Tag Archives: Conversion

Hibbeler Statics 14E P1.4 — Conversion from English units to Metric Units

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft3 to kN/m3, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

Statics of Rigid Bodies 14th by RC Hibbeler, Problem 1-4

Solution:

Part A

\begin{align*}
200\ \text{lb}\cdot \text{ft} & =\left( 200\ \text{lb}\cdot \text{ft} \right)\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\\
&=271\ \text{N}\cdot \text{m}
\end{align*}

Part B

\begin{align*}
350 \ \text{lb/ft}^3& = \left( \frac{350\ \text{lb}}{1\ \text{ft}^3} \right)\left( \frac{1\ \text{ft}}{0.3048\ \text{m}} \right)^3\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{1\ \text{kN}}{1000\ \text{N}} \right)\\
& = 55.0\ \text{kN/m}^3
\end{align*}

Part C

\begin{align*}
8 \ \text{ft/hr}& = \left( \frac{8\ \text{ft}}{1\ \text{hr}} \right)\left( \frac{1\ \text{hr}}{3600\ \text{s}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\left( \frac{1000 \ \text{mm}}{1 \ \text{m}} \right)\\
& = 0.677\ \text{mm/s}
\end{align*}
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Problem 1-10: The average speed of the earth in its orbit in kilometers per second and in meters per second

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PROBLEM:

a) The average distance between the Earth and the Sun is  108 km. Calculate the average speed of the Earth in its orbit in kilometers per second. 

b) What is this is meters per second?

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SOLUTION:

Part A

We assume that the earth revolves around the sun in a circular manner. Therefore, the distance between the earth and the sun will be the radius of its orbit.

The total distance traveled the earth in full revolution is

\begin{align*}
d & =2\pi r \\
& = 2\pi \left(10\right)^8\:\text{km} \\
\end{align*}

The total time of travel is

\begin{align*}
t & =365.25\:\text{days}\left(\frac{24\:\text{hr}}{1\:\text{day}}\right)\left(\frac{3600\:\text{s}}{1\:\text{hr}}\right) \\
& = 3.15576\times 10^7\:\text{sec} \\
\end{align*}

Therefore, the average speed is

\begin{align*}
\text{speed} & =\frac{\text{distance}}{\text{time}}=\frac{d}{t} \\
 & = \frac{2\pi \left(10\right)^8\:\text{km}}{3.15576\times 10^7\:\text{s}} \\
& = 19.91\:\text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the average speed of the earth is 19.91 km/s.

Part B

Convert 19.91 km/s to m/s.

\begin{align*}
s & =\left(19.91\:\frac{\text{km}}{\text{s}}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right) \\
& =19\,910\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Therefore, the velocity is 19 910 m/s. 

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Problem 1-9: The distance moved in 1 sec and the speed in kilometers per million years of tectonic plates

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PROBLEM:

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year.
(a) What distance does it move in 1.0 s at this speed?
(b) What is its speed in kilometers per million years?

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SOLUTION:

Part A

We know that the formula for the distance traveled, d, if we are given rate, r, and time, t is given by d=r\times t.

The rate in cm/s is 

\begin{align*}
r & =\left(4\:\frac{\text{cm}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{year}}}{365\:\frac{1}{4}\:\bcancel{\text{days}}}\times \frac{1\:\bcancel{\text{day}}}{24\:\bcancel{\text{hours}}}\times \frac{1\:\bcancel{\text{hour}}}{60\:\bcancel{\text{mins}}}\times \frac{1\:\bcancel{\text{min}}}{60\:\text{s}}\right) \\
& = 1.2675 \times 10^{-7} \  \text{cm/s}
\end{align*}

The distance traveled is

\begin{align*}
d & = r \times t \\
d & = \left( 1.2675\times 10^{-7} \ \text{cm/s} \right)\left( 1.0  \ \text{s} \right) \\
d & = 1.3\times 10^{-7} \ \text{cm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We need to convert 4.0 cm/year to km/My.

\begin{align*}
4.0 \ \frac{\text{cm}}{\text{year}} & = 4.0\:\frac{\bcancel{\text{cm}}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{m}}}{100\:\bcancel{\text{cm}}}\times \frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\times \frac{1\:000\:000\:\bcancel{\text{years}}}{1\:\text{My}} \\ \\
& =40\:\text{km/My} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the speed of the tectonic plates is 40 km/My. 

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Problem 1-8: The speed of sound in km/h

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PROBLEM:

The speed of sound is measured to be  342 m/s on a certain day. What is this in km/h?

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SOLUTION:

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

Convert 342 m/s.

\begin{align*}
342\:\text{m/s} & =\left(342\:\frac{\bcancel{\text{m}}}{\bcancel{\text{s}}}\right)\left(\frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\right)\left(\frac{3600\:\bcancel{\text{s}}}{1\:\text{hr}}\right) \\
& =1231.2\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the speed of sound in km/hr is 1231.2 km/hr.

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Problem 1-7: The height of Mount Everest in kilometers

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PROBLEM:

Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)

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SOLUTION:

Convert 29028 ft to km

29,028\:\text{ft}=\left(29\:028\:\text{ft}\right)\left(\frac{1\:\text{km}}{3281\:\text{ft}}\right)=8.847\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the height of Mount Everest is 8.847 kilometers.

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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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Problem 1-1: Converting 100 km/h to meters per second and miles per hour

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PROBLEM:

The speed limit on some interstate highways is roughly 100 km/h.
(a) What is this in meters per second?
(b) How many miles per hour is this?

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SOLUTION:

Part A

\begin{aligned}
100 \  \frac{ \text{km}}{\text{hour}} & =100 \ \frac{ \bcancel{\text{km}}}{\bcancel{\text{hour}}} \times \frac{1000 \ \text{m}}{1 \ \bcancel{\text{km}}} \times \frac{1 \ \bcancel{\text{hour}}}{3600 \ \text{sec}}\\
\\
&=27.7 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{aligned}

Part B

\begin{aligned}
100 \  \frac{ \text{km}}{\text{hour}} & =100 \ \frac{ \bcancel{\text{km}}}{\text{hour}} \times\frac{1 \ \text{mile}}{1.609\ \bcancel{\text{km}}} \\
\\
&=62.2 \ \text{mi/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{aligned}
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Video Solution:

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