Buoyancy and stability
- The nature of buoyancy
- Buoyant force
- Stability of completely submerged bodies
- Stability of floating bodies
- References
The nature of buoyancy
Submarines, boats, scuba diving, hot air balloons, and airships all rely on the principles of buoyancy and stability covered in these notes.
Any object submerged in a fluid experiences an upward buoyant force. The buoyant force is a direct consequence of the pressure-elevation relation that was discussed earlier. To see how this comes about let us consider a cube with side length $d$ submerged in water to a depth $h$. In the figure below the cube is shown head-on (so the front and back faces are not shown but they have the same pressure distribution as the left and right sides).
We now calculate the pressure distribution and corresponding force acting on each side of the cube. On the top surface the pressure is $p_0=\gamma_w h$ and produces a downward force of magnitude $F_{\rm top}=\gamma_w h d^2$ on the top face ($d^2$ is the area of the top of the cube).
The bottom of the cube is located at a depth $d+h$. The pressure on the bottom surface is therefore $\gamma_w (d+h)$ and produces an upward force $F_{\rm bottom}=\gamma_w (h+d) d^2$. The net vertical force on the cube is therefore
\[\begin{align} F_{\rm net}&=F_{\rm bottom}-F_{\rm top}=\gamma_w (h+d) d^2-\gamma_w h d^2\nonumber\\ &=\gamma_w d^3 \end{align}\]Since the pressure distribution on the left and right side of the cube are the same there is no net horizontal force. The same is true for the front and back faces which aren’t shown. So the only remaining force is the vertical one.
So for the case of the cube submerged in water the buoyance force is equal to the product of the specific weight of water, $\gamma_w$, times the volume of the cube, $d^3$.
Using calculus one can generalize this to any submerged object of arbitrary shape. It turns out that the buoyant force is always vertically upward and proportional to the volume of the object ($i.e.$ the displaced volume of water).
Buoyant force
Buoyancy is a force that opposes the weight of a submerged or partially submerged object. The buoyancy force acts vertically upward at the centroid of the displaced volume. It will be important later to differentiate between centroid and center of gravity. The centroid is the geometric center of the object. The center of gravity is the average location of the weight of an object. For objects of uniform density the centroid and center of gravity coincide. The magnitude of the buoyant force is given as;
\[\begin{align} \textrm{Buoyant force}&=\textrm{Specific weight of fluid}\times\textrm{Displaced volume of fluid}\nonumber\\ F_b&=\gamma_f\,\times\, V_d \label{eq:buoyancy} \end{align}\]An object will float if the buoyant force is larger than the force of gravity and will sink otherwise. The weight of a solid object is given by $w=\gamma V$, where $\gamma$ is the specific weight of the object (not the fluid) and $V$ is the volume of the object (which is different than $V_d$ if the object is partially submerged).
A cube $0.50~\m$ on a side is made of steel having a specific weight of $76.5~\kN/\m^3$. What force is required to hold the cube in equilibrium when completely submerged in water? How about if the cube was made of Styrofoam having a specific weight of $0.4~\kN/\m^3$?
We start by drawing a free-body diagram of the cube. The force of gravity acts at the center of gravity (assuming the cube has uniform density the center of gravity is the same as the centroid). It is labeled in the figure as $w$ and points downward. The buoyancy force (labeled $F_b$) acts at the centroid and points upward.
Our intuition tells us that the cube of steel will sink and that the external force must be upward. Our intuition is correct as an object with an average specific weight greater than that of surrounding fluid will tend to sink. We therefore draw the external force $F_{ext}$ as pointing upward in our free-body diagram.
In equilibrium the sum of the forces must equal zero,
\[\sum F_y=F_b+F_{ext}-w=0\]solving for $F_{ext}$
\[F_{ext}=w-F_b\]In evaluating the weight we will need the volume of the cube:
\[V=(0.5~\m)^3 = 0.125~\m^3\]The weight of the cube is therefore
\[w=\gamma_{\rm steel}\,\times\, V = 76.5~\frac{\kN}{\m^3}\times 0.125~\m^3 = 9.56~\kN\,.\]Since the cube is fully submerged the volume of water displaced by the cube is the same as the volume of the cube. The buoyancy force is therefore,
\[F_b=\gamma_f\,\times\, V_d = 9.81~\frac{\kN}{\m^3}\times 0.125~\m^3 =1.23~\kN\,.\]We therefore evaluation $F_{ext}$ to be
\[F_{ext}=w-F_b=9.56~\kN - 1.23~\kN = 8.33~\kN\]Since the external force evaluated to a positive value we chose the correct direction. That is, we correctly assumed that we would need to apply an upward force on the cube to prevent it from sinking. In this case the buoyancy force was not able to overcome the force of gravity.
We now consider the case when the cube is made of Styrofoam. Since the average specific weight of a Styrofoam cube is less than that of the surrounding water the cube will tend to float. However, let us wrongly assume that our external force is acting upwards so we can make use of the same free-body diagram and equations as before.
The weight of the Styrofoam cube is
\[w=\gamma_{\rm Styrofoam}\,\times\, V = 0.4~\frac{\kN}{\m^3}\times 0.125~\m^3 = 0.05~\kN\,.\]The buoyancy force is the same for Styrofoam and steel! The buoyancy force depends on the specific weight of the fluid (in this case water) and the volume displaced by the object. It doesn't matter what the object is made of!
We can now evaluate the external force and find $ F_{ext}=w-F_b=0.05~\kN - 1.23~\kN = -1.18~\kN $ The negative value of the external force resulted from the incorrect assumption about its direction. In order to keep the Styrofoam cube submerged we must supply a $1.18~\kN$ force downward.
Stability of completely submerged bodies
A completely submerged body is stable if the center of gravity $(cg)$ is below the center of buoyancy $(cb)$.
Stability of floating bodies
A partially submerged body is stable if the center of gravity $(cg)$ is below the metacenter $(mc)$.
| Symbol | name | description |
|---|---|---|
| cg | center of gravity | located at the center of mass of the entire volume |
| cb | center of buoyancy | located at the centroid of the displaced volume |
| mc | metacenter | see figure |
| MB | metacentric radius | distance between cg and mc |
References
https://maritime.org/doc/fleetsub/chap5.htm
https://maritime.org/doc/fleetsub/index.htm
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