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The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

  • diameter of the steel, d = 4 mm
  • the radius of the wire, r = 2mm = 0.002 m
  • original length of the wire, L₁
  • final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
  • extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
  • the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

[tex]A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2[/tex]

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

[tex]E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}[/tex]

[tex]F = \frac{200 \times 10^9\ \times\ 1.257\times 10^{-5}\ \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N[/tex]

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

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