Question

The effective cross sectional area of a horse's femur is 7×10^-4m^2 and the Young's modulus of this bone is 8.3×10^9pa.Calculate the strain that occurs in the femur when the horse of mass 600kg puts it's full weights on it

Answer 1

The Hooke's law states

$$\sigma=\varepsilon E$$

The stress is given by relation

$$\sigma=\frac{F}{A}$$

So, strain

$$\varepsilon=\frac{\sigma}{E}=\frac{F}{AE}=\frac{mg}{AE}$$

$$=\frac{600\:\rm{kg}\times 9.8\:\rm{N/kg}}{7\times 10^{-4}\:\rm{m^2}\times 8.3\times 10^9\:\rm{Pa}}$$

$$=0.001\:\rm{m}=1\:\rm{mm}$$