align*}
\dfrac{a}{h} \ge \sqrt {{6\,(1\alpha)\alpha}}
\label{static:eq:stabilityCritieriaCubic}
\end{align*}
Fig. 4.42 The maximum height reverse as a function of density ratio.
One of the interesting point for the above analysis is that there
is a point above where the ratio of the height to the body width
is not stable anymore.
In cylindrical shape equivalent to equation
qref{static:eq:stabilityCritieriaCubic} can be expressed.
For cylinder (circle) the moment of inertia is $I_{xx} = \pi\,b^4/64$.
The distance $\overline{BG}$ is the same as for the square shape
(cubic) (see above \eqref{static:eq:BGbar}).
Thus, the equation is
\begin{align*}
\dfrac{\overline{GM}}{h} = \dfrac{g}{64\,\alpha}
\left(\dfrac{b}{h}\right)^2
 \dfrac{1}{2} \left( 1  \alpha \right)
\end{align*}
And the condition for maximum height for stability is
\begin{align*}
\dfrac{b}{h} \ge \sqrt{{32\,(1\alpha)\,\alpha}}
\end{align*}
This kind of analysis can be carried for different shapes and
the results are shown for these two shapes in Figure 4.42.
It can be noticed that the square body is more stable than the circular body shape.
Principle Main Axises
Any body has infinite number of different axises around which moment of
inertia can be calculated.
For each of these axises, there is a different moment of inertia.
With the exception of the circular shape, every geometrical shape has
an axis in which the moment of inertia is without the
product of inertia.
This axis is where the main rotation of the body will occur.
Some analysis of floating bodies are done by breaking the
rotation of arbitrary axis to rotate around the two main axises.
For stability analysis, it is enough to find if the body is stable
around the smallest moment of inertia.
For example, a square shape body has larger moment of inertia
around diagonal.
The difference between the previous calculation and the moment of
inertia around the diagonal is
\[
\Delta I_{xx} =
\overbrace{\dfrac{\sqrt{2}\,a\left(
\dfrac{\sqrt{3}\,a}{2}\right)^3 }{6}}^{I\;diagonal\;axis} \;
\overbrace{\dfrac{a^4}{12}}^{``normal'' axis} \sim
0.07\,{a}^{4}
\]
Which show that if the body is stable at main axises,
it must be stable at the ``diagonal'' axis.
Thus, this problem is reduced to find the stability for principle axis.
Unstable Bodies
What happen when one increases the height ratio above
the maximum height ratio?
The body will flip into the side and turn to the next stable
point (angle).
This is not a hypothetical question, but rather practical.
This happens when a ship is overloaded with containers above the
maximum height.
In commercial ships, the fuel is stored at the bottom of the ship
and thus the mass center (point $G$) is changing during the voyage.
So, the ship that was stable (positive $\overline{GM}$)
leaving the initial port might became unstable
(negative $\overline{GM}$) before reaching the destination port.
Fig. 4.43 Stability of two triangles put tougher.
One way to make a ship to be a hydrodynamic
is by making the body as narrow as possible.
Suppose that two opposite sides triangle (prism) is attached to each other to create a long
``ship'' see Figure 4.43.
Supposed that $\mathbf{a/h}\longrightarrow \tilde 0$ the body will be unstable.
On the other side if the $\mathbf{a/h}\longrightarrow \tilde \infty$ the body is very stable.
What is the minimum ratio of $\mathbf{a/h}$ that keep the body stable at half of the volume in
liquid (water).
Assume that density ratio is $\rho_l / \rho_s = \bar{\rho}$.
Solution
The answer to the question is that the limiting case where $\overline{GM} = 0$.
To find this ratio equation terms in \eqref{static:eq:GMIntermidiate} have to
be found.
The Volume of the body is
\begin{align*}
V = 2\;\left( \dfrac {a^2 \, h} {2} \right) = a^2 \, h
\end{align*}
The moment of inertia is triangle (see explanation in example \eqref{mech:ex:triangleIxx} is
\begin{align*}
I_{xx} = \dfrac{a\,h^3}{2}
\end{align*}
And the volume is
\begin{align*}
V_{body} = a^2 \; \sqrt{h^2  \dfrac{a^2}{4} } = a^2\,h \; \sqrt{1  \dfrac{1}{4}\,
\dfrac{a^2}{h^2} }
\end{align*}
The point $\mathbf{B}$ is a function of the density ratio of the solid and liquid.
