Lecture #8
EGR 120 – Introduction to Engineering
Significant Digits and Systems of Units
Reading Assignment:
Read Chapters 5, 10, and 11 in Thinking Like An Engineer – An Active
Learning Approach, by Stephan
Unit Conversion Tables – Inside front cover and inside back cover of text
Homework Assignment:
Homework Assignment #4
1
Lecture #8
EGR 120 – Introduction to Engineering
Significant Digits
The use of significant digits in expressing measured quantities gives us the ability to
indicate the intended degree of precision. A different degree of precision is implied
by 5 gallons versus 5.000 gallons.
Three basic rules for significant digits
1. All non-zero digits are significant (Example: 54.87 has 4 significant digits)
2. Leading zeros are not significant (Example: 0.003 has 1 significant digit)
3. Trailing zeros are significant (except with whole numbers where their
significance may be uncertain) (Example: 2.20 has 3 significant digits)
(Example: 100 has an unclear number of significant digits (1, 2, or 3))
Example: How many significant digits are in each number below?
1. 241.692
2. 10.000
3. 0.000173
4. 0.0440
5. 5250
2
Lecture #8
EGR 120 – Introduction to Engineering
Significant Digits and Accuracy
When a measurement is recorded, the number should
reflect the accuracy of the measurement. A
measurement of 15.3 implies a certain accuracy: that
the true value of the quantity being measured is
between 15.25 and 15.35.
Illustration: Consider the three voltmeters shown to
the right. Why would the middle voltmeter read 15.3?
It must be that the actual value of the voltage is closer
to 15.3 than to 15.2 and is also closer to 15.3 than 15.4.
Since the midpoint between 15.2 and 15.3 is 15.25 and
the midpoint between 15.3 and 15.4 is 15.35, it follows
that: The reading 15.3 implies an accuracy of between
15.25 and 15.35.
1
5
.2
1
5
.3
1
5
.4
+
V
_
+
V
_
+
V
_
Example: Determine the accuracy for each measured quantity below.
1. 2.09
2. 4.80
3. 2.500
3
Lecture #8
EGR 120 – Introduction to Engineering
Significant Digits and Whole Numbers
Whole numbers can include trailing zeros that may or may not be significant.
For example, the number the zeros in the number 4000 may or may not be
significant. It is unclear.
Examples: How many significant digits do you think are implied by each of the
following?
1) A person states that the outside temperature is 100 degrees.
Probably all 3 digits are intended to be significant as since expressing
temperatures to the nearest degree is common.
2) A person states that his swimming pool holds 10000 gallons of water.
Probably all digits are not intended to be significant. If all 5 digits were
significant, this would imply an unusual accuracy.
So, how can the confusion over the number of significant digits be eliminated?
By using scientific notation.
4
Lecture #8 EGR 120 – Introduction to Engineering
Significant Digits and Scientific Notation
Using scientific notation is a sure way to clear up any possible confusion in the
number of significant digits. Trailing zeros are always significant when scientific
notation is used. The number 4000 has an unclear number of significant digits, but
4.000 x 103 clearly has 4 significant digits and 4.0 x 103 clearly has only 2 significant
digits.
Example: Recall the case of the 10,000 gallon swimming pool. The owner could have
used scientific notation to make the accuracy of the pool’s volume clear. The possible
choices are listed below.
Volume of water
in swimming pool
Number of
sig. digits
Implied
Accuracy
1. x 104 gallons
1.0 x 104 gallons
1.00 x 104 gallons
1.000 x 104 gallons
1.0000 x 104 gallons
5
Lecture #8
EGR 120 – Introduction to Engineering
Significant Digits and Percent Error
The number of significant digits used in expressing a measured quantity gives an
indication of the percent error in the measurement. To illustrate this, consider the case of
3 significant digits used to express a whole number. The smallest whole number that can
be expressed using 3 significant digits is 100 and the largest is 999.
Note that 100 implies an accuracy of 99.5 - 100.5 and 999 implies an accuracy of 998.5 999.5.
max value - min value
x 100%
nominal value
100.5 - 99.5
Then when the display shows the value of 100 the error is
x 100% = 1%
100
999.5 - 998.5
And when the display shows the value of 999 the error is
x 100%  0.1%
999
If error is defined as error =
So, a number expressed using 3 significant digits has a percent error somewhere between
0.1% and 1% or a maximum percent error of 1%. A similar process could be used in
other cases:
Number of significant digits
Maximum Error
1
2
3
4
5
6
6
Lecture #8
EGR 120 – Introduction to Engineering
Mathematical operations involving significant digits
Measured data is often used in calculations and the number of significant digits
used to express the result depends on the type of calculations performed. There
are 2 basic rules to consider:
1) When multiplying or dividing numbers, the results should be expressed using
the fewest number of significant digits contained in any of the numbers.
