Belt friction

Objective
To confirm the laws of belt friction and determine the co-efficient of friction of
the given belt and the pulley
Equipment
Steel cylinder with groves rotatable by handle.
Meter rule.
Thread.
Diagram

Spring balance
peg
steel cylinder with groove

belt
weight

Theory
T1 = tension in the belt where the weight is hanged
T2 = tension recorded by the spring balance on the other end
= the angle of lab or contact

= co-efficient between the belt and the pulley

handle

The ratio of the two tensions may be found by considering an elemental piece of the belt MN
subtending an angle

at the centre of the pulley B as shown below. Forces keeping MN in

equilibrium include:
Tension T in the belt at M acting tangentially,
Tension T +

in the belt at N acting tangentially,

Normal reaction R acting outward at P (p is at the middle of MN)

Friction force (F) = R acting at right angle to R and in the opposite direction of the motion
of the pulley.
Diagram of belt for flat

From the diagram above and by Pythagoras theorem,

Taking, T =(T + T ) ,

R=( T + T ) sin
sin

Since the angle is too small,

R= T

R=Tsin

=
2
2

+ T
T
2
2
2

)(

)(

( T ) + T

Since

is negligible, it implies that

)
R=T .. ( 1 )

Resolving vertical forces from diagram,

F=( T + T ) cos Tcos

2
2
Also, since is very small,
F=( T + T )T

It implies that,

But

F= R

Implying that, R= T
R=

T
.. ( 2 )

Equating (1) and (2),

T =
=

Therefore,

T
T

Integrating the above between limits T1 and T2,

T1

Implies that

T2

ln

Therefore,

T
=
T

T1
=
T2

T 1
=e
T2

T 1
For a flat belt, T 2 =e

cos

=1
thus, reduces to unity.
2

Diagram of the V-belt in the v groove

Let
RN = Normal reaction between belt

= angle of groove
= co-efficient of friction between belt and pulley

R = total reaction in the plane of groove

Resolving vertical forces on diagram,

R=Rnsin + Rnsin
2 Rnsin

Rn=

R
R
= cosec
2 sin 2

But friction 2 Rn
2

R
cosec
2

( Rn= R2 cosec )
R cosec

R ( cosec )

From the above comparing with flat belt,

T 1 (cosec )
=e
T2
e cosec
1

e
1=cosec
Therefore,

1=

sin

Procedure
With the rotating cylinder apparatus, the angle of groove

is selected by

moving the peg to which the spring balance is connected.

The angular distance is measured with a thread and its linear measurement
is taken by stretching the thread on a calibrated straight edge.
A known weight is hanged on the other end of the belt and its weight
registered on the spring balance is noted as T 1.the direction of rotation is
selected and with the help of the handle the steel groove is rotated slowly
and steadily and the new weight registered on the spring balance is noted as
T2.
The experiment is repeated for a range of values of the lap

and T1

by

setting the belt so that it rests on the flat surface.

Calculations and Results
FOR PEG ONE
Groove one
1 =

8.5
360=111.27
27.5

Groove two
2 =

7.5
360=91.53
29.5

FOR PEG TWO

Groove one
1 =

8.6
360=112.27
27.5

Groove two
2 =

9
360=109.83
29.5

FOR PEG THREE

Groove one
1 =

8.6
360=112.27
27.5

Groove two
2 =

9.3
360=113.49
29.5

Grove three
3 =

Grove three

7
360=84
30

3 =

Groove four
4 =

Grove three

9
360=108
30

3 =

Groove four

6.5
360=76.72
30.5

4 =

9.7
360=116.4
30

Groove four

9.3
360=109.77
30.5

4 =

9.8
360=115.67
30.5

PEG ONE

Load(
kg)
2.0
4.0
6.0
8.0
10.0

Groov
Groov
Groov
Groov
e1
e2
e3
e4
Log T1 Log T2 Log T1 Log T2 Log T1 Log T2 Log T1 Log T2
0.5051 0.778
2
0.7324 0.954
2
0.9444 1.079
2
0.9637 1.176
1
1.0863 1.255
3

