Axxxxx J. Dubrexxx
Mathematics Dept. Chairman
xxxxx (CT) High School
A. J. D.
7/98 for Historical Perspective (Section II
designed for overhead projector.)
Much effort spent to preserve original type-written pages.
1. The differentiation of measurable physical quantities.
2. The meaning and nature of Derived Quantities.
3. The preference of Mass over Force as a fundamental quantity.
4. The Uniformity of the Metric Prefixes.
5. The exactness of definition for the fundamental MKS units.
6. The relation of basic MKS units to their corresponding English units.
7. The overall simplicity of the MKS system.
1. The definitions for the remaining fundamental quantities.
2. Definitions for derived quantities such as Force, Work, Power, and Energy.
3. Common translations and everyday uses of the above
Prologue II p. i
Objectives I p. ii
Objectives II p. ii
Divisions of Physical Quantities p. 1
Systems of Measurement FPS p. 1
Systems of Measurement MKS p. 2
Metric Prefixes p. 3
Standard Fundamental Units Plus p. 3
Length p. 4
Mass p. 5
Time p. 5
Liquid p. 5
Area p. 6
Temperature p. 6
Conversion Summary I p. 7
Bibliography and Footnotes p. 8
Supplementary Lecture Notes p. 9
Pages are numbed in tradition format: ( ** # ** ).
The second division we refer to as the derived
quantities. These physical quantities provide rules through their definitions
for determining their values in terms of the measurable fundamental quantities.
Examples of derived quantities are:
velocity : length/time (kilometers/ hour)
acceleration : length/time/time (meter/sec/sec)
force : (via Newton's 2nd Law: F= m * a) - mass * length/time/time (1 Newton = 1- . kilogram* meter/sec/sec)
work : force * length (1 Joule = Newton meter)
power : work/time (1 Watt = Joule/sec)
Once having recognized the six fundamental
quantities, it remains to choose a standard for each with which unknowns
may be quantitatively compared (measured). In choosing a standard we must
consider its reproducibility and invariance, and in selecting a measuring
device we must consider its accuracy and dependability.
NOTE: * means "times", operation of multiplication
= = = = =
The FPS system is quite cumbersome for converting
from sub-units to units, to multiple-units since the multiples are not
consistent with-in the same unit and are very different for each fundamental
Length: 12 in = ft, 3 ft = 1 yd, 1760 yd = 1 mi.
Liquid: 8 oz = 1 cup, 2 cups = 1 pt, 2 pts = 1 qt, 4 qts = 1 gal.
The FPS system employs the pound (force) as
a fundamental unit. However, force is not invariant. The weight (gravitational
attractive force) of an object is quite dependent on its location. An object
has significantly different weight on the Moon than it does on the Earth
or in an orbiting space laboratory. It is the mass of an object, the amount
of matter it contains, which is invariant.
There are two basic subdivisions of the Metric System. The CGS system employs the centimeter (length), gram (mass, amount of matter), and second (time) to describe the phenomena of mechanics. It is essentially a laboratory-oriented system, particularly in used thermodynamics.
The MKS system employs the meter (length), kilogram (mass, amount of matter), and second (time) to describe the fundamental units of mechanics. The MKS system is best applicable to the macroscopic "real" world. This system's mechanics will concern us most.
NOTE: FPS short for North American United States Engineering
The subdivisions and multiples of the MKS units necessary for everyday use are relatively easy to learn. We further note that all are powers of ten.
The most often used subdivisions are identified by the Latin prefixes: milli (1/1,000), and centi (1/100).
The Greek prefixes required for multiple-units are: kilo (1,000), mega (1,000,000) and giga
Deci (1/10), deka (10) and hecto (100) are not commonly employed.
The milli-unit is familiar to us in the form
of the mill in our property tax rate. In financial circles, the centi-unit
is commonly referred to as a penny. Thus, we need only learn to replace
thousand by kilo, million by mega and billion by giga to complete our working
knowledge of the most useful metric sub-unit and multiple-unit prefixes.
These same prefixes apply to all the metric units.
The MKS unit of length Is the meter, 1 meter = 196,50,763-73 wavelengths of the orange-red light of krypton-86 as measured with a Michaelson interferometer. This definition is invariant, reproducible and sufficiently accurate.
The unit of mass (amount of matter things contain) in the MKS system is the kilogram (1,000 grams). Mass measurements are made with an equal arm balance. The standard kilogram is a platinum-iridium cylinder kept at the International Bureau of Weights and Measures at Serves, France.
