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By studying this chapter you will extend your knowledge and understanding of forces - in particular
the nature of forces exerted on solid bodies and small particles as they move through fluids. You
should also acquire a basic understanding of the processes of sedimentation, diffusion and
Brownian motion.
Minimum learning goals
When you have finished studying this chapter you should be able to do all of the following.
Explain, interpret and use the terms
drag force, lift force, terminal velocity, Brownian motion, diffusion, dynamic equilibrium,
Recognise and describe examples of situations involving forces exerted by fluids on moving
Draw force diagrams and analyse forces for systems subject to lift and drag forces.
Describe how the forces exerted by fluids on moving bodies vary with the speed of motion.
Describe and explain how fluids affect the motion of large objects - in particular, explain how
terminal velocity is reached in certain circumstances.
Describe the process of sedimentation.
Explain how fluids affect the motion of small particles and the nature of Brownian motion.
Describe and explain the important features of the diffusion process.
The laws governing the motion of an object, as given in the previous chapters, apply only when all
the forces acting on the object are taken into account. In the real world all objects move through
fluids so we must include the forces exerted by the extremely large number of fluid molecules. In
some situations the effect of these forces can be ignored, in others they play a crucial role in
determining the motion.
We have seen in chapter FE3 that, in the special case where a large object is at rest, the forces
exerted by the fluid molecules sum to give the buoyant force, a vertical force acting through the
centre of gravity of the displaced fluid, with magnitude equal to the weight of the displaced fluid.
The forces exerted by fluids on very small objects or on moving objects are, however, much more
complicated. This lecture will give a survey of these forces and explain qualitatively how living
organisms are affected by them. In addition to its intrinsic importance, this topic is a source of
models for non-mechanical phenomena.
Questions to think about
When a car is driven at constant speed along a level road, a constant force is exerted on the car
by the road surface. Why doesn't the car accelerate because of this force?
Oxygen gas is a denser fluid than nitrogen gas. Why doesn't the atmosphere stratify into a
layer of oxygen near the earth's surface and a layer of nitrogen above it?
Revise the section on buoyant forces and try question 3.10 in FE3.
FE4: Motion in Fluids
There are three important things to note about the force exerted by a fluid on a moving object which
is very much larger than a fluid molecule.
(i) The fluid force is not just the buoyant force. For example two pieces of cardboard with the
same mass and volume (and hence with the same weight and buoyant force acting on them) do not
fall at the same rate.
(ii) The force is not always vertical. For example the force exerted by the air on a ‘parachute’
moving horizontally at constant velocity must balance the other forces acting.
Hand force
Fluid force
Figure 4.1 Forces on a toy parachute
(iii) The force is not always along the line of motion. For example the fluid force on a marsupial
glider moving at constant velocity must be vertical, balancing the animal's weight.
For convenience, the fluid force is usually divided into three parts:
the usual buoyant force,
a lift force perpendicular to the line of motion, and
a drag force parallel to the line of motion, opposing the motion.
Figure 4.2
Forces on a glider exerted by the air
The glider's weight is not shown.
The lift and drag forces come into play only when the object is moving. Usually they increase
with speed and with the size of the object.
FE4: Motion in Fluids
The total fluid force does not necessarily act through the centre of gravity of the object. If it
does not, then the object may have a rotational acceleration. (See FE3.) This complication will be
ignored here.
There are some objects (such as non-rotating spheres) for which the lift force is small or zero.
Other objects produce large lift forces - aerofoils and birds' wings, for example.
There is always a drag force on an object moving in a fluid. For objects of diameter less than
about a millimetre moving in water or air, or larger objects moving in more viscous (sticky) fluids:
drag force ∝ speed .
For objects of diameter very much greater than a millimetre moving in water or air:
drag force ∝ (speed)2 .
A weighted balloon falls vertically through the air.
An object falling vertically in a fluid does not accelerate at a constant rate. The downward velocity
approaches a constant value known as the terminal velocity.
Downward velocity
without drag force
Terminal velocity
with drag force
Figure 4.3
Approach to terminal velocity
This behaviour is caused by the drag force. Consider the following force addition diagrams
(figure 4.4).
Total force
Total force
v =0
Figure 4.4
Drag force
B Buoyant
0 < v < vT
System of forces on a falling object
v = vT
FE4: Motion in Fluids
When the object is at rest, the forces on it are its weight and the buoyant force. The sum of
these forces (represented by the solid arrow in figure 4.4) produces a downward acceleration. As
the downward velocity increases, the drag force increases from zero, reducing the total downward
force. The downward acceleration gradually decreases to zero. The object then travels at its terminal
velocity, vT . (If there were no drag force, the total downward force and the downward acceleration
would have been constant.)
