11.1 Fluid Properties, Hydrostatic Forces & Buoyancy

Fluid Statics, Hydrostatic Forces & Buoyancy

In heavy industrial and mining operations—such as Coal India Limited (CIL) opencast pit dewatering, dense-medium coal washery cyclone circuits, tailings dam stability, and underground mine drainage—fluid statics forms the cornerstone of structural and hydraulic design. Fluid statics deals with fluids at rest, where shear stresses are identically zero ($\tau = 0$) and only normal compressive stresses (hydrostatic pressure) exist.

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1. Fundamental Fluid Properties

1.1 Density, Specific Weight, and Specific Gravity

• Mass Density ($\rho$): Mass per unit volume, $\rho = m/V$ (SI unit: $\text{kg/m}^3$). For pure water at $4^{\circ}\text{C}$ and standard atmospheric pressure, $\rho_{w} = 1000\text{ kg/m}^3$.
• Specific Weight / Weight Density ($\gamma$ or $w$): Weight per unit volume, $\gamma = \rho g$ (SI unit: $\text{N/m}^3$). For water, $\gamma_w = 1000 \times 9.81 = 9810\text{ N/m}^3 = 9.81\text{ kN/m}^3$.
• Specific Volume ($v$): Volume per unit mass, $v = 1/\rho$ (SI unit: $\text{m}^3\text{/kg}$).
• Specific Gravity / Relative Density ($S$ or $RD$): Dimensionless ratio of fluid density to standard reference fluid density: $S = \frac{\rho_{\text{fluid}}}{\rho_{\text{water at } 4^{\circ}\text{C}}} = \frac{\gamma_{\text{fluid}}}{\gamma_{\text{water}}}$ For mercury ($Hg$), $S_{Hg} = 13.6$, yielding $\rho_{Hg} = 13600\text{ kg/m}^3$ and $\gamma_{Hg} = 133.42\text{ kN/m}^3$.

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1.2 Viscosity: Dynamic and Kinematic

Viscosity is the property by virtue of which a fluid offers internal resistance to relative deformation or shear movement of adjacent fluid layers.

          y ^         Moving Plate (Velocity = U, Area = A)
            |        =======================================> Force F
            |         --> u(y) Velocity Profile
            |        ---->
            |       ------->
            |      ========== (Fixed Boundary, y = 0, u = 0)
            +--------------------------------------------> x

According to Newton's Law of Viscosity, the shear stress $\tau$ between adjacent fluid layers is directly proportional to the rate of shear strain (or velocity gradient perpendicular to the flow direction, $du/dy$):

$\tau = \mu \frac{du}{dy}$

Where:

• $\tau = \frac{F}{A}$ is the shear stress ($\text{N/m}^2$ or $\text{Pa}$)
• $\mu$ is the Dynamic Viscosity (or absolute viscosity)
• $\frac{du}{dy}$ is the velocity gradient or rate of angular deformation ($\text{s}^{-1}$)

Units and Conversions for Dynamic Viscosity ($\mu$):

• SI Unit: $\text{Pa}\cdot\text{s} = \frac{\text{N}\cdot\text{s}}{\text{m}^2} = \frac{\text{kg}}{\text{m}\cdot\text{s}}$
• CGS Unit: $\text{Poise} = \frac{\text{dyne}\cdot\text{s}}{\text{cm}^2} = \frac{\text{g}}{\text{cm}\cdot\text{s}}$
• Crucial PSU Conversion: $1\text{ Pa}\cdot\text{s} = 10\text{ Poise} = 1000\text{ cP (centipoise)}$ $1\text{ Poise} = 0.1\text{ Pa}\cdot\text{s} = 0.1\text{ N}\cdot\text{s/m}^2$ For water at $20^{\circ}\text{C}$, $\mu_w \approx 1.0\times 10^{-3}\text{ Pa}\cdot\text{s} = 1.0\text{ cP} = 0.01\text{ Poise}$.

