Tag Archives: Statics of Rigid Bodies

Problem 1-10| General Principles| Engineering Mechanics: Statics| RC Hibbeler

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (0.631 Mm)/(8.60kg)², (b) (35 mm)² (48 kg)³.

Solution:

a) \left(0.631\:Mm\right)/\left(8.60\:kg\right)^2=\left(\frac{0.631\left(10^6\right)m}{\left(8.60\right)^2kg^2}\right)=\frac{8532\:m}{kg^2}=8.53\left(10^3\right)m/kg^2=8.53\:km/kg

b) \left(35\:mm\right)^2\left(48\:kg\right)^3=\left[35\left(10^{-3}\right)m\right]^2\left(48\:kg\right)^3=135\:m^2\cdot kg^3

Problem 1-8| General Principles| Engineering Mechanics: Statics| RC Hibbeler

The specific weight (wt./vol.) of brass is 520 lb/ft³. Determine its density (mass/vol.) in SI units. Use an appropriate prefix. 

Solution:

First, we will convert 1 Pa to lb/ft².

520\:lb/ft^3=\left(\frac{520\:lb}{ft^3}\right)\left(\frac{1\:ft}{0.3048\:m}\right)^3\left(\frac{4.448\:N}{1\:lb}\right)\left(\frac{1\:kg}{9.81\:N}\right)=8.33\:Mg/m^3

Converting Measurement from Pounds Per Square Inch to Pascal

The pascal (Pa) is actually a very small unit of pressure. To show this, convert 1 Pa = 1N/m² to lb/ft². Atmospheric pressure at sea level is 14.7 lb/in². How many pascals is this?

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-7
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-16

Solution:

First, we will convert 1 Pa to lb/ft².

1 Pa=1 Nm2(1 lb4.4482 N)(0.30482 m21 ft2)=20.9(10−3) lb/ft2\begin{align*} 1\ \text{Pa} & =\frac{1\:\text{N}}{\text{m}^2}\left(\frac{1\:\text{lb}}{4.4482\:\text{N}}\right)\left(\frac{0.3048^2\:\text{m}^2}{1\:\text{ft}^2}\right)\\ &=20.9\left(10^{-3}\right)\:\text{lb/ft}^2 \end{align*}

Next, we convert 14.7 lb/in2 to Pa

14.7 lb/in2=14.7 lbin2(4.448 N1 lb)(144 in21 ft2)(1 ft20.30482 m2)=101.3(103) N/m2=101.3(103)Pa\begin{align*} 14.7 \ \text{lb/in}^2 & =\frac{14.7\:\text{lb}}{\text{in}^2}\left(\frac{4.448\:\text{N}}{1\:\text{lb}}\right)\left(\frac{144\:\text{in}^2}{1\:\text{ft}^2}\right)\left(\frac{1\:\text{ft}^2}{0.3048^2\:\text{m}^2}\right)\\ & =101.3\left(10^3\right)\:\text{N/m}^2\\ & =101.3\left(10^3\right) \text{Pa}\\ \end{align*}
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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*}
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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

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