Question #138581
What load in kilograms must be applied to a steel wire 6 m long and diameter 1.6 mm to produce an extension of 1 mm? (Young’s modulus for steel = 20 x 10^11 Nm^-2)
1
Expert's answer
2020-10-27T11:26:57-0400

Since Young's modulus is E=FSΔllE =\frac {\frac {F} {S}} {\frac {\Delta l} {l}} , where FF is the force of action on the body, SS is the cross-sectional area, ll is the body length ,Δl\Delta l is the extension, then

F=E×S×Δll=20×1011×π×(1.6×1032)2×1036670.21F=\frac{E\times S \times \Delta l}{l}=\frac{20\times10^{11}\times \pi\times (\frac{1.6\times 10^{-3}}{2})^2 \times 10^{-3}}{6}\approx670.21 N.


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