1. 
Viscosity
 
What is the cause of friction? It depends on what type of friction we discuss. 
Actually, friction is a certain force acting in the opposite direction of the body's 
motion, thus, decelerating it. If a solid body moves along a solid surface, the force 
of friction is just a force of the electromagnetic interaction between the atoms or 
molecules of a body and a surface. It has the same origin as elastic forces. The 
atoms are located at certain positions in their crystal lattices which they do not 
leave. The atoms of solid surface "pull" or "push" the atoms of the moving solid 
body in the direction opposite to the body's velocity. 
Contrary to solid bodies, fluids consist of randomly moving molecules flying at 
high speeds. They bump at a solid body inserted into fluid at various angles and 
at various speeds. During the collision, the electromagnetic forces of interaction 
are mainly repulsive, and their values and directions are random due to random 
character of the molecules' motion. Thus, it is useless to speak of the force's origin, 
but useful to note, that when bumping the molecules change their velocities, 
i.e. accelerate  
, and can produce a force ma = F according to Newton's law. If 
a body is motionless inside a fluid, all the molecules' bumps compensate each 
other and the total force is equal to zero. But if a body moves through a fluid, 
the accelerations of the molecules bumping from the front are larger than the 
accelerations of the molecules bumping from the back, and the resulting force F 
differs from zero. Its direction is opposite to the direction of body's motion, thus, 
it is effectively a force of resistance. The property of fluids to resist the motion of 
bodies inserted into them is called viscosity. The corresponding force of friction 
is called force of viscosity. 
What does the force of viscosity depend upon?
 
1.1. 
HandsOn: Viscosity
 
1) Take an exercise book and wave it in the air twice. First time wave it with the 
edge forward, and second time with the cover forward. 
Q: Can you feel the difference in resistance forces? Explain the role of cross- 
section area. 

2) Take two pieces of rough paper or hard plastic. One - resembling a ruler 
a feet long and an inch wide. Its area is, thus, 12 square inches. Another - a 
round one about 4 inches in diameter. Its area is about 12 square inches too. The 
weights of these pieces are, thus, also equal. Try to wave these objects in the air

with the same speed. Hold them with your fingers to become a more sensitive 
instrument. 
Q: Can you feel any difference in resistance forces? Does the form of the front 
cross-section make any difference? 
3) Take a sheet of paper and wave it in the air twice in such a way that the 
sheet is perpendicular to the motion. First time wave slowly, and second time 
fast. Use your fingers again. 
Q: Can you feel the difference in resistance forces? Explain the role of speed.  
"  
Summarizing the results of our HandsOn investigation, we see that the force of 
viscosity depends on the properties of a moving object: its form and dimensions, 
and on the speed of motion. Stokes analized theoretically the case of a sphere of 
radius r moving at constant speed v in a continuous fluid. He did not account 
for microscopic, i.e. molecular kinetic, reasons for the force of viscosity and found 
out that a sphere will suffer a force of viscosity equal to
 
FStokes 
= 6??vr
 
where ? (greek letter pronounced "e-ta") is the viscosity coefficient character- 
izing the viscous properties of fluid, and 6? is a form factor for sphere. The higher 
is the speed, the larger is the force of viscosity acting on a body in the direction 
opposite to the direction of its motion. The larger is the characteristic dimension 
of a body, the larger is the force of viscosity. What properties of fluid expressed 
by the viscosity coefficient will affect the body's motion? 
Turning back to the speculations about bumping molecules, we can guess that 
the force of viscosity which as we think originate from F = ma = m?v
?t 
must 
be proportional to the molecules' mass. Besides, the change of velocity for the 
same time interval ?t will be higher for higher values of velocity, i.e. for higher 
temperatures of the fluid.
 
1.2. 
SimuLab: Viscosity
 
1.2.1. 
Pre-Lab Discussion
 
In the Virtual Laboratory we can simulate an object's motion through a media 
consisting of randomly moving particles and find out from the molecular kinetic 
point of view the dependance of the viscosity coefficient on the parameters of the 
fluid. In the Post-Lab Discussion we will compare the experimental result with 
the theoretical one.
 
