\u0394 L n L mm Strain \u03b5 L L \u0394 N m 2 Stress \u03c3 A N m F 2 1 98 N 1323 mm 011mm 00084

Δ l n l mm strain ε l l δ n m 2 stress σ a n m f

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Δ = L n L 0 (mm) Strain ε) ( = L / L Δ N / m ) ( 2 Stress = σ) ( / A ( N / m ) F 2 1 9.8 N 13.23 mm 0.11mm 0.0084 3.141 2 19.6 N 13.35 mm 0.23 mm 0.0175 6.28 3 29.4 N 13.46 mm 0.34 mm 0.0259 9.42 4 39.2 N 13.56 mm 0.44 mm 0.0335 12.56 5 49 N 13.67 mm 0.55 mm 0.0419 15.71 6 58.8 N 13.79 mm 0.67 mm 0.0511 18.85 7 68.6 N 13.90 mm 0.78 mm 0.0594 21.98 Table II Weight (kg) Applied Force (N) (mm) L 0 L Δ = L n L 0 (mm) Strain ε) ( = L / L Δ N / m ) ( 2 Stress = σ) ( / A ( N / m ) F 2 7 68.6 N 13.90 mm 0.78 mm 0.0594 21.98 6 58.8 N 13.79 mm 0.67 mm 0.0511 18.85 5 49 N 13.69 mm 0.57 mm 0.0434 15.71 4 39.2 N 13.56 mm 0.44 mm 0.0335 12.56 3 29.4 N 13.46 mm 0.34 mm 0.0259 9.42 2 19.6 13.35 mm 0.23 0.0175 6.28 1 9.8 13.23 mm 0.11 0.0084 3.141 3.2 Calculations: L Δ = L n L 0 ε = L / L Δ / A σ = F L 3.23 3.12 Δ = 1 − 1 .11/13.12 ε = 0 .8/3.12 σ = 9 L .11 mm Δ = 0 .0084 N / m ε = 0 2 .141 N / m σ = 3 2 4. Analysis and Discussion
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By utilizing the equations pertaining to Hooke’s Law, we were able to calculate the Young Modulus of Elasticity. In order to get the modulus, we needed to find the stress and strain for each trial on the string. The ratio of stress over strain will give us the modulus, which was the main objective of the lab. Discussion and Questions 1. When the elastic limit is not exceeded, the deformation is recoverable so comparing both cases of increasing and decreasing the hanging weights, the strain should be the same for the same stress. That is the Young’s Modulus should be the same for both cases. Do you observe this?
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  • Spring '17
  • HalinaOprychal
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