2) When adding or subtracting numbers, the results should be expressed using
the smallest number of digits to the right of any common(*) decimal point
contained in any of the numbers.
* Note that using scientific notation allows the user to move the decimal point as
desired(such as 0.123 x 103 = 1.23 x 104 = 12.3 x 105 , etc) so be sure to use the
same exponent when determining the number of significant digits to the right of
any common decimal point.
Example: Express the result below using the proper number of significant digits.
78
x 87
7
Lecture #8
EGR 120 – Introduction to Engineering
Example: Express the result below using the proper number of significant digits.
3.729
x 1.6
Example: Express the result below using the proper number of significant digits.
0.0064  28.2 =
8
Lecture #8
EGR 120 – Introduction to Engineering
Example: Express the result below using the proper number of significant digits.
823.457
+ 438.9
Example: Express the result below using the proper number of significant digits.
1.67 x 103
+ 1.9 x 104
9
Lecture #8
EGR 120 – Introduction to Engineering
Example: Recall the example of the swimming pool that held 10000 gallons.
Suppose that someone spills a cup of water into the pool. What is the new
volume? (1 cup = 0.125 gallon)
10000.
+ 0.125
Rules for significant digits - Are they perfect?
No. Using the rules for significant digits is simply an attempt to express
answers with a reasonable degree of accuracy. There are other methods of
error analysis which give better results, using more complex techniques
involving calculus and statistics (such as worst-case analysis, Monte Carlo
analysis, and sensitivity analysis).
Using exact quantities in problems
Error is not introduced until measurement is made. So if a textbook
problem states that a triangle has a height of 10 inches, you can assume that
it is an exact value and the rules for significant digits are not an issue. If the
problem states that a person measured the height of the triangle to be 10.0
inches, then you can assume that the quantity uses 3 significant digits.
10
Lecture #8 EGR 120 – Introduction to Engineering
Dimensions and Systems of Units
Unit – a predetermined reference amount that is used to help us understand the
magnitude of a physical quantity
Dimension – a term used to describe a physical quantity, such as length,
temperature, time, etc.
Dimensions can be described using various units.
Example: Length is a dimension. It can be expressed using a variety of
different units, such as in the example below:
Length = 0.8333 ft = 10 in = 25.4 cm
Systems of Units
There are two systems of units:
1) SI Units (International System of Units)
2) US Customary Units
SI Units have some significant advantages over US Customary Units, but US
Customary Units are still used heavily in the US, so it is important to be very
familiar with both sets of units.
11
Lecture #8
EGR 120 – Introduction to Engineering
Advantages of SI Units
The SI system is better than the US Customary system because:
1) The SI system is a decimal system.
The SI system uses a system of prefixes. There is only one unit for each dimension
(such as the meter for length), whereas US units often have several units for each
dimension (such as inches, feet, yard, miles, etc., for length). Quantities are easily
expressed in the SI system with several different prefixes by simply moving the
decimal point. US Units, on the other hand, require the use of conversion factors
(such as 12 inches = 1 foot).
Example: Express the quantity below using other units (or prefixes) in its system of
units.
A) SI Units:
2500000 mm =
B) US Units:
2500000 in =
12
Lecture #8
EGR 120 – Introduction to Engineering
Advantages of SI Units (continued)
The SI system is better than the US Customary system because:
2) The SI system is a non-gravitational system.
SI units are based on mass, whereas US units are based on force (or weight). Mass is
simply a quantity of matter and does not depend on gravity. Force is the acceleration of
mass. Recall Newton’s 2nd law of physics:
F = ma
where a = acceleration
or if a = g = acceleration due to gravity, then force is called weight and Newton’s law
becomes
W = mg
The use of weight in the US system is awkward because weight varies slightly over the
surface of the earth (g is proportional to 1/R2, where R is the radius of the Earth).
Weight varies tremendously when we go outside of the arena of the Earth. Acceleration
due to gravity on the moon is only about 1/6 of the value on Earth, so a person weighing
180 lb on Earth only weighs about 30 lb on the moon.
Example: Suppose you are an engineer working for NASA and you are providing
specifications for a part on the space shuttle. Which of the following two choices would
you make?
