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• Level: GCSE
• Subject: Maths
• Word count: 1351

# Investigate the elastic properties of a strip of metal (hacksaw blade) and use the results to determine the value of Young's Modulus of the metal.

Extracts from this document...

Introduction

Baber Pervez 12K

Physics Coursework – Making sense of data

Aim

To investigate the elastic properties of a strip of metal (hacksaw blade) and use the results to determine the value of Young’s Modulus of the metal from the following experiment:

The Young’s Modulus, E, is given by:

a2  =         Ebd3   Cosθ        b = width of blade

6Mg        d = thickness of blade

g = acceleration due to gravity: 9.81ms-2

Young’s Modulus

For the description of the elastic properties of linear objects like wires, rods, columns which are either stretched or compressed, a convenient parameter is the ratio of the stress to the strain, a parameter called the Young's modulus of the material. Young's modulus can be used to predict the elongation or compression of an object as long as the stress is less than the yield strength of the material.

To obtain a suitable value for Young's Modulus, a graph needs to be plotted. We can see that the initial equation is in the form y = mx:

y = a2

m =          Ebd3

6Mg

x = Cosθ

From this we can say that the graph that is to be plotted will be a2

Middle

0.190

60

0.170

64

0.143

69.5

0.119

75.5

0.087

81

From this we needed to calculate the appropriate values which would be used in the graph. These are shown in the table below:

 a2 Cosθ 0.049729 0.71 0.042849 0.63 0.0361 0.50 0.0289 0.44 0.020449 0.35 0.014161 0.25 0.007569 0.16

From these calculations, the graph shown is over-leaf

As you can see from the graph three different lines have been drawn to represent the data: a line with maximum gradient, a best fit line and a line with minimum gradient.

The first line is the line with the maximum gradient according to the results. The following calculations were done in finding the maximum Young’s Modulus:

m        =        Δy

Δx

=        0.0361

0.5

=        0.0722

E  =        6Mgm

bd3

E        =        6 x 0.1 x 9.81 x 0.0722

0.01195 x 0.000683

=        1.131 x 1011 Pa

=        113.1 GPa

The second line is the line with the average gradient according to the results. The following calculations were done in finding the average Young’s Modulus:

m        =        0.0450

0.7

=        0.064285714

E        =        6 x 0.1 x 9.81 x 0.064285714

0.01195 x 0.000683

=        1.007 x 1011

Conclusion

θ taken would be incorrect as the temperature would have had considerable effect.

The results for the Young’s Modulus:

 Young’s Modulus (GPa) Maximum 113.1 Average 100.7 Minimum 74.1

From these results we can calculate the percentage error on the final values.

Percentage error        =        Average value – Minimum value         x       100

Average value

Percentage error        =         100.7 – 74.1         x       100

100.7

Percentage error        =        26.4151 %

This value is very similar compared with the errors in the gradient. This shows that the graph removed the errors which would be carried forward and multiplied. These percentage errors are due to human error in taking the measurements and plotting the graph.

In conclusion, the best fit line in the graph was a straight line through the origin which proves the initial conclusion that the graph will pass through the origin. The average value for the Young’s Modulus of the hacksaw blade was calculated to be 100.7 GPa which had an error of 26.4151 %. This error was accounted for by the reasons mentioned previously.

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

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