Answer to Question #138581 in Electric Circuits for Halyufer

Question

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)

Answer 1

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

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