Notes on Buoyancy, Stability, and Pressure Distribution (Pages 1–9)

Page 1

  • Buoyancy and Archimedes’ laws

    • The same hydrostatic principles used to compute forces on surfaces apply to the net pressure force on a submerged or floating body.
    • Archimedes’ two buoyancy laws (3rd century B.C.):
      1) A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces.
      2) A floating body displaces its own weight in the fluid in which it floats.
    • Archimedean derivations (illustrative): using vertical-force balance (Eq. (2.30)) for a body between an upper surface 1 and a lower surface 2 gives
      FB=F<em>y(2)−F</em>y(1)=(extfluidweightabove2)−(extfluidweightabove1)=extweightoffluidequivalenttothebodyvolumeag2.33FB = F<em>y(2) - F</em>y(1) = ( ext{fluid weight above } 2 ) - ( ext{fluid weight above } 1 ) = ext{weight of fluid equivalent to the body volume} ag{2.33}
    • Alternatively, summing vertical forces on elemental vertical slices yields
      FB =
      ho igl( V_{ ext{displaced}} igr) g = ext{weight of fluid displaced} ag{2.34}
    • Center of buoyancy (B): for uniform specific weight, the line of action of FB passes through the center of volume of the displaced fluid. This point B may not coincide with the actual center of mass of the body.
    • Generalization to layered fluids (LF): for layered fluids, sum the weights of each displaced layer:
      (FB)_{LF} =

    = extstyleigl(
    ho_i igr) g ( ext{displaced volume}) \
    (2.35)

    • Each displaced layer has its own center of volume; one sums the moments of incremental buoyant forces to get the center of buoyancy in LF.
    • Buoyancy in gases: even gases exert buoyancy. Example: an average person (~60 lbf/ft³) weighs ~180 lbf and has volume ~3.0 ft³ in air density ~0.0763 lbf/ft³; air buoyancy reduces apparent weight by about 0.23 lbf. In vacuum, the person would weigh ~0.23 lbf more.
    • For balloons/blimps, air buoyancy is the controlling factor in design.
    • Small buoyant forces can drive many flow phenomena (e.g., natural convection, vertical mixing in oceans).
    • Floating bodies: only a portion is submerged; the rest is above the surface. For a floating body, the buoyant force equals the body’s weight and acts collinearly with the weight (no net moment in static equilibrium). This yields
      FB =
      ho igl( ext{displaced volume}igr) g = W ag{2.36}

