Hydrostatics: Pressure, Manometry, Forces on Submerged Surfaces, Buoyancy and Stability

This chapter answers Section XE1.2 of XE1 Fluid Mechanics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam: “Buoyancy, Manometry, Forces on submerged bodies and their stability.” It teaches the pressure variation in a static fluid, absolute and gauge pressure, U-tube, differential and inclined manometers, the hydrostatic force and centre of pressure on plane surfaces, the horizontal and vertical components of force on curved surfaces, Archimedes’ principle, and the stability of floating and submerged bodies through the metacentric height. Every idea is used in a numerical: a gauge pressure at depth, a mercury manometer, a hinged gate, a quarter-circular gate, an apparent weight and a metacentric height. Throughout, g = 9.81 m/s² and the fluid densities are stated wherever they are needed.

1. Pressure in a static fluid

In a fluid at rest there is no shear stress, so the only surface force is the normal pressure p, and it acts equally in all directions at a point. Balancing the weight of a thin horizontal slab against the pressure difference gives dp/dz = −ρg (z measured upward). For a liquid of constant density this integrates to p = p₀ + ρgh, with h the depth below a reference surface at pressure p₀. Three consequences matter. Pressure depends only on depth, not on the shape or width of the vessel (the hydrostatic paradox). All points at the same level in a connected body of the same fluid are at the same pressure. A pressure applied to a confined static fluid is transmitted undiminished (Pascal’s law), which is how a hydraulic press gives a force ratio equal to the area ratio.

Absolute and gauge pressure. Absolute pressure is measured from perfect vacuum and can never be negative. Gauge pressure is measured from the local atmospheric pressure: p_gauge = p_abs − p_atm. A negative gauge pressure is a vacuum pressure, p_vac = p_atm − p_abs. The standard atmosphere is 101.325 kPa. For gases the variation over a few metres is negligible (ρgh for air over 10 m is about 0.1 kPa), but for the atmosphere over kilometres it is not constant-density.

Worked example 1. At a depth of 12 m in sea water (ρ = 1025 kg/m³) with p_atm = 101.3 kPa: p_gauge = ρgh = 1025 × 9.81 × 12 = 120.7 kPa and p_abs = 101.3 + 120.7 = 222.0 kPa. A gauge of −30 kPa in a room at 100 kPa means an absolute pressure of 70 kPa.

2. Manometers: U-tube, differential and inclined

A manometer measures pressure from the height of a liquid column. The method is one rule: start at a known pressure and walk through the liquid; add ρgh when going down in a fluid, subtract ρgh when going up, and stop when you reach the unknown. Never jump between two fluids at different levels unless you pass through the interface. A U-tube manometer joins a pressure source to a column of a heavy manometric liquid (mercury, ρ = 13600 kg/m³, or water) open to the atmosphere on the other side, and gives a gauge pressure. A differential manometer joins two tapping points and gives p_A − p_B directly.

Manometer rules
TypeResultNote
U-tube, pipe fluid ρ₁, manometric liquid ρ_m, reading hp_gauge = (ρ_m − ρ₁) g h + ρ₁ g (level corrections)Subtract the pipe fluid’s own column across the interface
Differential, same-level taps, same fluid ρ₁ in the pipesp_A − p_B = (ρ_m − ρ₁) g hh is the difference of the manometric levels
Inclined, reading L along a tube at angle θ to the horizontalΔp = ρ g L sin θVertical rise is L sin θ; small θ magnifies the scale

Worked example 2. Two water pipes at the same level are joined by a mercury U-tube (ρ_m = 13600, ρ_w = 1000 kg/m³) with the mercury levels differing by h = 0.30 m. Walking from A down to the lower mercury level, across to the higher level and up to B, all the water segments cancel except the water column of height h that replaces mercury: p_A − p_B = (13600 − 1000) × 9.81 × 0.30 = 37.08 kPa. Writing 13600 × 9.81 × 0.30 = 40.0 kPa forgets the water above the mercury and overstates the difference by 8 %. In an inclined manometer with θ = 30° the liquid moves twice as far along the tube as the vertical rise for the same pressure, so the scale is magnified two times.

