Upthrust In Fluids, Archimedes' Principle and Floatation Chapter Notes | Physics Class 9 ICSE PDF Download

Upthrust and Archimedes' Principle

Buoyancy and Upthrust

An upward force acts on a body when it is partially or fully immersed in a fluid, called upthrust or buoyant force, denoted by F_B. Buoyancy is the property of a fluid to exert this upward force on an immersed body.

Experiment 1: Pushing an empty can into water

• Place an airtight, empty can in a tub of water; it floats with most of its volume above water.
• Pushing the can deeper requires increasing force due to rising upthrust.
• When fully immersed, a constant force is needed to hold it; releasing it causes the can to float back up.

Experiment 2: Pushing a cork into water

• A cork floats with about 2/3 of its volume submerged.
• Pushing it underwater and releasing it results in the cork rising back to float.
• An upward force is felt when holding the cork submerged.

Explanation

• Two forces act: weight (W) downward and upthrust (F_B) upward.
• A body floats when W = F_B.
• Pushing the body deeper increases F_B, reaching maximum (F_B') when fully immersed.
• If released, F_B' > W causes the body to rise; an external force (F_B' - W) is needed to keep it submerged.

Note: Gases also exert buoyancy; e.g., a hydrogen-filled balloon rises because F_B from air exceeds its weight.

Conditions for floating or sinking

• Two forces: weight (W) downward, upthrust (F_B) upward.
• F_B depends on the submerged volume, maximized (F_B') when fully immersed.
• If F_B' > W or F_B' = W, the body floats.
  • F_B' > W: Floats partially submerged, with F_B = W.
  • F_B' = W: Floats fully submerged.
• If F_B' < W, the body sinks with acceleration a = (W - F_B') / m, ignoring viscosity.

Unit of upthrust: Newton (N) or kgf.

Characteristic Properties of Upthrust

Property 1: Greater submerged volume results in larger upthrust.

• Example: Pushing a can deeper into water increases upthrust.
• In air, a pebble falls faster than feathers of the same mass due to less upthrust on the smaller-volume pebble.

Property 2: For the same submerged volume, denser fluids exert greater upthrust.

• Example: A cork in glycerine submerges less than in water because glycerine’s higher density provides greater upthrust.

Property 3: Upthrust acts upward at the center of buoyancy, the center of gravity of the displaced fluid.

• For a fully immersed uniform body, the center of buoyancy aligns with the body’s center of gravity.
• For a partially submerged floating body, the center of buoyancy is below the body’s center of gravity.

Reason for Upthrust

• Fluid pressure is equal in all directions at a point and increases with depth.
• For a submerged block, pressure P₂ at the lower face (deeper) exceeds P₁ at the upper face (shallower).
• Net upward force (upthrust) = (P₂ - P₁) × area.
• Sidewall pressures cancel out as they are equal and opposite.
• Note: A thin lamina experiences negligible upthrust due to nearly equal pressure on both sides, causing it to sink.

Upthrust Equals Weight of Displaced Liquid (Mathematical Proof)

Upthrust on a body equals the weight of the fluid displaced by its submerged part.

Proof:

• Consider a cylindrical body of area A in a liquid of density ρ, with upper surface at depth h₁ and lower at h₂.
• Pressure on upper surface: P₁ = h₁ ρ g; downward thrust: F₁ = h₁ ρ g A.
• Pressure on lower surface: P₂ = h₂ ρ g; upward thrust: F₂ = h₂ ρ g A.
• Horizontal thrusts cancel out.
• Net upthrust: F_B = F₂ - F₁ = h₂ ρ g A - h₁ ρ g A = A (h₂ - h₁) ρ g.
• Since A (h₂ - h₁) = V (submerged volume), F_B = V ρ g.
• V ρ g = volume of displaced liquid × density × g = weight of displaced liquid.

Note:

• For a fully immersed body, displaced volume equals the body’s volume, maximizing upthrust (F_B').
• Valid for any shape.

Factors affecting upthrust:

• Volume of the submerged body.
• Density of the fluid.

Effect of upthrust:

• Reduces apparent weight; e.g., a bucket feels lighter in water than in air.
• A fish feels lighter in water; a body weighs slightly less in air than in a vacuum due to air’s upthrust.

Archimedes' Principle

• Statement: A body immersed partially or fully in a fluid experiences an upthrust equal to the weight of the fluid displaced.
• Applies to both liquids and gases.
• Apparent weight loss equals the upthrust.

Experimental Verification of Archimedes' Principle

Experiment 1:

• Use two cylinders (A solid, B hollow) of equal volume, suspended from a balance.
• Balance them in air with weights.
• Immerse cylinder A in water; the balance tips upward due to upthrust.
• Fill cylinder B with water until the balance levels, showing upthrust on A equals the weight of water displaced.

