Materials Tutorial 4  Fluid Materials
Contents 
A fluid material is one that adopts the shape of its container. It cannot resist any shear forces that are applied to it. It can be a liquid or a gas. The general properties of a fluid are:
It cannot resist permanent deformation;
It flows.
However there are significant differences between fluids, depending on the phase of the matter:
Liquids form a free surface that is not made by the container. This is on the top surface of the liquid. Liquids have a surface tension.
Liquids are almost impossible to compress;
Gases are not restricted by any top surface. Nor is there any surface tension;
Gases can be compressed easily.
Some apparently solid materials are actually very viscous (gooey) fluids. Glass is one such example:
Other examples include silicone putty. You can roll it into a ball. Then leave it a while and it flattens out into a disc.
Explain why glass can be considered as a supercooled liquid. 
You will have encountered pressure before. It is defined as:
perpendicular force per unit area
The equation is:
[ p  pressure (N m^{2}); F  force (N); A  area (m^{2}).]
The units for pressure are Pascals (Pa) or Newton per square metre (N m^{2}) where:
1 Pa = 1 N m^{2}
If the force is at an angle, we need to take the vertical component of the force.
So our equation is changed to:
A force of 250 N is applied at an angle of 40^{o} to the horizontal onto a surface of 25 cm in diameter.
Calculate the pressure. Give your answer to an appropriate number of significant figures. 
Pressure in a fluid acts equally in all directions. This is called Pascal's Law. This is why you are not crushed by atmospheric pressure.
Pressure and Hydraulic Systems
This has important implications in hydraulic systems. Consider a very simple hydraulic system consisting of a master cylinder of area A_{1} and a slave cylinder of area A_{2}. This is shown in the diagram below:
Force F_{1} is applied to the master cylinder. There is a resulting pressure, p. Since pressure is the same throughout the system, pressure p acts on the slave cylinder to give force F_{2}. So we can write:
Simple rearrangement gives:
A simple hydraulic system has a master cylinder of area 0.0015 m^{2}, while the slave cylinder has an area of 0.055 m^{2}. A force of 12 N is applied to the master cylinder. Calculate the force from the slave cylinder. 
The system above is a force multiplier. The ratio of the forces = ratio of the areas. This is shown below:
Show that your answer to Question 3 is consistent with the statement about force multipliers. 
SCUBA divers will tell you how when they dive, the pressure acting on them from the water increases. (SCUBA = Self Contained Underwater Breathing Apparatus).
Consider a cylinder of area A m^{2}, filled with fluid of density r kg m^{3} to a height of h m.
The volume of the cylinder = area × height = hA
The mass of the fluid in the cylinder = volume × density = hAr
Remember that weight is a force, NOT the mass.
The weight of the fluid = mass × gravitational field strength = hArg
Pressure = weight (force) ÷ area:
The areas cancel out to give:
p = hrg
Therefore the pressure is proportional to the depth in a liquid. This equation works for liquids, since they are incompressible, so the density of a liquid remains the same throughout the body, however deep we go. However this model does not work for a gas. Gases are easily compressed, so the density of a gas changes with the height, something that pilots of aeroplanes need to be aware of.
A SCUBA diver dives to a depth of 20 m. Calculate the pressure acting on the diver at that depth. Density of seawater = 1030 kg m^{3}. Gravitational field strength = 9.81 N kg^{1}. 
The Pascal is quite a small unit of pressure. Engineers tend to use the atmospheric pressure unit, bar.
1.0 bar = 1.0 × 10^{5 }Pa
Your answer should show that the pressure acting on the SCUBA diver is about 2 bar.
Any object that floats in a liquid has an upthrust that is equal to its weight. Since the weight acts vertically downwards, the upthrust acts vertically upwards. The idea is shown below:
The weight of water displaced is the same as the weight of the object. If the upthrust is less than the weight, the object will sink.
When an object is totally immersed in the water, the volume of water displaced is the volume of the object. You will have used this idea to find out the density of an irregular object. See Materials Tutorial 1.
Archimedes' principle states:
Any body wholly or partly immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced
We can explain Archimedes' Principle using the relationship between pressure and depth. Consider a uniform object of area A and thickness t at a depth of d in a liquid of density r:
The pressure on the top surface is:
Since force = pressure × area, we can work out the downwards force on the top surface:
Pressure on the bottom surface is:
Since force = pressure × area, we can work out the upwards force on the bottom surface:
The upwards force on the object is in the opposite direction to the downwards force, therefore:
The volume of the object is tA. This volume is the same as the volume of liquid that is displaced. The mass of liquid displaced is volume × density, so the mass of liquid displaced is tAr. Therefore the weight displaced is tArg. We have seen how the depth term has cancelled out. This is consistent with the density of the water being the same whatever the depth, due to liquids being incompressible.
Archimedes' principle can be applied to gases. However the upthrust from the air is small enough to be negligible.
