Chapter 1, Problem 1.104QP

Interpretation Introduction

Interpretation:

The given tire pressure is $\text{32}{\text{lb/in}}^{\text{2}}$.  It has to be converted into ${\text{g/cm}}^{\text{2}}$.

Explanation of Solution

Given, $\text{32}{\text{lb/in}}^{\text{2}}$.

The conversion factors are used to convert into ${\text{g/cm}}^{\text{2}}$.

The conversion factor used to convert pound into gram is

  $\begin{array}{l}\text{1}\text{lb}\text{=}\text{454}\text{g}\\ \text{=}\frac{\text{454}\text{g}}{\text{1}\text{lb}}\end{array}$

The conversion of inch to centimeter can be given as

  $\begin{array}{l}\text{12}\text{in}\text{=}\text{1}\text{ft;}\text{1}\text{yd}\text{=}\text{3}\text{ft;}\text{1}\text{yd}\text{=}\text{0}\text{.914}\text{m;}\text{1}\text{cm}\text{=}{\text{10}}^{\text{-2}}\text{m}\\ \text{conversion}\text{factor:}\\ \text{=}\frac{\text{12}\text{in}}{\text{1}\text{ft}}\text{×}\frac{\text{3}\text{ft}}{\text{1}\text{yd}}\text{×}\frac{\text{1}\text{yd}}{\text{0}\text{.914}\text{m}}\text{×}\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1}\text{cm}}\end{array}$

The conversion of $\text{32}{\text{lb/in}}^{\text{2}}$ into ${\text{g/cm}}^{\text{2}}$ is given as

  $\begin{array}{l}\text{=}\frac{\text{32lb}}{{\text{in}}^{\text{2}}}\text{×}\frac{\text{454}\text{g}}{\text{1}\text{lb}}\text{×}\frac{{\left(\text{12}\text{in}\right)}^{\text{2}}}{{\left(\text{1}\text{ft}\right)}^{\text{2}}}\text{×}\frac{{\left(\text{3}\text{ft}\right)}^{\text{2}}}{{\left(\text{1}\text{yd}\right)}^{\text{2}}}\text{×}\frac{{\left(\text{1}\text{yd}\right)}^{\text{2}}}{{\left(\text{0}\text{.914}\text{m}\right)}^{\text{2}}}\text{×}\frac{{\left({\text{10}}^{\text{-2}}\text{m}\right)}^{\text{2}}}{{\left(\text{1}\text{cm}\right)}^{\text{2}}}\\ \text{=}\frac{\text{32}\text{lb}}{{\text{in}}^{\text{2}}}\text{×}\frac{\text{454}\text{g}}{\text{1}\text{lb}}\text{×}\frac{\text{144}{\text{in}}^{\text{2}}}{\text{1}{\text{ft}}^{\text{2}}}\text{×}\frac{\text{9}{\text{ft}}^{\text{2}}}{\text{1}{\text{yd}}^{\text{2}}}\text{×}\frac{\text{1}{\text{yd}}^{\text{2}}}{\text{0}\text{.835}{\text{m}}^{\text{2}}}\text{×}\frac{{\text{10}}^{\text{-4}}{\text{m}}^{\text{2}}}{\text{1}{\text{cm}}^{\text{2}}}\\ \text{=}\frac{\text{1882}\text{.82}\text{g}}{\text{0}\text{.835}{\text{cm}}^{\text{2}}}\\ \text{=}\text{2254}\text{.8}{\text{g/cm}}^{\text{2}}\end{array}$

As the given value consists of two significant figures, the final converted value should be in two significant figures.  The final value is greater than five, so the second value is increased by one.  The rounded off value is $\text{2300}{\text{g/cm}}^{\text{2}}$.