Experiment 2:

• Suspend a solid from a spring balance; note its weight in air (e.g., 300 gf).
• Immerse it in a eureka can, collecting displaced water in a measuring cylinder (e.g., 100 cm³).
• Weight in water (e.g., 200 gf); loss in weight = 100 gf.
• Weight of displaced water (100 cm³ × 1 g cm^-3) = 100 gf, verifying upthrust equals weight of displaced water.

Solid Bodies’ Behavior Based on Density

• For a body of volume V, density ρ in a liquid of density ρ_L:
  • Weight: W = V ρ g.
  • Maximum upthrust: F_B' = V ρ_L g.
• Case 1: If ρ > ρ_L, W > F_B', body sinks with net force (W - F_B') downward.
• Case 2: If ρ = ρ_L, W = F_B', body floats fully submerged, net force zero.
• Case 3: If ρ < ρ_L, W < F_B', body floats partially submerged with volume v such that F_B = v ρ_L g = W, net force zero.
• Experiment 1:
  • An iron nail (ρ > ρ_water) sinks; a cork (ρ < ρ_water) floats.
• Experiment 2:
  • Test solids of varying densities; those with ρ ≤ ρ_water float, others sink.
• Example: An empty tin can floats due to low average density (including air), unlike a solid iron block.

Relative Density and Its Measurement by Archimedes' Principle

Density

Density (ρ) is the mass per unit volume of a substance: ρ = M / V.

Units:

• S.I.: kg m^-3.
• C.G.S.: g cm^-3.
• 1 g cm^-3 = 1000 kg m^-3.

Example: Iron’s density is 7.8 g cm^-3 or 7800 kg m^-3.

Effect of temperature:

• Most substances expand on heating, reducing density; contract on cooling, increasing density.
• Water’s density peaks at 4°C (1 g cm^-3 or 1000 kg m^-3), increasing from 0°C to 4°C, then decreasing above 4°C.

Relative Density

• Relative density (R.D.) is the ratio of a substance’s density (ρ_s) to water’s density at 4°C (ρ_w): R.D. = ρ_s / ρ_w.
• Also: R.D. = mass of substance / mass of equal volume of water at 4°C.
• Unit: No unit (pure ratio).
• Relationships:
  • In C.G.S.: R.D. = density in g cm^-3 (since ρ_w = 1 g cm^-3).
  • In S.I.: R.D. = density in kg m^-3 / 1000 (since ρ_w = 1000 kg m^-3).
• Examples:
  • Copper: density = 8.9 g cm^-3, R.D. = 8.9.
  • Mercury: density = 13.6 × 10³ kg m^-3, R.D. = 13.6.
  • Silver: R.D. = 10.8, density = 10.8 g cm^-3 or 10.8 × 10³ kg m^-3.
• Difference between density and R.D.:
  • Density: Mass per unit volume, units g cm^-3 or kg m^-3.
  • R.D.: Ratio of densities to water’s density, no unit.
• Densities and R.D. of substances:
  • Cork: 240 kg m^-3, 0.24 g cm^-3, R.D. = 0.24.
  • Wood (pine): 500 kg m^-3, 0.50 g cm^-3, R.D. = 0.50.
  • Petrol: 800 kg m^-3, 0.80 g cm^-3, R.D. = 0.80.
  • Turpentine: 870 kg m^-3, 0.87 g cm^-3, R.D. = 0.87.
  • Ice: 920 kg m^-3, 0.92 g cm^-3, R.D. = 0.92.
  • Water (4°C): 1000 kg m^-3, 1.00 g cm^-3, R.D. = 1.
  • Sea water: 1025 kg m^-3, 1.02 g cm^-3, R.D. = 1.02.
  • Glycerine: 1260 kg m^-3, 1.26 g cm^-3, R.D. = 1.26.
  • Glass: 2500 kg m^-3, 2.5 g cm^-3, R.D. = 2.5.
  • Aluminium: 2700 kg m^-3, 2.70 g cm^-3, R.D. = 2.70.
  • Iron: 7860 kg m^-3, 7.86 g cm^-3, R.D. = 7.86.
  • Copper: 8920 kg m^-3, 8.92 g cm^-3, R.D. = 8.92.
  • Silver: 10500 kg m^-3, 10.5 g cm^-3, R.D. = 10.5.
  • Mercury: 13600 kg m^-3, 13.6 g cm^-3, R.D. = 13.6.
  • Gold: 19300 kg m^-3, 19.3 g cm^-3, R.D. = 19.3.
  • Platinum: 21500 kg m^-3, 21.5 g cm^-3, R.D. = 21.5.

