• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28
  29. 29
    29
  30. 30
    30
  31. 31
    31
  32. 32
    32
  33. 33
    33
  34. 34
    34

Suspension Bridges. this extended essay is an investigation to study the variation in tension in the left segment of a relatively inelastic and an elastic string tied between two supports

Extracts from this document...

Introduction

image00.png


Abstract

Suspension bridges are made of a long roadway with cables that are anchored at both ends to pillars. Vehicles on the roadway exert a compression force on the pillars which in turn exert a tension in the cables. However, the tension is not the same at different points across the cable and it also varies as the length of the cable varies.Taking inspiration from the suspension bridge, this extended essay is an investigation to study the variation in tension in the left segment of a relatively inelastic and an elastic string tied between two supports with the: -

  • Point of Application of Force
  • Length of the string tied

Keeping in mind that the intrinsic property of the string plays a big role in determining the extent of deformation or change undergone due to the force applied, the research involves comparing the trends in the above variables for two strings of different strain values. Therefore, using their strain values, it would be safe to assume one string as relatively inelastic while the other as elastic.

The analyzed data showed that the tension in the relatively inelastic string increased moving horizontally across the string until it reached its maximum somewhere around the halfway distance between the two supports after which it decreased. For the elastic string, the tension increased slightly initially and the maximum was well before the halfway distance after which the tension started decreasing.  

Also, as the length of the string tied was increased, the tension decreased. However, the extent of decrease was not uniform for either string. The reason for the similar yet non-identical trend is attributed to the strain values of the strings and predictions have been made for strings with different strain values.

Table of Contents

Cover Page        1

Abstract        2

Table of Contents        3

List of figures, images and graphs        4

...read more.

Middle

12.2

1.8447 ± 0.04

19.2

1.1898 ± 0.02

101

11.5

1.6092 ± 0.02

17.5

1.0633 ± 0.01

117

9.2

1.3181 ± 0.02

14.4

0.8459 ± 0.01

126

7.2

1.0349 ± 0.01

11.6

0.6440 ± 0.005

129.4

6.1

0.8986 ± 0.01

10

0.5492 ± 0.005

image07.png

Horizontal Distance (x) from the Rigid Support 1

± 0.05 (in cm)

Length of the string = 148 cm

Length of the string = 152 cm

Vertical Distance (y1) from the Rigid Supports

± 0.05 (in cm)

T1

(in Newton)

Vertical Distance (y1) from the Rigid supports

± 0.05 (in cm)

T1

(in Newton)

9

18

0.7097 ± 0.002

19

0.7023 ± 0.002

18

22

0.7629 ± 0.004

23.5

0.7437 ± 0.004

27

24.8

0.8074 ± 0.004

26.5

0.7797 ± 0.005

38.5

27.6

0.8403 ± 0.005

29.4

0.8067 ± 0.005

66.3

30.1

0.8534 ± 0.006

32.2

0.8075 ± 0.006

84.3

29.6

0.7975 ± 0.004

31.6

0.7528 ± 0.004

101

27.4

0.6953 ± 0.001

29.2

0.6555 ± 0.002

117

23.1

0.5334 ± 0.001

24.9

0.4964 ± 0.001

126

19.5

0.3861 ± 0.001

21.2

0.3559 ± 0.001

129.4

17.8

0.3105 ± 0.001

19.5

0.2840 ± 0.001

String Type 2: - Elastic Band (Relatively Elastic)

Horizontal Distance (x) from the Rigid Support 1

± 0.05 (in cm)

Length of the string = 140 cm

Length of the string = 144 cm

Vertical Distance (y1) from the Rigid Supports

± 0.05 (in cm)

T1

(in Newton)

Vertical Distance (y1) from the Rigid supports

± 0.05 (in cm)

T1

(in Newton)

9

14.3

0.7500 ± 0.008

15.6

0.7328 ± 0.007

18

21.5

0.7700 ± 0.008

21.8

0.7657 ± 0.008

27

26.5

0.7797 ± 0.009

27.2

0.7695 ± 0.008

38.5

31.6

0.7717 ± 0.009

32.3

0.7617 ± 0.008

66.3

36.2

0.7362 ± 0.008

37.3

0.7195 ± 0.007

84.3

35.2

0.6857 ± 0.007

36.3

0.6681 ± 0.006

101

31.6

0.6097 ± 0.006

32.7

0.5910 ± 0.006

117

24.4

0.5061 ± 0.005

26.5

0.4677 ± 0.004

126

18.9

0.3980 ± 0.003

19.9

0.3785 ± 0.003

129.4

15.9

0.3513 ± 0.003

16.7

0.3306 ± 0.002

Horizontal Distance (x) from the Rigid Support 1

± 0.05 (in cm)

Length of the string = 148 cm

Length of the string = 152 cm

Vertical Distance (y1) from the Rigid Supports

± 0.05 (in cm)

T1

(in Newton)

Vertical Distance (y1) from the Rigid supports

± 0.05 (in cm)

T1

(in Newton)