2
We will drop a heavy object through the media consisting of light flying parti- 
cles and measure the speed of its fall. Three forces acting on an object - the force of 
gravity Fgrav 
, buoyant force Fbuo 
and the force of viscosity Fvisc 
- will compensate 
each other, when an object moves at constant speed. Thus, Fgrav 
?Fbuo 
?Fvisc 
= 0. 
For a gas system the buoyant force is much less than the other two forces, and 
the corresponding term can be omitted Fgrav 
? Fvisc 
= 0. The viscosity force we 
will write in the form analogous to the Stokes formula Fvisc 
= ?rv?, where ? is a 
constant form factor and r is the characteristic dimension of a body. Finally, we 
obtain
 
Mg ? ?rv? = 0
 
Here M is the mass of a body, g is the free fall acceleration. Since M = Nm, 
where m is mass of the body's molecule and N is the number of molecules of the 
body, we get
 
? = Nmg  
?rv
 
Thus, for the case of body falling at constant speed in a fluid, the viscosity 
coefficient appears to be proportional to the speed in the minus first power and 
to the body's characteristic dimension in the minus first power. Varying the 
parameters of the fluid and measuring speed of the falling body, we will judge of 
the viscosity coefficient dependance on these parameters. 
You will be able to 
a) Measure the speed of the body falling down in the gas of particles 
b) Find out the viscosity coefficient dependance on mass of the gas particles 
c) Find out the viscosity coefficient dependance on temperature of the gas 
d) Test the predicted dependance of the force of viscosity on the dimension of 
the body 
1) Open Visc.smd. Set Number of Hidden Steps to 1000. Click Options- 
Start Averaging-Graph updates-Average over hidden steps. Set Gravity 
to 0:04: 
You see a container which is 40 units high and a green object, a crystal, consist- 
ing of 7 particle of mass 1, surrounded by almost ideal gas of 193 blue particles.
 
3
The blue particles are comparatively light and have mass 0:1. The temperature is 
0:15, the interaction parameter for the gas particles is 0:05, and, thus, the blue 
particles remain in the gaseous phase and are practically unaffected by gravity. 
Choose a characteristic dimension of the crystal, e.g. its maximal width, mea- 
sure it with a ruler right from the screen and record the obtained value d. 
2) Press Start. Watch the y-concentration graph. 
You see that the two highest bins of the histogram contain green bars corre- 
sponding to the green crystal in the upper part of the screen. As time goes on, the 
crystal falls down, leaves the upper layers and the upper green bar vanishes. Your 
goal will be to measure the time it takes the ctystal to move from one bin to the 
next and calculate the average speed of the fall. 
3) Click Pause. Click Reset Experiment. Check the settings. 
4) Make a copy of the table below.
 
 
h  
90%  
80%  
70%  
60%  
50%  
40%  
30%  
20%  
ti          
?ti  
?         
vi  
?         
5) Click Start. Watch the y-concentration graph and averaging window. 
As soon as the green bar completely leaves the 90% level, record into the table 
the corresponding time value t9 
from the averaging window. Repeat the same for 
80% level and so forth until the table is filled in. 
Calculate ?t8 
= t9 
? t8 
;?t7 
= t8 
? t7 
;::: and v8 
= 40?0:1  
?t8 
;v7 
= 40?0:1  
?t7 
;:::and 
record the values into the table. Find v0:1 
= vavg 
= v8 
+v7 
+:::+v2  
7 
for temperature 
T = 0:15 
To be able to investigate the role of the gas molecules mass and preserve the 
fluid in the gas state, we will increse the gas molecule mass simultaneously with 
the decrease of gravity. 
6) Reset Experiment. Set Mass B to 0:2. Set Gravity to 0:02. 
7) Repeat steps 4-5. 
Get a new value of a constant speed free fall v0:2 
. 
8) Reset Experiment. Set Mass B to 0:4. Set Gravity to 0:01. 
9) Repeat steps 4-5. 
Get a new value of a constant speed free fall v0:4 
. 
According to ? = Nmg  
?rv 
, if ? were independant of gas molecule mass, the ob- 
tained values for v0:1 
;v0:2 
;v0:4 
must be proportional to g, i.e. v0:1 
: v0:2 
: v0:4 
= 4 : 
2 : 1. 
Q: Is it really strictly so?
 
