Buoyancy (DP IB Physics)

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Ashika

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Ashika

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Buoyancy

  • Buoyancy is experienced by a body which is partially or totally immersed in a fluid

    • The buoyancy force is exerted on a body due to the displacement of the fluid it is immersed in

  • Buoyancy keeps boats afloat and allows balloons to rise through the air

  • When a body travels through a fluid, it also experiences a buoyancy force (upthrust) due to the displacement of the fluid

  • Buoyancy is calculated using:

F subscript b space equals space rho V g

  • Where:

    • Fb = buoyancy force (N)

    • ρ = density of the fluid (kg m–3)

    • V = volume of the fluid displaced (m3)

    • g = acceleration of free fall (m s–2)

  • If you were to take a hollow ball and submerge it into a bucket of water, you would feel some resistance

  • Some water will flow out of the bucket as it is displaced by the ball

  • The buoyancy force, Fb of the water will push upward on the ball

  • When you let go of the ball, the buoyancy force of the water on the ball will cause the ball to accelerate to the surface

  • The ball will remain stationary floating on the surface of the water

  • A this point, the weight of the ball acting downward, Fg, is equal to the buoyancy force acting upwards, Fb

1-2-10-buoyancy

The ball floats when the buoyancy force and its weight are balanced

  • Notice that

F subscript g space end subscript equals space rho V g space equals fraction numerator space m over denominator V end fraction V g space equals space m g

  • Where:

    • m = mass of the ball (kg)

    • ρ = density of the ball (kg m–3)

    • V = volume of the ball (m3)

  • The buoyancy force and the weight force are equal

Drag Force at Terminal Speed

  • Terminal velocity, or terminal speed, is useful when working with Stoke’s Law

  • This is because, at terminal velocity, the forces in each direction are balanced

W subscript s space equals space F subscript d space plus thin space F subscript b (Equation 1)

  • Where:

    • Ws = weight of the sphere (N)

    • Fd = the drag force (N)

    • Fb = the buoyancy force / upthrust (N)

1-2-10-viscous-drag-force-ib-2025-physics

At terminal velocity, the forces on the sphere are balanced

  • The weight of the sphere is found using volume, density and gravitational field strength

W subscript s space equals space rho subscript s V subscript s g 

W subscript s space equals space 4 over 3 pi r cubed rho subscript s g (Equation 2)

  • Where

    • Vs = volume of the sphere (m3)

    • ρs = density of the sphere (kg m–3)

    • r = radius of the sphere (m)

    • g =  acceleration of free fall (m s−2)

  • Recall Stoke’s Law

F subscript d space equals space 6 pi eta r v (Equation 3)

  • Where

    • Fd = viscous drag force (N)

    • η = fluid viscosity (N s m−2 or Pa s)

    • r = radius of the sphere (m)

    • v = velocity of the sphere through the fluid (ms−1)

      • In this case, v is the terminal velocity

  • The buoyancy force equals the weight of the displaced fluid

    • The volume of displaced fluid is the same as the volume of the sphere

    • The weight of the fluid is found using volume, density and acceleration of free fall

F subscript b space equals space 4 over 3 pi r cubed rho subscript f g (Equation 4)

  • Substitute equations 2, 3 and 4 into equation 1

4 over 3 pi r cubed rho subscript s g italic space italic equals italic space italic 6 pi eta r v italic space italic plus italic space italic 4 over italic 3 pi r to the power of italic 3 rho subscript f g italic space

  • Rearrange to make terminal velocity the subject of the equation

v space equals space fraction numerator italic 4 over italic 3 pi r to the power of italic 3 g italic left parenthesis rho subscript s italic space italic minus italic space italic space rho subscript f italic right parenthesis over denominator italic 6 pi eta r end fraction space equals space fraction numerator 4 pi r to the power of italic 3 g italic left parenthesis rho subscript s italic space italic minus italic space italic space rho subscript f italic right parenthesis over denominator italic 18 pi eta r end fraction space

  • Finally, cancel out r from the top and bottom to find an expression for terminal velocity in terms of the radius of the sphere and the coefficient of viscosity

v space equals fraction numerator 2 pi r squared g italic left parenthesis rho subscript s italic space italic minus italic space italic space rho subscript f italic right parenthesis over denominator italic 9 pi eta end fraction space

 

  • This final equation shows that terminal velocity is:

    • directly proportional to the square of the radius of the sphere

    • inversely proportional to the viscosity of the fluid

Worked Example

Icebergs typically float with a large volume of ice beneath the water. Ice has a density of 917 kg m-3 and a volume of Vi.

The density of seawater is 1020 kg m-3.

What fraction of the iceberg is above the water?

A. 0.10 Vi          

B. 0.90 Vi          

C. 0.97 Vi          

D. 0.20 Vi

Worked example - Archimedes' principle iceberg (2), downloadable AS & A Level Physics revision notes

Examiner Tips and Tricks

Remember that ρ in the buoyancy force equation is the density of the fluid and not the object itself!

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.