Suspension Bridges Essay
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 exert 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 is relatively inelastic while the other is 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.
Writing service  Conditions  Website 
[Rated 96/100]  Prices start at $12
 
[Rated 94/100]  Prices start at $11
 
[Rated 91/100]  Prices start at $12

For the elastic string, the tension increased 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 nonidentical 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
 Introduction 5, 6
 Experiment Design 7, 8
 Controlling of Variables 7
 Hypothesis 9
 Experimental Procedure 10, 11
 Apparatus Utilized and their Specifications 10
 Procedure Followed 11
 Calculation of Tension 12
 Primary Data Collected and Tension Calculated 13, 14, 15
 Nylon String 14
 Elastic Band 15
 Strain Value of the strings 16
 Data Analysis 17, 18, 19, 20, 21, 22
 Variation of Tension with Point of Application for Nylon String of length 140 cm 17
 Variation of Tension with Point of Application for Elastic Band of length 140 cm 18
 Comparison of the graphs for nylon string and elastic band 19
 Axis of Symmetry 20
 Variation of Tension with Point of Application for Nylon String of lengths 140, 144, 148 and 152 cm 21
 Variation 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 of the string samples used 30
 List of Figures, Images, Tables and Graphs
 Figures
 Figure 1. It shows a bob tied to a rigid extended support 1
 Figure 2. Shows the experiment setup with the bob acting as the force 8
 Figure 3. The experiment setup with labelled measurements 12
 Figure 4. The experiment setup with labelled measurements 27
 Figure 5. The experiment setup showing sign convention used 28
 Images
 Image 1. The Severn Suspension Bridge 5
 Image 2. It shows the wooden board with the string hanging 10
 Image 3. Shows the bob hanging at some distance ‘x’ cm and the distance measured y1 cm 13
 Image 4. Nylon String used 30
 Image 5. Elastic Band used 30
 Tables
 Table 1 shows the point of application of force (x), y and tension calculated for lengths 140 cm and 144 cm 15
 Table 2 shows the point of application of force (x), y and tension calculated for lengths 148 cm and 152 cm 15
 Table 3 shows the point of application of force (x), y and tension calculated for lengths 140 cm and 144 cm 16
 Table 4 shows the point of application of force (x), y and tension calculated for lengths 148 cm and 152 cm 16
 Graphs
 Graph 1 Shows Tension vs Point of Application of Force for nylon string of length 140 cm 18
 Graph 2 Shows Tension vs Point of Application of Force for an elastic band of length 140 cm 19
 Graph 3 Shows Tension vs Point of Application of Force for the elastic band and nylon string of length 140 cm 20
 Graph 4 Shows Tension vs Point of Application of Force for nylon string of lengths 140, 144, 148 and 152 cm 22
 Graph 5 Shows Tension vs Point of Application of Force for an elastic band of length 140, 144, 148 and 152 cm 23
 Graph 6 Shows the curve for ? = 0.112 25
 Graph 7 Shows the curve for ?> 0.112 25
 Graph 8 Shows the curve for ?> 0.112 25
Introduction. According to any ordinary man, when he sees objects at rest, such as on top of a table or fixed to a wall, he concludes that there is no force acting on it because the object is at rest. However, that is a misconception that there is no force acting on the object. In fact, the forces act so that they cancel out the individual effect of the forces acting in different directions. Thus, a system in which the forces act only in two dimensions is said to be in equilibrium when the vector sum of all the forces, that is (horizontal direction),(vertical direction) is equal to zero. While studying Statics, I have often come across equilibrium in objects attached to rigid support by a string. Let us take an example of such an object.
In figure 1, we see that there is a force due to gravity experienced by the hanging bob, which is, in fact, its own weight acting vertically downwards. Even though the bob is not supported by a hand or any other rigid support beneath it, it does not fall but rather stays in equilibrium. This can only be possible if there is an equal and opposite force to the gravitational force, which acts upwards on the object (as). This force is actually exerted by the support at the top, thus generating tensile stress or tension throughout the material. However, this is a simple case of equilibrium when the string is attached to rigid support only at one end. Consider the case of a cable that is tied between two columns of a suspension bridge. A suspension bridge comprises a roadway that is supported by cables that are anchored on both ends, i.e., attached to rigid support on both ends.
