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Physics Bootcamp

Samuel J. Ling

Section 1.3 Metric and Other Units

In the last section, we introduced units of fundamental mechanical quantities, length, time, and mass. In the International System (SI system) fundamental units, also called metric units, are meter (m) for length, second (s) for time, and kilogram (kg) for mass. Although SI units are preferred in scientific circles, Imperial or English units are also commonly used, at least, in the United States. Therefore, it is important to note their relations.

Subsection 1.3.1 Imperial Units

The Imperial units now do not have their independent standards. Instead, we define them in relation to their metric units. First note that imperial unit of time is same as the metric unit of second. For length, the smallest imperial unit, i.e., an inch (in) is defined in terms of centimeter (cm), i.e., hundreth of a meter; inch is taken to be exactly 2.54 cm.
\begin{equation} 1\ \text{in} = 2.54\ \text{cm}\ \ \ \ \text{(Exactly)}.\tag{1.1} \end{equation}
The other units of length in the English/Imperial system have the following relations to inch; you can convert them in multiples of meters (i.e., into SI units) by first converting them into inches, and then using this exact conversion of inches into centimeters.
\begin{align*} \amp 1\ \text{ft} = 12\ \text{in} \\ \amp 1\ \text{yd}= 3\ \text{ft} \\ \amp 1\ \text{mi}= 1760\ \text{yd} \end{align*}
The English unit of mass is a pound (\(\text{lb}\)) or pound-mass, which is approximately
\begin{equation*} 1\ \text{lb} = 453.59237\ \text{grams} \approx 453.6\ \text{grams}. \end{equation*}
The unit of pound (lb) for mass is sometimes confused with a similar name for force, called pound force (lbf) which stands for the weight of a pound of something. The relation to the unit of force in SI unit, called Newton (N) turns out to be
\begin{equation*} 1\, \text{lbf} \approx 4.448\, \text{N}. \end{equation*}
This is obtained from the formula \(F=mg\) with \(m=1\,\text{lb}\approx 0.4536\,\text{kg}\) and \(g=9.81\,\text{m}/\text{s}^2\text{.}\) The unit of Newton stands for
\begin{equation*} \text{N} \equiv \text{kg}\, \frac{\text{m}}{\text{s}^2}. \end{equation*}

Subsection 1.3.2 Unit Prefixes

Often a meter, a kilogram, or a second is too large or too small for the physical phenomena under study. In the metric system, multiples of positive and negative powers of 10 are then used to simplify the numerical values to ordinary sizes and unit names are given to various multiples of the fundamental units meter, kilogram, and second in a uniform way by adding prefixes to the names of units (See Table 1.7. For instance, one hundredth of a meter is called a centimeter, one thousand meters is a kilometer, and a millionth of a second is called a microsecond.
Table 1.7. SI Prefixes
Factor Prefix Symbol Factor Prefix Symbol
\(10^{1}\) deka- da \(10^{-1}\) deci- d
\(10^{2}\) hecto- h \(10^{-2}\) centi- c
\(10^{3}\) kilo- k \(10^{-3}\) milli- m
\(10^{6}\) mega- M \(10^{-6}\) micro- \(\mu\)
\(10^{9}\) giga- G \(10^{-9}\) nano- n
\(10^{12}\) tera- T \(10^{-12}\) pico- p
\(10^{15}\) peta- P \(10^{-15}\) femto- f
\(10^{18}\) exa- E \(10^{-18}\) atto- a
\(10^{21}\) zetta- Z \(10^{-21}\) zepto- z
\(10^{24}\) yotta- Y \(10^{-24}\) yocto- y

Subsection 1.3.3 The SI AND CGS Systems of Units

There are two systems of units based on the metric units that are commonly used in science - the International System (SI) and the centimeter-gram-second (cgs) systems. In the SI system all units are expressed in meter, kilogram and second while in the cgs system they are in centimeter, gram and second. The SI system is dominant in textbooks while cgs is more convenient for laboratory experiments. The SI system for mechanical properties is also called MKS or meter-kilogram-second system.

