Tag Archives: Engineering Mechanics: Statics

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

(354 mg)(45 km)0.0356 kN=[354(10−3) g][45(103) m]0.0356 (103) N=0.447 (103)g⋅mN=0.447 kg⋅m/N\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

(0.00453 Mg)(201 ms)=[4.53(10−3)(103)kg][201 (10−3)s]=0.911 kg⋅s\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

435 MN/23.2 mm=435 (106) N23.2 (10−3) m=18.75(109) Nm=18.8 GN/m\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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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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Problem 1-5: Soccer field dimensions in feet and inchess

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PROBLEM:

Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches?

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SOLUTION:

The length in feet and inches are

115 m=115 m×1 ft0.3048 m=377.3 feet  (Answer)115 m=115 m×1 ft0.3048 m×12 inches1 ft=4528 inches  (Answer)\begin{aligned} 115\ \text{m} & = 115\ \bcancel{\text{m}} \times \frac{1\ \text{ft}}{0.3048\ \bcancel{\text{m}}} \\ \\ & =377.3\ \text{feet} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ 115\ \text{m} & = 115\ \bcancel{\text{m}} \times \frac{1\ \bcancel{\text{ft}}}{0.3048\ \bcancel{\text{m}}} \times \frac{12\ \text{inches}}{1\ \bcancel{\text{ft}}} \\ \\ & =4528\ \text{inches} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{aligned}

The width in feet and inches are

85 m=85 m×1 ft0.3048 m=278.9 ft  (Answer)85 m=85 m×1 ft0.3048 m×12 in1 ft=3346 inches  (Answer)\begin{aligned} 85\ \text{m} & = 85\ \bcancel{\text{m}} \times \frac{1\ \text{ft}}{0.3048\ \bcancel{\text{m}}}\\ \\ & =278.9\ \text{ft}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ 85\ \text{m} & = 85\ \bcancel{\text{m}} \times \frac{1\ \bcancel{\text{ft}}}{0.3048\ \bcancel{\text{m}}} \times \frac{12\ \text{in}}{1\ \bcancel{\text{ft}}}\\ \\ & =3346\ \text{inches} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
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Problem 1-4: The length of the American football field in meters

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PROBLEM:

American football is played on a 100-yard-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)

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SOLUTION:

100 yard=100 yard×3 feet1 yard×1 m3.281 feet=91.4 m  (Answer)\begin{aligned} 100 \ \text{yard} & = 100 \ \bcancel{\text{yard}} \times \frac{3\ \bcancel{\text{feet}}}{1 \ \bcancel{\text{yard}}}\times \frac{1 \ \text{m}}{3.281\ \bcancel{\text{feet}}} \\ \\ & =91.4 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
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Problem 1-2: Converting car speed of 33 m/s to kilometers per hour and determining if it exceeds the speed limit

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PROBLEM:

A car is traveling at a speed of 33 m/s.
(a) What is its speed in kilometers per hour?
(b) Is it exceeding the 90 km/h speed limit?

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SOLUTION:

Part A

33 m/s=33 ms×1 km1000 m×3600 s1 hr=118.8 km/hr  (Answer)\begin{aligned} 33 \ \text{m/s} & =33\ \frac{\text{m}}{\text{s}} \times \frac{1\ \text{km}}{1000 \ \text{m}} \times \frac{3600\ \text{s}}{1 \ \text{hr}} \\ \\ & =118.8 \ \text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}

Part B

At 118.8 km/h, the car is traveling faster than the speed limit of 90 km/h. (Answer)

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