Referring to Figure 1, wherein tensile stress is generated throughout the material by the support to which it is attached; we can infer that when the cable is tied between two rigid supports, it experiences tensile stress as exerted by supports on either side. At the point of application of force, the string gets divided into parts, that is, on either side of the point of application of the force. Thus, both the parts of the string will experience a tension force (same as tensile stress) which may be equal or different.
It is of utmost importance that a delicate balance is maintained in the cables to prevent any accident. Any external force applied to these cables may lead to them snapping or breaking down, which may be disastrous. The force acting on the cables in a suspension bridge can be due to the air pressure and wind velocity. This reallife practical setup propelled me to undertake 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 between the supports.
The first part involves studying the variation in tension force exerted as force is applied on different points across the tied string. Also, as in suspension bridges, the weight is felt by the roadway, which exerts a compression force in the pillars. This, in turn, pulls the cables tighter and keeping this factor in mind, the second part of my study is the variation in tension when different lengths of the same string are tied between two supports. At the end of the study, the trend between tension in the left segment and the point of application of force can be modelled by negative quadratic equations.
This goes on to explain that tension increases till the maximum point and then starts decreasing. Also, as the length of the string tied between is increased, the tension in the string decreases. The intrinsic property of strings, such as the extent of deformation when under force, determines the tension it experiences. This is indicated by the strain value, which has been subsequently calculated for the two strings in question; an inelastic and a relatively elastic string. At the end of the study, the strain value is related to the observations leaving scope for predictions about the behavior of strings of different properties.
Experimental Design. The practical application of my research question lies in suspension bridges and even guitar strings. However, for flexibility in my research and conforming to school laboratory apparatus, I have designed an experimental setup that comprises a wooden board with several nails hammered from top to bottom on either side. Also, a grid was drawn on the board to make out the point of application of the force easily. My preliminary view of the experiment comprised of two nails hammered into the wall. A string would then be attached between the two nails by tying the two ends of the string to the nails. However, the wooden board with a grid makes it easier to calculate and control variables. To keep a constant force, I chose to use a certain constant mass tied to the string and whose position could easily be varied.
The point of application of force, the horizontal distance from the rigid support, is assigned the independent variable, and the tension in the left segment becomes the dependent variable. Henceforth, all use of the word ‘tension’ would refer to tension in the left segment. The applied force is the controlled variable throughout the experiment. While comparing the variation of tension with the point of application of force in the two strings, the length of the string is also a controlled variable. However, the length of the string becomes an independent variable while studying the variation in tension with the length of the strings tied. Controlling of Variables.
1. Applied Force. The force applied must be constant at all points of application to prevent disparity in the tension exerted. This is done by employing a uniform spherical bob of mass 69.3 gms. Therefore, the force which is exerted in the form of the bob’s weight will be: –
 F = ma
 F = 0.0693 kg x 9.8 ms2
 = 0.679 N
2. Length of the string. The length of the string is taken to be constant when varying the point of application of force. This is because the tension exerted varies when the length of the string is different (This is also another part of the study, which is discussed later). To keep the length of the string constant, colored marks were made on the string and tied appropriately to the supports.
Firstly, at a length of string = 140 cm, data was collected for 10 points of application of forces. The force applied in the form of a spherical bob was movable. Thus, the readings were taken starting close to the first rigid support and moving away from such that by the time I reached the second rigid support; they were 10 points at which the vertical distance was measured. This was done for both the nylon string and the elastic band.
However, the experiment has another important observation attached to it: to study the variation in tension with differing lengths of strings. Thus, 4 different lengths of the string, which were greater than the least distance between the two nails (138 cm), were taken. I started with a length of 140 cm and measured the vertical distance for 10 points of application of force (horizontal distances). Then, the length of the string was increased by 4 cm, and the vertical distance was measured for the same 10 points of application of force. The same was done for a length of string equal to 148 cm and 152 cm. Both the elastic band and the nylon string followed the same procedures.
There were some initial obstacles that I came across but were overcome in due course. Firstly, I planned to use a length string equal to the distance between the two nails (i.e. 138 cm). However, data collection became difficult for a relatively nonelastic string as the force applied could not produce an appreciable change in the vertical distance of the string. Thus, I modified my plans and used lengths greater than the actual distance between the two nails, that is, 140 cm, 144 cm, 148 cm, and 152 cm.