⚠️ Trap
The pressure at the same level is equal only within one continuous fluid. Two points at the same height on the two sides of a U-tube are not at the same pressure if a different liquid, or a gas, separates them. Also, the density that multiplies h is that of the liquid in which you walked, not the pipe fluid for the whole path.

3. Force and centre of pressure on plane surfaces

For a plane surface of area A submerged with its centroid at vertical depth h_c, the resultant hydrostatic force is F = ρ g h_c A = p_c A, where p_c is the gauge pressure at the centroid. The force is normal to the surface. Because pressure rises with depth, the line of action passes below the centroid, at the centre of pressure. For a plane inclined at θ to the free surface, with y measured along the plane from the line where its extension meets the free surface, y_cp = y_c + I_c/(y_c A), where I_c is the second moment of area about the horizontal centroidal axis in the plane. The vertical depth of the centre of pressure is y_cp sin θ.

  • Rectangle b × d with its top edge at the free surface, vertical: I_c = bd³/12, h_c = d/2, so the centre of pressure is at 2d/3 below the surface.
  • Circle of diameter D: I_c = πD⁴/64. Triangle (base b, height h): I_c = bh³/36 about the centroidal axis parallel to the base.
  • As the plate is lowered, I_c/(y_c A) shrinks, so the centre of pressure approaches the centroid; for a horizontal plate they coincide.

Worked example 3 (hinged gate). A vertical rectangular gate, 2 m wide and 3 m high, has its top edge 1 m below the water surface (ρ = 1000 kg/m³) and is hinged along the top edge. Find the normal force P at the bottom edge that just holds it shut; neglect atmospheric pressure, which acts on both sides. Then h_c = 1 + 1.5 = 2.5 m and A = 6 m², so F = 9810 × 2.5 × 6 = 147.15 kN. With I_c = 2 × 3³/12 = 4.5 m⁴, y_cp = 2.5 + 4.5/(2.5 × 6) = 2.8 m below the surface, which is 1.8 m below the hinge. Taking moments about the hinge, P × 3 = 147.15 × 1.8, so P = 88.29 kN. Using the centroid (1.5 m below the hinge) would give 73.6 kN, which is too small.

⚠️ Trap
The centre of pressure depends only on geometry and depth, not on the liquid density or on g. Adding atmospheric pressure to the free surface raises F by p_atm A and moves the centre of pressure towards the centroid. If the atmosphere acts on the back of the plate as well, use gauge pressure and the formula above unchanged.

4. Force on curved surfaces

For a curved surface the pressure forces point in different directions, so resolve them. The horizontal component F_H equals the hydrostatic force on the vertical projection of the surface, and acts at the centre of pressure of that projection. The vertical component F_V equals the weight of the liquid vertically above the surface up to the free surface, real if the liquid lies above the surface and imaginary (the volume that would be filled) if the liquid is below it, and acts through the centroid of that volume. The resultant is F = √(F_H² + F_V²) at angle tan⁻¹(F_V/F_H) to the horizontal. For a circular arc every pressure force is normal to the surface and so passes through the centre of the circle, hence the resultant also passes through that centre.

Worked example 4 (quarter-circular gate). A gate is a quarter of a circular cylinder of radius R = 2 m and width w = 3 m. The centre O of the circle is on the water surface (ρ = 1000 kg/m³); the arc runs from the point R directly below O to the point R beside O on the surface, and the water lies on the concave side, above the arc. The vertical projection has height R and its centroid is R/2 deep, so F_H = ρg(R/2)(Rw) = 9810 × 1 × 6 = 58.86 kN. The water above the arc is a quarter-cylinder, so F_V = ρg(πR²/4)w = 9810 × 3.1416 × 3 = 92.46 kN. The resultant is √(58.86² + 92.46²) = 109.6 kN, passing through O, at tan⁻¹(92.46/58.86) = 57.5° below the horizontal.

⚠️ Trap
F_H uses the projected area on a vertical plane, not the curved area. F_V is the weight of liquid above the surface even where no liquid is actually there. On a closed submerged body the horizontal components on the two sides cancel and the net vertical component is the buoyant force, which links this section to Archimedes’ principle.

