Question #141131
A 5cm² cord has original length of 120cm is pulled by a 20N force. The change in length in the cord is 5mm. Find the strain, stress and elastic modulus.
1
Expert's answer
2020-10-30T13:19:30-0400

By definition, the stress is the ratio of a force to a cross-sectional area:


σ=FA=20N5×104m2=4×104Pa\sigma = \dfrac{F}{A} = \dfrac{20N}{5\times 10^{-4}m^2} = 4\times 10^4Pa

The strain is the ratio of a length change to the original length:


ε=ΔLL=0.5cm120cm4.17×103\varepsilon = \dfrac{\Delta L}{L} = \dfrac{0.5cm}{120cm} \approx 4.17\times 10^{-3}

The Hook's law states that:


σ=Eε\sigma = E\varepsilon

where EE is the elastic modulus. Thus:


E=σε=4×104Pa4.17×103=9.6×106Pa=9.6 GPaE = \dfrac{\sigma}{\varepsilon} = \dfrac{4\times 10^4Pa}{4.17\times 10^{-3}} = 9.6\times 10^6Pa = 9.6\space GPa

Answer. stress = 4×104Pa4\times 10^4Pa, strain = 4.17×1034.17\times 10^{-3}, elastic modulus = 9.6×106Pa9.6\times 10^6Pa.


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