Chapter 7, Problem 15P

To determine

Calculate the average pressure at the bottom of the women’s high-heeled dress shoe and a women’s athletic walking shoe.

Answer to Problem 15P

The average pressure at the bottom of the women’s high-heeled dress shoe is $\underset{\_}{2.17\text{psi}}$.

The average pressure at the bottom of the women’s athletic walking shoe is $\underset{\_}{1.67\text{psi}}$.

Explanation of Solution

Given information:

Weight of the women is $W=120\text{lb}$

Calculation:

The weight of the women is carried by both shoes. Hence, the weight (Force) acting on each shoe is as follows:

$\begin{array}{c}F=\frac{W}{2}\\ =\frac{120}{2}\\ =60\text{lb}\end{array}$

Sketch the profile of high-heeled dress shoe in inches as shown in Figure 1.

Refer to Figure 1.

The profile of contact area is divided into two equal parts and each part is divided into 8 trapezoids of equal heights.

Consider the area of top portion as $A_1$ and the bottom portion as $A_2$.

Apply trapezoidal rule as shown below.

$A=h\left[\frac{1}{2}{y}_0+y_1+y_2+\cdots +y_{n-2}+y_{n-1}+\frac{1}{2}{y}_n\right]$ (1)

Calculate the area of top portion $\left(A_1\right)$ using Equation (1) as shown below.

$\begin{array}{c}{A}_1=1\times \left[\begin{array}{l}\frac{1}{2}\left(0\right)+1.12+1.37+1.25+1.75\\ +2.12+2.12+2+1.87+\frac{1}{2}\left(0\right)\end{array}\right]\\ =13.6\text{in}{\text{.}}^2\end{array}$

Calculate the area of bottom portion $\left(A_2\right)$ using Equation (1) as shown below.

$\begin{array}{c}{A}_2=1\times \left[\begin{array}{l}\frac{1}{2}\left(0\right)+1.5+1.62+1.75+2.12\\ +2.12+2+1.62+1.25+\frac{1}{2}\left(0\right)\end{array}\right]\\ =13.98\\ \simeq 14\text{in}{\text{.}}^2\end{array}$

Calculate the total area of high-heeled dress shoe as shown below.

$A=A_1+A_2$

Substitute $13.6\text{in}{\text{.}}^2$ for $A_1$ and $14\text{in}{\text{.}}^2$ for $A_2$.

$\begin{array}{c}A=13.6+14\\ =27.6\text{in}{\text{.}}^2\end{array}$

Calculate the average pressure at the bottom of high-heeled dress shoe as shown below.

$\text{Pressure}=\frac{\text{Force}\left(F\right)}{\text{Area}\left(A\right)}$ (2)

Substitute $60\text{lb}$ for F and $27.6\text{in}{\text{.}}^2$ for A in Equation (2).

$\begin{array}{c}\text{Pressure}=\frac{60}{27.6}\\ =2.17\text{psi}\end{array}$

Hence, the average pressure at the bottom of the women’s high-heeled dress shoe is $\underset{\_}{2.17\text{psi}}$.

Sketch the profile of athletic walking shoe in inches as shown in Figure 2.

Refer to Figure 2.

The profile of contact area is divided into two equal parts and each part is divided into 12 trapezoids of equal heights.

Consider the area of top portion as $A_1$ and the bottom portion as $A_2$.

Calculate the area of top portion $\left(A_1\right)$ using Equation (1) as shown below.

$\begin{array}{c}{A}_1=1\times \left[\begin{array}{l}\frac{1}{2}\left(0\right)+1.12+1.37+1.25+1+0.87+1.12\\ +1.75+2.12+2.12+2+1.87+\frac{1}{2}\left(0\right)\end{array}\right]\\ =16.56\\ \simeq 16.6\text{in}{\text{.}}^2\end{array}$

Calculate the area of bottom portion $\left(A_2\right)$ using Equation (1) as shown below.

$\begin{array}{c}{A}_2=1\times \left[\begin{array}{l}\frac{1}{2}\left(0\right)+1.5+1.62+1.75+1.62+1.75+2\\ +2.12+2.12+2+1.62+1.25+\frac{1}{2}\left(0\right)\end{array}\right]\\ =19.35\\ \simeq 19.4\text{in}{\text{.}}^2\end{array}$

Calculate the total area of athletic walking shoe as shown below.

$A=A_1+A_2$

Substitute $16.6\text{in}{\text{.}}^2$ for $A_1$ and $19.4\text{in}{\text{.}}^2$ for $A_2$.

$\begin{array}{c}A=16.6+19.4\\ =36\text{in}{\text{.}}^2\end{array}$

Calculate the average pressure at the bottom of athletic walking shoe as shown below.

Substitute $60\text{lb}$ for F and $36\text{in}{\text{.}}^2$ for A in Equation (2).

$\begin{array}{c}\text{Pressure}=\frac{60}{36}\\ =1.67\text{psi}\end{array}$

Therefore, the average pressure at the bottom of the women’s athletic walking shoe is $\underset{\_}{1.67\text{psi}}$.