Tag Archives: represent number using appropriate prefix

Hibbeler Statics 14E P1.10 — Representing Combinations of Units in the Correct SI Form

_{Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-10}

Solution:

Part A

\begin{align*} \text{GN} \cdot \mu \text{m} & = \left( 10^9 \ \text{N} \right)\left( 10^{-6} \ \text{m} \right)\\ & = 10^3 \ \text{N} \cdot \text{m}\\ & = \text{kN} \cdot \text{m} \end{align*}

Part B

\begin{align*} \text{kg/}\mu\text{m} & = \frac{10^3 \ \text{g}}{10^{-6} \ \text{m}} \\ & = 10^9 \ \frac{\text{g}}{\text{m}} \\ & = \text{Gg/m} \end{align*}

Part C

\begin{align*} \text{N/ks}^2 & = \frac{\text{N}}{\left( 10^3 \ \text{s} \right)^2}\\ & = \frac{\text{N}}{10^6 \ \text{s}^2} \\ & = 10^{-6} \ \frac{\text{N}}{\text{s}^2} \\ & = \mu \text{N}/\text{s}^2 \end{align*}

Part D

\begin{align*} \text{kN}/ \mu\text{s} & = \frac{10^3 \ \text{N}}{10^{-6} \ \text{s}} \\ & = 10^9 \ \frac{\text{N}}{\text{s}}\\ & = \text{GN/s} \end{align*}

_{Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-8}

Solution:

Part A

\begin{align*} \text{Mg/mm} & = \frac{10^6 \ \text{g}}{10^{-3} \ \text{m}} \\ & = 10^9 \ \frac{\text{g}}{\text{m}}\\ & = \text{Gg/m} \end{align*}

Part B

\begin{align*} \text{mN/} \mu \text{s} & = \frac{10^{-3} \ \text{N}}{10^{-6} \ \text{s}} \\ & = 10^3 \ \frac{\text{N}}{\text{s}} \\ & = \text{kN/s} \end{align*}

Part C

\begin{align*} \mu \text{m}\cdot \text{Mg} & = \left( 10^{-6} \ \text{m} \right)\left( 10^6 \ \text{g} \right) \\ & = \text{m}\cdot \text{g}\\ \text{This can also be written as:} \\ & = \text{mm} \cdot \text{kg} \end{align*}

_{Statics of Rigid Bodies 14^th by RC Hibbeler, Problem 1-5}

Solution:

Part A

\begin{align*} 45 \ 320 \ \text{kN} & = 45.3 \times 10^3 \ \text{kN}\\ & = 45.3\times 10^3\times 10^3 \ \text{N}\\ & = 45.3 \times 10^6 \ \text{N}\\ & = 45.3 \ \text{MN} \end{align*}

Part B

\begin{align*} 568\left( 10^5 \right)\ \text{mm} & =568\left( 10^5 \right)\times 10^{-3} \ \text{m}\\ & = 568\times 10^2 \ \text{m} \\ & = 56.8 \times 10^3 \ \text{m}\\ & = 56.8 \ \text{km} \end{align*}

Part C

\begin{align*} 0.00563 \ \text{mg} & = 5.63\times 10^{-3} \ \text{mg}\\ & = 5.63\times 10^{-3}\times 10^{-3} \ \text{g}\\ & = 5.63\times 10^{-6} \ \text{g}\\ & = 5.63 \ \mu\text{g} \end{align*}