Tag Archives: RC Hibbeler

Problem 1-16| General Principles| Engineering Mechanics: Statics| RC Hibbeler

What is the weight in newtons of an object that has a mass of: (a) 10 kg, (b) 0.5 g, and (c) 4.50 Mg? Express the result to three significant figures. Use an appropriate prefix.

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Problem 1-15| General Principles| Engineering Mechanics: Statics| RC Hibbeler

Determine the mass of an object that has a weight of (a) 20 mN, (b) 150 kN, and (c) 60 MN. Express the answer to three significant figures.

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Problem 1-14| General Principles| Engineering Mechanics: Statics| RC Hibbeler

Evaluate each of the following and express with an appropriate prefix: (a) (430 kg) ², (b) (0.002 mg)², and (c) (230 m)³.

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Problem 1-13| General Principles| Engineering Mechanics: Statics| RC Hibbeler

Convert each of the following to three significant figures: (a) 20 lb·ft to N·m, (b) 450 lb/ft³ to kN/m³, and (c) 15 ft/h to mm/s.
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Problem 1-12| General Principles| Engineering Mechanics: Statics| RC Hibbeler

Convert each of the following and express the answer using an appropriate prefix: (a) 175 lb/ft³ to kN/m³, (b) 6 ft/h to mm/s, and (c) 835 lb·ft to kN·m. Continue reading →

Evaluation of Expressions to SI Units with Appropriate Prefix

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) 354 mg (45 km)/(0.0356 kN), (b) (0.00453 Mg)(201 ms), and (c) 435 MN/23.2 mm.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-11
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-18

Solution:

Part A

\begin{align*}
\frac{\left(354\:\text{mg}\right)\left(45\:\text{km}\right)}{0.0356\:\text{kN}} & = \frac{\left[354\left(10^{-3}\right)\:\text{g}\right]\left[45\left(10^3\right)\:\text{m}\right]}{0.0356\:\left(10^3\right)\:\text{N}}\\
& = \frac{0.447\:\left(10^3\right)\text{g}\cdot \text{m}}{\text{N}}\\
& = 0.447\:\text{kg}\cdot \text{m/N}
\end{align*}

Part B

\begin{align*}
\left(0.00453\:\text{Mg}\right)\left(201\:\text{ms}\right) & =\left[4.53\left(10^{-3}\right)\left(10^3\right)\text{kg}\right]\left[201\:\left(10^{-3}\right)\text{s}\right]\\
& =0.911\:\text{kg}\cdot \text{s}\\
\end{align*}

Part C

\begin{align*}
435\:\text{MN}/23.2\:\text{mm} & =\frac{435\:\left(10^6\right)\:\text{N}}{23.2\:\left(10^{-3}\right)\:\text{m}}\\
& = \frac{18.75\left(10^9\right)\:\text{N}}{\text{m}}\\
& =18.8\:\text{GN/m}
\end{align*}
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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 1-9| General Principles| Engineering Mechanics: Statics| RC Hibbeler

A rocket has a mass of 250(10³) slugs on earth. Specify (a) its mass in SI units and (b) its weight in SI units. If the rocket is on the moon, where the acceleration due to gravity is g_m=5.30\:ft/s^2, determine to three significant figures (c) its weight in SI units and (d) its mass in SI units.

Solution:

a) 250\left(10^3\right)slugs=\left[250\left(10^3\right)\:slugs\right]\left(\frac{14.59\:kg}{1\:slug}\right)=3.6457\left(10^6\right)\:kg=3.65\:Gg

b) W_e=mg=\left[3.6475\left(10^6\right)kg\right]\left(9.81\:m/s^2\right)=35.792\left(10^6\right)kg\cdot m/s^2=35.8\:MN

c) W_m=mg_m=\left[250\left(10^3\right)slugs\right]\left(5.30\:ft/s^2\right)=\left[1.325\left(10^6\right)lb\right]\left(\frac{4.448\:N}{1\:lb}\right)=5.894\left(10^6\right)N=5.89\:MN

d) m_m=m_e=3.65\left(10^6\right)kg=3.65\:Gg

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.

\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

\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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