A diver is salvaging a spherical cannon ball from the wreck of an ancient ship. The cannon ball has a diameter of 10 cm and is made of iron of density 7900 kg m^{3}. The density of seawater is 1030 kg m^{3}. What is the force needed to lift the cannon ball: (a) on land; (b) under the water?
Does the depth of the wreck matter?
Acceleration due to gravity = 9.81 m s^{2}. 
This is a special application of the Archimedes Principle. Suppose we have a block of iron which has a density of 7900 kg m^{3}. If placed in water of density 1000 kg m^{3}, it will, of course, sink. The sunken block will displace its own volume in the water. When iron ships first started to be built, many thought they would sink like a stone. But they didn't.
Worked Example A barge has a mass of 750 tonnes (1 tonne = 1000 kg). (a) What is the volume of steel used in its construction? (Density of steel is 7800 kg m^{3}) (b) The barge is built so that it is approximately a box 30 m long, 10 m wide, and 5 m high. Calculate the volume of the box. (c) What is the weight of water this barge will displace? Hence what volume of water will it displace? (g = 9.8 N kg^{1}) (d) What depth will the barge float at? 
Answer (a) Volume = mass ÷ density = 750 000 kg ÷ 7800 kg m^{3} = 96.2 m^{3}. (b) Volume of barge = 30 m × 10 m × 5 m = 1500 m^{3}. (c) Weight of water = 750 000 kg × 9.8 N kg^{1} = 7.35 × 10^{6} N. (Who was going to write 750 000 kg?) The mass of water displaced is 750 000 kg. Since the density of water is 1000 kg m^{3}, the volume is 750 m^{3}. (d) The area of the bottom of the barge is 300 m^{2}. Therefore the depth is 750 m^{3} ÷ 300 m^{2} = 2.5 m. This depth is called the draught of the barge. 
The barge will float like this.
If the barge went out to sea, where the density of sea water is about 1050 kg m^{3}, the draught would be reduced, and the barge would float higher in the water.
The way floating objects float higher in liquids of higher density is put to use in the hydrometer. This is a glass tube with a weight at the bottom, as shown.
It is the home winemaker's best friend. Water with sugar dissolved has a higher density. The more sugar dissolved, the higher the density. A typical home made wine will have enough sugar dissolved in it to give it a density of 1085 kg m^{3} (1.085 g cm^{3}). The hydrometer floats high in the unfermented wine. As the yeast feeds on the sugar, converting it to carbon dioxide and alcohol, the density gets lower, so the hydrometer floats lower. When the density reaches 1.000 (g cm^{3}), the wine is considered to be fermented out, and can be transferred to glass jars called demijohns or jorams to mature.
Terminal Velocity in Liquids
It is easy to measure terminal velocity in a liquid, since the terminal speed is low. This experiment is one of the core practicals:
When a ball bearing is dropped into a viscous liquid, it almost immediately reaches its terminal speed, and measuring it is simply a matter of timing the motion between two fixed points a known distance apart. You will also have to:
Note the mass, m, of each ball.
Use a micrometer to measure the diameter, d.
When the ball bearing drops into the liquid (usually glycerine), there are two forces:
the weight;
the upthrust.
Since the weight is greater than the upthrust, the ball bearing will accelerate. The velocitytime graph will look like this:
At terminal speed, the upwards forces of upthrust and drag and the downwards force of the weight are balanced.
The upthrust is the same as the weight of fluid displaced by Archimedes' principle. Therefore, if the weight is greater than the upthrust, the object will accelerate downwards until the drag balances the difference between the weight and the upthrust. This is true of all fluids, for example air, or water, or chocolate.
A ball bearing of diameter 2.50 mm is made of steel. It is released into glycerine and falls. (a) What is the upthrust? (b) Show that the acceleration is about 8 ms^{2}.
The density of this grade of steel is 8050 kg m^{3}; The density of glycerine is 1260 kg m^{3}; Acceleration due to gravity is 9.81 m s^{2}. 
You will see that the initial acceleration is not 9.81 m s^{2}. This is because upthrust is a force acting in the opposite direction to the weight. However you would not be able to time the ball bearing as it fell through the glycerine if its acceleration were 8 m s^{2}. Some other force is acting. This is the drag force.
Viscosity and Drag Force
Viscosity has a working definition of:
the quantity that describes a fluid's resistance to flow.
There are more formal definitions, but we won't use them here. Viscosity is temperature dependent. If you heat up a viscous material like oil, it becomes less viscous, i.e. more runny. Cold car engines use more petrol because the oil is more viscous, so more energy is needed to circulate it.
When the ball bearing is falling at terminal velocity, the resultant force = 0 (Newton I). The weight is balanced by the upthrust and the drag force.
Weight  (Upthrust + Drag) = 0
Use answers from Question 7 to work out what the drag force is at terminal velocity. What do you notice? 