Denote the liquid density as $\rho_l$ and solid density as $\rho_s$.
The point $\mathbf{B}$ can be expressed as
\begin{align*}
B = \dfrac {a\, \rho_s} {2\, \rho_l}
\end{align*}
And thus the distance $\overline{BG}$ is
\begin{align*}
\overline{BG} = \dfrac{a}{2} \left( 1  \dfrac{\rho_s}{\rho_l} \right)
\end{align*}
The limiting condition requires that $ \overline{GM} = 0$ so that
\begin{align*}
\dfrac{\rho_l\,I_{xx}}{\rho_{s} V_{body}} = \overline{BG}
\end{align*}
Or explicitly
\begin{align*}
\dfrac{\rho_l \,\dfrac{a\,h^3}{2}} { \rho_s \,a^2\,h \; \sqrt{1  \dfrac{1}{4}\,
\dfrac{a^2}{h^2} } } = \dfrac{a}{2} \left( 1  \dfrac{\rho_s}{\rho_l} \right)
\end{align*}
After rearrangement and using the definitions of $\xi= h/a$ $\bar{\rho}
\rho_l/\rho_s$ results in
\begin{align*}
\dfrac{\bar{\rho} \,\xi^2 }{ \sqrt{1  \dfrac{\xi^2}{4} } }
= \left( 1  \dfrac{1}{\bar{\rho}} \right)
\end{align*}
The solution of the above solution is obtained by squaring both sides and defining a
new variable such as $x=\xi^2$.
After the above manipulation and selecting the positive value and to keep stability as
\begin{align*}
x < \dfrac{\sqrt{\dfrac{
\sqrt{64\,{\bar{\rho}}^{4}64\,{\bar{\rho}}^{3}+{\bar{\rho}}^{2}2\,\bar{\rho}+1}
}{\bar{\rho}}+\dfrac{1}{\bar{\rho}}1}}{2\,\sqrt{2}\,\bar{\rho}}
\end{align*}
4.6.1.1 Stability of Body with Shifting Mass Centroid
Fig. 4.44 The effects of liquid movement on the $\overline{GM}$.
Ships and other floating bodies carry liquid or have a load which
changes the mass location during tilting of the floating body.
For example, a ship that carries wheat grains where the cargo is
not properly secured to the ship.
The movement of the load (grains, furniture, and/or liquid) does
not occur in the same speed as the body itself or the displaced
outside liquid.
Sometimes, the slow reaction of the load, for stability
analysis, is enough to be ignored.
Exact analysis requires taking into account these shifting
mass speeds.
However, here, the extreme case where the load reacts in the
same speed as the tilting of the ship/floating body is examined.
For practical purposes, it is used as a limit for the
stability analysis.
There are situations where the real case approaches to this
extreme.
These situations involve liquid with a low viscosity (like water,
alcohol) and ship with low natural frequency (later on the
frequency of the ships).
Moreover, in this analysis, the dynamics are ignored and only the
statics is examined (see Figure 4.44).
A body is loaded with liquid ``B'' and is floating in a liquid ``A''
as shown in Figure 4.44.
When the body is given a tilting position the body displaces
the liquid on the outside.
At the same time, the liquid inside is changing its mass
centroid.
The moment created by the inside displaced liquid is
\begin{align}
M_{in} = g\, {\rho_l}_B \beta {I_{xx}}_B
\label{static:eq:momentL}
\end{align}
Note that ${I_{xx}}_B$ isn't the same as the moment of inertia
of the outside liquid interface.
The change in the mass centroid of the liquid ``A'' then is
\begin{align}
\overline{G_{1}G_{1}'}
= \dfrac{\cancel{g}\, \cancel{{\rho_l}_B} \beta {I_{xx}}_B}
{\underbrace{\cancel{g}\,V_B\,\cancel{{\rho_l}_B}}
_{\text{Inside liquid weight }}}
= \dfrac{{I_{xx}}_B}{V_B}
\label{static:eq:GG'}
\end{align}
Equation \eqref{static:eq:GG'} shows that $\overline{GG^{'}}$ is only a function of the geometry.
This quantity, $\overline{G_1G_1'}$, is similar for all liquid tanks on the floating body.
