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    A phone line 2.0 m in diameter and 0.5 km long is stretched 0.5 m by a force of 4000 N. Determine:
    1. the stress on the wire
    2. the strain on the wire
    3.Young's modulus for the cable material.

    © BrainMass Inc. brainmass.com December 24, 2021, 5:15 pm ad1c9bdddf
    https://brainmass.com/physics/classical-mechanics/solving-an-elasticity-problem-38321

    SOLUTION This solution is FREE courtesy of BrainMass!

    d=diameter of line= 2 m
    Therefore cross sectional Area of line =pi d^2/4= 3.1416 m^2

    1. the stress on the wire
    Force= 4000 N
    Stress= Force/Area= 1273.2366 N/m^2 =4000/ 3.1416

    2. the strain on the wire

    strain = elongation / original length
    elongation= 0.5 m
    length = 0.5 km= 500 m

    Strain= 0.001 =0.5/ 500

    3.Young's modulus for the cable material.

    Young's modulus= Stress /Strain= 1,273,237 N/m^2 (or Pascal Pa)
    This can also be written as 1.27 Mpa

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:15 pm ad1c9bdddf>
    https://brainmass.com/physics/classical-mechanics/solving-an-elasticity-problem-38321

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