9

18

0.7097 ± 0.007

23

0.6816 ± 0.007

18

24.3

0.7348 ± 0.008

28.7

0.6970 ± 0.007

27

28.8

0.7486 ± 0.008

33.5

0.7015 ± 0.007

38.5

33

0.7523 ± 0.008

36.7

0.7100 ± 0.007

66.3

38.5

0.7025 ± 0.007

41.5

0.6649 ± 0.006

84.3

38.1

0.6415 ± 0.006

41.2

0.6017 ± 0.006

101

34.5

0.5632 ± 0.005

38

0.5170 ± 0.005

117

27.9

0.4455 ± 0.004

32

0.3917 ± 0.004

126

23.1

0.3274 ± 0.002

26.8

0.2838 ± 0.003

129.4

20.2

0.2743 ± 0.002

24.2

0.2302 ± 0.002

image08.png

Strain in the strings

To quantitatively draw a comparison between the elasticity of the two strings, it is better to calculate their strain values. This is done by taking a fixed length of both the strings and applying a constant force. The ratio of change in the length of the strings to the actual length indicates their responsiveness to force.

Strain, δ = image58.pngimage58.png

For this, a controlled initial length (44 cm) was taken for both the strings and also the same bob was employed as the constant force (0.679 N).

Nylon String                                                                Elastic Band

Initial Length = 44 cm                                                Initial Length = 44 cm

Length after force applied = 49 cm                                Length after force applied = 54 cm

Change in length = 49 – 44 = 5 cm                                Change in length = 54 – 44 = 10 cm

δnylon = image59.pngimage59.png = 0.114                                                δband = image60.pngimage60.png = 0.227

Thus, the elasticity of the elastic band is almost twice that of the nylon string. These strain values shall be used in the analysis to analyze the reason for the variation in tension in the analysis.

Data Analysis

  • To compare the tension in a nylon and elastic band with the point of application of force for a constant length of 140 cm

String Type: - Nylon String (Relatively Non Elastic)

image61.pngimage10.png

image11.png

As we move horizontally from left to right across the nylon string while applying force, the tension experienced is highest at a point which is approximately the mid-point of the distance between the two rigid supports (nails). From that point onwards, the tension starts decreasing. The minimum tension experienced will be at a point of application of force which is the farthest from the first rigid support.  Hence, the curve is roughly like a wave which is extended towards the end.

String Type: - Elastic Band

image62.png

image12.png

Here we see that the tension is approximately the highest initially itself. From the first point of application onwards, there is a very slight increasing trend observed and the tension is highest at a point of application of force which is before the halfway distance between the rigid supports. From the maximum point onwards, the tension keeps on reducing and the minimum point lies when the point of application of force is the furthest from the first rigid support

Comparing the occurrence of the maximum tension in a nylon string and an elastic band

Both the nylon and the elastic band can be modeled by quadratic equations. This enables us to study the symmetry in the tension trend for the different strings.

image63.pngimage14.pngimage13.png

image15.png

Axis of Symmetry

The axis of symmetry of the above graph is an important indication of the trend in tension as it divides all tension values in two parts gives the point of application of force wherein the tension is expected to be maximum.

The general quadratic equation y = ax2 + bx + cimage16.png

Axis of symmetry = image64.pngimage64.png

Nylon String                                                                Elastic Band

y = -2.413x2 + 3.099x + 0.952                                                y = -0.504x2 + 0.392x + 0.711

Axis of symmetry = image65.pngimage65.png                                                                            Axis of symmetry = image67.pngimage67.png

Axis of symmetry = 0.6421 m                                                 Axis of symmetry = 0.3889 m

From the values calculated above, we can infer that for a relatively non-elastic string the tension force is maximum further away from the rigid support and occurs somewhat halfway the distance (d/2 = 0.69 m) between the two supports.

However, that is not the case with a relatively elastic string wherein the tension is maximum and the trend is symmetrical quite nearer to the rigid support.  

  • To study the variation in tension with different lengths of the string

String Type: - Nylon String

image68.png

image17.png

The graph makes it clear that as the length of the string tied between the two supports decreases, the tension values increases. It is not only the maximum tension, but in fact tension at all points of application of force is more when the length of the string is 140 cm. The tension starts decreasing with increasing length. However, the decrease in the tension values is not at all linear or equal. Rather, the area between the curve of 140 cm and 144 cm is larger compared to that between 144 cm and 148 cm. This indicates that the tension values decrease but not uniformly.

String Type: - Elastic Band

image69.png

image19.png

Due to the close proximity of the four series, it is difficult to display the polynomial equation on the graph. Here are the polynomial equations for all the four series: -

Length of the String (in cm)

Polynomial Equation

R2

152

y = -0.566x2 + 0.436x + 0.635

0.9948

148

y = -0.574x2 + 0.456x + 0.668

0.9927

144

y = -0.510x2 + 0.386x + 0.702

0.9948

140

y = -0.504x2 + 0.392x + 0.711

0.9913

...read more.

Conclusion

T1’ and ‘T2’, respectively.

