Tag Archives: Engineering Mechanics: Statics and Dynamics 13th Edition by RC Hibbeler Solution Manual

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 to correct SI form using appropriate prefix

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m/ms, (b) μkm, (c) ks/mg, and (d) km·μN.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-9

Solution:

Part A

\displaystyle \text{m/ms}=\left(\frac{\text{m}}{10^{-3}\:\text{s}}\right)=10^3\:\text{m/s}=\:\text{km/s}

Part B

\displaystyle \mu \text{km}=10^{-6}\cdot 10^3\:\text{m}=10^{-3}\:\text{m}=\text{mm}

Part C

\displaystyle \text{ks/mg}=\frac{10^3\:s}{10^{-6}\:\text{kg}}=10^9\:\frac{\text{s}}{\text{kg}}=\text{Gs/kg}

Part D

\displaystyle \text{km} \cdot \mu \text{N}=10^3\:\text{m}\cdot 10^{-6}\:\text{N}\:=10^{-3}\:\text{m}\cdot \text{N}=\text{mm}\cdot \text{N}

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*}
Advertisements

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*}