Question #220292

A 3.00 m long steel wire is fixed in a ceiling and a mass of 12.3 kg is hanging from the wire. The wire has a diameter d = 0.80 mm, a Young’s modulus 2.00 GPa and mass density 7860 kg/m3 .

(a) Calculate the stress in the wire.

(b) Calculate the strain.

(c) Calculate the elongation of the wire.

(d) Calculate the transverse speed of pulse.

(e) Calculate the longitudinal sound speed.


1
Expert's answer
2021-07-25T09:18:38-0400

Explanations & Calculations


  • Assume the thread's mass to be negligible compared to the hanging mass hence a light string.
  • Load applied to the wire = mg=12kg×9.8ms2=120.54N\small mg = 12kg\times9.8ms^{-2} = 120.54\,N
  • Area of the wire = πd24=π×(0.80×103m)24=5.03×107m2\small \pi \frac{d^2}{4} = \pi \times \frac{(0.80\times10^{-3}m)^2}{4}=5.03\times10^{-7}m^2

(a)

  • Strain is given by

σ=FA=120.54N5.03×107m2=2.396×1082.40×108Pa\qquad\qquad \begin{aligned} \small \sigma&=\small \frac{F}{A}\\ &=\small \frac{120.54\,N}{5.03\times10^{-7}m^2}\\ &=\small 2.396\times10^8\\ & \approx\small 2.40\times10^8\,Pa \end{aligned}

(b)

  • Using the stress-strain equation, the strain can be found

σ=Yϵϵ=2.40×108Pa2.00×109Pa=0.12\qquad\qquad \begin{aligned} \small \sigma&=\small Y\epsilon\\ \small \epsilon&=\small \frac{2.40\times10^8Pa}{2.00\times10^9Pa}\\ &=\small \bold{0.12} \end{aligned}

(c)

  • Once the strain is known, elongation can be calculated

ϵ=eLe=0.12×3.00m=0.36m\qquad\qquad \begin{aligned} \small \epsilon&=\small \frac{e}{L}\\ \small e&=\small 0.12\times3.00m\\ &=\small \bold{0.36 m} \end{aligned}

(d)

  • Transverse speed is given by

v=Tμ=Fμ=FAρ=120.54N5.03×107m2×7860kgm3=174.61ms1\qquad\qquad \begin{aligned} \small v&=\small \sqrt{\frac{T}{\mu}}=\sqrt{\frac{F}{\mu}}=\sqrt{\frac{F}{A\rho}}\\ &=\small \sqrt{\frac{120.54N}{5.03\times10^{-7}m^2\times7860kgm^{-3}}}\\ &=\small \bold{174.61\,ms^{-1}} \end{aligned}

(e)

  • Normally longitudinal speeds are described for stiff rod-like objects, and this thread could be thought similar to such a situation as it has some mass(own mass) & thickness.
  • Longitudinal speed is given by

vL=Eρ=2.00×109Pa7860kgm3=504.43ms1\qquad\qquad \begin{aligned} \small v_L&=\small \sqrt{\frac{E}{\rho}}\\ &=\small \sqrt{\frac{2.00\times10^9Pa}{7860kgm^{-3}}}\\ &=\small \bold{504.43\,ms^{-1}} \end{aligned}


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