Chapter 9, Problem 77GP

(a)

To determine

To Find: The stress in the tibia of a leg.

Answer to Problem 77GP

The stress in the tibia of the leg is $\text{ 7 }\times {\text{ 10}}^{\text{8}}{\text{ N/m}}^{\text{2}}$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{Impulse force = }-\text{540 N}\cdot \text{s}\\ \text{Average force = 2}\text{.1}\times {\text{10}}^{\text{5}}\text{ N}\\ \text{Leg bone area A = 3}\times {\text{10}}^{\text{-4}}{\text{ m}}^{\text{2}}\end{array}$

Formula used:

Stress in the tibia of the leg is determined by the following formula,

  $\text{Stress = }\frac{\text{Force(Average)}}{\text{Area}}$

Calculation:

Substitute the given values,

  $\begin{array}{l}\text{Stress = }\frac{\text{2}\text{.1}\times {\text{10}}^{\text{5}}\text{ N}}{\text{3}\times {\text{10}}^{\text{-4}}{\text{ m}}^{\text{2}}}\\ \text{Stress = 7}\times {\text{10}}^{\text{8}}{\text{ N/m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the stress in the tibia of the leg is $\text{7}\times {\text{10}}^{\text{8}}{\text{ N/m}}^{\text{2}}$ .

(b)

To determine

To explain: Whether the bone will break or not.

Explanation of Solution

Introduction:

The strength of the bone of the leg is basically depends on the stress that the tibia of the leg received.

The ultimate (maximum) strength of the bone is equal to $1.7\times {10}^8{\text{Nm}}^{\text{-2}}$ Which means that, bone can sustained the stress up to $1.7\times {10}^8{\text{Nm}}^{\text{-2}}$ . But, when the stress in the tibia of the leg is higher than the ultimate strength of the bone then the bone cannot sustained the load and it will be break.

Conclusion:

Here, the stress in the tibia of the leg is $7\times {10}^8{\text{Nm}}^{\text{-2}}$ which is higher than the ultimate strength of the bone. So, in this case bone will break.

(c)

To determine

To find: The stress in the tibia of a leg.

To explain: Whether the bone will break or not.

Answer to Problem 77GP

The stress in the tibia of the leg is $\text{1}\text{.6 }\times {\text{10}}^7{\text{ Nm}}^{\text{-2}}$ .

The bone will not break.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Impulse force = -540 N}\text{.s}\\ \text{Average force = 4}\text{.9}\times {\text{10}}^{\text{3}}\text{ N}\\ \text{Leg bone area A = 3}\times {\text{10}}^{\text{-4}}{\text{ m}}^{\text{2}}\\ \text{d = 50 cm = 0}\text{.5 m}\end{array}$

Formula used:

Stress in the tibia of the leg is determined by the following formula,

  $\text{Stress = }\frac{\text{Force(Average)}}{\text{Area}}$

Calculation:

Substitute the given values,

  $\begin{array}{l}\text{Stress = }\frac{\text{4}\text{.9}\times {\text{10}}^{\text{3}}\text{ N}}{\text{3}\times {\text{10}}^{\text{-4}}{\text{ m}}^{\text{2}}}\\ \text{Stress = 1}\text{.6}\times {\text{10}}^{\text{7}}{\text{ N/m}}^{\text{2}}\end{array}$

The strength of the bone of the leg is basically depends on the stress that the tibia of the leg received. The ultimate (maximum) strength of the bone is equal to $1.7\times {10}^8{\text{Nm}}^{\text{-2}}$ . Which means that, bone can sustained the stress up to $1.7\times {10}^8{\text{Nm}}^{\text{-2}}$ . But, when the stress in the tibia of the leg is smaller than the ultimate strength of the bone then the bone can sustained the load and it will not break.

Conclusion:

Here, the stress in the tibia of the leg is $\text{1}\text{.6 }\times {\text{10}}^7{\text{ Nm}}^{\text{-2}}$ which is lower than the ultimate strength of the bone. So, in this case bone will not break.