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 ]

Answer

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]

Answer:

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:
Card 8:

Title 8

This is the card content

Card 9:

Title 9

This is the card content

Card 10:

Title 10

This is the card content

Click here to see out online tuition demo