0.5441 1.000
0
0.6990 1.021
2
0.9138 1.130
3
1.0294 1.217
5
1.0792 1.290
0

0.5441 0.875
1
0.7782 1.021
2
0.8573 1.146
1
1.0212 1.204
1
1.0969 1.278
8

0.5441 0.7782

Groov
e3
Log T1

Groov
e4
Log T1

0.7782 0.9294
0.8921 1.0792
0.9542 1.1523
1.0531 1.2304

PEG TWO

Load(
kg)
2.0

Groov
e1
Log T1
0.6021

4.0

0.7243

6.0

0.8751

8.0

1.0000

10.0

1.0969

Log
T2
0.903
1
1.041
4
1.130
3
1.243
0
1.311

Groov
e2
Log T1

Log T2

0.6232 0.929
4
0.6990 1.079
2
0.9294 1.209
5
1.0212 1.290
0
1.1139 1.352

0.6021
0.7782
0.9294
1.0000
1.0864

Log
T2
0.929
4
1.079
2
1.204
1
1.278
8
1.342

Log T2

0.5798 0.8129
0.7404 1.0000
0.8751 1.1614
0.9395 1.2175
1.0682 1.2900

PEG THREE

Load(
kg)
2.0
4.0
6.0
8.0
10.0

Groov
e1
Log T1 Log
T2
0.4914 1.000
0
0.7782 1.146
1
0.9294 1.255
3
0.9542 1.301
0
1.1139 1.352
2

Groove

0.6021
0.7782
0.9031
1.0000
1.0414

Log
T2
1.000
0
1.146
1
1.255
3
1.332
4
1.361
7

Peg1
111.27
091.53
084.00
076.72

0.6021
0.6990
0.9294
0.9777
1.0792

Peg2
112.58
109.83
108.00
109.77
PEG ONE

Groove1

Groov
e3
Log T1

Log
T2
1.000
0
1.176
1
1.267
2
1.342
4
1.361
7

Groov
e4
Log T1

Log T2

0.5441

0.9542

0.7404

1.1139

0.9031

1.2041

1.0212

1.2788

1.0414

1.3324

Linear circumferential
distances(cm)
29.00
31.00
34.00
36.00

1
2
3
4
Groove
1
2
3
4

Groov
e2
Log T1

Peg3
136.15
113.49
116.40
115.67

a graph of LogT against logT

1.2
1

f(x) = 0.91x - 0.19

0.8

Log T2

0.6

Linear (Log T2)

0.4
0.2
0
0.95

1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45

Groove2
A graph of LogT against logT
1.2
1

f(x) = 0.98x - 0.3

0.8
0.6
0.4
0.2
0
0.95

1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45

Groove3
A graph of LogT against logT
1.2
1

f(x) = 1.36x - 0.7

0.8
0.6
0.4
0.2
0
0.95

Groove4

1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.2
1

f(x) = 1.24x - 0.55

0.8
0.6
0.4
0.2
0
0.9

0.95

1.05

1.1

1.15

1.2

1.25

1.3

1.35

PEG TWO
Groove1
1.2
1

f(x) = 1.23x - 0.67

0.8
0.6
0.4
0.2
0
1

1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45

Groove2
1.2
1

f(x) = 0.89x - 0.17

0.8
0.6
0.4
0.2
0
1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

Groove3
1.2
1

f(x) = 1.1x - 0.37

0.8
0.6
0.4
0.2
0
0.9 0.95

1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4

Groove4
1.2
1

f(x) = 0.89x - 0.17

0.8
0.6
0.4
0.2
0
1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

PEG THREE
Groove1

1.45

1.2
1

f(x) = 1.1x - 0.51

0.8
0.6
0.4
0.2
0
1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

Groove2
1.2
1

f(x) = 1.42x - 0.9

0.8
0.6
0.4
0.2
0
1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.3

1.35

1.45

Groove3
1.2
1

f(x) = 1.37x - 0.77

0.8
0.6
0.4
0.2
0
1.05

Groove4

1.1

1.15

1.2

1.25

1.4

1.2
f(x) = 0.97x - 0.27

1
0.8
0.6
0.4
0.2

0
0.9

1.1

1.2

1.3

1.4

1.5

From the equation, taking log of both sides we have

log

T1
=log e
T2

log T 1 log T 2=
log T 1 =log T 2 +
By comparison of this equation to the standard equation of a straight line
y=mx +c