The velocity-time graph (figure 4.3) is characterised by the terminal velocity and the "rise
time" - the time taken to reach terminal velocity (or more precisely some specified fraction of the
terminal velocity). Both these quantities can be found from the equation of motion:
m dt = total force.
An object travelling at terminal velocity is in equilibrium. The terminal velocity is the velocity
at which the drag force balances the other forces acting;
total force = 0.
An approach to a constant velocity will always occur whenever there is a constant driving force
and an opposing force which increases with velocity until it is as large as the driving force.
Birds of prey, initially travelling faster than their terminal velocity are slowed down to terminal
velocity by the drag force.
Particles of different sizes and densities have different terminal velocities in water. (See the
section on sedimentation in the POST-LECTURE.)
We now consider the motion of very small particles in fluids.
Examples include the circulation of nutrients and waste materials in living organisms. Particles
move independently of the flow of blood in animals or the flow of sap in plants. Particles of matter
move through the fluids, rather than being carried along with the fluid. What are the forces exerted
by fluids on the particles in these examples?
Here we are looking at very small objects in a fluid which are continually undergoing random
collisions with fluid molecules. This situation is in contrast with that previously discussed, in which
the objects had diameters very much larger than 10-6 m and masses more than 1012 times the mass
of a fluid molecule. In those cases only the effects of "averaged out" forces (buoyant, lift and drag)
were detectable.
For objects smaller than 10-6 m the accelerations caused by individual collisions are larger.
Such small objects also travel further between collisions (about 10-10 m to 10-7 m) since they are
less likely to collide with fluid molecules because of their smaller sizes. The objects therefore make
a series of random movements called Brownian motion. (See figure 4.5.)
A large collection of objects, each undergoing a random Brownian motion, can move as a
whole in an organised way. This process is called diffusion.
A one dimensional mechanical analogue of diffusion is presented. It is shown how objects can
spread out by a sequence of purely random motions. The people move randomly using a code
based on the last digits from a page of telephone numbers: 0-3 means stand still; 4-6, move right: 79, move left.
FE4: Motion in Fluids
Figure 4.5
Brownian motion
Two important features of a diffusion process
(i) There is an overwhelming probability that the random movements will cause objects to diffuse
from regions of high concentration (of objects) to those of low concentration.
(ii) The average time for an object to move a distance d from its starting point is proportional to
d 2.
For example, glucose molecules (relative molecular mass 240) in water (relative molecular
mass 18) take 10-3 s to diffuse 10-6 m but 109 s (about 20 y) for 1 m.
Diffusion is an effective mechanism of transport of materials over typical cell dimensions (less
than about 10-5 m) in water. For example jellyfish without circulatory systems can obtain oxygen
and other nutrients from the surrounding water and lose carbon dioxide. Larger organisms require a
circulatory system and perhaps a respiratory system to bring the nutrients to the cells .
Diffusion occurs much faster in gases than in liquids because of the lower density of the fluid.
To illustrate this, the upward diffusion of bromine gas into air, is demonstrated. A state of dynamic
equilibrium is eventually reached where, at each point in space, the net rate of upward diffusion to
regions of lower bromine concentration is balanced by the rate of falling.
The final demonstration shows people participating in a two-dimensional mechanical analogue
of diffusion.
Consider a small particle of density d and volume V falling with velocity v in a viscous fluid of
density ρL. The forces acting on the particle are shown in figure 4.6.
Buoyant force = ρ L Vg
Drag force = λ v
Weight = ρ Vg
Figure 4.6
Forces on a particle during sedimentation
FE4: Motion in Fluids
The constant λ in the expression for the drag force depends on the viscosity of the fluid as
well as the size and the shape of the particle.
When the particle is travelling at its terminal velocity (or sedimentation rate) vT, these three
forces are in balance.
ρ LV g + λ vT = ρ V g ;
vT = λ (ρ - ρL) g .
Since the value of V/λ , and hence the terminal velocity or sedimentation rate, depends on the
size of the particle, sedimentation is a useful method for separating particles of different sizes. It is
used, for example, in agricultural laboratories to determine the amounts of various types of soils in a
The sedimentation rate depends on the difference in the densities of the particles and the
surrounding fluid. It is often quite slow for biological materials in water since their density is almost
equal to that of the surrounding water.
For a spherical particle of radius r falling in a viscous fluid characterized by a coefficient of viscosity η ,
the terminal velocity is
vT =
(ρ - ρL) g .
Find the terminal velocity or sedimentation rate of a soil particle of density 2.6 × 103 kg.m-3 and
radius 0.50 × 10-6 m in water.
Viscosity of water η = 1.0 × 10-3 kg.m-l.s-l .
Density of water ρL = 1.0 × 103 kg.m-3.
A man with a parachute is falling vertically through the atmosphere with a velocity v . The system, man
plus parachute, experiences a drag force described by λv2 where λ is a constant.
