Tag Archives: Concept of Statics

Expressing the Density of Water in SI Units

Water has a density of 1.94 slug/ft³. What is the density expressed in SI units? Express the answer to three significant figures.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-19
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-17

Solution:

\begin{align*}
\rho _w & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =999.6\:\frac{\text{kg}}{\text{m}^3}\\
& =1.00\:\text{Mg/m}^3\\
\end{align*}
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Showing How an Equation is Dimensionally Homogeneous

Using the SI system of units, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-18
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-15

Solution:

To prove that F is in Newtons, we have

\begin{align*}
\text{F} & =\text{G}\cdot \frac{\text{m}_1\text{m}_2}{\text{r}^2}\\
& =\left(\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}\right)\left(\frac{\text{kg}\cdot \text{kg}}{\text{m}^2}\right)\\
& =\frac{\text{kg}\cdot \text{m}}{\text{s}^2}\\
& =\text{N}
\end{align*}

Now, if we substitute the given values into the equation

\begin{align*}
\text{F} & = 66.73\left(10^{-12}\right)\left[\frac{200\left(200\right)}{0.6^2}\right]\\
& = 7.41\left(10^{-6}\right) \text{N}\\
& =7.41\ \mu  \text{N}\\
\end{align*}
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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 1-5: 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

\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

\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:

\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 1-3: Converting 1.0 m/s to 3.6 km/h

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PROBLEM:

Show that 1.0 m/s=3.6 km/h.

Hint: Show the explicit steps involved in converting 1.0 m/s=3.6 km/h.

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SOLUTION:

We know that 1 hr = 3600 s, and 1 km = 1000 m. 

\begin{aligned}
1.0 \ \text{m/s} & = 1.0 \ \frac{\bcancel{\text{m}}}{ \bcancel{\text{s}}} \times \frac{3600 \ \bcancel{\text{s}}}{1 \ \text{hr}}\times \frac{1 \ \text{km}}{1000 \ \bcancel{\text{m}}} \\
\\
& =3.6 \ \text{km/hr} \ \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

\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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