The MKS unit of time is the second. 1
second = 9,192,631, 770 oscillations of cesium-133 atom. An atomic
clock device is employed to record time.
The unit of liquid measure is a derived unit,
the liter ( l ). it has a volume of exactly 1,000 cubic centimeters.
1 kilometer (km) = 1,000 meters (1,000+ yd) while 1 mile (mi) = 1760 yards. More to the point, and to two significant figures, 1 km = .62 mi and 1 mi = 1.6 km. Thus, distances that would be appropriately expressed in miles may be converted to kilometers by multiplying the number of units by 8/5ths. Therefore, a distance of 5 miles becomes 8 kilometers (5*8/5 = 8), 25 miles becomes 40 kilometers (25 * 8/5 = 40) and 100 mi becomes 160 km (100 * 8/5 = 160 or 100 * 1.60 = 160).
1 meter = 100 centimeters (cm) and 1 meter
= 39..37 inches combine to yield the relation 1 in = 2.54 cm. Thus
distances that would be appropriately expressed in inches may be roughly
converted to centimeters by doubling the number of units and adding half
the number to this result. Therefore:
6 in .ap=. 15 cm (2*6 + .5*6 = 15 or 6*2.54 = 15.24) ,
12 in .ap=. 30 cm (2*12 + .5*12 = 30 or 12*2.54 = 30.48).
A 15-cm shoe size would be appropriate for a woman whereas 25 cm would usually be for a man.
NOTE .ap=. means "approximately
A football playing field plus one end zone .ap=. 100m (100.58)
l is the symbol for liter
1 in = 2.54 cm by U. S. definition
Thus the weight of a kilogram mass is approximately
2.2 pounds and a 1 pound weight has the mass of about .4536 kilograms.
For a rough approximation of mass given the weight, divide the number of units in half.
For example, 1 lb .ap=. ½ kgm, 2 lbs .ap=. 1 kgm.
And. reversing direction 1 kgm .ap=. 2.2 lb, 1.5 kgm .ap=. 3.3 lb, 2 kgm .ap=. 4.4 lb, 2.5 kgm .ap=. 5.5 lb.
---------------------- ----------------- ---------
NOTE .ap=. still means approximately equal.
in^2 means in * in or square inches
The hectare is a square hectometer (1 hm =
100 m = 328.1 ft). Thus, 1 hectare = 107, 650 square feet.
Therefore 1 hectare = 2.471 (2.5) acres and 1 acre = .4047 (2/5) hectare.
We will remember that our "usual" temperature scale, the Fahrenheit scale, assigns the arbitrary number 32 to waters freezing point and the arbitrary number 212 to the boiling point of the same. Thus, each Fahrenheit degree (deg-F) is exactly 1/180 the temperature difference between water's boiling and freezing points.
Conversion from one arbitrary scale to another
is "simple" once we have established the relationship between them.
The first thing to be done is establish the relationship of their "zero"
points. The second thing to be done is establish the relationship
the relative size of each unit.
As we have seen, the Fahrenheit scale starts off with a +32 at water's freezing point (O deg-C). Thus, when converting from Fahrenheit to centigrade, we must make the zero adjustment first by subtracting 32 from the Fahrenheit reading, then we may complete the unit conversion. When converting from centigrade to Fahrenheit, we must make the zero adjustment after the unit conversion by adding 32 to the result.
Since there are fewer centigrade degrees between the two fixed points than there are Fahrenheit degrees, the centigrade degree must be larger than the Fahrenheit degree. We note that this ratio is 180/100 or 9/5. This means it requires only 5/9ths as many centigrade degrees as Fahrenheit degrees to express the same temperature difference. In a like manner, it requires 9/5ths times more Fahrenheit degrees as centigrade degrees to express equivalent temperature differences.
The above discussion leads naturally to the
following conversion formulas:
C = 5/9*(F - 32) and F = 9/5*C + 32.