A) specify its weight - If its weight is 100 lb on Earth then it weighs only about 17 lb on
the moon and its weight is approximately zero in space. This is not a good choice!
B) specify its mass - If it has a mass of 100 kg on Earth then its mass is also 100 kg on
13
the moon or in space. This is a much better choice!
14
Lecture #8
EGR 120 – Introduction to Engineering
Advantages of SI Units (continued)
The SI system is better than the US Customary system because:
3) The SI system is a worldwide system.
The United States is the only country still using US Customary Units. The rest
of the world uses SI Units. Obviously it would be better for the United States
to switch to the system used by everyone else.
Will the United States ever switch to the SI system?
Efforts in the past
In the early 1970’s the United States attempted to switch to the SI system. Speed
limit signs on interstates were all changed to include both mph and km/h.
Speedometers on cars had to show speeds using both types of units. Schools began
teaching children the SI system. One of the most common statics and dynamics
textbooks used by engineering students in college is by the authors Beers & Johnston.
In the early 1970’s their text switched from using US units to using SI units.
Unfortunately, special interests such as the building and manufacturing industries,
derailed the efforts. The speed limit signs were eventually changed back to only mph.
Beer & Johnston’s engineering textbook soon switched to teach 50% US units and
50% SI units. It still uses that format today (as you will see in EGR 140 and EGR
245 at TCC).
Lecture #8
EGR 120 – Introduction to Engineering
Canada
In the early 1970’s Canada joined the US in our efforts to switch to the SI
system, but Canada didn’t give up! They now use only the SI system.
(Instructor’s note: A Canadian student was in my Engineering Statics class
(EGR 140) a few years ago. Canada made the switch to SI units when he was
in kindergarten. He had almost no knowledge of US units and struggled a bit
with the US units in Statics. In one generation the switch was complete!)
Efforts today
There is reason for hope. We have much more of a worldwide economy today.
US Engineering companies often sell products and services to other countries
so they must work in their system of units. Many engineering companies
report that a significant portion of their work is now done in SI units. In 1992
the federal government began requiring engineering work done on federal
facilities to use SI units. This has had a positive effect. For example, suppose a
local a engineering company designs a building for the Norfolk Naval base.
Not only is the design done in SI units, but the local subcontractors that do the
plumbing, electrical, HVAC (heating, ventilation, and air-conditioning),
15
landscaping, etc., must all deal with SI specifications.
Lecture #8
EGR 120 – Introduction to Engineering
Shown below is an article from the local newspaper announcing that the federal
government began requiring work in SI units in 1992. It wasn’t exactly front page
news (it appeared on page A10), but it will hopefully bring us closer to someday
switching to the SI system.
A10
THE VIRGINIAN-PILOT
THURSDAY OCTOBER 1, 1992
----------------------------------------------------------------------------------COLORADO SPRINGS, Colo. –
All federal agencies, if possible, are
supposed to begin conducting business today like the rest of the world
does. By the metric system.
In reality, leaving behind a system in which Americans measure
their beer in pints and their binges
in pounds will probably happen at a
glacial pace.
Even Dr. Gary Carver, chief of
the Metric Program Office, a division of the Commerce Department,
realizes the government will have to
move as slowly as an inchworm –
or a centipede.
16
17
Lecture #8
EGR 120 – Introduction to Engineering
The SI System Units
The SI system is built on 7 precisely defined base units. All other units (called
derived units) are built on these 7 units. Before examining each unit in detail, a
couple of points should be made.
1) The definitions of the units have changed over time. In recent times
attempts have been made to define the units in terms of quantities in nature that
are constant.
2) The units are man-made. They do not occur naturally. Any advanced
civilization would discover the value of  (the ratio of the circumference to the
diameter of a circle), e (the base of the natural log), or many other constants.
However, there is nothing in nature that would lead them to discover the meter,
second, kilogram, or any other SI unit (or US unit).
18
Lecture #8
EGR 120 – Introduction to Engineering
Table 2: SI Base Units
Quantity
Length
Mass
Time
Electric current
Thermodynamic temperature
Amount of substance
Luminous intensity
Name
meter
kilogram
second
ampere
kelvin
mole
candela
Symbol
m
kg
s
A
K
mol
cd
Suggestion: Know the SI
Base Units for Test #1.
Notes
1) Note that the base unit for mass is the kilogram, not the gram. It is an odd feature of
the SI system that this unit includes a prefix. This sometimes causes some confusion.