  • Examples and implications

    • If a body’s weight and displaced volume correspond to the fluid’s density, the body is neutrally buoyant (no net vertical force when fully submerged or at a given submerged volume).
    • Neutral buoyancy is exploited in practice (e.g., Swallow float for flow visualization; submarines adjust ballast tanks for positive/neutral/negative buoyancy).
    • Floating stability: a floating body may overturn if disturbed; stability is tested by small perturbations to see if a restoring moment arises.
Page 2
  • Stability and metacenter concepts
    • For a floating body, static stability is analyzed by first finding the basic floating position via (2.36) and locating the centers G (center of gravity) and B (center of buoyancy).
    • Tilt the body by a small angle A0. The new buoyancy center moves to B′. A vertical line through B′ intersects the line of symmetry at the metacenter M. For small A0, M is effectively fixed.
    • Stability criterion: if M is above G (MG > 0), a restoring moment exists and the body is stable; if M is below G (MG < 0), the body is unstable and tends to overturn. Stability generally increases with a larger metacentric height MG.
  • Equation for metacentric height (stability criterion)
    • The general, elegant relationship linking key points (center of gravity G, center of buoyancy B, metacenter M) and waterline geometry is
      MG = rac{Io}{V{ ext{sub}}} - BG ag{2.37}
      where:
    • I_o is the area moment of inertia of the waterline area about the tilt axis,
    • V_sub is the submerged volume,
    • BG is the distance between G and B.
    • This formula applies with the simplifying assumption of a smooth waterline shape; it underpins naval design and stability testing.
  • Practical notes
    • The metacentric height MG is a property of the cross-section and draft; bigger MG implies greater stability.
    • Naval architects use the waterline area’s area moment of inertia (I_o) to compute MG efficiently, especially for vessels with varying cross-sections.
    • Stability analysis must consider real-world constraints (e.g., grounding, maneuvering limits); the Costa Concordia example (Fig. 2.19) illustrates potential failures if stability is not properly accounted for.
  • Example: neutral buoyancy and stability concepts explained further (lead into Example 2.12, to be detailed in the next page)
Page 3
  • Example 2.11: determining a block’s average specific weight from immersion data
    • Given: a concrete block weighs 100 lbf in air and 60 lbf when submerged in freshwater (density 62.4 lbf/ft³).
    • Free-body balance in immersion:
      extapparentweight+F<em>B−W=0ext{apparent weight} + F<em>B - W = 0 or equivalently 60extlbf+F</em>B−100extlbf=0 <br/>ightarrowFB=40extlbf60 ext{ lbf} + F</em>B - 100 ext{ lbf} = 0 \ <br /> ightarrow F_B = 40 ext{ lbf}.
    • Buoyant force equals weight of displaced fluid, so
      FB = ho{ ext{water}} g imes V{ ext{block}} \ ightarrow 40 = (62.4 rac{ ext{lbf}}{ ext{ft}^3}) imes V{ ext{block}} \
      ightarrow V_{ ext{block}} = rac{40}{62.4} ext{ ft}^3 = 0.641 ext{ ft}^3.
    • Specific weight of block
      ar{
      ho}{ ext{block}} = rac{W}{V{ ext{block}}} = rac{100 ext{ lbf}}{0.641 ext{ ft}^3} \
      ightarrow ar{
      ho}_{ ext{block}} o 156 rac{ ext{lbf}}{ ext{ft}^3}.
    • Neutral buoyancy: if the body’s weight equals the buoyant force for some displaced volume, the body remains at rest at that level. Neutrally buoyant particles are used for flow visualization; a Swallow float is a neutrally buoyant device used to track ocean currents. Submarines adjust buoyancy by ballast tanks.
  • Stability concept recap
    • Stable: a small disturbance yields a restoring moment; unstable: overturns rather than returns.
    • Naval and marine design rely on MG > 0; otherwise a vessel may tip over under perturbations.
Page 4
  • Static stability calculation steps for a symmetric floating body
    1) Compute the basic floating position using Eq. (2.36) to locate G and B.
    2) Tilt the body by a small angle A0; determine the new waterline and the new center of buoyancy B′.
    3) Draw a vertical line through B′ to locate the metacenter M; M’s location is effectively independent of small A0 for small angles.
    4) Evaluate MG = MI/VSUB − BG; if MG > 0, the body is stable; if MG < 0, unstable.
  • Stability and waterline area relation (Naval architecture result)
    • A compact form relates the stability parameter MG to the geometry of the waterline area via the area moment of inertia Io:
      MG = rac{Io}{V{ ext{sub}}} - BG ag{2.37}
    • Io is the area moment of inertia of the waterline area about the tilt axis, and V_sub is the submerged volume.
    • The engineer computes G and B from body geometry and then uses Io and V_sub to determine MG.
  • Real-world note
    • Stability analysis can be complicated for irregular shapes; the example of icebergs and large ships shows why robust design and operational practices are essential.
  • Example: a barge with rectangular cross section (to be detailed next)
Page 5
  • Example 2.12: metacentric height for a barge with a uniform rectangular cross section
    • Geometry: barge with width 2L, vertical draft H, length into the page b.
    • Waterline area (relative to tilt axis O): base b, height 2L ⇒
      I_o = rac{b (2L)^3}{12} = rac{8 b L^3}{12} = rac{2}{3} b L^3.
    • Submerged volume: for a barge with cross-section 2L by b and draft H,
      Vextsub=2LimesbimesH.V_{ ext{sub}} = 2 L imes b imes H./
    • If G is exactly at the waterline, then BG = H/2 (center of buoyancy B lies at mid-depth of the submerged portion).
    • Using (2.37):
      MG = rac{Io}{V{ ext{sub}}} - BG = rac{ rac{2}{3} b L^3}{2 L b H} - rac{H}{2} = rac{L^2}{3H} - rac{H}{2} = rac{2L^2 - 3H^2}{6H}.
    • Stability condition: MG > 0 ⇒ 2L^2 > 3H^2 ⇒ L^2 > frac{3}{2} H^2 ⇒ 2L > 2.45 H (approximately).
    • Interpretation: the wider the barge relative to its draft, the more stable it is; lowering the center of gravity G also improves stability.
  • Remarks
    • The result demonstrates how geometry dictates stability: as L increases relative to H, MG increases, enhancing restoration after tilt.
Page 6
  • Pressure distribution in rigid-body motion
    • In rigid-body motion, all particles translate and rotate together with no internal deformation; there are no viscous effects in the balance, so the pressure field satisfies