Subsection 1.3.4 Base Units

All mechanical properties can be expressed in terms of length, mass and time. Therefore, their units can be related to the units of length, mass and time. For this reason, the units meter, kilogram and second are called fundamental units while other units are said to be derived from them and are called derived units.
There are other properties that are not related to motion of objects, such as an electric charge or temperature of an object and therefore, the set of fundamental units or base units contains additional quantities. Presently, there are seven base standards in the SI system of units as shown in Table 1.8. In the initial part of the book we will be using only the units of length, time and mass.
Table 1.8. Base SI units
Quantity SI Unit Name Unit Symbol
Length Meter m
Time Second s
Mass Kilogram kg
Electric current Ampere A
Temperature Kelvin K
Amount of substance Mole Mol
Luminous intensity Candela Cd

Subsection 1.3.5 Conversion of Units

Conversion of units is frequently needed in physics and you should make sure you have a method of calculation that works consistently for you. A particularly useful method is to multiply the given number by a fraction whose value is \(1\) but has different units on the numerator and denominator. The fraction is chosen so that numerator and denominator are different units for the same physical quantity. For instance, if we need to change minutes to hours, we will need a fraction of hours to minutes or minutes to hours depending on whether minute to be changed is in the numerator or the denominator respectively.
\begin{equation*} \frac{1\, \text{hr}}{60\,\text{min}}= 1 = \frac{60\,\text{min}}{1\, \text{hr}}. \end{equation*}

Example 1.9. Changing One Unit Only.

How many seconds are in 20 minutes?
Answer.
\(1200\, \text{sec}\text{.}\)
Solution.
You can start with 20 minutes and then multiply with a fraction of second [the unit desired] to minute [the unit to be converted] whose value is 1. The fraction is obtained by the conversion of 60 seconds in 1 minute.
\begin{align*} 20\, \text{min}\ \amp = 20\, \text{min} \times \frac{60\, \text{sec}}{1\, \text{min}}\\ \amp = 20\, \cancel{\text{min}} \times \frac{60\ \text{sec}}{1\, \cancel{\text{min}}}\\ \amp = 1200\, \text{sec}. \end{align*}

Example 1.10. Changing Product of Units.

A rectangular board is \(5\text{ in}\) by \(2\text{ in}\text{.}\) The are of the board is \(10\) square inches. How many square centimeters in \(10\) square inch?
Answer.
\(64.5\ \text{cm}^2.\)
Solution.
A square inch is a unit of area. Note that square inches stands for inches times inches. It is usually helpful to write out square inches this way before become proficient at conversions and can skip this step. We need a fraction for each inch to be converted. The fraction needed has \(\text{cm}\) at the numerator and \(\text{in}\) at the denominator, which we can construct from the fact that there are \(2.54\text{ cm}\) in \(1\text{ in}\text{.}\)
The calculation is presented below. You should note that when I used the calculator I got way too many digits than the three digits in \(2.54\) that went into the calculation, so I put in another step where the final answer was rounded off to three digits. This happens a lot in calculations and you should learn to round off numbers to correct number of digits. I will show you later a consistent way of rounding off that is commonly practised.
\begin{align*} 10\ \text{in}^2\ \amp = 10\ \text{in} \times \text{in}\\ \amp = 10\ \cancel{\text{in}} \times \cancel{\text{in}}\times \frac{2.54\ \text{cm}}{1\ \cancel{\text{in}}}\times \frac{2.54\ \text{cm}}{1\ \cancel{\text{in}}}\\ \amp \ \ \text{(two factors needed)}\\ \amp = 10\times2.54\text{ cm}\ \times2.54\text{ cm}\\ \amp = 64.516\ \text{[from calculator]}\ \text{cm}^2\\ \amp = 64.5\ \text{cm}^2. \end{align*}
Figure 1.11.

