Category Archives: Engineering Mechanics: Statics

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

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.


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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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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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}