      abla p =
      ho (oldsymbol{g} - oldsymbol{a}) ag{2.38}
      utfull
    • Here, g is gravitational acceleration vector, a is the rigid-body acceleration (translation + rotation effects). The pressure gradient aligns with the direction of g − a, and surfaces of constant pressure are perpendicular to this direction.
    • The general case for combined translation and rotation will be discussed in Chap. 3 (Fig. 3.11).
Page 7
  • Uniform linear acceleration (translational acceleration only)
    • Consider a tank of water in a car that accelerates with ax (horizontal) and az (vertical, if any).
    • The direction of the greatest increase of pressure is given by the vector sum g − a; the angle ϕ that the pressure-gradient lines make with the horizontal satisfies
      heta = an^{-1} rac{ax}{g + az}. ag{2.39} (Note: ϕ is the tilt of the constant-pressure lines; the free surface tilts accordingly.)
    • The magnitude of the pressure gradient along the steepest ascent is
      G = igl(ax^2 + (g + az)^2igr)^{1/2}. ag{2.40}
    • These results hold independent of container size/shape as long as the fluid remains connected.
  • Example 2.13 (slosh and corner pressure)
    • Problem: A drag racer accelerates at 7 m/s²; mug depth 10 cm, coffee depth 7 cm at rest; density 1010 kg/m³.
    • (a) Spill check: tilt angle θ from ax/g:
      heta = an^{-1} rac{a_x}{g} = an^{-1} rac{7}{9.81}

ightarrow heta
obreak = 35.5^ deg.
- The free surface rises on the rear side by Az = (3 cm) tan θ ≈ 2.14 cm. Since 2.14 cm < 3 cm (the remaining clearance to the rim, 10 cm − 7 cm = 3 cm), there is no spill.

  • (b) Pressure at corner A: at rest, PA ≈ ρ g h_rest = (1010)(9.81)(0.07) ≈ 694 Pa.
    • Under acceleration, the pressure at A increases to about 906 Pa (about 31% higher). An alternative projection along the slant yields the same result when accounting for the tilt and the spatial location of A relative to the slanted free surface.
Page 8
  • Rotating fluids (rigid-body rotation about the vertical z-axis)
    • Consider rotation with angular velocity Ω about the z-axis, with no translation. The velocity field is Ω × r, and the acceleration is the centripetal term a = −r Ω² i (radial inward).
    • The force balance becomes