An ideal elastic string would be a onepiece rubber band. However, another issue was that I could not find a rubber band having a length of more than 1 m. Initially, I decided to join several rubber bands and assumed them to be acting as one string. But, this restricted the movement of the mass, which acted as the force and the properties of the joined rubber band varied due to the knots. Thus, I settled to use an elastic band similar to the kind used in garments.
Hypothesis. I predicted the result of my experiment that as the distance from the rigid support increases, the tension force in that part of the string (irrespective of whether it is elastic or inelastic), tied to the corresponding rigid support, should increase. Therefore, I envisaged a linear curve. Also, I believed that the tension in the elastic string would be correspondingly less than the nonelastic string. For the second part of my observation, I expected the tension to be more when the length is relatively less, both for elastic and nonelastic strings. Also, I had predicted that the tension is maximum at a point that was approximately half the distance between the rigid supports. Experimental Procedure. Apparatus Utilized and their Specifications
 Plywood Board of dimensions 1.5 m x 1 m
 18 nails (9 nails hammered vertically on either side at a vertical distance of 10 cm from each other and perpendicular horizontal distance of 138 cm)
 Nylon String (Length ? 1.6 m)
 Elastic Band (Length ? 1.6 m)
 A string of negligible weight (50 cm)
 Uniform Circular Metal Bob (Mass – 69.3 gms)
 Wooden Ruler (1 m)
 Measuring Tape (1 m)
 Stools to balance the wooden board (2 nos.)
Procedure Followed. The procedure described here is for the relatively nonelastic nylon string. The same procedure is also adopted for the elastic band.
 Firstly, 140 cm of the nylon string was measured using a measuring tape, and both ends were tied to the nails on either side.
 Next, a string of negligible weight was attached to the circular bob and tied to the nylon string. Thus, the nylon string became rigid.
 A horizontal distance of 9 cm from the nail on the left was measured using a ruler, and the string attached to the bob was moved to exactly that point.
 The bob acted like a force downwards, the vertical distance until the nails were measured using the ruler. This became the first reading.
 Steps 3 and 4 were followed for horizontal distances – 18 cm, 27 cm, 38.5 cm, 66.3 cm, 84.3 cm, 101 cm, 117 cm, 126 cm and 129.4 cm.
 The second part of the study involved varying the length of the string. Thus, the length of the string was increased by 4 cm so that it was 144 cm.
 Steps 2, 3, 4 and 5 were again followed.
 Readings were taken by increasing 4 cm each time such that the lengths of the string for which readings were taken were 140 cm, 144 cm, 148 cm and 152 cm.
Calculation of Tension. I have restricted myself to find out the tension T1. As the system is in Static Equilibrium, the sum of all the vector forces must be equal to 0. Three main forces acting on the system: the bob’s weight, the tension T1 existing in the left part of the string, and the tension T2, which exists in the right part of the string. The bob’s weight acts vertically downwards, and the two tensions can be resolved into their horizontal and vertical components. The summation of all vertical forces and also the horizontal forces. Using this property, the formula to calculate the tension T1 is found out to be
 T1 =
 Moreover, as we have taken y1 = y2
 T1 =
For the complete derivation of the tension T1, refer to Appendix 1. Primary Data Collected and Tension Calculated. My data collection involves measuring the y1 distance for the corresponding application of force which is the horizontal distance from the first rigid support. Using the formula for Tension as derived in the last section, I have also calculated the tension T1 existing in the left part of the string.
