Answer:
[tex]F_{max}=2.95*10^{5}N[/tex]
[tex]\Delta l=0.25cm[/tex]
Explanation:
E=1.50x10^10 N/m2 Young's modulus of bone
σmax=1.50x10^8 N/m2 Max stress tolerated by the bone
Relation between stress and Force:
[tex]\sigma=\frac{F}{A}=\frac{F}{\pi*d^{2}/4}[/tex]
[tex]F_{max}=\sigma_{max}*\pi*d^{2}/4}=1.50*10^{8}*\pi*(2.5*10^{-2})^{2}=2.95*10^{5}N[/tex]
Relation between stress and strain:
Young's modulus is defined by the ratio of longitudinal stress σ , to the longitudinal strain ε:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]\epsilon=\frac{\Delta l}{l}[/tex]
We solve these equations to find the bone compression:
[tex]\Delta l=l*\frac{\sigma}{E}=25*\frac{1.50*10^{8}}{1.50*10^{10}}=0.25cm[/tex]
