Upthrust

Buoyancy is experienced by a body partially or totally immersed in a fluid, such as a water
- The size of the force that produces this is equal to the weight of the fluid displacement (sometimes referred to as Archimedes principle)
This force, called upthrust, 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

$F_{b} = ρ V g$

Where:
- F_b = buoyancy force (N)
- ρ = density of the fluid (kg m^–3)
- V_g = volume of the fluid displaced (m³)
- g = gravitational field strength (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 will act upwards on the ball to bring it to the surface
The ball will remain stationary floating when its weight acting downwards, F_g equals the buoyancy force acting upwards, F_b

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The ball floats when it is balanced by the buoyancy force and its weight

Notice that

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

which is the weight of the object submerged in the fluid

Where:
- m = mass of the ball (kg)
- ρ = density of the ball (kg m^–3)
- V = volume of the ball (m³)

Drag Force at Terminal Velocity

Terminal velocity is useful when working with Stoke’s Law since 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
- F_d = the drag force (N)
- F_b = the buoyancy force / upthrust (N)

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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 = gravitational field strength (N kg⁻¹)

Recall Stoke’s Law

$F_{d} = 6 π η r v$ (equation 3)

Where 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 from volume, density and gravitational

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

Exam Tip

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

Upthrust (HL IB Physics)

Revision Note

Upthrust

Drag Force at Terminal Velocity

Worked example

Exam Tip

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Author: Ashika