Kinematic Viscosity ($\nu$):

Kinematic viscosity represents the ratio of absolute dynamic viscosity to mass density, measuring momentum diffusivity:

$\nu = \frac{\mu}{\rho}$

• SI Unit: $\text{m}^2/\text{s}$
• CGS Unit: $\text{Stokes} = \text{cm}^2/\text{s}$
• Crucial PSU Conversion: $1\text{ Stoke} = 10^{-4}\text{ m}^2/\text{s}$ $1\text{ cSt (centistoke)} = 10^{-6}\text{ m}^2/\text{s} = 1\text{ mm}^2/\text{s}$

Temperature Dependence of Viscosity:

1. Liquids: Viscosity is dominated by intermolecular cohesive forces. As temperature increases, thermal agitation expands intermolecular spacing, reducing cohesion. Consequently, liquid viscosity decreases with increasing temperature: $\mu_L(T) = \mu_0 e^{-b T}$
2. Gases: Viscosity is dominated by molecular momentum transfer across streamlines. As temperature rises, average molecular speed $\bar{v}_{rms} \propto \sqrt{T}$ increases, intensifying collisions and momentum exchange. Consequently, gas viscosity increases with increasing temperature: $\mu_G \propto \sqrt{T}$

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2. Rheological Classification of Fluids

Real engineering fluids, coal-water slurries, drill polymer muds, and lubricants are categorized by their constitutive shear stress vs. strain rate relation, formalized by the Ostwald-de Waele Power Law Model:

$\tau = \tau_0 + k \left(\frac{du}{dy}\right)^n$

Where $\tau_0$ is the yield stress, $k$ is the consistency index, and $n$ is the flow behavior index. The apparent dynamic viscosity is $\mu_a = \frac{\tau}{du/dy} = \tau_0 \left(\frac{du}{dy}\right)^{-1} + k \left(\frac{du}{dy}\right)^{n-1}$.

   Shear Stress (tau) ^
                      |        / Bingham Plastic (tau_0 > 0, n = 1)
                      |       /  (Toothpaste, Sewage Sludge, Drilling Mud)
                      |      / 
              tau_0 --+-----+--------------------------------
                      |    /     / Dilatant (n > 1, shear-thickening)
                      |   /     /  (Quicksand, Starch/Water, Butter)
                      |  /     / 
                      | /     /-- Newtonian (n = 1, tau_0 = 0)
                      |/     /    (Water, Air, Gasoline, Thin Oils)
                      |     / 
                      |    / Pseudoplastic (n < 1, shear-thinning)
                      |   /  (Blood, Paint, Milk, Coal Slurry)
                      +-------------------------------------------> du/dy

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3. Surface Tension, Droplet Pressure & Capillarity

3.1 Surface Tension ($\sigma$)

Surface tension arises from the unbalanced cohesive attractive forces acting on molecules at the liquid-gas interface, resulting in an inward tensile membrane force. Unit: $\text{N/m}$ or $\text{J/m}^2$.

Internal Excess Pressure ($\Delta P = P_{int} - P_{ext}$):

1. Liquid Droplet (1 Interface): $\Delta P \cdot \left(\frac{\pi d^2}{4}\right) = \sigma \cdot (\pi d) \implies \Delta P = \frac{4\sigma}{d} = \frac{2\sigma}{R}$
2. Hollow Soap Bubble (2 Interfaces - Inner & Outer): $\Delta P \cdot \left(\frac{\pi d^2}{4}\right) = 2 \cdot \sigma \cdot (\pi d) \implies \Delta P = \frac{8\sigma}{d} = \frac{4\sigma}{R}$
3. Liquid Jet (Cylindrical Interface of length $L$): $\Delta P \cdot (L \cdot d) = \sigma \cdot (2L) \implies \Delta P = \frac{2\sigma}{d} = \frac{\sigma}{R}$

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3.2 Capillarity

Capillarity is the rise or depression of a liquid meniscus in a small-bore tube (diameter $d$) caused by the competition between adhesion (liquid-to-solid attraction) and cohesion (liquid-to-liquid attraction).