What other vertical forces are acting on the system? At terminal velocity, the vertical forces add to
give zero total force.
Find an expression for the terminal velocity in terms of these forces.
Assume that the man and parachute together weigh 850 N and that the buoyant force on them is
negligible. (See question 3.10 in FE3.)
The constant λ depends on the shape of the falling object. When the man is falling in the sky-diving
position (on his stomach with arms extended) with his parachute unopened, λ is 0.22 kg.m-l and when he
has his parachute opened up, λ is 26 kg.m-1.
iii) Find the terminal velocity of the man (and parachute) in these two positions.
If the man pulls his ripcord after sky-diving, he decelerates from the first of these terminal velocities to
the second in about 1 s .
iv) What would be the average acceleration of the man during this l s interval? (Compare this with the
acceleration due to gravity.)
Suppose you are driving a car along a straight level road at a constant speed of about 80 km.h-l by keeping
the accelerator pedal in a fixed position. At a high speed like this, the drag force, which is proportional to
the square of the speed, is quite large. If you depress the accelerator pedal a bit more and keep it in this
new position, the driving force on the car increases to a new constant value.
i) Explain, in terms of the drag and driving forces, what happens to the speed of the car.
ii) What happens if you then let the accelerator pedal come back to its original position? Sketch a
velocity-time graph for the complete motion.
In the absence of air resistance, an object thrown vertically upwards would take the same time to rise to its
maximum height as it does to return from it. With a drag force present this is no longer the case. Use
graphs of acceleration and velocity as functions of time to determine whether the object takes longer to rise
or to fall. You may neglect the buoyant force here.
(Hint: Use slopes and the areas between these graphs and the time axis where appropriate.)
FE4: Motion in Fluids
The mathematical equations used to describe the motion of a small object falling in a fluid, which
exerts a drag force proportional to the velocity, can also be used to describe currents in certain
electrical circuits and nuclei undergoing radioactive decay. Try question 4.5 if you have the
necessary mathematical expertise.
If the buoyant force is neglected, a small object falling vertically through the air has a downward
component of velocity, vx , at time t given by
- t
mg 
vx =
1− e m  .
λ 
What is the limit of the velocity (i.e. the terminal velocity) as t → ∞ ?
What is the starting velocity at t = 0 ?
iii) After what time (the rise time) is the difference between the terminal velocity and the actual velocity equal
to l/e of the terminal velocity ?
iv) Sketch the velocity component as a function of time.
Show by differentiating the expression for v that the downward component of acceleration, ax , at time
t is given by
a x = ge m .
vi) Sketch this acceleration component as a function of time.
Hence verify that the velocity component satisfies the equation of motion
m x = mg − λ vx
viii) Does the terminal velocity obtained directly from the equation of motion agree with that found in (i)?
ix) Is the terminal velocity ever reached in this model ? How does the real world differ from the model in this
Show by differentiating the result of (v) that the acceleration component obeys the equation
m x = −λa x
Insoluble particles small enough to undergo Brownian motion will not fall in a fluid. They form
what is known as a colloidal suspension which may be thought of as something intermediate
between a true suspension and a solution. Common examples of colloids include smoke particles,
the fat in homogenised milk and the protein molecules and protoplasm of cells. The particles are
kept from clumping together, usually by some kind of electrostatic repulsion. If the surrounding
fluid is altered in some way, for example by changing its acidity, temperature or chemical properties,
the colloidal particles can often aggregate and settle, as for example in the setting of jellies and
clotting of blood.
In the two dimensional mechanical analogue of diffusion in the video lecture, you may have noticed two
people attempting to sabotage the demonstration by making precisely the same move as one another at
each stage. At each stage, each person could move in 9 distinct ways, each with equal probability.
What is the probability that two people at the same position at the same time will make precisely the
same next move, purely by chance?
What is the probability that they will make the same next ten moves?
Carbon dioxide gas molecules take 25 seconds to diffuse about a centimetre in air. How long would it take
them to diffuse through l metre?
Interlude 4 - Energies
Energy equivalent* of Sun's mass
Solar energy received by Earth annually
Earthquake, cyclone, H bomb
8 hours hard labour
KE of a rifle bullet
Energy of a nuclear bond
Ionization energy
Energy of a chemical bond
105 0 __
10 __
Sun's annual output
103 0 __
102 0 __ Annual world use of energy
(Energy equivalent* of 1 g of matter
...(Energy from fission of 1 kg of U235
(Burning 7000 tonnes of coal
10 __
1 kilowatt hour (3.6 × 106 J)
Energy equivalent of a proton
Energy equivalent of an electron
10 __
1 electron volt (1.6 × 10-19 J)
* Energy equivalent of mass is calculated using E = m c 2 , where c is the speed of light in vacuum.