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:

$F_{b} = ρ V g$

Where:
- F_b = 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, F_b 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, F_g, is equal to the buoyancy force acting upwards, F_b

1-2-10-buoyancy

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

Notice that

$F_{g} = ρ V g = \frac{m}{V} V g = m g$

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

$W_{s} = F_{d} + F_{b}$ (Equation 1)

Where:
- W_s = weight of the sphere (N)
- F_d = the drag force (N)
- F_b = 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

$W_{s} = ρ_{s} V_{s} g$

$W_{s} = \frac{4}{3} π r^{3} ρ_{s} g$ (Equation 2)

Where
- V_s = volume of the sphere (m³)
- ρ_s = 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

$F_{d} = 6 π η r v$ (Equation 3)

Where
- F_d = 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

$F_{b} = \frac{4}{3} π r^{3} ρ_{f} g$ (Equation 4)

Substitute equations 2, 3 and 4 into equation 1

$\frac{4}{3} π r^{3} ρ_{s} g = 6 π η r v + \frac{4}{3} π r^{3} ρ_{f} g$

Rearrange to make terminal velocity the subject of the equation

$v = \frac{\frac{4}{3} π r^{3} g (ρ_{s} - ρ_{f})}{6 π η r} = \frac{4 π r^{3} g (ρ_{s} - ρ_{f})}{18 π η r}$

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

$v = \frac{2 π r^{2} g (ρ_{s} - ρ_{f})}{9 π η}$

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 V_i.

The density of seawater is 1020 kg m^-3.

What fraction of the iceberg is above the water?

A. 0.10 V_i B. 0.90 V_i C. 0.97 V_i D. 0.20 V_i

Examiner Tip

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

Buoyancy (HL IB Physics)

Revision Note

Buoyancy

Drag Force at Terminal Speed

Worked example

Examiner Tip

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Author: Ashika