Determination of Relative Density of a Solid by Archimedes' Principle

R.D. = Weight of body / Weight of water displaced = W₁ / (W₁ - W₂).

Case 1: Solid denser than water, insoluble

• Procedure:
  • Weigh the solid in air using a physical balance: W₁ gf.
  • Place a beaker of water on a wooden bridge over the balance, ensuring no contact with the pan.
  • Immerse the solid fully in water without touching the beaker’s walls or bottom; weigh: W₂ gf.
• Calculations:
  • Loss in weight = W₁ - W₂ gf.
  • R.D. = W₁ / (W₁ - W₂).

Case 2: Solid denser than water, soluble

• Use a liquid of known R.D. where the solid is insoluble and sinks.
• Repeat the procedure, weighing in air (W₁) and in the liquid (W₂).
• R.D. = [W₁ / (W₁ - W₂)] × R.D. of liquid.

Determination of Relative Density of a Liquid by Archimedes' Principle

• R.D. = Weight of liquid displaced / Weight of water displaced = (W₁ - W₂) / (W₁ - W₃).
• Procedure:
  • Use a solid heavier than and insoluble in both the liquid and water.
  • Weigh the solid in air (W₁ gf), in the liquid (W₂ gf), and after washing and drying, in water (W₃ gf).
  • Calculate R.D. = (W₁ - W₂) / (W₁ - W₃).

Floatation

Principle of Floatation

Forces on a floating body:

• Weight (W) downward at the center of gravity (G).
• Upthrust (F_B) upward at the center of buoyancy (B), equal to the weight of displaced liquid.

Principle: Weight of a floating body equals the weight of the liquid displaced by its submerged part: W = F_B. 
Apparent weight of a floating body is zero.

Cases:

• Case 1: W > F_B' (ρ > ρ_L), body sinks; apparent weight = W - F_B'.
• Case 2: W = F_B' (ρ = ρ_L), body floats fully submerged; apparent weight = 0.
• Case 3: W < F_B' (ρ < ρ_L), body floats partially submerged; W = F_B, apparent weight = 0.

Relation Between Volume of Submerged Part, Densities

For a body of volume V, density ρ_s, floating with volume v submerged in a liquid of density ρ_L:

• Weight: W = V ρ_s g.
• Upthrust: F_B = v ρ_L g.
• For floatation: W = F_B, so V ρ_s g = v ρ_L g.
• Thus: v / V = ρ_s / ρ_L.

Examples:

• Cork (ρ_s = 0.75 g cm^-3) in water (ρ_L = 1 g cm^-3): v / V = 0.75, so 3/4 submerged.
• Ice (ρ_s = 0.9 g cm^-3) in water: v / V = 0.9, so 90% submerged.

Applications of the Principle of Floatation

(i) Floatation of iron ship

• An iron nail sinks (ρ > ρ_water), but a ship floats due to low average density from air-filled hollow spaces.
• Loaded ships submerge more to displace enough water to balance increased weight.
• Ships submerge more in river water (lower density) than sea water.
• Plimsoll line: Marks safe loading limit for water density of 1000 kg m^-3.
• Unloaded ships use ballast (sand) to lower the center of gravity for stability.

(ii) Floatation of human body

• Average human density: 1.07 g cm^-3 (empty lungs), 1.00 g cm^-3 (full lungs).
• Swimmers float with lungs full, displacing water equal to their weight.
• Easier to swim in sea water (ρ = 1.026 g cm^-3) than river water (ρ = 1.0 g cm^-3) due to higher upthrust.
• In the Dead Sea (ρ = 1.16 g cm^-3), less submersion is needed, making swimming easier.

(iii) Floatation of submarines

• Submarines use ballast tanks to control density.
• To dive: Fill tanks with water, increasing density above sea water’s.
• To rise: Expel water with compressed air, reducing density below sea water’s.

(iv) Floatation of iceberg

• Ice (ρ = 0.917 g cm^-3) floats on water (ρ = 1 g cm^-3) with 91.7% submerged.
• In sea water (ρ = 1.026 g cm^-3), 89.3% submerged.
• Icebergs are dangerous as ~90% is submerged, making size estimation difficult.
• Melting ice doesn’t change water level, as melted water equals the volume above water.

(v) Floatation of fish

• Fish use swim bladders like ballast tanks.
• To rise: Fill bladder with gas, reducing density.
• To sink: Empty bladder, increasing density.

(vi) Rising of balloons

• Helium-filled balloons (ρ < ρ_air) rise due to upthrust exceeding weight.
• They stop rising when upthrust equals weight, as air density decreases with altitude.