5. Buoyancy and Archimedes’ principle

Archimedes’ principle: a body wholly or partly immersed in a fluid at rest experiences an upward buoyant force equal to the weight of the fluid it displaces, F_B = ρ_f g V_disp, acting through the centre of buoyancy B, the centroid of the displaced volume. A body floats in equilibrium when its weight equals the buoyant force, so for a homogeneous floating body the submerged fraction is V_disp/V = ρ_body/ρ_fluid. The same body floats lower in a lighter fluid (the principle of the hydrometer), and its weight does not change on moving between fluids.

Worked example 5 (apparent weight). A stone weighs 50 N in air and 30 N when fully immersed in water (ρ_w = 1000 kg/m³). The buoyant force is 20 N, so V = 20/(1000 × 9.81) = 2.04 × 10⁻³ m³. Its mass is 50/9.81 = 5.10 kg, so ρ = 5.10/2.04 × 10⁻³ = 2500 kg/m³, which is also SG = 50/20 × 1 = 2.5, and g cancels. A wooden block of density 600 kg/m³ floating in water sinks to 60 % of its height; in a liquid of density 800 kg/m³ it sinks to 75 %.

🎯 Floating ice
A floating body displaces its own weight of fluid. Ice floating in water displaces exactly the volume of water that it will become on melting, so for pure ice in pure water the level does not change on melting. If the ice held a stone that would sink on melting, the level would fall; in a denser liquid such as brine the melt water adds more volume than the ice displaced, and the level rises.

6. Stability of floating and submerged bodies

Stability asks what happens after a small tilt. Let G be the centre of gravity and B the centre of buoyancy. Fully submerged body: the body is stable if G lies below B (the buoyant force then gives a restoring couple), neutral if G and B coincide, unstable if G is above B. Floating body: when it tilts the shape of the displaced volume changes and B moves sideways. The metacentre M is the point where the line of the buoyant force meets the original vertical axis for small tilts. The body is stable if M lies above G, which is the same as a positive metacentric height GM = BM − BG (BG positive when G is above B). Here BM = I/V_disp, with I the second moment of the waterplane area about the axis of tilt and V_disp the displaced volume. The restoring moment at a small angle φ is W · GM · sin φ.

Worked example 6 (barge). A rectangular barge 10 m long, 4 m wide, floats at a draught of 1.5 m in water, with its centre of gravity G at 1.2 m above the keel, and rolls about its longitudinal axis. V_disp = 10 × 4 × 1.5 = 60 m³ and B is 0.75 m above the keel, so BG = 1.2 − 0.75 = 0.45 m. The waterplane has I = 10 × 4³/12 = 53.33 m⁴ about the roll axis, so BM = 53.33/60 = 0.889 m and GM = 0.889 − 0.45 = 0.439 m > 0: stable. Rolling about the axis that gives the smaller I (the longer dimension along the axis) is the critical one, so check pitch only if it is smaller; here pitch has I = 4 × 10³/12, which is far larger.

⚠️ Trap
Do not use the submerged-body rule (G below B) for a floating body. A floating ship has G above B and is still stable if BM exceeds BG, because the waterplane gives a wide BM. Raising G (loading on deck) lowers GM, and widening the beam raises the waterplane I in proportion to the cube of the width, so BM rises steeply (as the square of the beam at the same draught). A rigid-body tank in uniform acceleration has p increasing along the effective gravity g − a, which tilts the free surface, but that is only a passing item for this section.

Key takeaways

  • In a static fluid dp/dz = −ρg, so p = p₀ + ρgh for constant ρ; pressure depends on depth only, and p_abs = p_gauge + p_atm with p_abs never negative.
  • For manometers walk through the liquid, adding ρgh going down and subtracting going up; a same-level differential reading gives Δp = (ρ_m − ρ₁) g h, and an inclined reading gives Δp = ρ g L sin θ.
  • On a plane surface F = ρ g h_c A acts at y_cp = y_c + I_c/(y_c A), always below the centroid for a non-horizontal surface, and independent of ρ.
  • On a curved surface F_H is the force on the vertical projection and F_V is the weight of the liquid above the surface (real or imaginary); a circular arc’s resultant passes through its centre.
  • Archimedes: F_B = ρ_f g V_disp through the centroid of the displaced volume; a homogeneous floater has submerged fraction ρ_body/ρ_fluid, and apparent weight in water equals true weight minus F_B.
  • A floating body is stable if GM = I/V_disp − BG > 0; a fully submerged body is stable only if G is below B.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Two open vessels of very different shapes, one narrow and one wide with sloping walls, are filled with the same liquid to the same height and have equal base areas. Which statement about the pressure on their bases is correct?

