Chapter 13, Problem 26SP

The sole of a man’s size-10 shoe is around 11.0 in. by 4.00 in. Determine the gauge pressure under the feet of a 200-lb man standing upright. Give your answer in both $\text{lb/in}{\text{.}}^{\text{2}}$ and Pa. [Hint: $1.00{\text{ lb/in}}^{\text{2}}=6895\text{ Pa}$. Check your work using $1.00\text{ in}{\text{.}}^{\text{2}}=6.45\times {10}^{–4}{\text{ m}}^{\text{2}}$ and $1.00\text{ lb}=4.448\text{ N}$.]

To determine

The gauge pressure under the feet of a 200-lb man standing upright if the size of the man’s sole is around 11.0 in. by 4.00 in.

Answer to Problem 26SP

Solution:

$2.27{\text{ lb/in}}^{\text{2}}$, $15.7\text{ kPa}$

Explanation of Solution

Given data:

The weight of the person is $200\text{ lb}$.

The length of each foot is $11\text{ in}$

The breadth of each foot is $\text{4 in}$

Formula used:

Write the expression for pressure exerted by the shoe:

$P=\frac{F}{A}$

Here, $P$ is the pressure exerted, $F$ is the force applied, and $A$ is the area on which the force is applied.

Write the expression of the area of one shoe:

$A=lb$

Here, $A$ is the area, $l$ is the length, and $b$ is the breadth.

Explanation:

Recall the expression of the area of one shoe:

$A=lb$

Here, $A$ is the area of one shoe, $l$ is the length of one shoe, and $b$ is the breadth of one shoe.

Substitute $11\text{ in}$ for $l$ and $4\text{ in}$ for $b$

$\begin{array}{l}A=\left(11\text{ in}\right)\left(4\text{ in}\right)\\ =44i{n}^2\end{array}$

Here, the force applied by both feet is equal to the weight of the person which is equally divided on both feet. Hence, the force on one foot will be $\frac{F}{2}$.

Thus, the expression of gauge pressure under the feet of the man standing upright will be:

$P=\frac{F}{2A}$

Here, $P$ is the gauge pressure exerted by the standing man, $F$ is the weight of the man, and $A$ is the area of one foot.

Substitute $200\text{ lb}$ for $F$ and $44{\text{ in}}^{\text{2}}$ for $A$

$\begin{array}{c}P=\frac{200\text{ lb}}{2\left(44{\text{ in}}^2\right)}\\ =2.27{\text{ lb/in}}^{\text{2}}\end{array}$

Convert the pressure from ${\text{lb/in}}^{\text{2}}$ to $\text{Pa}$

$\begin{array}{c}P=\left(2.27{\text{ lb/in}}^2\right)\left(\frac{6895\text{ Pa}}{1{\text{ lb/in}}^2}\right)\\ =\left(15,651.65\text{ Pa}\right)\left(\frac{1\text{ kPa}}{1000\text{ Pa}}\right)\\ =15.65\text{ kPa}\\ \simeq 15.7\text{ kPa}\end{array}$

Conclusion:

Therefore, the gauge pressure under the feet of a 200-lb man standing upright is $2.27{\text{ lb/in}}^{\text{2}}$ or $15.7\text{ kPa}$.