Category Archives: Engineering Mathematics Blog

Expressing units in the correct SI form using an appropriate prefix

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) kN/μs, (b) Mg/mN, and (c) MN/(kg•ms).

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-2

Solution:

Part A

\begin{align*}
\text{KN}/\mu\text{s} & = \frac{\left( 10 \right)^3\ \text{N}}{\left( 10 \right)^{-6}\ \text{s}} \\
& =\left( 10 \right)^9 \ \text{N/s}\\
& = \text{GN/s}
\end{align*}

Part B

\begin{align*}
\text{Mg/mN} & =\frac{\left(10^6\right)\text{g}}{\left(10^{-3}\right)\text{N}}\\
& = 10^9\:\text{g/N}\\
& =\text{Gg/N}
\end{align*}

Part C

\begin{align*}
\text{MN}/\left(\text{kg}\cdot \text{ms}\right) & =\frac{10^6\:\text{N}}{\text{kg}\cdot \left(10^{-3}\right)\text{s}}\\
& =10^9\:\frac{\text{N}}{\text{kg}\cdot \text{s}}\\
& =\text{GN}/\left(\text{kg}\cdot \text{s}\right)
\end{align*}
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Problem 1.2 – Expressing density to SI units

Wood has a density of 4.70 slug/ft3. What is its density expressed in SI units?

Solution:

\displaystyle 4.70\:\text{slug/ft}^3\times \left[\frac{\left(1\:\text{ft}^3\right)\left(14.59\:\text{kg}\right)}{\left(0.3048\:\text{m}\right)^3\:\left(1\:\text{slug}\right)}\right]=2.42\:\text{Mg/m}^3

Rounding off to 3 significant figures

Round off the following numbers to three significant figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, and (d) 7555 kg.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-6

Solution:

Part A

\begin{align*}
58 \ 342 \ \text{m} & = 58.342\times 10^{3} \ \text{m}\\
& = 58.342 \ \text{km}\\
& = 58.3 \ \text{km}
\end{align*}

Part B

\begin{align*}
68.534 \ \text{s} & = 68.5 \ \text{s}
\end{align*}

Part C

\begin{align*}
2553 \ \text{N} & = 2.553 \ \text{kN}\\
& = 2.55 \ \text{kN}
\end{align*}

Part D

\begin{align*}
7555 \ \text{kg} & = 7.555\times 10^3 \ \text{kg}\\
& = 7.555 \times 10^3 \times 10^3 \ \text{g}\\
& = 7.555 \times 10^6 \ \text{g} \\
& = 7.555 \ \text{Mg}\\
& = 7.56 \ \text{Mg}
\end{align*}

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 5

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \infty }\left(\frac{8x-5}{\sqrt{4x^2+3}}\right)

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SOLUTION:

Divide by the highest denominator power

\begin{align*}

\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8x-5}{\sqrt{4x^2+3}}\right) & =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8x-5}{\sqrt{4x^2+3}}\cdot \displaystyle \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\right) \\
\\

& =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{8x}{x}-\displaystyle \frac{5}{x}}{\sqrt{\displaystyle \frac{4x^2}{x^2}+\displaystyle \frac{3}{x^2}}}\right)\\
\\
& =\displaystyle \lim\limits_{x\to \infty }\left(\displaystyle \frac{8-\displaystyle \frac{5}{x}}{\displaystyle \sqrt{4+\displaystyle \frac{3}{x^2}}}\right) \\

\\
& =\displaystyle \frac{8-0}{\sqrt{4+0}} \\
\\
& =\displaystyle \frac{8}{2} \\
\\
& =\displaystyle 4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}
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