    1. The pressure is greater in the wide vessel, because it holds more liquid and so a greater weight
    2. The gauge pressure on the two bases is the same, because it depends only on the liquid density and the depth
    3. The pressure is greater in the narrow vessel, because the same liquid is squeezed into a smaller width
    4. The pressure is greater in the vessel with outward-sloping walls, because the walls push the liquid downward
    Show answer

    Answer: B — The gauge pressure on the two bases is the same, because it depends only on the liquid density and the depth

    With the same liquid and the same depth, p = ρgh is the same at both bases whatever the shape (the hydrostatic paradox); the extra weight of liquid in the wide vessel is carried by its sloping walls. The tempting reply that more liquid means more pressure confuses total weight with pressure.

  2. A vacuum gauge on a closed vessel reads 30 kPa of vacuum while the local atmospheric pressure is 100 kPa. What is the absolute pressure in the vessel?

    1. 130 kPa
    2. −30 kPa
    3. 70 kPa
    4. 30 kPa
    Show answer

    Answer: C — 70 kPa

    Vacuum pressure is p_atm − p_abs, so p_abs = 100 − 30 = 70 kPa. Adding the vacuum reading to atmospheric pressure (130 kPa) treats it as a positive gauge pressure, and an absolute pressure can never be negative, so −30 kPa is a gauge value and not an absolute one.

  3. A vertical plane plate is fully submerged in a liquid and is slowly lowered, staying vertical. How do the hydrostatic force F and the distance between the centre of pressure and the centroid change?

    1. F increases in proportion to the centroid depth, while the centre of pressure moves closer to the centroid
    2. F increases, and the centre of pressure moves farther below the centroid because the pressure is larger
    3. F stays the same, because the plate area is fixed, and the centre of pressure stays at the same distance below the centroid
    4. F increases, and the centre of pressure moves above the centroid for large depths
    Show answer

    Answer: A — F increases in proportion to the centroid depth, while the centre of pressure moves closer to the centroid

    F = ρ g h_c A grows linearly with h_c. The offset y_cp − y_c = I_c/(y_c A) decreases as y_c increases, so the centre of pressure approaches the centroid but stays below it. Thinking that a larger pressure pushes the centre of pressure further down mixes up the pressure level with its relative variation across the plate.

  4. A curved surface of a tank has liquid on its lower side only, and no liquid lies vertically above it. The vertical component of the hydrostatic force on it is directed upward. How is its magnitude found?

    1. It is zero, because there is no liquid above the surface to press on it
    2. It equals the weight of the liquid actually present in the tank on the lower side of the surface
    3. It equals the hydrostatic force on the vertical projection of the surface
    4. It equals the weight of the imaginary liquid that would fill the volume between the surface and the level of the free surface
    Show answer

    Answer: D — It equals the weight of the imaginary liquid that would fill the volume between the surface and the level of the free surface

    F_V is the net vertical pressure force and equals the weight of the volume of liquid, real or imaginary, lying vertically between the surface and the free-surface level, directed upward when the liquid is below. The force on the vertical projection is the horizontal component F_H, not F_V, and an absence of liquid above does not make F_V zero.

  5. A long horizontal circular cylinder of radius 1 m and length 2 m floats in water of density 1000 kg/m³ with exactly its lower half immersed. Taking g = 9.81 m/s², what is the net vertical force exerted on it by the water?

    1. 30.8 kN upward
    2. 61.6 kN upward, the buoyant force for the whole cylinder submerged
    3. 15.4 kN upward, equal to ρg times the volume of a full cylinder of diameter 1 m and length 2 m
    4. 19.6 kN upward, equal to ρg times the volume of a rectangular block 1 m × 1 m × 2 m
    Show answer

    Answer: A — 30.8 kN upward

    The displaced volume is half the cylinder: (π × 1²/2) × 2 = π m³ = 3.1416 m³, so F_B = 1000 × 9.81 × 3.1416 = 30.8 kN. The 61.6 kN figure doubles the displaced volume (it is the full-submersion value), and the other two use volumes (a smaller full cylinder, a rectangular block) that are not the displaced half-cylinder.