Whenever performing calculations in the SI system involving mass, be sure to express
mass in kilograms, not grams. For example,
Example: Calculate force using F = ma
where m = 2 kg and a = 4 m/s2
F = (2 kg)(4 m/s2) = 8 N (not 8 kN)
Similarly, if m = 4 g and a = 3 m/s2
F = (4 g)(3 m/s2) = (0.004 kg)(3 m/s2) = 0.012 N = 12 mN (not 12 N)
2) It is said that all of the base units except the kilogram can be reproduced in a “wellequipped lab.” This would take an incredible “lab” as will be apparent after the
19
definitions are presented! The kilogram will probably be redefined at some point.
Lecture #8
EGR 120 – Introduction to Engineering
Definitions of units in Table 2
Length - The meter is a length equal to the distance traveled by light in a vacuum during
1/299792458 s. The meter was defined by the CGPM that met in 1983.
Time - The second is the duration of 9,192,631,770 periods of radiation corresponding to the
transition between the two hyperfine levels of the ground state of the cesium-133 atom. The
second was adopted by the Thirteenth CGPM in 1967.
Mass - The standard for the unit of mass, the kilogram, is cylinder of platinum-iridium alloy kept
by the International Bureau of Weights and Measures in France. A duplicate copy is maintained in
the United States. The unit of mass is the only base unit nonreproducible in a properly equipped
lab.
Electric Current - The ampere is a constant current which, if maintained in two straight parallel
conductors of infinite length and of negligible circular cross sections and placed one meter apart
in vacuum, would produce between these conductors a force equal to 2 x 10 -7 newton per meter of
length. The ampere was adopted by the Ninth CGPM in 1948.
Temperature - The kelvin, a unit of thermodynamic temperature, is the fraction 1/273.16 of the
thermodynamic triple point of water. The kelvin was adopted by the Thirteenth CGPM in 1967.
Amount of substance - The mole is the amount of substance of a system that contains as many
elementary entities as there are atoms in 0.012 kilogram of carbon-12. The mole was defined by
the Fourteenth CGPM in 1971.
Luminous intensity - The base unit candela is the luminous intensity in a given direction of a
source that emits monochromatic radiation of frequency 540 x 10 12 hertz and has a radiant
intensity in that direction of 1/683 watts per steradian.
20
Scientists close to new definition of the kilogram
By Caroline Copley
LONDON | Sun Jan 23, 2011
LONDON (Reuters Life!) - Scientists say they are close to achieving a 200-year-old goal of
creating a universal system of measurements based on stable quantities, as they progress toward
changing how the kilogram is defined.
The kilogram is the only base unit in the International System of Units (SI) that is still defined by a physical object -a prototype of platinum-iridium kept in the vaults of the International Bureau of Measurements (BIPM) in France.
The stability of the kilogram is crucial as it forms the basis from which many other units are derived.
But measurements made over more than 100 years suggest that the mass of the international prototype may have
changed by about 50 micrograms -- the size of a small grain of sand -- prompting the BIPM to try to develop a new
definition based on a fundamental physical property.
Scientists will gather at the Royal Society, Britain's national academy of science, on Monday to present their progress
on redefining the kilogram according to something called "the Planck constant," a fundamental constant of quantum
physics. "International consensus has been achieved, that in the near future the kilogram shall be redefined,
based on a fixed value of the Planck constant," Michael Stock, a physicist at the International Bureau of Weights
and Measurements (BIPM), said in a statement.
Stock said researchers have been conducting experiments that establish a link between mass and the Planck constant
by comparing measurements of electrical and mechanical power. But the new definition of the kilogram can't take
place until the results of tests, conducted in laboratories across the world, are in agreement, he explained.
The International System of Units is the world's most widely used system of measurements for commerce and
science. It is made up of seven base units -- meter, kilogram, second, ampere, kelvin, candela and mole -- each of
which represents a different physical quantity. Its origins can be traced back to 18th century France and it has been
recognized internationally as the standard metric system since the 1960s. The meter was once defined as the distance
between two lines on a platinum-iridium prototype, but is now defined by the speed of light.