Some solved examples:
-10 deg-F by C = 5/9*(-10 - 32) = 5/9*(-42) = -23-33 yields -23.3 deg-C
+10 deg-F by C = 5/9*(+10 - 32) = 5/9*(-22) = -12.22 yields -12.2 deg-C
30 deg-F by C = 5/9*( 30 - 32) = 519*(- 2) = - 1.11 yields - 1.1 deg-C
50 deg-F by C = 5/9*( 50 - 32) = 5/9*(+18) = +10.00 yields +10.0 deg-C
70 deg-F by C = 5/9*( 70 - 32) = 5/9*( 38) = 21.11 yields 21.1 deg-C
90 deg-F by C = 5/9*( 90 - 32) = 5/9*( 58) = 32.22 yields 32.2 deg-C
- - - - - - - - - - - - - - - --
NOTE: centigrade is used throughout to emphasize
the metric prefix centi
The key to the whole thing is to attain a conceptual comprehension of the basic metric units in terms of the more familiar English units. This simply means to relate the meter to 1+ yards the kilogram mass to increments of 2+ pounds, the second to itself, the liter to 1+ quarts, the hectare to 2.5 acres, and the double relation of 0 deg-C to 32 deg-F and 100 deg-C to 212 deg-F. Possibly, a brief summary table, as listed above, for the more "common" temperatures (or any other unit needed) may aid in learning the "new metric units.
Once we have achieved this qualitative relationship
for the basic metric units and learned the meanings of the metric prefixes,
we will be better able to understand the specific metric sub-units and
multiple units for themselves. We will NOT attempt to memorize the
conversion decimals for each English unit (1 in = 2..54 cm, 1 ft = 30.48
cm, etc.) as these are useless units in the metric system (and may be found
in tables if needed). Such an approach would require more memorization
and effort than our present elementary and secondary students are willing
to put forth. And, would be a momentous task for our adult population.
-1- Black, N. H. (Harvard), Little, E. P. (Wayne); An
Introductory Course In COLLEGE
PHYSICS, 4th ed., The MacMillan Co., N. Y., 1956.
-2- Dull, C. E., Metcalfe, H. C., Williams J E., MODERN
PHYSICS, Holt, Rinehart &
Wilson, Inc., N. Y., 1960.
-3- Sears, F. W. (Dartmouth), Zemansky M. W. (City
College, N. Y.), UNIVERSITY
PHYSICS, 4th ed., Addison-Wesley Pub. Co., Inc., 1970
-4- Williams J E, Metcalfe H. C., Trinklein, F. E., Lefler,
R. W., MODERN PHYSICS, Holt,
Rinehart & Wilson, Inc., New York, 1968
A. IN LANGUAGE, A STATEMENT
(DECLARATIVE SENTENCE) CONTAINS TWO
ESSENTIAL PARTS: 1) a SUBJECT and 2) a VERB.
B. OFTEN, STATEMENTS ARE SO
CONSTRUCTED AS TO HAVE TWO, OR MORE,
INTERPRETATIONS. THAT IS, THEY ARE JUST SO MUCH RHETORIC.
C. IN MATHEMATICS AND SCIENCE,
A STATEMENT MUST MEET THE SAME
CRITERIA AS IN "A" ABOVE,
IN CONTRAST TO "B" ABOVE,
A STATEMENT CAN HAVE ONE AND ONLY ONE INTERPRETATION !!!
TO UNDERSTAND AND APPRECIATE THE BEAUTY OF THE METRIC SYSTEM, AND INDEED THE SCIENCE FROM WHICH IT SPRUNG. WE MUST KEEP THIS RELATION CONSTANTLY IN MIND.
r^2 => r * r \\ * => multiplication \\ => means implies
Newtons 2nd Law \\ Ns L of Universal Gravitation \\ Ns 1st Law
Ns 2nd L : F = m*a or F*dt = d(m*v) => F = d(m*v)/dt
Ns L of UG : F = G*(m*M)/r^2
m = MASS ( m or M )
a = ACCELERATION
v = VELOCITY
r = RADIAL DISTANCE
t = time
G = GRAVITATIONAL CONSTANT (a number)
IGNORE "d" \\ Use s for
displacement and r for radius
2. MASS ( m or M ) IS NEVER ALONE ON THE
RIGHT SIDE OF THESE EQUATIONS.
THIS TELLS US :
1. THERE IS A RELATION BETWEEN THE FORCE F AND THE MASS ( m or M ).
2. BUT!!! THEY ARE NOT EQUAL! THEY ARE DIFFERENT QUANTITIES!!!
A. N's lst L : MOVING OBJECTS (MASSES)
MUST MAINTAIN A STRAIGHT
LINE PATH UNLESS ACTED UPON BY AN EXTERNAL FORCE.
B. K's lst L : THE PLANETS MOVE IN ELLIPTICAL (CURVED) ORBITS.
C. THUS, THE PLANETS REVOLVING ABOUT THE SUN,
RVOLVING ABOUT THEIR RESPECTIVE PLANETS, OUR ARTIFICIAL
SATELLITES REVOLVING ABOUT THE EARTH, MOON AND MARS
MUST EXPERIENCE AN EXTERNAL FORCE.
