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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 13^th Edition by RC Hibbeler, Problem 1-11}
_{Engineering Mechanics: Statics 14^th 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*}

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 13^th Edition by RC Hibbeler, Problem 1-7}
_{Engineering Mechanics: Statics 14^th 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*}

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

\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

\begin{align*} 35.3\left( 10^3 \right) \ \text{N} & = 35.3 \ \text{kN} \end{align*}

Part C

\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 14^th Edition by RC Hibbeler, Problem 1-2}

Solution:

Part A

\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

\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

\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*}

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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Tag Archives: engineering sciences

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.

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:

Problem 1-9| General Principles| Engineering Mechanics: Statics| RC Hibbeler

Solution:

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:

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?

Solution:

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:

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.

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.