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Deformation of Solids: Question 3

Syllabus 6.1

Structured AS 8 marks

A technician investigates a metal wire of original length 2.50 m2.50\text{ m} and uniform diameter 0.80 mm0.80\text{ mm}. One end of the wire is clamped and the wire hangs vertically. A load of 45 N45\text{ N} is hung from the free end, producing an extension of 1.5 mm1.5\text{ mm}. The wire does not exceed its limit of proportionality.

(a) Calculate the cross-sectional area of the wire. [2]

(b) Calculate the stress in the wire when the 45 N45\text{ N} load is applied. [2]

(c) Calculate the strain in the wire when the 45 N45\text{ N} load is applied. [2]

(d) Determine the Young modulus of the material of the wire. [2]

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Worked solution

Part (a): Cross-sectional area

The wire has diameter d=0.80 mmd = 0.80\text{ mm}, so its radius is: r=d2=0.80×10−32=0.40×10−3 m=4.0×10−4 mr = \frac{d}{2} = \frac{0.80\times10^{-3}}{2} = 0.40\times10^{-3}\text{ m} = 4.0\times10^{-4}\text{ m}

The cross-sectional area of a wire (a circle) is: A=πr2=π×(4.0×10−4)2=π×1.6×10−7A = \pi r^2 = \pi \times (4.0\times10^{-4})^2 = \pi \times 1.6\times10^{-7} A=5.03×10−7 m2A = 5.03\times10^{-7}\text{ m}^2

Part (b): Stress

Stress is defined as force per unit cross-sectional area: σ=FA=455.03×10−7\sigma = \frac{F}{A} = \frac{45}{5.03\times10^{-7}} σ=8.95×107 Pa\sigma = 8.95\times10^{7}\text{ Pa}

Part (c): Strain

Strain is defined as extension per unit original length. First convert the extension to metres: x=1.5 mm=1.5×10−3 mx = 1.5\text{ mm} = 1.5\times10^{-3}\text{ m}

Then: ε=xL=1.5×10−32.50\varepsilon = \frac{x}{L} = \frac{1.5\times10^{-3}}{2.50} ε=6.0×10−4\varepsilon = 6.0\times10^{-4}

(Strain is a ratio of two lengths, so it has no units.)

Part (d): Young modulus

The Young modulus is the ratio of stress to strain, within the limit of proportionality: E=σε=8.95×1076.0×10−4E = \frac{\sigma}{\varepsilon} = \frac{8.95\times10^{7}}{6.0\times10^{-4}} E=1.49×1011 PaE = 1.49\times10^{11}\text{ Pa}

As a check, using E=FLAxE = \dfrac{FL}{Ax} directly: E=45×2.505.03×10−7×1.5×10−3=112.57.54×10−10=1.49×1011 PaE = \frac{45 \times 2.50}{5.03\times10^{-7} \times 1.5\times10^{-3}} = \frac{112.5}{7.54\times10^{-10}} = 1.49\times10^{11}\text{ Pa} Both methods agree.

Final answers

  • (a) Cross-sectional area =5.03×10−7 m2= \boxed{5.03\times10^{-7}}\text{ m}^2
  • (b) Stress =8.95×107 Pa= \boxed{8.95\times10^{7}}\text{ Pa}
  • (c) Strain =6.0×10−4= \boxed{6.0\times10^{-4}} (no units)
  • (d) Young modulus =1.49×1011 Pa= \boxed{1.49\times10^{11}}\text{ Pa}