 Sample Calculation of Tension T1
 When the length of the string = 140 cm and the point of application of force ‘x’ cm = 9 cm
 T1 = = = = 1.1863 ï¿½ 0.02 N
 String Type 1: – Nylon String (Relatively inelastic)
 Horizontal Distance (x) from the Rigid Support 1 ï¿½ 0.05 (in cm)
 Length of the string = 140 cm
 Length of the string = 144 cm
 Vertical Distance (y1) from the Rigid Supports ï¿½ 0.05 (in cm)
 T1 (in Newton)
 Vertical Distance (y1) from the Rigid supports ï¿½ 0.05 (in cm)
 T1 (in Newton)
 9
 5.7
 1.1863 ï¿½ 0.02
 10.1
 0.8502 ï¿½ 0.01
 18
 8
 1.4538 ï¿½ 0.02
 13.5
 0.9841 ï¿½ 0.01
 27
 9.6
 1.6303 ï¿½ 0.02
 15.7
 1.0865 ï¿½ 0.01
 38.5
 11
 1.7821 ï¿½ 0.03
 17.6
 1.1775 ï¿½ 0.02
 66.3
 12.3
 1.9340 ï¿½ 0.03
 19.5
 1.2503 ï¿½ 0.02
 84.3
 12.2
 1.8447 ï¿½ 0.04
 19.2
 1.1898 ï¿½ 0.02
 101
 11.5
 1.6092 ï¿½ 0.02
 17.5
 1.0633 ï¿½ 0.01
 117
 9.2
 1.3181 ï¿½ 0.02
 14.4
 0.8459 ï¿½ 0.01
 126
 7.2
 1.0349 ï¿½ 0.01
 11.6
 0.6440 ï¿½ 0.005
 129.4
 6.1
 0.8986 ï¿½ 0.01
 10
 0.5492 ï¿½ 0.005
 Horizontal Distance (x) from the Rigid Support 1 ï¿½ 0.05 (in cm)
 Length of the string = 148 cm
 Length of the string = 152 cm
 Vertical Distance (y1) from the Rigid Supports ï¿½ 0.05 (in cm)
 T1 (in Newton) Vertical Distance (y1) from the Rigid supports ï¿½ 0.05 (in cm)
 T1
 (in Newton)
 9
 18
 0.7097 ï¿½ 0.002
 19
 0.7023 ï¿½ 0.002
 18
 22
 0.7629 ï¿½ 0.004
 23.5
 0.7437 ï¿½ 0.004
 27
 24.8
 0.8074 ï¿½ 0.004
 26.5
 0.7797 ï¿½ 0.005
 38.5
 27.6
 0.8403 ï¿½ 0.005
 29.4
 0.8067 ï¿½ 0.005
 66.3
 30.1
 0.8534 ï¿½ 0.006
 32.2
 0.8075 ï¿½ 0.006
 84.3
 29.6
 0.7975 ï¿½ 0.004
 31.6
 0.7528 ï¿½ 0.004
 101
 27.4
 0.6953 ï¿½ 0.001
 29.2
 0.6555 ï¿½ 0.002
 117
 23.1
 0.5334 ï¿½ 0.001
 24.9
 0.4964 ï¿½ 0.001
 126
 19.5
 0.3861 ï¿½ 0.001
 21.2
 0.3559 ï¿½ 0.001
 129.4
 17.8
 0.3105 ï¿½ 0.001
 19.5
 0.2840 ï¿½ 0.001
 String Type 2: – Elastic Band (Relatively Elastic)
 Horizontal Distance (x) from the Rigid Support 1 ï¿½ 0.05 (in cm)
 Length of the string = 140 cm
 Length of the string = 144 cm
 Vertical Distance (y1) from the Rigid Supports ï¿½ 0.05 (in cm)
 T1 (in Newton)
 Vertical Distance (y1) from the Rigid supports ï¿½ 0.05 (in cm)
 T1 (in Newton)
 9
 14.3
 0.7500 ï¿½ 0.008
 15.6
 0.7328 ï¿½ 0.007
 18
 21.5
 0.7700 ï¿½ 0.008
 21.8
 0.7657 ï¿½ 0.008
 27
 26.5
 0.7797 ï¿½ 0.009
 27.2
 0.7695 ï¿½ 0.008
 38.5
 31.6
 0.7717 ï¿½ 0.009
 32.3
 0.7617 ï¿½ 0.008
 66.3
 36.2
 0.7362 ï¿½ 0.008
 37.3
 0.7195 ï¿½ 0.007
 84.3
 35.2
 0.6857 ï¿½ 0.007
 36.3
 0.6681 ï¿½ 0.006
 101
 31.6
 0.6097 ï¿½ 0.006
 32.7
 0.5910 ï¿½ 0.006
 117
 24.4
 0.5061 ï¿½ 0.005
 26.5
 0.4677 ï¿½ 0.004
 126
 18.9
 0.3980 ï¿½ 0.003
 19.9
 0.3785 ï¿½ 0.003
 129.4
 15.9
 0.3513 ï¿½ 0.003
 16.7
 0.3306 ï¿½ 0.002
 Horizontal Distance (x) from the Rigid Support 1
 ï¿½ 0.05 (in cm)
 Length of the string = 148 cm
 Length of the string = 152 cm
 Vertical Distance (y1) from the Rigid Supports ï¿½ 0.05 (in cm)
 T1 (in Newton) Vertical Distance (y1) from the Rigid supports ï¿½ 0.05 (in cm)
 T1 (in Newton)
 9
 18
 0.7097 ï¿½ 0.007
 23
 0.6816 ï¿½ 0.007
 18
 24.3
 0.7348 ï¿½ 0.008
 28.7
 0.6970 ï¿½ 0.007
 27
 28.8
 0.7486 ï¿½ 0.008
 33.5
 0.7015 ï¿½ 0.007
 38.5
 33
 0.7523 ï¿½ 0.008
 36.7
 0.7100 ï¿½ 0.007
 66.3
 38.5
 0.7025 ï¿½ 0.007
 41.5
 0.6649 ï¿½ 0.006
 84.3
 38.1
 0.6415 ï¿½ 0.006
 41.2
 0.6017 ï¿½ 0.006
 101
 34.5
 0.5632 ï¿½ 0.005
 38
 0.5170 ï¿½ 0.005
 117
 27.9
 0.4455 ï¿½ 0.004
 32
 0.3917 ï¿½ 0.004
 126
 23.1
 0.3274 ï¿½ 0.002
 26.8
 0.2838 ï¿½ 0.003
 129.4
 20.2
 0.2743 ï¿½ 0.002
 24.2
 0.2302 ï¿½ 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,? = 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
 ?nylon = = 0.114 ?band = = 0.227
