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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

Mg/ms=103 kg10−3 s=106kg/s=Gg/s\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

N/mm=1 N10−3 m=103 N/m=kN/m\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

mN(kg⋅μs)=10−3 N10−6 kg⋅s=103 N/(kg⋅s)=kN/(kg⋅s)\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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Engineering Mechanics: Statics 14th Edition by RC Hibbeler Solution Manual

You can purchase the complete solution manual of the Engineering Mechanics: Statics 14th Edition by Russell Hibbeler in this page.

General Principles: Statics of Rigid Bodies 14th Edition by RC Hibbeler

General Principles

Force Vectors: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Force Vectors

Equilibrium of a Particle: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Equilibrium of a Particle

Force System Resultants: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Force System Resultants

Equilibrium of a Rigid Body: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Equilibrium of a Rigid Body

Structural Analysis: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Structural Analysis

Internal Forces: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Internal Forces

Friction: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Friction

Center of Gravity and Centroid: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Center of Gravity and Centroid

Moment of Inertia: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Moment of Inertia

Virtual Work: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Virtual Work

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Problem 1-6| General Principles| Engineering Mechanics: Statics| RC Hibbeler

If a car is traveling at 55 mi/h, determine its speed in kilometers per hour and meters per second.

Solution:

First, convert 55 mi/h to km/h. 

55\:\frac{mi}{h}=\left(\frac{55\:mi}{1\:h}\right)\left(\frac{5280\:ft}{1\:mi}\right)\left(\frac{0.3048\:m}{1\:ft}\right)\left(\frac{1\:km}{1000\:m}\right)=88.5\:\frac{km}{h}

Next. we convert 88.5 km/h to m/s.

88.5\:\frac{km}{h}=\left(\frac{88.5\:km}{1\:h}\right)\left(\frac{1000\:m}{1\:km}\right)\left(\frac{1\:h}{3600\:s}\right)=24.6\:\frac{m}{s}

Representing combinations of units in correct SI Form

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) 0.000431 kg, (b) 35.3(10³) N, and (c) 0.00532 km.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-7

Solution:

Part A

0.000 431 kg=0.431×10−3 kg=0.431×10−3×103 g=0.431 g\begin{align*} 0.000 \ 431 \ \text{kg} & = 0.431 \times 10^{-3} \ \text{kg}\\ & = 0.431\times 10^{-3} \times 10^3 \ \text{g} \\ & = 0.431 \ \text{g} \end{align*}

Part B

35.3(103) N=35.3 kN\begin{align*} 35.3\left( 10^3 \right) \ \text{N} & = 35.3 \ \text{kN} \end{align*}

Part C

0.00532 km=0.00532×103 m=5.32×10−3×103 m=5.32 m\begin{align*} 0.00532 \ \text{km} & = 0.00532 \times 10^3 \ \text{m} \\ & = 5.32 \times 10^{-3} \times 10^3 \ \text{m} \\ & = 5.32 \ \text{m} \end{align*}

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(10−3)N=109 g/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/(kg⋅ms)=106 Nkg⋅(10−3)s=109 Nkg⋅s=GN/(kg⋅s)\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

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*}