       Water in Glass Tube (Wetting)          Mercury in Glass Tube (Non-Wetting)
           theta < 90 deg (theta ~ 0)               theta > 90 deg (theta ~ 130 deg)
               |   \_____/   |                          |    ____    |
               |      |      |                          |   /    \   |
             --+------|------+-- Free Surface         --+--+------+--+-- Free Surface
               |      h (Rise)                          |  |      |  |
               |      |      |                          |  |  h   |  | (Depression)
               |             |                          |  v      v  |

Equating upward vertical surface tension component to downward weight of the liquid column:

$(\sigma \cos\theta) \cdot (\pi d) = \rho g \left(\frac{\pi d^2}{4}\right) h$

$h = \frac{4\sigma \cos\theta}{\rho g d} = \frac{4\sigma \cos\theta}{\gamma d}$

• Wetting Liquid (e.g., Clean Water & Glass): Adhesion $>$ Cohesion, contact angle $\theta < 90^{\circ}$ (for pure water-clean glass, $\theta \approx 0^{\circ} \implies \cos\theta = 1$). Results in capillary rise with a concave meniscus ($h > 0$).
• Non-Wetting Liquid (e.g., Mercury & Glass): Cohesion $>$ Adhesion, contact angle $\theta > 90^{\circ}$ (for mercury-glass, $\theta \approx 128^{\circ} - 130^{\circ} \implies \cos\theta < 0$). Results in capillary depression with a convex meniscus ($h < 0$).
• Limit of Manometer Tube Diameter: To avoid capillary errors exceeding $1%$, manometer tube inner diameters must be $d \ge 6\text{ mm}$ (preferably $d \ge 10\text{ mm}$).

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3.3 Compressibility and Bulk Modulus

• Bulk Modulus of Elasticity ($K$): $K = -V \frac{dP}{dV} = \rho \frac{dP}{d\rho} = \frac{1}{\beta}$ Where $\beta$ is fluid compressibility ($\text{Pa}^{-1}$). For water at ambient temperature, $K \approx 2.06\times 10^9\text{ Pa} = 2.06\text{ GPa}$, rendering it practically incompressible except in hydraulic water hammer calculations.
• Speed of Sound in Fluid Medium ($c$): $c = \sqrt{\frac{K}{\rho}}$

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4. Hydrostatic Pressure & Pascal's Law

4.1 Pascal's Law

Pascal's Law states that the intensity of pressure at any point in a static fluid is equal in all spatial directions:

$P_x = P_y = P_z$

4.2 Hydrostatic Law (Variation of Pressure with Depth)

In a static fluid under gravity, the vertical pressure gradient balances specific weight:

$\frac{dP}{dz} = -\rho g = -\gamma \implies \frac{dP}{dh} = \rho g = \gamma$

Integrating for an incompressible homogeneous fluid ($\rho = \text{constant}$):

$P = P_0 + \rho g h = P_{\text{atm}} + \gamma h$

• Pressure Head ($h$): Height of equivalent column of liquid of density $\rho$ generating pressure $P$: $h = \frac{P}{\rho g} = \frac{P}{\gamma}$
• Absolute vs. Gauge Pressure: $P_{\text{abs}} = P_{\text{atm}} + P_{\text{gauge}}$ $P_{\text{vacuum}} = P_{\text{atm}} - P_{\text{abs}} = -P_{\text{gauge}}$ Standard atmospheric pressure: $P_{\text{atm}} = 101.325\text{ kPa} = 1.01325\text{ bar} = 760\text{ mm Hg} = 10.33\text{ m of } \text{H}_2\text{O}$.

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5. Manometry and Pressure Measurement

Manometers measure static pressure differences by balancing fluid columns.

5.1 Simple U-Tube Manometer

For measuring gauge pressure $P_A$ of a fluid of density $\rho_1$ with manometric fluid of density $\rho_m$ (where $\rho_m > \rho_1$):

$P_A + \rho_1 g h_1 = P_{\text{atm}} + \rho_m g h_2$

$P_{A,\text{gauge}} = \rho_m g h_2 - \rho_1 g h_1$

5.2 Differential U-Tube Manometer

Connecting two pipe points $A$ and $B$ containing fluids $\rho_1$ and $\rho_2$ at levels $h_1$ and $h_2$ above a datum, with manometric liquid $\rho_m$ showing deflection $x$:

$P_A - P_B = \rho_m g x + \rho_2 g h_2 - \rho_1 g (h_1 + x)$

If both pipes carry the same fluid ($\rho_1 = \rho_2 = \rho$) at the same center level ($h_1 = h_2$):

$P_A - P_B = g x (\rho_m - \rho) = \rho g x \left(\frac{S_m}{S} - 1\right)$

$\Delta h = \frac{P_A - P_B}{\rho g} = x \left(\frac{S_m}{S} - 1\right)$

5.3 Inverted Differential U-Tube Manometer

Used for measuring tiny pressure differences between liquids. The top is filled with a lighter manometric fluid (air or light oil of density $\rho_m < \rho$):

