Chapter 9, Problem 78QAP

To determine

(a)

To identify about who will suffer more transverse break in a bone in human or a cow.

Answer to Problem 78QAP

The human suffers more transverse break in a bone than a cow

Explanation of Solution

Given:

Ultimate stress in the transverse direction for human bone $=133\times {10}^{6}N/m^2$

Ultimate stress in the transverse direction for cow bone $=178\times {10}^{6}N/m^2$

Formula used:

  $Ultimatestrength=\frac{F_{\max }}{A_{\min }}$

Here, all alphabets are in their usual meanings.

Calculation:

Since, $Ultimatestrength\propto \frac{1}{A_{\min }}$

Here, human has less area of the cross-section than cow and their ultimate stresses in the transverse direction are nearly equal as compare to their area of the cross-sections.

Hence, human suffers more transverse break in a bone than a cow

Conclusion:

Thus, the human suffers more transverse break in a bone than a cow.

To determine

(b)

To explain, how cow's bones are much more capable of supporting their extreme weight than human.

Answer to Problem 78QAP

The cow's bones are much more capable of supporting their extreme weight than human because of high longitudinal ultimate stress and wider bone.

Explanation of Solution

Given:

Longitudinal yield stress for human bone $=182\times {10}^{6}N/m^2$

Longitudinal ultimate stress for human bone $=195\times {10}^{6}N/m^2$

Longitudinal yield stress for cow bone $=196\times {10}^{6}N/m^2$

Longitudinal ultimate stress for cow bone $=237\times {10}^{6}N/m^2$

Formula used:

  $Ultimatestrength=\frac{F_{\max }}{A_{\min }}$

Here, all alphabets are in their usual meanings.

Calculation:

Since, $Ultimatestrength\propto \frac{1}{A_{\min }}$

Here, cow has more area of the cross-section and wider bone (or thicker) than man but longitudinal ultimate stress for cow bone $is237\times {10}^{6}N/m^2$ which is more than the longitudinal ultimate stress for human bone $\left(195\times {10}^{6}N/m^2\right).$

Hence, the cow's bones are much more capable of supporting their extreme weight than human.

Conclusion:

Thus, the cow's bones are much more capable of supporting their extreme weight than human because of high longitudinal ultimate stress and wider bone.

To determine

(c)

The compressive force before the breaking the bone.

Compressed length of the bone.

Answer to Problem 78QAP

The compressive force before the breaking the bone is $4.0\times {10}^{4}N$.

Compressed length of the bone is $0.5mm$.

Explanation of Solution

Given:

Compressive ultimate stress for human bone $=133\times {10}^{6}N/m^2$

Longitudinal elastic modulus of human bone $Y=9.6\times {10}^{9}N/m^2$

Area of cross-section of woman's leg $A=3.00\times {10}^{-4}{m}^2$

Length of bone $L=35cm=35\times {10}^{-2}m$

Let $\Delta L$ be the compressed length and Let $F_{\max }$ be the compressed force

Formula used:

  $Ultimatecompressivestrength=\frac{compressedforce}{areaofcross-\sec tion}=\frac{F_{\max }}{A}------(i)$

Longitudinal elastic modulus $Y=\frac{\left(\frac{1}{10}{F}_{\max }\right)/A}{\Delta L/L}-------(ii)$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in above formula,

  $Ultimatecompressivestrength=\frac{F_{\max }}{A}$

  $\begin{array}{l}or,F_{\max }=\left(Ultimatecompressivestrength\right)\times A\\ or,F_{\max }=\left(133\times {10}^{6}N/m^2\right)\times \left(3.00\times {10}^{-4}{m}^2\right)\\ or,F_{\max }=4.0\times {10}^{4}N\end{array}$

Now, using equation (ii),

  $\text{Y}=\frac{\left(\frac{1}{10}{F}_{\max }\right)/A}{\Delta L/L}$

  $\begin{array}{l}or,\Delta L=\left(\frac{F_{\max }}{10A}\right)\times \frac{L}{\text{Y}}\\ or,\Delta L=\left(\frac{4.0\times {10}^{4}N}{10\times \left(3.00\times {10}^{-4}{m}^2\right)}\right)\times \frac{\left(35\times {10}^{-2}m\right)}{\left(9.6\times {10}^{9}N/m^2\right)}\\ or,\Delta L=4.9\times {10}^{-4}m=0.5mm\end{array}$

Hence, the compressive force before the breaking the bone is $4.0\times {10}^{4}N$.

Compressed length of the bone is $0.5mm$.

Conclusion:

Thus, the compressive force before the breaking the bone is $4.0\times {10}^{4}N$.

Compressed length of the bone is $0.5mm$.