Question

A $40kg$ boy whose leg bones are $4c{m}^2$ in the area and $50cm$ long falls through a height of $50cm$ without breaking his leg bones. If the bones can stand a stress of $0.9\times {10}^{8}N{m}^{-2}$, calculate Young's modulus for the material of the bone. Take $g=10m{s}^{-2}$.

Answer 1

Given: $Mass=m=40kg$

$Area=A=4cm²=0.0004m²$

$l=50cm=0.5m$

$h=50cm=0.5m$

Now

$\text{Young's modulus}=Y=\frac{Stress}{Strain}$

$Y=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$ (1)

As

$mgh=\frac{1}{2}F\Delta L$ where F is force applied

$\Delta L=\frac{2mgh}{F}$ (2)

Substituting (2) in (1)

$Y=\frac{\frac{F}{A}}{\frac{\frac{2mgh}{F}}{L}}$

$Y=\frac{F^{2}L}{2mghA}$

$Y=\frac{(0.9\ast {10}^8{)}^2\ast 0.5}{2\ast 40\ast 10\ast 0.5\ast 0.0004}=2.5\times {10}^{6}N/m^2$