Author Archives: Engineering Math

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

KN/μs=(10)3 N(10)6 s=(10)9 N/s=GN/s\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

Mg/mN=(106)g(103)N=109g/N=Gg/N\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

MN/(kgms)=106Nkg(103)s=109Nkgs=GN/(kgs)\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

58 342 m=58.342×103 m=58.342 km=58.3 km\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

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

Part C

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

Part D

7555 kg=7.555×103 kg=7.555×103×103 g=7.555×106 g=7.555 Mg=7.56 Mg\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 limx(8x54x2+3)\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

limx(8x54x2+3)=limx(8x54x2+31x1x)=limx(8xx5x4x2x2+3x2)=limx(85x4+3x2)=804+0=82=4  (Answer)\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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