(Writing by Caroline Copley; editing by Paul Casciato)
Lecture #8
EGR 120 – Introduction to Engineering
Derived Units
There are hundreds of other units in the SI system; however, there are no more definitions. All
other SI units (called derived units) are built on the 7 base units. Several derived units are
shown in table 6.3 in the text and they fall into two categories:
1) Derived units that are given new names
Example: Force, F = ma = (1 kg)(1 m/s2) = 1 kg.m/s2 = 1 newton (1 N)
(so the new derived unit name of newton was assigned to the kg.m/s2 )
Example: Work = (Force)(distance) = (1 N)(1 m) = 1 N.m = 1 joule (1 J)
(so the new derived unit name of joule was assigned to the N.m = kg.m2/s2 )
So the newton and the joule don’t need to be scientifically defined. They are based on the
definitions of the kilogram, meter, and second.
2) Derived units that are NOT given new names
Example: Velocity = Distance/Time = (1 m)/(1 s) = 1 m/s
(but no new name was given to the m/s. Why not? Who knows? Most new unit names are
given in honor of a pioneer in the field, such as Newton, Pascal, and Faraday (unit of farads).
Perhaps the m/s will some day be called the Andretti in honor of the race car driver!)
Example: Acceleration = Velocity/Time = (1 m/s)/(1 s) = 1 m/s/s = 1 m/s2
(but again no new name was given to the m/s2)
22
Lecture #8
EGR 120 – Introduction to Engineering
SI Prefixes
Engineers should be
familiar with all SI
prefixes since they are
commonly used.
Table 6 in the textbook
shows the SI prefixes.
They are also shown to the
right.
Suggestion: Know this
table for Test #1.
Multiplier Prefix name Symbol
10
+18
exa
E
10
+15
peta
P
10+12
tera
T
10+9
giga
G
10
+6
mega
M
10
+3
kilo
k
10+2
hecto
h
10+1
deka
da
10
-1
deci
d
10
-2
centi
c
10
-3
milli
m
10-6
micro
m
10-9
nano
n
10
-12
pico
p
10
-15
femto
f
atto
a
10-18
23
Lecture #8
EGR 120 – Introduction to Engineering
Rules for SI Prefixes
1) Do not use scientific notation with prefixes.
Example: F = 2.75 x 106 N or F = 2.75 MN are acceptable, but F = 2.75 x 103 kN is
usually considered to be poor style.
2) Prefixes are not used with units of temperature.
Example: 4.00 (103) C, not 4.00 kC
1.25 (103) K, not 1.25 kK
3) Choose the correct prefix such that the number lies in the range 0.1 to 1000.
Example: Express each of the following quantities with the correct prefix.
A)
154.0 x 104 N
B)
0.000070 mm
C)
0.0139 x 10-8 g
D)
12752 GN
E)
0.0552 mg
24
Lecture #8
EGR 120 – Introduction to Engineering
Rules for SI Prefixes
4) For tables it is generally preferable to use a common prefix (an exception to
rule 3)
Example:
Example:
Poor style (mixed prefixes)
Preferred style (common prefixes)
x
0.1
0.2
0.3
0.4
0.5
F
97 N
12.65 kN
2.95 MN
550 kN
80 N
x
0.1
0.2
0.3
0.4
0.5
F (kN)
0.097
12.65
2950
550
0.08
25
Lecture #8
EGR 120 – Introduction to Engineering
Rules for Units Names and Symbols
Just as doctors and lawyers should know proper terminology and notation within
their profession, engineers should know the correct way to write and use units.
Several rules are presented below. You will probably recognize how some of the
rules are ignored in everyday use.
1) Use proper unit symbols, not abbreviations.
Example: t = 10 s,
not t = 10 sec. (see Table 2 - the symbol for the unit
second is s)
I = 15 A, not
I = 15 Amps
(I represents electric current)
2) Do not use periods with unit symbols.
Example: W = 10 lb,
not
W = 10 lb.
3) Never make unit symbols plural.
Example: W = 10 lb, not W = 10 lbs
Distance = x = 15 m, not
x = 15 ms
(it looks like milliseconds instead of
meters!)
26
27
Lecture #8 EGR 120 – Introduction to Engineering
Rules for Units Names and Symbols (continued)
4) Unit symbols use lower case letters unless they are proper names and then only
the first letter is in upper case.
Examples:
P = 10 kPa (Pa is the unit symbol for pressure in pascals
from the proper name Pascal)
f = 10 kHz
(Hz is the unit symbol for frequency in hertz
from the proper name Hertz)
F = 10 kN
(N is the unit symbol for force in newtons
from the proper name Newton)
t = 10 ms
(s is the unit symbol for time in seconds - not a proper name)
G = 10 mS
(S is the unit symbol for conductance in siemens
from the proper name Siemen)
5) Full symbol names can be made plural and should be in all lower case.