$P_A - P_B = \rho g (h_1 - h_2) - \rho_m g x = g x (\rho - \rho_m)$

$\Delta h = x \left(1 - \frac{S_m}{S}\right)$

5.4 Inclined Micro-Manometer

An inclined leg at angle $\theta$ to the horizontal magnifies meniscus travel $L$ along the tube:

$h = L \sin\theta$

$\text{Magnification / Sensitivity Factor} = \frac{L}{h} = \frac{1}{\sin\theta}$

For $\theta = 30^{\circ}$, $\sin 30^{\circ} = 0.5$, giving a magnification of $2.0\times$; for $\theta = 5.74^{\circ}$, $\sin\theta \approx 0.1$, giving a $10\times$ magnification.

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6. Hydrostatic Forces on Submerged Plane & Curved Surfaces

                      Free Water Surface (h = 0)
   ===============================================================
          \  theta (Angle of inclination)
           \ 
            \      Centroid G (at depth h_bar, slant distance y_bar)
             \------* G
              \      \ 
               \      * CP (Center of Pressure at depth h_cp, slant y_cp)
                \======

6.1 Total Hydrostatic Force on Plane Surfaces

For any plane surface of area $A$ inclined at angle $\theta$ to the free surface with its area centroid at depth $\bar{h}$:

$F = \int P \, dA = \int (\rho g h) \, dA = \rho g \sin\theta \int y \, dA = \rho g \sin\theta (A \bar{y}) = \rho g A \bar{h}$

$F = \gamma A \bar{h}$

Key Rule: Total hydrostatic thrust depends solely on fluid density $\rho$, surface area $A$, and centroidal depth $\bar{h}$. It is entirely independent of the angle of inclination $\theta$.

6.2 Center of Pressure ($h_{cp}$ or $y_{cp}$)

The Center of Pressure is the point on the submerged surface where the resultant hydrostatic force acts. By taking moments about the free surface axis ($F \cdot y_{cp} = \int y \cdot dF$):

$y_{cp} = \bar{y} + \frac{I_G}{A \bar{y}}$

Since $h = y \sin\theta$, $\bar{h} = \bar{y} \sin\theta$, and $h_{cp} = y_{cp} \sin\theta$:

$h_{cp} = \bar{h} + \frac{I_G \sin^2\theta}{A \bar{h}}$

Where $I_G$ is the second moment of area (area moment of inertia) of the plane figure about its centroidal axis parallel to the free liquid surface.

Fundamental Properties of Center of Pressure:

1. Since $\frac{I_G \sin^2\theta}{A \bar{h}} > 0$, the center of pressure $h_{cp}$ always lies strictly below the centroid $\bar{h}$.
2. For a vertical plane ($\theta = 90^{\circ} \implies \sin^2\theta = 1$): $h_{cp} = \bar{h} + \frac{I_G}{A \bar{h}}$
3. For a horizontal plane ($\theta = 0^{\circ} \implies \sin^2\theta = 0$): $h_{cp} = \bar{h}$
4. As depth increases ($\bar{h} \to \infty$), $\frac{I_G \sin^2\theta}{A \bar{h}} \to 0$, meaning $h_{cp}$ approaches $\bar{h}$.

Geometric Properties of Common Submerged Surfaces:

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6.3 Hydrostatic Forces on Curved Submerged Surfaces

On a curved surface, pressure vectors vary continuously in direction. The resultant force is resolved into orthogonal components:

            Free Water Surface
    ================================
    |              |               |
    |  Fluid       | Vertical      |
    |  Column      | Projected     |
    |  (Weight =   | Area A_x      |
    |   F_y)       |               |
    |              v               |
    +--------------+ <--- F_x (Horizontal Component = rho*g*A_proj*h_bar)
     \            /
      \  Curved  /
       \ Surface/
        +------+
           |
           v F_y (Vertical Component = Weight of fluid column above surface)

1. Horizontal Component ($F_x$): Equal to the hydrostatic force exerted on the vertical projection of the curved surface ($A_{\text{proj}}$): $F_x = \rho g A_{\text{proj}} \bar{h}_{\text{proj}}$ Acts at the center of pressure of the projected vertical area.
2. Vertical Component ($F_y$): Equal to the weight of the liquid column vertically above the curved surface extending up to the free liquid surface: $F_y = \rho g V_{\text{column}} = \gamma V_{\text{column}}$ Acts through the center of gravity of this imaginary or real liquid volume.
3. Resultant Force ($F_R$) and Line of Action: $F_R = \sqrt{F_x^2 + F_y^2}$ $\tan\alpha = \frac{F_y}{F_x}$ For circular-arc gates/surfaces, $F_R$ always passes through the center of curvature because every differential hydrostatic pressure vector $dP = P , dA$ acts normal to the surface (radially through the center).

