7.3 Buoyancy, Flotation, and Accelerated Fluid Masses

7.3 Buoyancy, Flotation, and Accelerated Fluid Masses

This section addresses the mechanics of floating and submerged bodies under static conditions, as well as the dynamic state of fluids subjected to constant linear acceleration or uniform rotation—a condition known in engineering mechanics as relative equilibrium of liquids.

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1. Archimedes' Principle & Buoyant Force

Archimedes' Principle states that any body completely or partially submerged in a fluid experiences an upward vertical force equal to the weight of the fluid displaced by the body.

Mathematical Formulation

Consider an elemental vertical prism of height $h = h_2 - h_1$ and cross-sectional area $dA$ submerged in a fluid of specific weight $\gamma_f$: $dF_b = (P_{\text{bottom}} - P_{\text{top}}) dA = (\gamma_f h_2 - \gamma_f h_1) dA = \gamma_f (h_2 - h_1) dA = \gamma_f dV$ Integrating across the entire submerged body: $F_b = \gamma_f V_d = \rho_f g V_d$ where:

• $F_b$ is the buoyant force ($\text{kN}$ or $\text{N}$), acting vertically upward.
• $V_d$ is the displaced fluid volume (submerged volume of the body).
• The line of action of $F_b$ passes directly through the centroid of the displaced volume, designated as the Center of Buoyancy ($CB$ or $B$).

Principle of Flotation

For any body floating in static equilibrium in a single liquid, vertical force equilibrium requires: $\sum F_y = 0 \implies W = F_b \implies \gamma_{\text{body}} V_{\text{total}} = \gamma_f V_d$ $\frac{V_d}{V_{\text{total}}} = \frac{\gamma_{\text{body}}}{\gamma_f} = \frac{SG_{\text{body}}}{SG_f}$ For a prismatic floating body of uniform horizontal cross-section and total vertical height $H$, the draft (submerged depth $D$) is: $D = H \left(\frac{SG_{\text{body}}}{SG_f}\right)$

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2. Stability of Submerged vs. Floating Bodies

Submerged Bodies (Submarines, Torpedoes)

For completely submerged bodies, the displaced shape does not change as the body rotates, so the Center of Buoyancy ($B$) remains fixed relative to the body.

1. Stable Equilibrium: Center of Gravity ($G$) lies vertically below the Center of Buoyancy ($B$). Any angular tilt creates a restoring righting couple ($W$ downward at $G$, $F_b$ upward at $B$).
2. Unstable Equilibrium: $G$ lies vertically above $B$. Any slight tilt creates an overturning couple that causes the body to capsize until $G$ settles below $B$.
3. Neutral Equilibrium: $G$ and $B$ coincide.

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Floating Bodies and the Metacenter ($M$)

When a floating body heels (tilts) by an angle $\theta$, the submerged geometry changes, shifting the Center of Buoyancy from $B$ to a new position $B'$. The vertical line of action through $B'$ intersects the original vertical axis of symmetry at the Metacenter ($M$).

Metacentric Radius ($MB$)

By equating the moment of the submerged volume shift (wedges of immersion and emersion) to the moment of the buoyant force: $MB = \frac{I}{V_d}$ where:

• $I$ is the second moment of area (moment of inertia) of the waterline horizontal plane about the tilting axis (longitudinal axis for rolling, transverse axis for pitching).
• $V_d$ is the displaced volume of fluid.

Metacentric Height ($MG$)

The metacentric height ($MG$) is the primary measure of floating stability: $MG = MB \pm GB$

• If $G$ is located above $B$: $MG = MB - GB = \frac{I}{V_d} - GB$.
• If $G$ is located below $B$: $MG = MB + GB = \frac{I}{V_d} + GB$.

Stability Conditions for Floating Bodies:

• $MG > 0$ ($M$ above $G$): Stable equilibrium. A restoring couple forms to return the vessel to upright.
• $MG = 0$ ($M$ at $G$): Neutral equilibrium. The vessel remains in the tilted position without restoring or capsizing forces.
• $MG < 0$ ($M$ below $G$): Unstable equilibrium. An overturning couple accelerates the heel, causing capsizing.

Righting and Overturning Moments

For small angles of heel (typically $\theta \le 10^\circ\text{–}12^\circ$): $RM = W \cdot \overline{GZ} = W \cdot MG \sin \theta$ where $\overline{GZ} = MG \sin \theta$ is the righting arm.

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3. Relative Equilibrium: Uniform Linear Acceleration

When a container of liquid accelerates uniformly, no relative motion occurs between fluid particles; the liquid behaves as a rigid body with zero internal shear stresses.

Governing Differential Equations

Summing forces on a fluid element of mass $dm = \rho dx dy dz$ undergoing acceleration components $a_x$ (horizontal) and $a_y$ (vertical): $\frac{\partial P}{\partial x} = -\rho a_x, \quad \frac{\partial P}{\partial y} = -\rho (g \pm a_y)$

Inclination of Isobaric Surfaces (Free Surface Angle $\theta$)

Surfaces of constant pressure (isobars), including the liquid free surface, tilt at an angle $\theta$ relative to the horizontal: $\tan \theta = \frac{a_x}{g \pm a_y}$

• Vertical Acceleration Convention: Use $+a_y$ for upward acceleration and $-a_y$ for downward acceleration.
• If a container is in free fall ($a_y = -g$), $\tan \theta \to \infty$ and hydrostatic pressure drops to zero everywhere ($P = 0$).