The total change of the vessel is then calculated similarly to center
area calculations.
\begin{align}
\cancel{g}\,m_{total}\, \overline{GG'} =
\cancelto{0}{{g}\,m_{body}} +
\cancel{g}\, m_f \overline{G_1G_1'}
\label{static:eq:GGtotal}
\end{align}
For more than one tank, it can be written as
\begin{align}
\overline{GG'} =
\dfrac{g}{W_{total}} \sum_{i=1}^n \overline{G_iG_i}
{\rho_l}_i V_i
=
\dfrac{g}{W_{total}} \sum_{i=1}^n \dfrac{{{I_{xx}}_b}_i}
{{V_b}_i}
\label{static:eq:totalGG}
\end{align}
A new point can be defined as $G_c$.
This point is the intersection of the center line with the
vertical line from $G'$.
\begin{align}
\overline{G\,G_c} = \dfrac{\overline{GG'}} {\sin\beta}
\label{static:eq:GGc}
\end{align}
The distance that was used before $\overline{GM}$ is replaced by
the criterion for stability by $\overline{G_c\,M}$ and is expressed as
\begin{align}
\overline{G_c\,M} =
{\dfrac{g\, \rho_A\,{I_{xx}}_A}{\rho_{s} V_{body}}}
\overline{BG}
 \dfrac{1}{m_{total}}\, \dfrac{{I_{xx}}_b}{V_b}
\label{static:eq:GcM}
\end{align}
If there are more than one tank partially filled with liquid, the general formula is
\begin{align}
\overline{G_c\,M} =
{\dfrac{g\, \rho_A\,{I_{xx}}_A}{\rho_{s} V_{body}}}
\overline{BG}
 \dfrac{1}{m_{total}}
\sum_{i=1}^{n}\dfrac{{{I_{xx}}_b}_i}{{V_b}_i}
\label{static:eq:GcMg}
\end{align}
Fig. 4.45 Measurement of GM of floating body.
One way to reduce the effect of the moving mass center
due to liquid is done by substituting a single tank with
several tanks.
The moment of inertial of the combine two tanks is smaller than
the moment of inertial of a single tank.
Increasing the number of tanks reduces the moment of inertia.
The engineer could design the tanks in
such a way that the moment of inertia is operationally changed.
This control of the stability, $\overline{GM}$, can be achieved by
having some tanks spanning across the entire body with
tanks spanning on parts of the body.
Movement of the liquid (mostly the fuel and water) provides way to
control the stability, $GM$, of the ship.
Metacentric Height, $\overline{GM}$, Measurement
The metacentric height can be measured by finding the change in
the angle when a weight is moved on the floating body.
Moving the weight, $T$ a distance, $d$ then
the moment created is
\begin{align}
M_{weight} = T\,d
\label{static:eq:Td}
\end{align}
This moment is balanced by
\begin{align}
M_{righting} = W_{total} \overline{GM}_{new} \,\theta
\label{static:eq:TdR}
\end{align}
Where, $W_{total}$, is the total weight of the floating
body including measuring weight.
The angle, $\theta$, is measured as the difference in the
orientation of the floating body.
The metacentric height is
\begin{align}
\overline{GM}_{new} = \dfrac{T\,d}{W_{total} \,\theta}
\label{static:eq:GMmessured}
\end{align}
If the change in the $\overline{GM}$ can be neglected,
equation \eqref{static:eq:GMmessured} provides the solution.
The calculation of $\overline{GM}$ can be improved
by taking into account the effect of the measuring weight.
The change in height of $G$ is
\begin{align}
\cancel{g}\, m_{total}\, G_{new} =
\cancel{g}\, m_{ship}\, G_{actual}
+ \cancel{g}\,T\,h
\label{static:eq:deltaGMR}
\end{align}
Combining equation \eqref{static:eq:deltaGMR} with
equation \eqref{static:eq:GMmessured} results in
\begin{align}
\overline{GM}_actual =
\overline{GM}_{new}\, \dfrac{m_{total}}{m_{ship}} 
h \, \dfrac{T}{m_{ship}}
\label{static:eq:GMactual}
\end{align}
The weight of the ship is obtained from looking at the ship
depth.
4.6.1.3 Stability of Submerged Bodies
The analysis of submerged bodied is different from the stability
when the body lays between two fluid layers with different density.
When the body is submerged in a single fluid layer, then none of the
changes of buoyant centroid occurs.
Thus, the mass centroid must be below than buoyant centroid
in order to have stable condition.
However, all fluids have density varied in some degree.
In cases where the density changes significantly,
it must be taken into account.