Example: F = 10 N or F = ten newtons (not Newtons)
P = 10 W or P - ten watts (not Watts)
6) Use a space between the number and the unit.
Example: W = 10 lb,
not W = 10lb
Lecture #8
EGR 120 – Introduction to Engineering
Performing calculations involving prefixes
The clearest method is:
1) Convert each quantity from using prefixes to using scientific notation
2) Convert any units to base units if necessary (Note: the base unit for mass in
the kilogram, not the gram.)
2) Perform the calculation using scientific notation
3) Express the final result using a prefix.
Example: Using F = ma calculate F in each case below. Express the result with
the correct SI prefix.
A) Find F if m = 6.00 Gg, a = 8.50 cm/s2
Step 1) F = ma = (6.00 Gg)(8.50 cm/s2) = (6.00 x 109 g)(8.50 x 10-2 m/s2)
Step 2) F = (6.00 x 106 kg)(8.50 x 10-2 m/s2)
Step 3) F = 5.10 x 105 N (note that 1 kg.m/s2 = 1 N)
Step 4) F = 0.510 x 106 N = 0.510 MN
B) Find F if m = 35.0 Pg, a = 625 mm/s2
F = ma = (35.0 Pg)(625 mm/s2) = (35.0 x 1015 g)(625 x 10-6 m/s2) = (35.0 x 1012
kg)(625 x 10-6 m/s2)
28
F = 2.1875 x 1010 N = 21.875 x 109 N = 21.9 GN
Lecture #8
EGR 120 – Introduction to Engineering
Examples: Using the formula F = ma, calculate F in each case below. Express the
result with the correct SI prefix.
1. m = 175 mg, a = 255 m/s2
2. m = 875.0 pg, a = 425.5 km/s2
3. m = 125 g, a = 7.00 m/s2
4. m = 0.625 Pg, a = 9.25 km/min2
29
Lecture #8 EGR 120 – Introduction to Engineering
US Customary Units
The following US Customary Units are three of the key base units:
Quantity
Length
Time
Force
Unit name
feet
second
pound
Unit symbol
ft
s
lb
US Customary units, unfortunately, also rely on numerous other units.
Examples:
Length might be expressed in terms of inches, feet, yards, or miles.
Time might be expressed in terms of seconds, minutes, or hours.
Force might be expressed in terms of ounces, pounds, or tons.
30
Lecture #8
EGR 120 – Introduction to Engineering
Mass versus Force
Note that the kilogram, a unit of mass, is a base unit in the SI system, whereas the pound, a
unit of force, is a base unit in the US system. Using F = ma it can be seen that the newton, a
unit of force, is a derived unit in the SI system and there is a derived unit for mass in the US
system - called the slug. These relationships are shown below.
Using SI Units
Using US Units
If m = 1 kg and a = 1 m/s2 (all base units),
then
If F = 1 lb and a = 1 ft/s2 (all base units
then
F = ma = (1 kg)(1 m/s2 )
m = F/a = (1 lb)/(1 ft/s2 )
F = 1 kg.m/s2 = 1 newton = 1 N (a derived unit)
F = 1 slug (a derived unit)
So the slug is the derived unit for mass in the US system and
1 slug = (1 lb)/(1 ft/s2 ) = 1 lb.s2/ft
lb  s 2
1 slug  1
ft
31
Lecture #8
EGR 120 – Introduction to Engineering
Calculating Weight
A special case of
F = ma
(Newton’s 2nd Law of Physics)
occurs when a = g = acceleration due to gravity.
F = ma is then often written as:
W = mg.
G varies slightly over the surface of the area, but the following average values are commonly
used:
g = 32.174  32.2 ft/s2 = acceleration due to gravity (using the average radius of the Earth)
g = 9.806650 m/s2  9.81 m/s2
It is recommended that Engineering students memorize the following constants:
g = 32.2 ft/s2 = 9.81 m/s2
Example: A man weighs 180 lb. Determine his mass in slugs.
32
Lecture #8
EGR 120 – Introduction to Engineering
An alternate unit for mass in the US system: pounds-mass (lbm)
The unit slug is unfamiliar to the general public in the US. As discussed earlier, providing
engineering specifications using pounds (weight) can cause problems since weight varies
slightly with acceleration due to gravity on the Earth and weight varies tremendously
outside of the Earth. Providing specifications using mass is a better approach since mass
does not vary over the surface of the Earth or outside of the Earth.
