# 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

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Introduction

## 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

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

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 |

### 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 =

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 = = 0.114 δband = = 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)

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

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.

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 + c

Axis of symmetry =

Nylon String Elastic Band

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

Axis of symmetry = Axis of symmetry =

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

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

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 |

Conclusion

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: -

- Tension, T1 in the part of the string tied to the point of suspension on the left
- Tension, T2 in the part of the string tied to the point of suspension on the right
- 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: -

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: -

As Σ F x = 0

-T1 cosӨ1 + T2cosӨ2 = 0

Substituting cosӨ1 = and cosӨ2 = above

-T1 + T2 = 0

T1 = T2 ---------------------------------------- (1)

As Σ F y = 0

T1 sinӨ1 + T2 sinӨ2 – W = 0

Substituting sinӨ1= and sinӨ2= above

T1 + T2 - W = 0 ---------------------------------------- (2)

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

T1 + T1 - W = 0

T1 + T1- W = 0

= W

Or, = W

= W

= W

Or, T1 = -------------------------- (3)

When y1= y2

T1 = -------------------------- (4)

## Appendix 2 – String Samples used

Candidate Name: Ladha, Sanchit

Candidate Session Number: 001070 - 006 Page

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

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