For an example of such a case is an object floating in a solar pond
where the upper layer is made of water with lower salinity than
the bottom layer(change up to 20% of the density).
When the floating object is immersed into two layers, the stability
analysis must take into account the changes of the displaced liquids
of the two liquid layers.
The calculations for such cases are a bit more complicated but based
on the similar principles.
Generally, this density change helps to increase the
stability of the floating bodies.
This analysis is out of the scope of this book (for now).
4.6.1.4 Stability of None Systematical or ``Strange'' Bodies
Fig. 4.46 Calculations of $\overline{GM}$ for abrupt shape body.
While most floating bodies are symmetrical or semi–symmetrical,
there are situations where the body has a ``strange'' and/or
unsymmetrical body.
Consider the first strange body that has an abrupt step change
as shown in Figure 4.46.
The body weight doesn't change during the rotation
that the green area on the left and the green area on right are
the same (see Figure 4.46).
There are two situations that can occur.
After the tilting, the upper part of the body is above the liquid
or part of the body is submerged under the water.
The mathematical condition for the border is
when $b=3\,a$.
For the case of $b< 3\,a$ the calculation of moment of inertia
are similar to the previous case.
The moment created by change in the displaced liquid (area)
act in the same fashion as the before.
The center of the moment is needed to be found.
This point is the intersection of the liquid line with the brown middle line.
The moment of inertia should be calculated around this axis.
For the case where $b < 3\,a$ $x$ some part is under the liquid.
The amount of area under the liquid section depends on the tilting
angle.
These calculations are done as if none of the body under the
liquid.
This point is intersection point liquid with lower body
and it is needed to be calculated.
The moment of inertia is calculated around this point
(note the body is ``ended'' at end of the upper body).
However, the moment to return the body is larger
than actually was calculated and the bodies tend to be more
stable (also for other reasons).
4.6.1.5 Neutral frequency of Floating Bodies
This case is similar to pendulum (or mass attached to spring).
The governing equation for the pendulum is
\begin{align}
ll \ddot{\beta}  g\,\beta = 0
\label{static:eq:govPendulum}
\end{align}
Where here $ll$ is length of the rode (or the line/wire) connecting
the mass with the rotation point.
Thus, the frequency of pendulum is
$\dfrac{1}{2\,\pi}\sqrt{\dfrac{g}{ll}}$ which measured in $Hz$.
The period of the cycle is $2\,\pi\,\sqrt{ll/g}$.
Similar situation exists in the case of floating bodies.
The basic differential equation is used to balance
and is
\begin{align}
\overbrace{I\ddot{\beta}}^{rotation} 
\overbrace{V\,\rho_s\,\overline{GM}\,\beta}^{rotating\;moment}=0
\label{static:eq:govFloat}
\end{align}
In the same fashion the frequency of the floating body is
\begin{align}
\dfrac{1}{2\,\pi} \sqrt{\dfrac
{V\,\rho_s\,\overline{GM}}{I_{body}}}
\label{static:eq:floatFreq}
\end{align}
and the period time is
\begin{align}
2\,\pi
\sqrt{\dfrac{I_{body}} {V\,\rho_s\,\overline{GM}}}
\label{static:eq:periodFreq}
\end{align}
In general, the larger $\overline{GM}$ the more stable the floating
body is.
Increase in $\overline{GM}$ increases the frequency of the
floating body.
If the floating body is used to transport humans and/or other
creatures or sensitive cargo it requires to reduce the $\overline{GM}$
so that the traveling will be smoother.
4.6.2 Surface Tension
The surface tension is one of the mathematically complex topic and
related to many phenomena like boiling, coating, etc.
In this section, only simplified topics like constant value will be discussed.
In one of the early studies of the surface tension/pressure was done by
Torricelli.
In this study he suggest construction of the early barometer.
In barometer is made from a tube sealed on one side.
The tube is filled with a liquid and turned upside down into the liquid container.
The main effect is the pressure difference between the two surfaces (in the tube
and out side the tune).
However, the surface tension affects the high.
This effect is large for very small diameters.
In interaction of the molecules shown in Figure ? describe the existence of surface tension.
Explain why this description is erroneous?
Solution
The upper layer of the molecules have unbalanced force towards
the liquid phase.
Newton's law states when there is unbalanced force, the body should
be accelerate.
However, in this case, the liquid is not in motion.
Thus, the common explanation is wrong.