Thus, the elasticity of the elastic band is almost twice that of the nylon string. These strain values shall be used to analyze the reason for the variation in tension in the analysis. Data Analysis. To compare the tension in nylon and elastic band with the point of application of force for a constant length of 140 cm String Type: – Nylon String (Relatively NonElastic) 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 midpoint 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 that 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 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 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. Quadratic equations can model both the nylon and the elastic band. 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. It divides all tension values into two parts, giving 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 = 2.413×2 + 3.099x + 0.952 y = 0.504×2 + 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 nonelastic 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 with maximum tension, 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 than that between 144 cm and 148 cm. This indicates that the tension values decrease but not uniformly. String Type: – Elastic Band. Due to the proximity of the four series, it is difficult to display the polynomial equation on the graph. Here are the polynomial equations for all four series: – Length of the String (in cm)
 Polynomial Equation
 R2
 152
 y = 0.566×2 + 0.436x + 0.635
 0.9948
 148
 y = 0.574×2 + 0.456x + 0.668
 0.9927
 144
 y = 0.510×2 + 0.386x + 0.702
 0.9948
 140
 y = 0.504×2 + 0.392x + 0.711
 0.9913
Although quadratic equations also model the trends in tension for an elastic band, it is not similar to the nylon string. Firstly, the maximum tension occurs closer to the rigid support, as observed in Part 1 already. Secondly, the difference in the range of tension in the elastic band when taking different lengths is not much. This is why the curves for all four lengths are very close to each other, unlike the nylon string.
Conclusion. Variation in tension with the point of application of force. My hypothesis of a linear increasing trend for both the strings was proved wrong by Graphs 1 and 2. The inelastic string is in the form of a wave, with the maximum tension occurring approximately half the distance between the two supports. However, the same is not reciprocated by the elastic band. Firstly, the magnitude and the range of the tension experienced is less compared to the inelastic string. This is because it has a higher strain value, which means a greater stretch than the inelastic string when the same force is applied. Thus, the same force is spread over a larger elastic band area than the nylon string. Secondly, the maximum tension occurs nearer to the rigid support as depicted by the axis of symmetry.
Variation in tension concerning the length of the string used. Graphs 4 and 5 both confirm my hypothesis that as the length of the string increases, the tension in the string decreases. Thus, we can come upon the inference that, Tension in a nonelastic string? However, the decrease in the tension is not uniform in the case of the inelastic string. As the elasticity increases, the difference in tension due to the length of the string is relatively less, which is shown by the tension trend of the elastic string. This happens because as the elasticity of the string increases, it stretches. Therefore, the tension force is also spread about a wider area—relation of tension to the strain values of strings.
The strain values have been calculated in the previous section, and it is found out that? nylon < ?band. By observing the changes in the trends of the two strings, is it possible to predict the tension behavior of strings having others? Values also. First, we shall take the change in the trend which takes place as we move from? Nylon to? Band (0.114 to 0.227)Using the observations made earlier, it would be safe to assume that as the strain value increases, that is to say, the string becomes more elastic, the tension values decrease. This is shown in the transition from graph 6 to graph 7.