Using all these variables and applying the conditions for Static Equilibrium, I have found a general formula as follows to calculate ‘T1 and ‘T2’. The derivation of the formula is as follows: -

As the system is in equilibrium, the sum of all the vector forces acting must be equal to 0. In the system, there are three main forces which are acting: -

  1. Tension, T1 in the part of the string tied to the point of suspension on the left
  2. Tension, T2 in the part of the string tied to the point of suspension on the right
  3. Force acting downwards, which in this case, the weight ‘W’ of the bob

We shall resolve all the forces acting on the system into two parts and since the system is in equilibrium, the sum of all the forces in their respective components will also be 0: -

Σ F x = 0 (Sum of all horizontal forces is equal to 0)

Σ F y = 0 (Sum of all vertical forces is equal to 0)

It is also important to define the sign convention as we are dealing with a vector quantity (Force). The following figure gives the sign convention:  -

image42.png

image43.png

Using the Cartesian plane, I have assigned the point of application of force as the point of origin. Therefore, all forces acting from the positive x and y direction shall be positive whereas those acting from the negative x and y direction shall be negative. To find out the formula, all the x forces are resolved as follows: - image44.png

As Σ F x = 0

-T1 cosӨ1 + T2cosӨ2 = 0

Substituting cosӨ1 = image71.pngimage71.pngand cosӨ2 = image72.pngimage72.pngabove

-T1 image71.pngimage71.png + T2 image72.pngimage72.png = 0

T1 image71.pngimage71.png = T2 image72.pngimage72.png                                ---------------------------------------- (1)image46.png

As Σ F y = 0

T1 sinӨ1 + T2 sinӨ2 – W = 0                

Substituting sinӨ1= image74.pngimage74.pngand sinӨ2= image75.pngimage75.pngabove

T1image74.pngimage74.png + T2image75.pngimage75.png  - W = 0                          ---------------------------------------- (2)

Substituting the value for T2 in Equation (2) from Equation (1)

T1 image74.pngimage74.png + T1 image76.pngimage76.pngimage77.pngimage77.png  - W = 0image47.pngimage47.png

T1 image74.pngimage74.png + T1image78.pngimage78.png- W = 0

image79.pngimage79.pngimage80.pngimage80.png= W

Or, image82.pngimage82.png = W

image79.pngimage79.pngimage83.pngimage83.png= W

image79.pngimage79.pngimage84.pngimage84.png = W

Or, T1 = image81.pngimage81.png                                  -------------------------- (3)

When y1= y2

T1 =  image54.pngimage54.png                                        -------------------------- (4)

Appendix 2 – String Samples used

image85.pngimage48.png

image86.pngimage49.png

Candidate Name: Ladha, Sanchit

Candidate Session Number: 001070 - 006        Page

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Physics essays

  1. Physics Extended essay

    of Tension with Point of Application for Elastic Band of lengths 140, 144, 148 and 152 cm 22 Conclusion 23, 24 Relation of tension and strain values of the string 24 Evaluation 25 Acknowledgment and Bibliography 26 Appendix 1 - Derivation of Tension 27, 28, 29 Appendix 2 - Images

  2. Analyzing Uniform Circular Motion

    The frequency, as a result, decreases. However, a linear graph is then attained from using the formula;. The graph is a straight line because f2 is indirectly proportional to r. The original formula does not have any y intercept, however for the equation gained from the graph above there is a definite y intercept or b value.

  1. Investigating the Breaking Distance of a Cart

    The dependent variable was chosen to be the distance traveled by the cart, meaning it's braking distance. Whilst the independent variable was the initial velocity obtained by the cart. There were two controlled variables within this experiment, which are as follows: frictional force of the track surface and mass of the metal crate itself.

  2. Physics Wave revision question

    be reflected and be inverted. D. be reflected and not be inverted. (1) 20. A source produces water waves of frequency 10 Hz. The graph shows the variation with horizontal position of the vertical displacement of the surface of water at one instant in time.

  1. In this extended essay, I will be investigating projectile motion via studying the movement ...

    Therefore, the vertical acceleration of the metal ball will be considered as constant throughout the entire trajectory and the vertical motion of metal ball is considerded to be under free fall after projection. Note that the vertical component of projection velocity of metal ball is zero as the metal ball

  2. Investigate the Size of Craters in Sand Due to Dropped Object.

    Calculation of Uncertainties for the Volume of Craters, cm � Uncertainties = � 0.05 Uncertainties = � 0.05 2.00 0.50 � 0.72 4.00 0.60 � 0.75 6.00 0.70 � 0.78 8.00 0.90 � 0.84 10.00 1.00 � 0.87 12.00 1.10 � 0.90 * Fourth Reading Calculation Height of Dropped, cm Depth, cm (d)

  1. Forces Lab. I decide to investigate the relationship between the propelling force exerted on ...

    Collecting the raw data: The experiment will be repeated 3 times for each of 4 different masses of the hanging weights. (100g, 150g, 200g, 250g) I select 100g to be my minimum value of the mass of the hanging weights because 50g is too light to propel the weighted margarine tub and make it move.

  2. HL Physics Revision Notes

    superposition: The effect of two separate causes is equal to the sum of the separate causes. Constructive interference occurs when two waves are in phase with eachother. The resultant displacement is the sum of both displacements. Destructive interference occurs when two waves are out of phase.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work