      abla p =
      ho ( oldsymbol{g} - oldsymbol{a}) =
      ho(-g oldsymbol{k} + r ext{-dependent terms}) ag{2.41, 2.42}
      abla p =
      ho( -g oldsymbol{k} + oldsymbol{r} imes oldsymbol{Ω}^2 )
      abla p equations.
      abla p ext{ components give }
      abla p/
      abla r =
      ho Ω^2 r,
      abla p/
      abla z = -
      ho g.
  • Solving (by integrating with respect to r and z) yields p = rac{1}{2} ho Ω^2 r^2 + f(z). ag{2.44}
    • The remaining equation ∂p/∂z = −ρ g implies f′(z) = −ρ g, so f(z) = −ρ g z + C. Therefore
      p(r,z) = rac{1}{2}
      ho Ω^2 r^2 -
      ho g z + C. ag{2.44'}
    • The constant C is determined by the condition on the free surface (where p = 0). The resulting constant-pressure surfaces are paraboloids (not exponential curves as sometimes informally described). The still-water level is a reference where p = Pa, and the surface in rotation bows outward with increasing r.
  • Figure and interpretation
    • Fig. 2.22 shows the development of paraboloid constant-pressure surfaces in a fluid in rigid-body rotation; the dashed line indicates the direction of maximum pressure increase.
    • This analysis highlights how rotation modifies the hydrostatic pressure field in a rotating container.
Page 9
  • Completion of rotating-fluid analysis
    • The parabolic shape of the free surface under rotation is a classic result from p = 1/2 ρ Ω^2 r^2 − ρ g z + C with the free-surface boundary condition p = 0 along the surface.
    • For small Ω, the surface remains nearly flat near the axis; for large Ω, the surface bulges outward more strongly. This has practical implications for rotating tanks, partially filled spinning spacecraft, and geophysical flows in rotating systems.
Connections and significance
  • Archimedes’ buoyancy: foundational concept tying submerged volume to buoyant force; center of buoyancy and the alignment of FB with displaced fluid volume are central to stability analyses.
  • Metacentric height (MG) and stability: a compact design criterion for ships and floating bodies; Io, V_sub, BG connect geometry, weight distribution, and stability.
  • Stability metrics inform design choices: waterline geometry, center-of-gravity location, and draft all influence stability margins.
  • Pressure distribution in non-inertial frames: rigid-body translation and rotation modify the hydrostatic pressure field, with practical implications for accelerating vehicles, sensors, and industrial equipment.
  • Real-world relevance: neutrally buoyant systems, submarines, icebergs (density contrasts and submergence fractions), and extreme events (capsizing risks) are all anchored in these principles.
Key equations (summary)
  • Buoyant force (Archimedes’ law):
    FB =
    ho{ ext{fluid}} g igl( V{ ext{displaced}} igr) = W_{ ext{displaced}} ag{2.34}
  • Floating body buoyancy (full-submersion case):
    FB =
    ho g igl( V_{ ext{displaced}} igr) = W ag{2.36}
  • Center of buoyancy and neutral buoyancy concepts:
    • Center of buoyancy B is the centroid of the displaced fluid volume.
    • Neutral buoyancy occurs when W = FB for each submerged condition.
  • Metacentric height (stability):
    MG = rac{Io}{V{ ext{sub}}} - BG ag{2.37}
  • Example geometry (barge stability condition):
    • Io = rac{b(2L)^3}{12}, \, V{ ext{sub}} = 2LbH, \, BG = rac{H}{2}
    • Hence, MG = rac{L^2}{3H} - rac{H}{2} = rac{2L^2 - 3H^2}{6H} \ ext{Stable if } MG > 0 \ 2L^2 > 3H^2
      ightarrow 2L > 2.45 H
  • Uniform rigid-body acceleration (translation):

    abla p =
    ho(oldsymbol{g} - oldsymbol{a}) ag{2.38}
  • Tilt angle under horizontal acceleration:
    heta = an^{-1} rac{ax}{g + az} ag{2.39}
  • Magnitude of greatest pressure gradient:
    G = igl(ax^2 + (g + az)^2igr)^{1/2} ag{2.40}
  • Pressure field under rotation about z-axis:
    p = rac{1}{2}
    ho Ω^2 r^2 -
    ho g z + C ag{2.44'}