Category Archives: Sciences
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
Structural Analysis
Internal Forces
Friction
Center of Gravity and Centroid
Moment of Inertia
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)
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](https://s0.wp.com/latex.php?latex=%5Cleft%2835%5C%3Amm%5Cright%29%5E2%5Cleft%2848%5C%3Akg%5Cright%29%5E3%3D%5Cleft%5B35%5Cleft%2810%5E%7B-3%7D%5Cright%29m%5Cright%5D%5E2%5Cleft%2848%5C%3Akg%5Cright%29%5E3%3D135%5C%3Am%5E2%5Ccdot+kg%5E3&bg=ffffff&fg=000000&s=1&c=20201002)
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 
, 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](https://s0.wp.com/latex.php?latex=250%5Cleft%2810%5E3%5Cright%29slugs%3D%5Cleft%5B250%5Cleft%2810%5E3%5Cright%29%5C%3Aslugs%5Cright%5D%5Cleft%28%5Cfrac%7B14.59%5C%3Akg%7D%7B1%5C%3Aslug%7D%5Cright%29%3D3.6457%5Cleft%2810%5E6%5Cright%29%5C%3Akg%3D3.65%5C%3AGg&bg=ffffff&fg=000000&s=1&c=20201002)
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](https://s0.wp.com/latex.php?latex=W_e%3Dmg%3D%5Cleft%5B3.6475%5Cleft%2810%5E6%5Cright%29kg%5Cright%5D%5Cleft%289.81%5C%3Am%2Fs%5E2%5Cright%29%3D35.792%5Cleft%2810%5E6%5Cright%29kg%5Ccdot+m%2Fs%5E2%3D35.8%5C%3AMN&bg=ffffff&fg=000000&s=1&c=20201002)
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](https://s0.wp.com/latex.php?latex=W_m%3Dmg_m%3D%5Cleft%5B250%5Cleft%2810%5E3%5Cright%29slugs%5Cright%5D%5Cleft%285.30%5C%3Aft%2Fs%5E2%5Cright%29%3D%5Cleft%5B1.325%5Cleft%2810%5E6%5Cright%29lb%5Cright%5D%5Cleft%28%5Cfrac%7B4.448%5C%3AN%7D%7B1%5C%3Alb%7D%5Cright%29%3D5.894%5Cleft%2810%5E6%5Cright%29N%3D5.89%5C%3AMN&bg=ffffff&fg=000000&s=1&c=20201002)
d)
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.
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.
Next. we convert 88.5 km/h to m/s.
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