Fig. 4.47 A heavy needle is floating on a liquid.
Needle is made of steel and is heavier than water and many
other liquids. However, the surface tension between the needle and
the liquid hold the needle above the liquid.
After certain diameter, the needle cannot be held by the liquid.
Calculate the maximum diameter needle that can be inserted into liquid
without drowning.
Solution
Under Construction
4.7 Rayleigh–Taylor Instability
Rayleigh–Taylor instability (or RT instability) is named after
Lord Rayleigh and G. I. Taylor.
There are situations where a heavy liquid layer is placed over a lighter fluid layer.
This situation has engineering implications in several industries.
For example in die casting, liquid metal is injected in a cavity filled with air.
In poor designs or other situations, some air is not evacuated
and stay in small cavity on the edges of the shape to be casted.
Thus, it can create a situation where the liquid metal is above
the air but cannot penetrate into the cavity because of instability.
This instability deals with a dense, heavy fluid that is being placed above a
lighter fluid in a gravity field perpendicular to interface.
Example for such systems are dense water over oil (liquid–liquid), or water over air(gas–liquid).
The original Rayleigh's paper deals with the dynamics and density variations.
For example, density variations according to the bulk modulus
(see section 4.3.3.2) are always stable but
unstable of the density is in the reversed order.
Supposed that a liquid density is arbitrary function of the
height.
This distortion can be as a result of heavy fluid above the lighter liquid.
This analysis asks the question of what happen when a small amount of
liquid from the above layer enter into the lower layer?
Whether this liquid continue and will grow or will it return to its
original conditions?
The surface tension is the opposite mechanism that will returns the liquid to its
original place.
This analysis is referred to the case of infinite or very large surface.
The simplified case is the two different uniform
densities.
For example a heavy fluid density, $\rho_L$, above lower fluid with lower
density, $\rho_G$.
For perfectly straight interface, the heavy fluid will stay above
the lighter fluid.
If the surface will be disturbed, some of heavy liquid moves down.
This disturbance can grow or returned to its original situation.
This condition is determined by competing forces, the surface density, and
the buoyancy forces.
The fluid above the depression is in equilibrium with the sounding pressure
since the material is extending to infinity.
Thus, the force that acting to get the above fluid down is the buoyancy
force of the fluid in the depression.
Fig. 4.48 Description of depression to explain the Rayleigh–Taylor instability.
The depression is returned to its original position if the surface forces
are large enough.
In that case, this situation is considered to be stable.
On the other hand, if the surface forces (surface tension) are not sufficient,
the situation is unstable and the heavy liquid enters into the liquid
fluid zone and vice versa.
As usual there is the neutral stable when the forces are equal.
Any continues function can be expanded in series of cosines.
Thus, example of a cosine function will be examined.
The conditions that required from this function will be required from all
the other functions.
The disturbance is of the following
\begin{align}
h = h_{max} \cos \dfrac{2\,\pi\,x} {L}
\label{static:eq:disturbance}
\end{align}
where $h_{max}$ is the maximum depression and $L$ is the characteristic
length of the depression.
The depression has different radius as a function of distance from the center
of the depression, $x$.
The weakest point is at $x=0$ because symmetrical reasons the surface
tension does not act against the gravity as shown in Figure
qref{static:fig:cos}.
Thus, if the center point of the depression can ``hold'' the intrusive
fluid then the whole system is stable.
The radius of any equation is expressed by equation \eqref{intro:eq:radius}.
The first derivative of $\cos$ around zero is $\sin$ which is approaching zero
or equal to zero.
Thus, equation \eqref{intro:eq:radius} can be approximated as
\begin{align}
\dfrac{1}{R} = \dfrac{d^2h}{dx^2}
\label{static:eq:radiusAppx}
\end{align}
For equation \eqref{static:eq:disturbance} the radius is
\begin{align}
\dfrac{1}{R} =\dfrac{4\,\pi^2\,h_{max}}{L^2}
\label{static:eq:sinR}
\end{align}
According to equation \eqref{intro:eq:STbcylinder} the pressure difference
or the pressure jump is due to the surface tension at this point
must be
\begin{align}
P_H  P_L = \dfrac{4\,h_{max}\,\sigma\,\pi^2}{L^2}
\label{static:eq:deltaP}
\end{align}
The pressure difference due to the gravity
at the edge of the disturbance is then
\begin{align}
P_H  P_L = g\,\left( \rho_H\rho_L \right) h_{max}
\label{static:eq:detlaPe}
\end{align}
Comparing equations \eqref{static:eq:deltaP} and qref{static:eq:detlaPe}
show that if the relationship is
\begin{align}
\dfrac{4\,\sigma\,\pi^2}{L^2} > g \,\left( \rho_H\rho_L \right)
\label{static:eq:noEq}
\end{align}
It should be noted that $h_{max}$ is irrelevant for this analysis
as it is canceled.
