When the mass is removed, the length of the cable is found to be l0 = 4.76 m. After the mass is added, the length is remeasured and found to

Question

When the mass is removed, the length of the cable is found to be l0 = 4.76 m. After the mass is added, the length is remeasured and found to be l1 = 5.43 m. Determine Young’s Modulus Y in N/m2 for the steel cable if the weight has a mass m = 35 kg and the cable has a radius r = 0.025 m.

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Gerda 5 years 2021-08-22T03:01:39+00:00 2 Answers 11 views 0

Answers ( )

  1. thienthi
    0
    2021-08-22T03:03:13+00:00
    Reply

    Answer:

    1242337 N/m^2

    Explanation:

    The cross section area of the cable is

    A = \pi r^2 = \pi * 0.025^2 = 0.002 m^2

    Let g = 9.81m/s2. The stress acting on the cable when mass is added is

    \sigma = F/A = mg/A = 35*9.81/0.002 = 174867 Pa

    The strain when the cable is stretched from 4.76 to 5.43 m is

    \epsilon = \frac{\Delta L}{L} = \frac{5.43 - 4.76}{4.76} = 0.14

    So the young modulus of the cable is

    E = \sigma / \epsilon = 174867 / 0.14 = 1242337 Pa = 1242337 N/m^2

  2. Eirian
    0
    2021-08-22T03:03:25+00:00
    Reply

    Answer:1265841.097N/m^2

    Explanation:LO = 4.76M

    L1 = 5.43M, young’s modulus (Y) = ?

    Mass (M) = 35kg, radius(R) = 0.025

    Young’s modulus = tensile stress/tensile strain

    But tensile stress = force(F)/area(A)

    A = the area of the circle of the cable = πr^2

    Weight of mass 35kg = force.  Hence W = my = ma

    g = 10m/s^2

    Force = 35*10 = 350N

    area = 3.142*0.025^2 = 3.142*0.000625 = 0.00196375m^2

    Tensile stress = 350/0.00196375 = 178230.4265N/m^2

    Tensile strain = extension/L0(original length)

    Extension = L1 – L0 = 5.43-4.76 = 0.67m

    Tensile strain = 0.67/4.76 = 0.1408

    Hence young modulus = 178230.4265/0.1408 = 1265841.097N/m^2

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