  6. A homogeneous block floats in water of density 1000 kg/m³ with 60 % of its volume submerged. It is moved to oil of density 800 kg/m³, in which it also floats. What fraction of its volume is now submerged?

    1. 48 %
    2. 60 %
    3. 80 %
    4. 75 %
    Show answer

    Answer: D — 75 %

    The block density is 600 kg/m³. In the oil the submerged fraction is ρ_block/ρ_oil = 600/800 = 0.75. Multiplying 0.6 by 0.8 gives 48 %, which moves the wrong way (a lighter liquid must be sunk further), and 60 % would be right only if the liquid had not changed.

  7. A piece of pure ice floats in a glass of pure water at 0 °C, with the water open to the air. The ice melts completely to water at 0 °C. What happens to the water level in the glass?

    1. It falls, because the ice has a lower density than liquid water and the density of the water formed is higher
    2. It stays the same, because the ice displaced exactly its own weight of water, which is the volume of water it becomes
    3. It rises, because the ice expands on melting into liquid water
    4. It rises by the volume of the part of the ice that was above the surface
    Show answer

    Answer: B — It stays the same, because the ice displaced exactly its own weight of water, which is the volume of water it becomes

    A floating body displaces its own weight of liquid, so the ice displaced a mass of water equal to its own mass, and on melting it becomes exactly that mass of water, filling the same volume. The level is unchanged. The belief that the level falls uses the lower density of ice but ignores that only the submerged part counts, and ice contracts, not expands, on melting.

  8. A ship has its centre of gravity G above its centre of buoyancy B, with BG = 0.3 m. For small rolling about its longitudinal axis the waterplane gives I/V_disp = 0.9 m. Which statement is correct?

    1. It is stable, because the metacentre lies above G by GM = 0.9 − 0.3 = 0.6 m
    2. It is unstable, because G lies above B and a floating body needs G below B
    3. It is stable only if the ship is fully submerged, because a floating ship has no metacentre
    4. It is neutrally stable, because BM and BG have the same sign
    Show answer

    Answer: A — It is stable, because the metacentre lies above G by GM = 0.9 − 0.3 = 0.6 m

    For a floating body the criterion is GM = BM − BG > 0, with BM = I/V_disp. Here GM = 0.6 m, so the metacentre is above G and the body is stable. The rule 'G below B' applies only to a fully submerged body; applied to a floating ship it wrongly condemns every ordinary vessel.

  9. An inclined-tube manometer is set up with its tube at 30° to the horizontal. For the same pressure difference, how does the distance moved by the liquid along the tube compare with the vertical rise it would have in a vertical tube?

    1. It is half as large, because the tube is tilted away from the vertical
    2. It is twice as large, because the vertical rise is L sin 30° = L/2
    3. It is 1.73 times as large, because the horizontal span is L cos 30°
    4. It is the same, because pressure depends only on the liquid density and not on the tube angle
    Show answer

    Answer: B — It is twice as large, because the vertical rise is L sin 30° = L/2

    Pressure difference corresponds to a vertical height h = L sin θ, so L = h/sin θ. At 30°, L = 2h, which is why inclined manometers magnify small readings. Pressure depends on the vertical height, not the length along the tube, so the same-reading answer ignores that the length changes while the height does not.

  10. Statements about hydrostatic force are given.
    P: On a submerged plane surface the resultant force is ρ g h_c A, with h_c the depth of the centroid.
    Q: The centre of pressure of a submerged vertical plane surface lies above its centroid.
    R: The horizontal component of the force on a curved surface equals the force on its vertical projection.
    S: The vertical component of the force on a curved surface equals the weight of liquid above it, up to the free surface, real or imaginary.
    Select all correct statements.

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; C — R; D — S

    P, R and S are the standard results. Q is wrong: pressure increases with depth, so the resultant acts below the centroid, and the two coincide only for a horizontal surface.

  11. Statements about stability are given.
    P: A floating body is stable for small tilts if its metacentre lies above its centre of gravity.
    Q: A fully submerged body of fixed shape is stable if its centre of gravity lies below its centre of buoyancy.
    R: Raising the centre of gravity of a floating body increases its metacentric height.
    S: The distance BM equals V_disp divided by the second moment of area of the waterplane about the axis of tilt.
    Select all correct statements.

