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Hibbeler Statics 14E P1.3 — Representing a combination of units in the correct SI form

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

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-3

Solution:

Part A

\begin{align*}
\text{Mg/ms} & = \frac{10^3 \ \text{kg}}{10^{-3} \ \text{s}} \\
& = 10^6 \text{kg/s}\\
& = \text{Gg/s}
\end{align*}

Part B

\begin{align*}
\text{N/mm} & = \frac{1\ \text{N}}{10^{-3} \ \text{m}}\\
& = 10^3 \ \text{N/m}\\
& = \text{kN/m}
\end{align*}

Part C

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

College Physics by Openstax Chapter 4 Problem 1

A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s2. What is the net external force on him?

Solution:

So, we are given mass, m = 63.0 \ \text{kg} , and acceleration, a = 4.20 \ \text{m/s}^2.

The net force has a formula 

\text{F}=\text{m}a

Substituting the given values, we have

\begin{align*}
F & = \left( 63.0 \ \text{kg} \right)\left( 4.20 \ \text{m/s}^2 \right) \\
F & = 265 \  \text{kg}\cdot \text{m/s}^2 \\
F & = 265 \ \text{N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
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