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7. Archimedes' Principle, Buoyancy & Floatation

7.1 Archimedes' Principle and Center of Buoyancy

• Buoyant Force ($F_B$): When a body is completely or partially submerged in a fluid, it is buoyed up by a net upward vertical force equal to the weight of the fluid displaced by the body: $F_B = \rho_{\text{fluid}} \cdot g \cdot V_{\text{displaced}} = \gamma_{\text{fluid}} \cdot V_{\text{displaced}}$
• Center of Buoyancy ($B$): The point of application of the buoyant force $F_B$. It coincides exactly with the centroid of the displaced liquid volume.
• Principle of Floatation: For a floating body in static equilibrium, total weight $W$ equals buoyant force $F_B$: $W = F_B \implies m_{\text{body}} g = \rho_{\text{fluid}} g V_{\text{submerged}}$ $\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}} = \frac{S_{\text{body}}}{S_{\text{fluid}}}$

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7.2 Metacenter and Metacentric Height ($GM$)

For a floating body given a small angular tilt (heel angle $\theta$):

• The submerged shape changes, causing the centroid of displaced volume to shift from $B$ to $B'$.
• The line of action of the new buoyant force through $B'$ intersects the original vertical center line of the body at the Metacenter ($M$).

              Original Axis        Tilted Axis (Angle theta)
                   |                    /
                   |                   / M (Metacenter)
                   |                  /  |
                   |                 /   | GM (Metacentric Height)
                   |                * G  |
                   |               /     | Restoring Couple =
                   |              * B    | W * GM * sin(theta)
                   |             /       v
                   |            * B'

• Metacentric Height ($GM$): The distance between the center of gravity $G$ and the metacenter $M$: $GM = BM - BG$ Where:
  • $BM$ is the Metacentric Radius (distance from $B$ to $M$): $BM = \frac{I_{xx}}{V_{\text{displaced}}}$ where $I_{xx}$ is the second moment of area of the waterplane section (plane of flotation at liquid surface) about the tilting/longitudinal axis.
  • $BG$ is the vertical distance between the center of buoyancy $B$ and center of gravity $G$ ($BG = y_G - y_B$).
  • Therefore: $GM = \frac{I}{V} - BG$

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7.3 Stability Conditions for Submerged vs. Floating Bodies

A. Completely Submerged Bodies (e.g., Submarines, Submerged Mine Sensors):

| State of Equilibrium | Relative Positions of $B$ and $G$ | Physical Behavior upon Small Angular Tilt | | :--- | :--- | :--- | :--- | | Stable Equilibrium | $B$ is strictly above $G$ ($BG > 0$) | $F_B$ and $W$ produce a restoring couple that returns body to upright position | | Unstable Equilibrium | $B$ is strictly below $G$ | $F_B$ and $W$ produce an overturning couple that increases tilt | | Neutral Equilibrium | $B$ and $G$ coincide ($B = G$) | No couple is generated; body remains in its new tilted position |

B. Partially Submerged / Floating Bodies (e.g., Dredgers, Barges, Pontoons):

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7.4 Time Period of Rolling (Oscillation) of a Floating Vessel

Treating the rolling floating vessel as an angular harmonic oscillator:

$I_{\text{mass}} \frac{d^2\theta}{dt^2} + W \cdot GM \cdot \sin\theta = 0$

For small angular displacements ($\sin\theta \approx \theta$ and $I_{\text{mass}} = \frac{W}{g} k^2$, where $k$ is the radius of gyration of the body about the longitudinal axis of oscillation):

$T = 2\pi \sqrt{\frac{k^2}{g \cdot GM}}$

Engineering Trade-Off in Vessel Design:

• Large $GM$ (High Stability): Very safe against capsizing, but results in a small time period $T$ (fast, violent rolling with high angular accelerations, causing severe passenger discomfort and cargo shift).
• Small $GM$ (Passenger / Cargo Comfort): Large time period $T$ (smooth, gentle rolling), but reduces safety margin against overturning.
• Typical $GM$ values: Merchant cargo ships $\approx 0.3 - 1.0\text{ m}$; Warships $\approx 1.0 - 1.8\text{ m}$; River barges $\approx 1.5 - 3.0\text{ m}$.