Also, along with the decrease in the tension values, the curve stretches out (depicted by graph 8), which indicates that the range of tension (maximumminimum) decreases. Thus, we can say that the tension for a highly elastic string (? >0.227) would be expected to be roughly uniform irrespective of the point of application of force and the length of the string. Conversely, for highly inelastic strings (? < 0.114), the range of tension values is expected to be high, and also, there would be an exponential increase in the tension values with reducing lengths of the strings.
Evaluation. Error Analysis. The systematic error in the experiment is very insignificant. Compared to the tension calculated, the uncertainty values are shallow, which is why they are not visible in the graphs as error bars. The uncertainty for the tension values calculated varies with every observation. The minimum uncertainty is ï¿½0.001 N, and the maximum is ï¿½ 0.05 N. Besides, there are some random errors in the design of the experiment are as follows: –
 The spherical bob, which was used as a constant force, was not fully suspended in the air. Instead, some part of it rested on the wooden board, which may have slightly reduced the effective weight of the bob and thus the force. Thus, what I have taken as a force in my calculations may actually be a little greater than the actual force.
 The elastic band was wider than the string with which the bob was tied. Thus, the elastic band became a little twisted at the point of application of force. Yet care was taken to try and measure the vertical distance as far as possible, keeping the center point of the band as the reference point.
 The nylon string used was not straight, but it tended to incur folds at certain points. Although it was kept under force for quite some time to become uniform, all complete uniformity could not be achieved. This slight nonuniformity may have led to variation in the strain value of the string.
Acknowledgement. With the fulfillment of a project, especially from deep within the heart, reflecting the inner ambitions, there comes a joy unfathomable. Physics has and will always fascinate me with its simple intricacies. To sustain my interest in the pursuit of physical knowledge, the extended essay being a part of that pursuit, there are a few people I would like to thank: –
 My Physics teacher and Extended Essay supervisor, Mr. Venkataragavaraj, for being a mentor and the light to guide me on my way throughout my research and otherwise.
 The Lab Assistant, Mr. Sivakumar, for his utter sincerity and dedication towards helping me carry out the research effectively.
 My willpower, for I, had considered it an entity henceforth, for keeping me going when things got tough.
Bibliography. Books Referred
 Physics for Scientists and Engineers by Douglas Giancoli; PrenticeHall Publications. Third International Edition. Chapter 12, Page 300
 Advanced Physics by Tom Duncan and John Murray Fifth International Edition
Websites Referred
 Image 1. Severn Suspension Bridge was taken from http://en.wikipedia.org/wiki/Image:Suspension_bridgepanoramic.jpg Accessed on 30th August 2007
 http://www.matsuobridge.co.jp/english/bridges/basics/suspension.shtm Accessed on 15th May 2007
Appendix 1 – Calculation of Tension. Figure 3 indicates that the bob is at a certain random distance ‘x cm’ from the suspension point (nail) on the left. If the distance between the twopoint of suspensions on either side is taken to be ‘l,’ therefore, the distance between the point of application of force and the point of suspension on the right would be ‘l – x ‘. We can see that due to the application of force, the string gets divided into two parts. The vertical distance to both the left and the right point of suspensions is assumed to be ‘y1 cm’ and ‘y2 cm’, respectively. If we assume a horizontal line to exist at the point of application of force, we can see that an angle is subtended at the point of application of force on both sides.
The angle subtended by the string on the point of application of force to the point of suspension on the left is assumed to be ‘?1’ and to the point of suspension on the right is assumed to be ‘?2’. Finally, the Tension force exerted by the two parts of the string to their respective rigid supports is assumed to be ‘T1’ and ‘T2’, respectively. Using all these variables and applying the conditions for Static Equilibrium, I have found a general formula 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, three main forces 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: –
 F x = 0 (Sum of all horizontal forces is equal to 0)
 F y = 0 (Sum of all vertical forces is equal to 0)
It is also important to define the sign convention as we deal 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 ydirection shall be positive, whereas those acting from the negative x and ydirection shall be negative. To find out the formula, all the x forces are resolved as follows: –
 As ? F x = 0
 T1 cos?1 + T2cos?2 = 0
 Substituting cos?1 = and cos?2 = above
 T1 + T2 = 0
 T1 = T2 ————————————— (1)
 As ? F y = 0
 T1 sin?1 + T2 sin?2 – W = 0
 Substituting sin?1 = and sin?2 = 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