After a fall, a 75 kg rock climber finds himself dangling from the end of a rope that had been 18 m long and 11 mm in diameter but has stret

Question

After a fall, a 75 kg rock climber finds himself dangling from the end of a rope that had been 18 m long and 11 mm in diameter but has stretched by 2.1 cm. For the rope, calculate (a) the strain, (b) the stress, and (c) the Young’s modulus.

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Trung Dũng 4 years 2021-07-25T13:46:25+00:00 1 Answers 22 views 0

Answers ( )

  1. Bo
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    2021-07-25T13:47:33+00:00
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    Answer:

    (A) Strain = 0.0012

    (b) Stress =77.44\times 10^5N/m^2

    (c) Young’s modulus =6.45\times 10^9N/m^2    

    Explanation:

    Mass of the rock m = 75 kg

    So weight of the rock F=mg=75\times 9.8=735N

    Length of the rope l = 18 m

    Diameter of the rope d = 11 mm

    Change in length of rope \Delta l=2.1cm =0.021m

    So radius r = 5.5 mm = 0.0055 m

    Cross sectional area A=\pi r^2

    A=3.14\times 0.0055^2=9.49\times 10^{-5}m^2

    (a) Strain is equal to ratio change in length to original length

    So strain =\frac{\Delta l}{l}=\frac{0.021}{18}=0.0012

    (b) Stress =\frac{Weight}{area}

    =\frac{735}{9.49\times 10^{-5}}=77.44\times 10^5N/m^2

    (c) Young’s modulus is equal to ratio of stress and strain

    So young’s modulus =\frac{77.44\times 10^5}{0.0012}=6.45\times 10^9N/m^2

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