We can calculate the drag force by Stokes' Law. It was worked out by George Gabriel Stokes (1819  1903). Consider a ball falling through a viscous liquid at terminal velocity. There are the three forces acting on it:
The downwards force of weight, which is balanced by the drag and the upthrust;
The upwards force of upthrust which can be worked out using Archimedes;
The upwards force of drag, which is worked out by Stokes' Law.
Stokes' Law is given by:
[F_{d}_{ } drag force (N); r  radius of the sphere (m); v  terminal speed (m s^{1} ).]
The strange looking symbol, h, is "eta", a Greek lower case letter long 'ē', the Physics Code for the coefficient of the viscosity of a fluid.
The units for h are N s m^{2}. An alternative SI unit is Pascal seconds (Pa s).
For air, h = 1.8 × 10^{5} N s m^{2}. For glycerine, h = 0.950 N s m^{2}.
This relationship only works when the fluid flow around the object is laminar (smooth). It does not work if the flow is turbulent.
Use your answer to Question 8 to calculate the terminal velocity of the ball bearing in Question 7, as it falls through the glycerine. Comment on your answer. For glycerine, h = 0.950 N s m^{2}. 

A student carelessly omits the upthrust in answering Questions 7, 8, and 9. What is the effect on the answer for the terminal velocity? 
When you are measuring the terminal velocity of ball bearings in a viscous fluid, you need to have a start point some distance below the surface of the glycerine.
Use your answer to Question 9 to give an estimate of the distance travelled by the ball bearing as it accelerates to terminal velocity. What is your assumption? 
So the start line a couple of centimetres below the surface will ensure that the ball bearing is travelling at the terminal velocity. However the Stokes' Law equation does not lend itself easily to graphical analysis. Let's look at this more closely:
The temptation is to say that:
There is a problem in that the drag force is not constant. If we use a ball bearing of half the diameter, the weight goes down by 8 times. Using the answers to the questions above, the weight of a ball bearing of diameter 1.25 mm will give these results:
Diameter / × 10^{3 }m 
Weight / N 
Upthrust / N 
Drag Force / N 
2.50 
6.46 × 10^{4} N 
1.01 × 10^{4} N 
5.45 × 10^{4} N 
1.25 
8.08 × 10^{5} N 
1.27 × 10^{5} N 
6.81 × 10^{5} N 
Therefore the drag force will be eight times less. So instead of the terminal velocity being doubled when the radius is halved, it is reduced by four times. Let's look at this further.
The 6ph is a constant, which we will call k. So the relationship between drag force and terminal velocity becomes:
If the radius is halved, the force goes down eight times. So the new terminal velocity v' is given by:
Therefore we need to look at a new relationship between the radius and the terminal velocity.
Relationship between Radius and Terminal Velocity
I have written two arguments here to establish the relationship between the radius of a ball bearing and its terminal velocity. The first does not take into account the upthrust (which is not on the syllabus for some boards). The second does take into account the upthrust.
Argument without upthrust
Upthrust is not on the syllabus for some of the boards. The derivation of this equation is an approximation, as we are not taking into account the upthrust. The resulting uncertainty is relatively low, compared with the uncertainties of the other measurements made in the experiment. If you want to see the effect of upthrust on the argument, please go to the next section.
Consider a ball bearing of mass m made of material of density r being dropped into a viscous fluid of viscosity h. The gravity field is g (= 9.81 m s^{2} as you aren't likely to take the apparatus elsewhere).
Weight = density × volume × g
We can write this as:
So we can bring in the Stokes' Law equation in by writing:
Cancelling out gives us:
And rearranging gives:
Then we take the square root to show the equation in the form it's usually presented:
We can rearrange to make
A ball bearing of diameter 2.50 mm is made of steel. It is released into glycerine and falls.
Calculate the terminal velocity.
The density of this grade of steel is 8050 kg m^{3}; For glycerine, h = 0.950 N s m^{2}; Acceleration due to gravity is 9.81 m s^{2}. 
Argument with upthrust
The derivation of this equation takes into account the upthrust.
Consider a ball bearing of mass
m
made of material of density r
being dropped into a viscous fluid of viscosity
h, and
density s.
The gravity field is g
(= 9.81 m s^{2} as you aren't
likely to take the apparatus elsewhere).
The strange looking symbol s, is sigma, a Greek lower case letter 's'.
Weight = r × V × g
Upthrust = s × V × g
We can write the weight as:
We can write the upthrust as:
We know that the drag force = weight  upthrust:
We can rewrite this by substituting:
We can write this as:
So we can bring in the Stokes' Law equation in by writing:
Cancelling out gives us:
And rearranging gives:
This rearranges to make v the subject:
A ball bearing of diameter 2.50 mm is made of steel. It is released into glycerine and falls.
Calculate the terminal velocity.
The density of this grade of steel is 8050 kg m^{3}; The density of glycerine is 1260 kg m^{3}; For glycerine, h = 0.950 N s m^{2}; Acceleration due to gravity is 9.81 m s^{2}.
