Buoyancy (DP IB Physics)
Revision Note
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:
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
The ball floats when the buoyancy force and its weight are balanced
Notice that
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
(Equation 1)
Where:
Ws = weight of the sphere (N)
Fd = the drag force (N)
Fb = the buoyancy force / upthrust (N)
At terminal velocity, the forces on the sphere are balanced
The weight of the sphere is found using volume, density and gravitational field strength
(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
(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
(Equation 4)
Substitute equations 2, 3 and 4 into equation 1
Rearrange to make terminal velocity the subject of the equation
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
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
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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