Answer to Question #125880 in Physics for Hum

Question

3. An iron rod 4 m long and 0.5 cm2 in cross-section stretches 1 mm when a mass of 225 kg is hung from its lower end. Compute Young's modulus for the iron.

Answer 1

The relation between stretch and load is given by the Hooke's law:

$\dfrac{mg}{A} = Y\dfrac{\Delta L}{L}$

where $m = 225kg$ is the mass of the load, $g = 9.81m/s^2$ is the constant, $\Delta L = 1mm = 10^{-3}m$ is the strech, $L = 4m$ is the length of the rod, $A = 0.5cm^2 = 0.5\cdot 10^{-4}m^2$ and $Y$ is the Young's modulus.

Expressing the Young's modulus, get:

$Y = \dfrac{mgL}{A\Delta L} = \dfrac{225\cdot 9.81\cdot4}{0.5\cdot 10^{-4}\cdot 10^{-3}} = 176.58\cdot 10^9Pa = 176.58GPa$

Answer. 176.58 GPa.