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; B — Q

    P and Q are the two standard criteria. R is wrong because GM = BM − BG and raising G increases BG, lowering GM. S inverts the formula: BM = I/V_disp, not V_disp/I.

  12. Statements about pressure in a fluid at rest are given.
    P: In a single connected body of one liquid, points at the same level are at the same pressure.
    Q: A negative gauge pressure is impossible.
    R: In a hydraulic press, the ratio of the output force to the input force equals the ratio of the piston areas, neglecting friction and weight.
    S: At a point in a static fluid the pressure is greater in the vertical direction than in the horizontal direction.
    Select all correct statements.

    1. P
    2. Q
    3. R
    4. S
    Show answer

    Answer: A — P; C — R

    P and R are correct (R is Pascal’s law applied to a press). Q is wrong: a vacuum is a negative gauge pressure. S is wrong: with no shear stress the pressure at a point is the same in all directions.

  13. Two horizontal water pipes A and B at the same elevation are connected by a differential U-tube manometer containing mercury, and the rest of the tubing is filled with water. The mercury levels in the two limbs differ by 0.30 m. Take ρ_mercury = 13600 kg/m³, ρ_water = 1000 kg/m³ and g = 9.81 m/s². The pressure difference between the two pipes is ____ kPa (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 37.08

    Δp = (ρ_m − ρ_w) g h = (13600 − 1000) × 9.81 × 0.30 = 12600 × 2.943 = 37 081.8 Pa = 37.08 kPa. Using 13600 alone gives 40.02 kPa, which forgets the water that fills the other limb.

  14. A vertical rectangular gate, 2 m wide and 3 m high, has its top edge 1 m below the surface of still water of density 1000 kg/m³ and is hinged along its top edge. Atmospheric pressure acts on both sides, g = 9.81 m/s². The force that must be applied normal to the gate at its bottom edge to hold it closed is ____ kN (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 88.29

    h_c = 2.5 m, A = 6 m², F = 9810 × 2.5 × 6 = 147 150 N. I_c = 2 × 27/12 = 4.5 m⁴, y_cp = 2.5 + 4.5/15 = 2.8 m, which is 1.8 m below the hinge. Moments about the hinge: P × 3 = 147 150 × 1.8, so P = 88 290 N = 88.29 kN.

  15. A gate is a quarter of a circular cylinder of radius 2 m and width 3 m. The centre of the circle lies on the free surface of water of density 1000 kg/m³; the arc runs from the point 2 m directly below the centre to the point 2 m beside the centre on the surface, and the water lies on the concave side of the arc. Take g = 9.81 m/s². The magnitude of the resultant hydrostatic force on the gate is ____ kN (to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 109.6

    F_H = ρg (R/2)(R w) = 9810 × 1 × 6 = 58 860 N. F_V = weight of the quarter-cylinder of water above the arc = 9810 × (π × 4/4) × 3 = 92 457 N. Resultant = √(58 860² + 92 457²) = 109 603 N = 109.6 kN.

  16. A stone weighs 50 N in air and 30 N when fully immersed in water of density 1000 kg/m³. Take g = 9.81 m/s². The density of the stone is ____ kg/m³ (to the nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 2500

    Buoyant force = 50 − 30 = 20 N = ρ_w g V, so V = 20/9810 = 2.039 × 10⁻³ m³. Mass = 50/9.81 = 5.097 kg. ρ = 5.097/2.039 × 10⁻³ = 2500 kg/m³. Equivalently ρ = ρ_w × 50/20, and g cancels.

  17. A rectangular barge 10 m long and 4 m wide floats in water at a draught of 1.5 m. Its centre of gravity is 1.2 m above the keel. For rolling about the longitudinal axis, the metacentric height GM is ____ m (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.44

    V_disp = 10 × 4 × 1.5 = 60 m³; B is at 0.75 m above the keel, so BG = 1.2 − 0.75 = 0.45 m. I = L b³/12 = 10 × 64/12 = 53.33 m⁴, BM = 53.33/60 = 0.8889 m. GM = 0.8889 − 0.45 = 0.4389, which is 0.44 m.