A poor attempt to reconcile this problem of using weight versus using the unfamiliar unit of
the slug was to introduce the unit of pounds-mass (lbm). The mass in pounds-mass is equal
to the weight in pounds (lb) or pounds-force (lbf) only when g = 32.174 049 ft/s2 (exactly).
We will typically round this value to g = 32.2 ft/s2.
Example: Recall the example above where a man weighs 180 lb = 180 lbf.
Determine his mass in pounds-mass.
Solution:
Since g = 32.2 ft/s2 , his mass is 180 lbm
(This essentially means that he has a mass such that when g = 32.2 ft/s2 his weight will be 180
lbf.)
33
Lecture #8
EGR 120 – Introduction to Engineering
The problem with using the unit of pounds-mass
The unit lbm is not generally used in formulas involving mass. The slug is the proper unit for
calculations in the US system, so it is generally necessary to convert mass to slugs before
performing calculations involving mass. The following conversion can be used:
1 slug = 32.2 lbm
Example : A man has a mass of 200 lbm. If he travels to the
moon where g = 5.37 ft/s2 , determine his weight.
Other alternate (informal) units
The kilogram is the base unit for mass in the SI system.
One kilogram could also be expressed as one kilogram-mass (so 1 kg = 1 kgm).
The kilogram force (kgf) is an informal unit of force equal to the force exerted on one
kilogram of mass by g = 9.80665 m/s2 (exactly). Since the official unit for force in the
SI system is N, care should be taken to convert from kgf to N when doing calculations.
The following conversion factor can be used:
1 kgf = 9.80665 N
34
Lecture #8
EGR 120 – Introduction to Engineering
Unit conversions
Engineers must be able to work with a wide variety of units in both the SI and the US
systems. A systematic approach to unit conversions will help the engineer to avoid errors.
Unit conversions can be accomplished by the following methods:
1) Dimension analysis
2) Calculators
3) Software programs (such as MATLAB or MathCAD) or online conversion programs
(such as DigitalDutch)
Note: Conversion factors are available in an appendix to the textbook
35
Lecture #8
EGR 120 – Introduction to Engineering
Dimension Analysis
Dimension analysis is a method of balancing units during the unit conversion process.
When someone is converting from inches to feet and knows that 12 in = 1 ft, sometimes
the person mistakenly divides by 12 when he or she should have multiplied by 12, or vice
versa. Dimension analysis will eliminate this type of error.
 12 in 
 1 ft 
or

 12 in 
 1 ft 


When using dimension analysis, the person would multiply by the fraction 
- whichever one resulted in canceling the unit to be eliminated. Note that since 12 in = 1 ft,
multiplying by one of these fractions is equivalent to multiplying by 1, so the value of the
quantity does not change, but the units do change.
Example: Convert the velocity V = 50 in/s to miles per hour (mi/h or mph).
36
Lecture #8
EGR 120 – Introduction to Engineering
Example: Convert the pressure P = 32.5 lb/in 2 to N/m2 .
Example: Convert the flow rate of 500 in3/min to m3/s
37
Lecture #8
EGR 120 – Introduction to Engineering
Example: Convert the torque of 55.2 ft-lb to N-m
Example: Convert the mass-density of 1.60 slugs /ft3 to g/cm3
38
Lecture #8
EGR 120 – Introduction to Engineering
Conversions using online conversion programs
You can easily search online for a unit conversion program. , such as Digital Dutch shown
below (http://www.digitaldutch.com/unitconverter/). Online programs may not include all
possible dimensions or combinations of units. For example, the flow rate conversion
example recently covered is not easily performed using Digital Dutch.
39
Lecture #8
EGR 120 – Introduction to Engineering
Conversions using calculators
Engineering calculators typically have built in unit conversion capabilities. Some
calculators, such as the TI-89/92, or HP-48/49, attach an assigned unit to a number and use
it in subsequent calculations, generating derived or compound units if necessary. Other
calculators, such as the TI-85/86, simply convert a number into another value, but do not
retain the unit information in any form. Whichever method is used, the user still needs to
have a knowledge of appropriate units, prefixes, and dimension analysis because the
calculators only give a selection of units. They will not handle all possible unit combinations
without some intelligent assistance.
TI-85/86 Examples:
Example: Convert a temperature of 78F to Celsius using the TI-85/86.