The point where the situation is neutral stable
\begin{align}
L_c = \sqrt {\dfrac{4\,\pi^2 \sigma}{g\left( \rho_H\rho_L \right)} }
\label{phase:eq:notEq}
\end{align}
An alternative approach to analyze this instability is suggested here.
Consider the situation described in Figure 4.49.
If all the heavy liquid ``attempts'' to move straight down, the lighter
liquid will ``prevent'' it.
The lighter liquid needs to move up at the same time but in a different
place.
The heavier liquid needs to move in one side and the lighter
liquid in another location.
In this process the heavier liquid ``enter'' the lighter liquid in one point
and creates a depression as shown in Figure 4.49.
Fig. 4.49 Description of depression to explain the instability.
To analyze it, considered two control volumes bounded by the blue lines
in Figure 4.49.
The first control volume is made of a cylinder with a radius $r$ and the
second is the depression below it.
The ``extra'' lines of the depression should be ignored,
they are not part of the control volume.
The horizontal forces around the control volume are canceling each other.
At the top, the force is atmospheric pressure times the area.
At the cylinder bottom, the force is $\rho\,g\,h\times A$.
This acts against the gravity force which make the cylinder to be
in equilibrium with its surroundings if the pressure at bottom is
indeed $\rho\,g\,h$.
For the depression, the force at the top is the same force at the
bottom of the cylinder.
At the bottom, the force is the integral around the depression.
It can be approximated as a flat cylinder that has depth of $r\,\pi/4$
(read the explanation in the example 4.23)
This value is exact if the shape is a perfect half sphere.
In reality, the error is not significant.
Additionally when the depression occurs, the liquid level
is reduced a bit and the lighter liquid is filling the missing portion.
Thus, the force at the bottom is
\begin{align}
F_{bottom} \sim \pi\,r^2
\left[
\left( \dfrac{\pi\,r}{4} + h \right)
\,\left( \rho_L  \rho_G\right) \,g + P_{atmos}
\right]
\label{static:eq:bottomDepression1}
\end{align}
The net force is then
\begin{align}
F_{bottom} \sim \pi\,r^2
\left(\dfrac{\pi\,r}{4} \right)
\,\left( \rho_L  \rho_G\right) \,g
\label{static:eq:bottomDepression}
\end{align}
The force that hold this column is the surface tension.
As shown in Figure 4.49, the total force is then
\begin{align}
F_{\sigma} = 2\,\pi \, r \,\sigma\, \cos\theta
\label{static:eq:sigmaF}
\end{align}
The forces balance on the depression is then
\begin{align}
2\,\pi \, r \,\sigma \cos\theta \sim \pi\,r^2
\left(\dfrac{\pi\,r}{4} \right)
\,\left( \rho_L  \rho_G\right) \,g
\label{static:eq:balance}
\end{align}
The radius is obtained by
\begin{align}
r \sim \sqrt{ \dfrac { 2\,\pi\,\sigma\cos\theta}{
\,\left( \rho_L  \rho_G\right) \,g }}
\label{static:eq:radiusMinTubeExI}
\end{align}
The maximum surface tension is when the angle, $\theta=\pi/2$.
At that case, the radius is
\begin{align}
r \sim \sqrt{ \dfrac { 2\,\pi\,\sigma}{
\,\left( \rho_L  \rho_G\right) \,g }}
\label{static:eq:radiusMinTubeEx}
\end{align}
Fig. 4.50 The cross section of the interface. The purple color represents the
maximum heavy liquid raising area. The yellow color represents
the maximum lighter liquid that ``goes down.''
The maximum possible radius of the depression depends on the geometry
of the container.
For the cylindrical geometry, the maximum depression radius is about half
for the container radius (see Figure 4.50).
This radius is limited because the lighter liquid has to enter at the
same time into the heavier liquid zone.
Since the ``exchange'' volumes of these two process are the same,
the specific radius is limited.