, the graph gives a straight line with its gradient being 1(one) and an

intercept
c= so the co-efficient of friction can be determined from the graph each co-

efficient of friction can be calculated as

=

1
Therefore, = sin

For V-belt
T1
=e
T2
1

Grove1

Grove2

1=0.188111.27=1.68910-3

0.554

91.53=6.053 10

=0.297112.27=2.64510-3

0.672 109.83=6.119 10

=0.703112.27=6.26210-3

0.171 112.27=1.55067 10
Grove3

Flat

7=0.368 84=4.381 103

8 0.499 108=4.62

10 0.895 76.72=1.2 10

103

11

0.766 109.77=6.978 103

9 0.505 116.4=4.338

12

103

0.267 115.67=2.30 103

Angle of lap/rads

Groove1
log

1.46607
1.597499
1.942

T
T

1
2

-0.4624
-0.3912
-0.4087

Groove2
log

T
T

1
2

-0.4437
-0.3553
-0.3232

Groove3
log

T
T

Flat
log

-0.426
-0.3485
-0.3011

T
T

1
2

-0.426
-0.3485
-0.3011

A graph of log(T1/T2) against groove1

-0.34
1.4
1.5
-0.36 f(x) = - 0.25x

1.6

1.7

1.8

1.9

-0.38
-0.4
-0.42
-0.44
-0.46
-0.48

A graph of log(T1/T2) against for groove2

0
1.4
-0.1

1.5

1.6

-0.2
-0.3
-0.4
-0.5

f(x) = -f(x)
0.22x
= - 0.22x

1.7

1.8

1.9

A graph of log(T1/T2) against for groove3

0
-0.051.4

1.5

1.6

1.7

1.8

1.9

-0.1
-0.15
-0.2
-0.25
-0.3
-0.35

f(x) = - 0.21x

-0.4
-0.45

A graph of log(T1/T2) against for flat

0
1.4
-0.05

1.5

1.6

1.7

1.8

1.9

-0.1
-0.15
-0.2
-0.25
-0.3

f(x) = - 0.19x

-0.35
-0.4

Groove

Slope ()

1
2
3
Flat

-0.247
-0.208
-0.217
-0.187

Conclusion

The limitations were mainly due to the fact that the rate of the rotation
of the cylinder whether fast or slow could change the spring balance reading.
Much more accurate results can be obtained if motor is used to rotate or turn
the cylinder to maintain constant speed and make reading of the values
easier.
In cases where the belt is not slipping will provide the belt with maximum torque. This is
because the force generated from the wheel on which the belt rotates will not be used in directing
the belt to a particular position. If such an experiment is performed the expected outcome must
be close to about 90% of the theoretical calculations.
This could be tested by the use of a belt drive. This can be used to
drive a pulley of the same diameter (diameter of driven pulley and driver
pulley). The ratio of torque in the driven pulley to the power of the driver
pulley gives .
This is due to the fact that belt drives use the friction between the pulley and the belt around the
arc of the contact to transmit torque. The belt is initially stretched with an initial tension between
the two pulleys.
Comparing obtained data from experiment with printed data differ slightly.
Printed Data
Flat belt: 10x2.2mm, leather/polyamide
V-belt: 5x3mm, rubber/fabric
References
Mechanics of Machines by James N. Asante
Design of Machinery by Robert L. Norton
College Physics by Buffa and Wilson