Chapter 9, Problem 51P

Part (a)

To determine

Whether the bond would break or not.

Answer to Problem 51P

Solution: Bone will not break.

Explanation of Solution

To know if the bone is going to break we need to find if the stress is bigger than the ultimate compressive strength of bone.

Given: Compressive force, $F=3.3\times {10}^4\text{N}$

Length of the bone, $=L=22\text{cm}=0.22\text{m}$

Cross-sectional area of the bone, $A=3.6{\text{ cm}}^{\text{2}}=3.6\times {10}^{-4}{\text{m}}^{\text{2}}$

Formula Used:

  ${\sigma }_{comp}=\frac{F}{A}$

Here, ${\sigma }_{comp}$ is the compressive stress, F is the applied force andA is the area of cross section.

Calculation:

  $\begin{array}{l}{\sigma }_{comp}=\frac{3.3\times 10^4}{3.6\times 10^{-4}}\\ {\sigma }_{comp}=0.916\times 10^{8}N/m^2\end{array}$

The compressive strength of bone is $1.70\times {10}^8{\text{N/ m}}^{\text{2}}$ . Above strength is less than compressive strength of bone.

Conclusion:Thus, bone will not break.

Part (b)

To determine

The amount by which the bone shortens.

Answer to Problem 51P

Solution: Bone is shortened by 1.34 mm.

Explanation of Solution

Explanation: Given: Compressive force, $F=3.3\times {10}^4\text{N}$

Length of the bone, $=L=22\text{cm}=0.22\text{m}$

Cross-sectional area of the bone, $A=3.6{\text{ cm}}^{\text{2}}=3.6\times {10}^{-4}{\text{m}}^{\text{2}}$

Formula Used:

Change in length can be found out by determining the deformation

  $\delta l=\frac{FL}{AE}$

  $\delta l$ is the deformation, F is the applied force, L is initial length, A is the area of cross section, and E is the Young’s modulus.Calculation: Where $\delta l$ is the change in length, F is the force, L is the length of bone, and E is the young modulus of bone which is 15 ×10⁹ N/ m²

  $\begin{array}{l}\delta l=\frac{FL}{AE}\\ \delta l=\frac{3.3\times 10^4\times 0.22}{3.6\times 10^{-4}\times 15\times 10^9}\\ \delta l=0.00134m\\ \delta l=1.34mm\end{array}$

Conclusion: Bone will not break. It is shortened by 1.34 mm.