Solution:
(Enter the value)
78
(this appears on the screen)
(Select 2nd - CONV)
(Select F5 for TEMP on the screen menu)
(Select F and then C on the screen menu)
78 F  C
(this appears on the screen)
(Press the ENTER key)
25.55555556 (the final result appears on the screen)
40
Lecture #8
EGR 120 – Introduction to Engineering
Example: Convert a pressure of 32.5 lb/in 2 to N/m 2 using the TI-85/86.
Solution:
(Enter the value)
32.5
(this appears on the screen)
(Select 2nd - CONV)
(Select MORE until PRESS appears)
(Select F3 for PRESS on the screen menu)
(Select F4 (lb/in 2 ) and then F3 (N/m2 ) on the screen menu)
32.5 lb/in 2  N/m 2
(this appears on the screen)
(Press the ENTER key)
2.240796 E5
(the final result appears on the screen)
41
EGR 120 – Introduction to Engineering
Lecture #8
Example: Convert a mass density of 1.60 slug/ft 3 to g/cm3 using the TI-85/86.
The problem is that the TI-85 does not have a unit category for mass density.
However, treating 1.60 slug/ft 3 as (1.60 slugs)/(1 ft 3 ), the numerator (mass)
and the denominator (volume) can be converted separately.
Solution:
(Enter the numerator value)
1.60
(this appears on the screen)
(Select 2nd - CONV)
(Select MORE and then F1 for MASS)
(Select F5 (slug) and then F1 (gm))
1.60 slug  gm
More difficult calculator conversions
Suppose that the calculator does not
have the desired units. Sometimes this
can be handled by breaking the
numerator and denominator into
separate parts and converting each part.
Then divide the final numerator by the
final denominator for the final result.
(this appears on the screen)
(Press the ENTER key)
2.33503 E4 (the final result appears on the screen)
(Press STO  and then the letter A and then ENTER to save the result as variable A)
Ans  A
2.33503 E4
(these two lines appear on the screen)
(Enter the denominator value)
1
(this appears on the screen)
(Select 2nd - CONV)
(Select F3 for VOL)
(Select MORE than F3 (ft 3 ) and then F1 (cm3 ))
1 ft 3  cm3
(this appears on the screen)
(Press the ENTER key)
2.83168 E4 (the final result appears on the screen)
(Press STO  and then the letter B and then ENTER to save the result as variable B)
Ans  B
2.83168 E4
(these two lines appear on the screen)
(Select ALPHA then A then  then ALPHA then B then ENTER)
A/B
0.824606
(these two lines appear on the screen)
42
Lecture #8
EGR 120 – Introduction to Engineering
TI-89 Example: (the TI-92 Plus is similar)
Example: Convert a pressure of 32.5 lb/in 2 to N/m 2 using the TI-89.
Solution:
(Enter the value)
32.5
(this appears on the screen)
(Select 2nd - UNITS)
(A menu appears. Scroll down to Pressure)
(Use the right arrow to view unit choices for Pressure)
(Select psi (pounds per square inch or lb/in 2 ))
32.5 psi
(this appears on the screen)
(Press 2nd  )
(Select 2nd - UNITS)
(Scroll down to Pressure and select Pa (1 Pa = 1 N/m 2 ))
32.5 psi  Pa
(this appears on the screen)
(Press the ENTER key)
224080 Pa
(the final result appears on the screen)
43
Lecture #8
EGR 120 – Introduction to Engineering
Combinations of units on the TI-89/92 Plus
Units can be multiplied, divided, or raised to powers using the TI-89/92 Plus. Note that the
unit ft2 is not listed under area, but the user should instead use the unit of length (ft) and
square it. Combinations of units can be easily converted into combinations of compatible
units as shown in the example below.
Example: Convert a mass density of 1.60 slug/ft3 to g/cm3 using the TI-89.
Solution:
(Enter the value)
1.6
(this appears on the screen)
(Select 2nd - UNITS, scroll down to Mass, press the right arrow, and select slug)
Select 
(Select 2nd - UNITS, scroll down to Length, press the right arrow, and select ft)
(Select ^3 to cube the unit ft)
(Press the ENTER key (optional, but this will display the units perhaps more clearly))
1.6
_slug
_ft 3
(this appears on the screen)
(Select 2nd  )
(Select 2nd - UNITS, scroll down to Mass, press the right arrow, and select gm)
Select 
(Select 2nd - UNITS, scroll down to Length, press the right arrow, and select cm)
(Select ^3 to cube the unit cm)
(Press the ENTER key)
0.824606
_gm
_cm 3
(the final result appears on the screen)
44