Archimedes’ Principle

Card 1: Archimedes’ Principle

Archimedes’ Principle Archimedes Principle states that when a body is wholly or partially immersed in a fluid it experiences an upthrust equal to the weight of the fluid displaced.

Example 1

The density and mass of a metal block are   3.2 ×103 kg m-3 and 5.0 kg respectively. Find the upthrust that act on the metal block when it is fully immersed in water. [ Density of water = 1000kg m-3 ]

The upthrust can be found from the formula F = ρVg. From the question, we know that

Density of the water, ρ = 1000kg m-3
Gravitational field strength, g = 10 ms-2

Since the metal block is fully immersed in water, the volume of the displaced water = volume of the block.

The volume of the block, V can be found be using the equation of density, ρ = m/V.

$\begin{gathered} {\text{Density of Block, }}\rho_{metal} {\text{ = 3200kgm}}^{{\text{ - 3}}} \hfill \\ {\text{Mass of Block, }}m = 5kg \hfill \\ {\text{Hence, Volume of Block,}} \hfill \\ V = \frac{m} {\rho } = \frac{{(5)}} {{(3200)}} = 0.0015625m^3 \hfill \\ \end{gathered}$

Upthrust exerted on the block,

$\begin{gathered} F = \rho Vg \hfill \\ F = (1000)(0.0015625)(10) \hfill \\ F = 14.625N \hfill \\ \end{gathered}$
Card 2: Buoyant force

Buoyant force

Buoyant force is an upward force exerted by a fluid on an object immersed in it.

Card 3: Principle of Floatation:

Principle of Floatation:

• Displaced volume of fluid = volume of the object that immerse in the fluid.
• If weight of the object > upthrust, the object will sink into the fluid.
• If weight of the object = upthrust, the object is in balance and therefore float on the surface of the fluid.

Card 4: Case 1 - Density of Object < Density of water: Partially Immerse

Density of Object < Density of water: Partially Immerse Upthrust = Weight

Upthrust, F = ρliquidVliquidg
Weight, W = mg =ρobjectVobjectg

Card 5: Case 2 - Density of Object < Density of water: Fully Immerse

Density of Object < Density of water: Fully Immerse F = T + W

F = Upthrust
T = Tension of the string
W = Weight

Upthrust, F = ρliquidVliquidg
Weight, W = mg =ρobjectVobjectg

Card 6: Case 3 - Density of Object > Density of water: Fully Immerse

Density of Object > Density of water: Fully Immerse T + F = W

F = Upthrust
T = Tension of the string
W = Weight

Upthrust, F = ρliquidVliquidg
Weight, W = mg =ρobjectVobjectg

Example A block that has volume of 0.2 m3 is hanging in a water tank as shown in the figure above. Find the tension of the string? [ Density of the metal = 8 × 103 kg m-3, density of water = 1 × 103 kg m-3]

Volume of metal block, Vblock = 0.2 m3
Density of metal block, ρblock = 8 × 103 kg m-3
Density of metal block, ρwater = 1 × 103 kg m-3

The system is in equilibrium, hence

Upthrust + Tension = Weight
ρwaterVwaterg + T = ρblockVblockg
T = ρblockVblockg - ρwaterVwaterg
T = (8000)(0.2)(10) - (1000)(0.2)(g)
T = 14000 N

Card 7: Applications of Archemedes Principle

Application of Archemedes Principle

• Submarine
• Hot air balloon
• Hydrometer
• Hydrometer is an instrument used to measure the relative density of liquids.
• Ship – Plimsoil line:
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