Pressure Variation in Linearly Accelerated Fluids

$P_2 - P_1 = \rho a_x (x_1 - x_2) + \rho (g \pm a_y) (y_1 - y_2)$ For purely vertical depth $h$ below the free surface: $P = \gamma \left(1 \pm \frac{a_y}{g}\right) h$

Surface Geometry in Open and Closed Rectangular Tanks

• Open Tanks (No Spill): The liquid volume is conserved, so the free surface pivots about the centroid of the original static surface. If length is $L$, the liquid rises at the trailing wall and drops at the leading wall by: $\Delta y = \frac{L}{2} \tan \theta$
• Open Tanks (Spilling): If $\Delta y$ exceeds the original freeboard, liquid spills over the edge, and the remaining liquid volume determines the new free surface plane.

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4. Relative Equilibrium: Uniform Rotation (Forced Vortex)

When a cylindrical tank of liquid rotates about its vertical centerline at constant angular velocity $\omega$ ($\text{rad/s}$), centripetal acceleration ($a_r = -\omega^2 r$) drives the liquid outwards, creating a forced vortex.

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Equation of the Parabolic Free Surface

From the balance of radial pressure gradient and centrifugal force $\frac{\partial P}{\partial r} = \rho \omega^2 r$ and vertical hydrostatic balance $\frac{\partial P}{\partial z} = -\rho g$: $\frac{dz}{dr} = \frac{\omega^2 r}{g} \implies z = \frac{\omega^2 r^2}{2g} + C$ Measuring the vertical coordinate $y$ from the lowest point (vertex) of the paraboloid: $y = \frac{\omega^2 r^2}{2g}$ At the outer container wall ($r = R$): $H = \frac{\omega^2 R^2}{2g}$ where $H$ is the total height of the paraboloid of revolution.

Critical Geometric Properties of the Paraboloid

1. Volume of the Paraboloid: $V_{\text{paraboloid}} = \frac{1}{2} \pi R^2 H$ The volume of a paraboloid of revolution equals exactly one-half the volume of its circumscribing cylinder of radius $R$ and height $H$.
2. Unspilled Liquid Symmetry Rule: If no liquid spills from the open cylinder, the volume of air space before rotation equals the volume of the paraboloid above the vertex. Consequently, the liquid rises at the perimeter wall by $H/2$ and depresses at the center axis by $H/2$ relative to the original static liquid level ($d_0$): $d_{\text{rim}} = d_0 + \frac{H}{2}, \quad d_{\text{center}} = d_0 - \frac{H}{2}$
3. Base Pressure Distribution: Pressure at the bottom of the container varies with radial position $r$: $P(r) = \gamma h(r) = \gamma \left(d_{\text{center}} + \frac{\omega^2 r^2}{2g}\right)$
   • Minimum pressure occurs at the center ($r = 0$): $P_{\text{center}} = \gamma d_{\text{center}}$.
   • Maximum pressure occurs at the outer rim ($r = R$): $P_{\text{rim}} = \gamma d_{\text{rim}}$.

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5. Worked Example: Floating Timber Scow Stability

Problem Statement: A rectangular timber scow has a width of $B = 6.0\text{ m}$, a length of $L = 12.0\text{ m}$, and a total height of $3.0\text{ m}$. The total weight of the scow and its cargo is $W = 1,271.4\text{ kN}$. It floats in seawater with a specific weight of $\gamma_{sw} = 10.05\text{ kN/m}^3$. The composite center of gravity ($G$) is located along the vertical centerline at $KG = 1.60\text{ m}$ above the flat bottom.

1. Determine the draft $D$ of the scow.
2. Determine the metacentric height $MG$ for rolling.
3. Determine the righting moment $RM$ when the scow heels by an angle of $\theta = 6^\circ$.

Step-by-Step Solution:

1. Determine draft $D$:

   • By Archimedes' principle: $W = F_b = \gamma_{sw} V_d$ $V_d = \frac{W}{\gamma_{sw}} = \frac{1,271.4\text{ kN}}{10.05\text{ kN/m}^3} = 126.51\text{ m}^3$
   • Displaced volume is $V_d = L \times B \times D$: $D = \frac{V_d}{L \times B} = \frac{126.51\text{ m}^3}{12.0\text{ m} \times 6.0\text{ m}} = \frac{126.51}{72.0} = 1.757\text{ m}$

2. Determine Metacentric Height ($MG$):

   • Center of Buoyancy ($B$) from the bottom: $KB = \frac{D}{2} = \frac{1.757\text{ m}}{2} = 0.879\text{ m}$
   • Distance from $B$ to $G$ ($G$ is above $B$): $GB = KG - KB = 1.600\text{ m} - 0.879\text{ m} = 0.721\text{ m}$
   • Metacentric radius $MB$ for roll (rotation about longitudinal centerline): $MB = \frac{B^2}{12 D} = \frac{(6.0\text{ m})^2}{12 (1.757\text{ m})} = \frac{36.0}{21.084} = 1.707\text{ m}$
   • Metacentric height: $MG = MB - GB = 1.707\text{ m} - 0.721\text{ m} = +0.986\text{ m}$
   • Because $MG > 0$, the vessel is in stable equilibrium.

3. Determine Righting Moment ($RM$ at $\theta = 6^\circ$): $RM = W \cdot MG \sin \theta = (1,271.4\text{ kN})(0.986\text{ m}) \sin 6^\circ$ $RM = (1,253.6\text{ kN}\cdot\text{m})(0.10453) = 131.04\text{ kN}\cdot\text{m}$

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6. CELE Exam Traps & Common Computational Errors