4
Q: How does fluid molecule's mass affect the speed of the body? 
Q: How does fluid molecule's mass affect the viscosity coefficient? 
To investigate the viscosity coefficient dependance on temperature 
10) Reset Experiment. Set Mass B to 0:1. Set Gravity to 0:04. Set 
Temperature to 0:2: 
11) Repeat steps 4-5. Find v0:1 
for T = 0:2 
12) Reset Experiment. Set Mass B to 0:1. Set Gravity to 0:04. Set 
Temperature to 0:25 
13) Repeat steps 4-5. Find v0:1 
for T = 0:25 
Q: How does fluid's temperature affect the speed of the body? 
Q: How does fluid' temperature affect the viscosity coefficient? 
14) Open Visc1a.smd. Set Number of Hidden Steps to 1000. Click 
Options-Start Averaging-Graph updates-Average over hidden steps. 
Set Gravity to 0:04: 
You see a container which is 40 units high and a green crystal which has a 
larger characteristic dimension, than the previous one, because it consists of 19 
particle of mass 1: The temperature is 0:15 again and the interaction parameter 
for the gas particles is the same as before, i.e. equal to 0:05: 
Choose the similar characteristic dimension of this crystal as before, measure 
it with a ruler right from the screen and record the obtained value D. 
15) Repeat steps 4-5. 
Find v0:1 
(19) for 19 particle crystal. 
Now we will test if the formula ? = Nmg  
?rv 
reflects the dependance of the force of 
viscosity on an object's dimension in the right way. Since all the fluids parameters 
are the same in both cases, the viscosity coefficient ? must also remain the same. 
Then 7mg 
?dv0:1 
(7) 
must be equal to 19mg 
?Dv0:1 
(19) 
and, thus,
 
v0:1 
(7) 
v0:1 
(19) 
= 7D 
19d
 
Q: Do your results support this equality?  
--- 
 
1.3. 
Post-Lab In-Depth Discussion
 
To recognize the meaning of the obtained experimental results let's regard the 
microscopic dynamics of viscosity in a more detailed way. Imagine a solid surface 
parallel to the y-axis. When the surface is motionless, the gas molecules that are to
 
5
the right of the surface move chaotically, their average speed being vavg 
. But when 
the surface starts to move parallel to y-axis, the gas molecules obtain a regular 
velocity component parallel to the surface motion direction due to collisions with 
the surface. Subsequently, the molecules collide with other molecules and transfer 
the obtained regular component of the momentum to the layers that are further 
and further from the surface. We see that a momentum transfer takes place. 
To estimate the rate of this transfer, take a plane region of area ?A also parallel 
to the y-axis at a certain distance x from the surface. Every time interval ?t; 
the molecules crossing this region from right to left and from left to right transfer 
certain momenta with them. Let a molecule flying from left to right transfer 
momentum mv1 
, where v1 
is the regular component of the molecule's velocity at 
a distance x ? l from ?A. Here l is the mean free path. Then the decrease of 
the momentum of the layer to the left of ?A will be equal to mv1 
N+ 
, where N+
 
is the number of molecules that crossed ?A by ?t. Similarly, the momentum 
transfered by the molecules through ?A from right to left by the same ?t will be 
equal to mv2 
N? 
, where v2 
is the regular component of the molecule's velocity at 
a distance x + l from ?A, and N? 
is the number of molecules that crossed ?A 
by ?t from right to left. 
How many molecules cross ?A by ?t from each side? Only those molecules 
can reach ?A that are located in the cylinder with ?A base and vavg 
?t height. 
Their number is equal to number density of the molecules times volume of this 
cylinder, i.e. nvavg 
?A?t. But not all of the molecules located there actually reach 
?A because they fly in various directions. On the average only 1 
6 
of the molecules 
fly towards ?A. Thus, the numbers of molecules that cross ?A by ?t from each 
side are equal and equal to N+ 
= N? 
= 1 
6 
nvavg 
?A?t. But the corresponding 
momenta mv1 
N+ 
and mv2 
N? 
transferred by them are not equal. According to 
Newton's law, the difference between them is equal to the product of the force of 
viscosity Fvisc 
times time interval ?t. Thus, we obtain
 
Fvisc 
?t = 1 
6 
nvavg 
m(v1 
? v2 
)?A?t
 
Dividing both sides of this equation by ?A we get a force of viscosity acting 
over unit area
 
fvisc 
= Fvisc  
?A 
= ?1
3 
nmvavg 
v2 
? v1  
2 
= ?1
3 
nmvavg 
lv2 
? v1  
2l
 
Note now that v2 
? v1 
is the velocity change ?v at the distance ?x = 2l, and 
the expression v2 
?v1  
2l 
= ?v 
?x 
is the velocity gradient. The combination of the rest 
parameters is called the viscosity coefficient ? = 1 
3 
nmvavg 
l. Finally, we get
 
6
fvisc 
= ???v
?x
 
We see that our experimental results are consistent with the above said. Really, 
it was found out that the larger is the molecule mass the larger is the viscosity 
coefficient. And since vavg 
? 
p 
T 
, the experimental dependance of the viscosity 
coefficient on temperature is also in accordance with theoretical speculations. 
Note, that the mean free path is inversly proportional to number density, l ? 1 
n 
, 
and the viscosity coefficient does not depend on number density of the fluid.
 
7