Ns L OF U G DESCRIBES THIS FORCE:
F = G * m * M/ r^2
Me O < -------------------------- r ------------------------ > o ms
Mass earth => M, in equation above.
SPACE LABORATORY (ms) mass space-lab => m, in equation above
mv o < -------------------------- r ------------------------ > O Mm
VIKING ORBITER (mv)
mass viking-orbiter => m, in equation above.
MARS (Mm) Mass mars => M, in equation above.
r = THE RADIAL DISTANCE
OF SEPARATION BETWEEN EACH SATELLITE (ms) AND ITS REFERENCE MASS (Me or
Mm ) DIVIDES THE NUMERATOR (GmM) BY ITS SQUARE.
FOR A GIVEN SYSTEMS, THE PRODUCT GmM REMAINS CONSTANT. THUS DIVIDING BY r^2, AS "ri" BECOMES LARGER RESULTS IN A SMALLER VALUE FOR F. IF WE LET G * m * M = 1000, THEN (for study purposes only)
F = 1000/ r ^2
r = 1 F = 1000/ 1^2 = 1000/1 = 1000
r = 1 F = 1000/10^2 = 1000/100 = 10
r = 100 F = 1000/100^2 = 1000/10000 = .1
r = 1000
F = 1000/ 1000^2 =
1000/(1000* 1000) =
THIS CLEARLY INDICATES THAT FORCE (F) IS NOT INVARIANT, IT CHANGES FROM PLACE TO PLACE. FORCE (OR WEIGHT) DEPENDS ON WHERE WE REFERENCE IT FROM.
THE QUANTITY OF MATTER AN OBJECT CONTAINS,
ITS MASS, IS INVARIANT.
M = AMOUNT OF MATTER AN OBJECT CONTAINS. (( MASS ))
F = THE PRODUCT
OF AN OBJECTS MASS AND THE ACCELERATI0N IT
EXPERIENCES : (( FORCE )) ( via N's 2nd L: F = m * a )
WEIGHT = THE GRAVITATIONAL ATTRACTIVE FORCE BETWEEN AN OBJECT
AND SOME CELESTIAL REFERENCE BODY SUCH AS THE EARTH, MOON,
MARS, VENUS, SUN, ETC.
THIS NOW LEADS US TO CONSIDER THE FUNDAMENTAL QUANTITIES OF OUR UNIVERSE. THE AFORE MENTIONED EQUATIONS (OF MECHANICS) CAN ALL BE EXPRESSED IN TERM OF ONLY THREE ELEMENTARY CONCEPTS.
QUANTITY M K S UNIT F P S UNIT ---1---
LENGTH (s) METER (m) FOOT (ft)
MASS (m or M) KILOGRAM (kgm) ((SLUG))
North American U. S. Engineering System or English Gravitational System.
- - - - - - - - - - - - - - -
s is used for displacement or distance.
ALL THE REMAING QUANTITIES (OF MECHANICS), VIA THEIR DEFINITIONS, MAY BE EXPRESSED IN TERMS OF JUST THREE FUNDAMETAL METRIC UNITS ( LENGTH, MASS & TIME ).
QUANTITY DEFINITION EQUATION MKS UNIT FPS UNIT
VELOCITY (v) TIME (t) RATE
v = s/t
CHANGE OF LENGTH (s)
ACCELERATION (a) TIME (t) RATE OF
a = v/t
CHANGE OF VELOCITY (v)
MOMENTUM (p) MASS TIMES VELOCITY p =
= N = lb
WORK (W) FORCE TIMES LENGTH (s) W = F*s N*m = Joule ft*lb
POWER (P) WORK DIVIDED BY TIME P = W/t J/sec = Watt ft*lb/sec
NEWTON (N) JOULE (J) SLUG
= Weight/32ft/sec^2 (the FPS unit of mass)
TERA T 10^12
GIGA G 10^9 1,000,000,000 BILLION
MEGA M 10^6 1,000,000 MILLION
KILO k 10^3 1,000 THOUSAND
HECTO h 10^2 100 HUNDRED
DEKA D 10^1 10 TEN
UNIT * 10 ^0 1 ONE
CENTI c 10^-2 1/100 HUNDREDTH
MILLI m 10^-3 1/1,000 THOUSANDTH
MICRO 'mu' 10^-6 1/1,000,000 MILLIONTH
NANO n 10^-9 1/1,000,000,000 BILLIONTH
PICA p 10^-1 2 1/1,000,000,000,000 TRILLIONTH
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