Thus, it can be written that the minimum radius is
\begin{align}
{r_{min}}_{tube} = 2\,\sqrt{\dfrac{2\,\pi\,\sigma}{
g\,{\left(\rho_L\rho_G\right)} }}
\label{static:eq:solutionD2}
\end{align}
The actual radius will be much larger.
The heavier liquid can stay on top of the lighter liquid without
being turned upside down when the radius is smaller than the
equation
ef{static:eq:solutionD2}.
This analysis introduces a new dimensional number that will be discussed in
a greater length in the Dimensionless chapter.
In equation \eqref{static:eq:solutionD2} the angle was assumed to be 90
degrees.
However, this angle is never can be obtained.
The actual value of this angle is about $\pi/4$ to $\pi/3$
and in only extreme cases the angle exceed this value (considering dynamics).
In Figure 4.50, it was shown that the depression and
the raised area are the same.
The actual area of the depression is only a fraction of the interfacial cross section and is a function.
For example,the depression is larger for square area.
These two scenarios should be inserting into equation 4.168 by introducing
experimental coefficient.
Estimate the minimum radius to insert liquid aluminum into
represent tube at temperature of 600$[K]$.
Assume that the surface tension is $400[mN/m]$.
The density of the aluminum is $2400 kg/m^3$.
Solution
The depression radius is assume to be significantly smaller and thus
equation \eqref{static:eq:radiusMinTubeEx} can be used.
The density of air is negligible as can be seen from the temperature
compare to the aluminum density.
\[
r \sim \sqrt{\dfrac{8\,\pi\,\overbrace{0.4}^{\sigma}}{ 2400\times 9.81}}
\]
The minimum radius is $r \sim 0.02 [m]$ which demonstrates the assumption
of $h>>r$ was appropriate.
Fig. 4.51 Three liquids layers under rotation with various critical situations.
Open Question by April 15, 2010
The best solution of the following question will win 18 U.S. dollars and your name will be associated
with the solution in this book.
A canister shown in Figure 4.51 has three layers of different fluids
with different densities.
Assume that the fluids do not mix.
The canister is rotate with circular velocity, $\omega$.
Describe the interface of the fluids consider all the limiting cases.
Is there any difference if the fluids are compressible?
Where is the maximum pressure points?
For the case that the fluids are compressible, the canister top center is connected to another tank
with equal pressure to the canister before the rotation (the connection point).
What happen after the canister start to be rotated?
Calculated the volume that will enter or leave, for known geometries of the fluids.
Use the ideal gas model.
You can assume that the process is isothermal.
Is there any difference if the process is isentropic?
If so, what is the difference?
4.8 Qualitative questions
These qualitative questions are for advance students and for those who would like
to prepare themselves preliminary examination (Ph. D. examinations).
Additional Question
 1.
The atmosphere has different thickness in different locations.
Where will be atmosphere thickness larger in the equator or the north pole?
Explain your reasoning for the difference.
How would you estimate the difference between the two locations.
 2.
The author's daughter (8 years old) stated that fluid mechanics make no sense.
For example, she points out that warm air raise and therefor the warm spot
in a house is the top floor (that is correct in a 4 story home).
So why when there is snow on high mountains?
It must be that the temperature is below freezing point on the top of the mountain
(see for example Mount Kilimanjaro, Kenya).
How would you explain this situation?
Hint, you should explain this phenomenon using only concepts that where developed in this chapter.
 3.
The surface of the ocean has spherical shape.
The stability analysis that was discussed in this chapter was based
on the assumption that surface is straight.
How in your opinion the effect of the surface curvature affects the stability
analysis.
 4.
If the gravity was changing due to the surface curvature what is the effect on the stability.
 5.
A car is accelerated (increase of velocity) in an include surface upwards.
Draw the constant pressure line.
What will constant pressure lines if the car will be driven downwards.
 6.
A symmetrical cylinder filled with liquid is rotating around its center.
What are the directions of the forces that acting on cylinder.
What are the direction of the force if the cylinder is not symmetrical?
 7.
A body with a constant area is floating in the liquid.
The body is pushed down of the equilibrium state into the liquid by a distance $ll$.
Assume that the body is not totally immersed in the liquid.
What are simple harmonic frequency of the body.
Assume the body mass is $m$ its volume is, $V$.
Additionally assume that the only body motion is purely vertical and
neglect the add mass and liquid resistance.
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