Exercises Exercises

1. Changing Imperial Units to SI Units.
Convert \(1.5\ \text{lb mi/h}^2\) into \(\text{kg m/s}^2\text{.}\)
Hint.
Be careful with two factors of hour.
Answer.
\(8.4\times10^{-5} \text{kg m/s}^2.\)
Solution.
This example is definitely messier than the previous ones. We will need four fractions, one for each factor of the units; we will combine the two factor of hours into one to save space. We will also need to round off the final answer. Here is the gory details of the calculation.
\begin{align*} 1.5\ \frac{\text{lb mi}}{\text{h}^2}\ \amp = 1.5\ \frac{\text{lb mi}}{\text{h}^2}\times \frac{0.4536\ \text{kg}}{1\ \text{lb}} \times \frac{1.6093\ \text{km}}{1\ \text{mi}}\times \frac{1\ \text{h}^2}{3600^2\ \text{s}^2}\\ \amp = 8.4488...\ \text{[from calculator]}\ \text{kg m/s}^2\\ \amp = 8.4\times10^{-5} \text{kg m/s}^2. \ \text{[Rounding off to two digits.]} \end{align*}
Note that I rounded off the final number to two digits - I did that because the least precise number was 1.5 which has two significant digits.
2. Unit Conversion from MPH to Meters per Second.
A car travels at a speed of 65.0 miles per hour (mph). What is the speed in meters/second?
Hint.
Convert both units.
Answer.
\(29.1\,\text{m/s}.\)
Solution.
\begin{align*} 65.0\ \frac{\text{mi}}{\text{h}}\ \amp = 65.0\ \frac{1609.344\text{m}}{3600\text{s}}\\ \amp = 29.0576\,\frac{\text{m}}{\text{s}}\ \text{(from calculator)}\\ \amp = 29.1\,\text{m/s}. \ \text{[Rounding off to three digits.]} \end{align*}
Note that I rounded off the final number to three digits - I did that because the least precise number was 65.0 which has three significant digits.
3. Unit Conversion from Light Years to Meters.
In astronomy, a common unit of distance is a light-year (ly), which is equal to the distance that light travels in one year, which is taken to be 365.25 days. How many meters are one ly?
Hint.
Use \(d = v\,t\text{.}\)
Answer.
\(9,460,730,472,580,800\, \text{m}.\)
Solution.
\begin{align*} d \amp = v\, t \\ \amp = 299792458\, \text{m/s}\times 365.25\times24\times3600\,\text{s}\\ \amp = 9,460,730,472,580,800\, \text{m}. \end{align*}
4. Unit of Density in Imperial Units.
The density of aluminum is 2.7 grams per cubic-centimeter. (a) What is the density in kilogram/cubic meter? (b) What is the density in \(\text{lb/ft}^3\text{.}\)
Hint.
Beware of three factors of \(cm\text{,}\) each needs to be converted!
Answer.
(a) \(2.7\times 10^3\,\text{kg/m}^3\text{,}\) (b) \(168.8\, \text{lb/ft}^3\text{.}\)
Solution 1. (a)
(a)
\begin{equation*} 2.7\,\frac{\text{g}}{\text{cm}^3} = 2.7\,\frac{10^{-3}\text{kg}}{\left(10^{-2}\right)^3\,\text{m}^3} = 2.7\times 10^3\,\text{kg/m}^3. \end{equation*}
Solution 2. (b)
(b)
\begin{align*} 2.7\,\frac{\text{g}}{\text{cm}^3} \amp = 2.7\,\frac{(1/453)\text{lb}}{\left((1/2.54)\times(1/12)\right)^3\,\text{ft}^3} \\ \amp = 168.8\, \text{lb/ft}^3. \end{align*}
5. Average Distance Between Sun and Mercury.
An astronomical unit (AU) is equal to the average distance between Earth and Sun. It is approximately equal to 150 million kilometers. Evaluate the average distances (a) from Sun to Mercury, (b) from Sun to Mars, and (c) Sun to Jupiter in AU if they are \(5.79 \times 10^{10}\ \text{m}\text{,}\) \(2.28 \times 10^{11}\ \text{m}\) and \(7.78 \times 10^{11}\ \text{m}\) respectively.
Hint.
In AU, the distance from the Sun to Earth is \(1.0\, \text{AU}\text{.}\)
Answer.
(a) \(d_\text{S-Mer}=0.386\, AU\text{,}\) (b) \(d_\text{S-Mar}=1.52\ AU\text{,}\) (c) \(d_\text{S-J}=5.19\ AU\) .
Solution 1. (a)
\begin{equation*} d_\text{S-Mer} = 5.79 \times 10^{10}\ \text{m}\times \frac{1\, \text{AU}}{1.50\times 10^{11}\,\text{m} } = 0.386\,\text{AU}. \end{equation*}
Solution 2. (b)
\begin{equation*} d_\text{S-Mar} = 2.28 \times 10^{11}\ \text{m}\times \frac{1\, \text{AU}}{1.50\times 10^{11}\,\text{m} } = 1.52\,\text{AU}. \end{equation*}
Solution 3. (c)
\begin{equation*} d_\text{S-J} = 7.78 \times 10^{11}\ \text{m}\times \frac{1\, \text{AU}}{1.50\times 10^{11}\,\text{m} } = 5.19\,\text{AU}. \end{equation*}
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