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8. Worked Engineering Examples

Example 1: Hydrostatic Force and Center of Pressure on an Inclined Sluice Gate

Problem: A rectangular sluice gate of width $b = 2\text{ m}$ and height $d = 3\text{ m}$ in a coal washery reservoir is hinged at its top edge. The gate is inclined at an angle of $\theta = 60^{\circ}$ to the horizontal. The top edge of the gate is at a vertical depth of $h_1 = 2\text{ m}$ below the free water surface ($\rho = 1000\text{ kg/m}^3, g = 9.81\text{ m/s}^2$). Determine:

1. The total hydrostatic thrust on the gate.
2. The vertical depth to the center of pressure ($h_{cp}$).

Solution:

1. Centroid Depth ($\bar{h}$): $\bar{h} = h_1 + \left(\frac{d}{2}\right) \sin\theta = 2.0 + \left(\frac{3.0}{2}\right) \sin 60^{\circ} = 2.0 + 1.5 \times 0.8660 = 2.0 + 1.299 = 3.299\text{ m}$
2. Total Hydrostatic Thrust ($F$): $A = b \times d = 2.0 \times 3.0 = 6.0\text{ m}^2$ $F = \rho g A \bar{h} = 1000 \times 9.81 \times 6.0 \times 3.299 = 194,179\text{ N} = 194.18\text{ kN}$
3. Center of Pressure Depth ($h_{cp}$): $I_G = \frac{b d^3}{12} = \frac{2.0 \times 3.0^3}{12} = \frac{54.0}{12} = 4.5\text{ m}^4$ $h_{cp} = \bar{h} + \frac{I_G \sin^2\theta}{A \bar{h}} = 3.299 + \frac{4.5 \times (\sin 60^{\circ})^2}{6.0 \times 3.299}$ $(\sin 60^{\circ})^2 = 0.75$ $h_{cp} = 3.299 + \frac{4.5 \times 0.75}{19.794} = 3.299 + \frac{3.375}{19.794} = 3.299 + 0.1705 = 3.4695\text{ m} \approx 3.47\text{ m}$

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Example 2: Metacentric Height of a Rectangular Pontoon

Problem: A rectangular flat-bottomed pontoon barge used for mine sump dredging has a width $B = 6\text{ m}$, length $L = 10\text{ m}$, and draught $D = 2\text{ m}$ in fresh water. The center of gravity $G$ of the loaded barge is on the vertical centerline at $y_G = 1.6\text{ m}$ above the base. Determine the metacentric height ($GM$) and state whether the barge is in stable equilibrium.

Solution:

1. Center of Buoyancy ($y_B$): For a uniform rectangular hull, the centroid of displaced volume is at half the draught: $y_B = \frac{D}{2} = \frac{2.0}{2} = 1.0\text{ m above the base}$
2. Distance $BG$: $BG = y_G - y_B = 1.6 - 1.0 = 0.6\text{ m}$
3. Metacentric Radius ($BM = I/V$):
   • Waterplane area moment of inertia about the tilting (longitudinal) axis: $I_{xx} = \frac{L B^3}{12} = \frac{10.0 \times 6.0^3}{12} = \frac{10.0 \times 216.0}{12} = 180.0\text{ m}^4$
   • Displaced volume of water: $V_{\text{disp}} = L \times B \times D = 10.0 \times 6.0 \times 2.0 = 120.0\text{ m}^3$
   • Metacentric radius: $BM = \frac{I_{xx}}{V_{\text{disp}}} = \frac{180.0}{120.0} = 1.5\text{ m}$
4. Metacentric Height ($GM$): $GM = BM - BG = 1.5 - 0.6 = +0.9\text{ m}$ Since $GM = +0.9\text{ m} > 0$ ($M$ lies $0.9\text{ m}$ above $G$), the pontoon is in stable equilibrium.