Transcript 02~chapter_2

Exploring Engineering
Chapter 2
Key elements in
Engineering Analysis
What we are going to learn
 Maybe the most important single lecture in this course
(which you should have already read ahead).
 Engineering is about units as well as numbers.
 How to deal with units and dimensions
 Newton’s 2nd law of motion
 SI and Engineering English units
 “gc” and “g”
 Significant figures
Prefix
pico
nano
micro
mili
centi
deci
deca
hecto
kilo
mega
giga
tera
10
1
10
12
10
23
10
36
10
69
Abbreviation
m
kg
s
Q
10
SI Unit
meter
kilogram
second
Coulomb
9
Symbol
L, l
M, m
T, t
Q,q
10
Quantity
length
mass
time
charge
21
10
32
10
63
10
96
10
9
Basic SI Units and Prefixes
Symbol
P
n
µ
m
c
d
da
h
k
M
G
T
Value
10 12
10 9
10 6
10 3
10 2
10 1
101
10 2
10 3
10 6
10 9
1012
Derived SI Units
Quantity
Symbol
SI Unit
Abbreviatio
n
electric charge
Q, q
coulomb
C
electric potential
V, v
volt
V
resistance
R
ohm

conductance
G
siemens
S
inductance
L
henry
H
capacitance
C
farad
F
frequency
f
hertz
Hz
force
F,f
newton
N
energy, work
W, w
joule
J
power
P, p
watt
W

weber
Wb
B
tesla
T
magnetic flux
magnetic flux density
Units and Dimensions
 In this course we will require the units to be manipulated
in square brackets […] in each problem.
 While easy to get the previous solutions without this method,
many engineering problems are much harder than this & need this
apparently clumsy methodology.
 Computerized unit conversions are available in free
software on the Internet (for example at:
http://joshmadison.com/software/convert-for-windows)
More Conversion Examples:
 These use conversion factors you can paste from Convert.exe
 800 m/s to mph
 800 [m/s][3.28 ft/m][1/5280 miles/ft][3600 s/hr]
 800 x 2.236 = 1790 [mph]
 2,000 hp to kW
 2,000 [hp][0.7457 kW/hp] = 1492 kW
 9.81 x 104 N m to ft lbf
 9.81 x 104 [N m][1/4.448 lbf/N][3.28 ft/m]
  9.81 x 104 x 0.737 = 7.23 x 104 ft lbf
Newton’s 2nd Law and Units
 What Newton discovered was not “may the force be with
you”, nor “may the mass  acceleration be with you” but that
force is proportional to the acceleration that it produces
on a given mass.
F   mass  accelerat ion 
or F  ma
Force, Weight, and Mass
 In high school you learned F = ma but there’s more to
it
 Newton said that force was proportional to mass x
acceleration (not equal to it) because the equation also defines
force
 So an undefined force is given by F  ma and in some also
undefined unit system F1  m1a1 (e.g., Force in units of
wiggles, mass in carats and acceleration in
furlongs/fortnight2)
 Eliminate the proportionality,
 F1 
F
ma
ma

and  F  
F1 m1a1
 m1a1 
g c  32.174
lbm ×ft
lbf × s 2
Force, Weight, and Mass
 The ratio (F1/m1a1) is arbitrary. Picking it defines the
unit of force.
 SI system: F1  1 Newton when m1 = 1 kg and a1 = 1
m/s2
 Then you can use F = ma
 English system: F1  1 lb force when m1 = 1 lb mass
and a1 = 32.174 ft/s2

ma
 lbm  ft 
Define g c  32.174 
and F 
lbf
2 
gc
 lbf  s 
Example 1
 What would the SI force on a body if its mass were 856
grams?
Need: Force on a body of mass 856 g (= 0.856 kg)
accelerated at 9.81 m/s2
 Know: Newton’s Law of Motion, F = ma
 How: F in N, m in kg and a in m/s2.
 Solve: F = ma = 0.856  9.81 [kg] [m/s2 ] = 8.397 = 8.40 N

Example 2
 What would the lbf force on a body if its mass were 3.25 lb
mass?
 Need: lbf on a body of 3.25 lbm accelerated at 32.2 ft/s2
 Know: Newton’s Law of Motion, F = ma/gc
 How: gc = 32.2 lbm ft/lbf s2
 Solve: F = ma/gc = 3.25  32.2 /32.2[lbm] [ft/s2 ][lbf s2]/[lbm
ft] = 3.25 lbf
 Weight is W = mg/gc – a special familiar force.
Example 3
 What would the lbf force on a body located on the moon (g
= 5.37 ft/s2) if its mass were 3.25 lbm?
 Know: Newton’s Law of Motion, F = ma/gc
 How: gc = 32.2 lbm ft/lbf s2 unchanged
 Solve: F = ma/gc = 3.25  5.37 /32.2[lbm] [ft/s2 ][lbf
s2]/[lbm ft] = 0.542 lbf
Newton’s 2nd Law and Units
 It bears repeating: SI system is far superior and simpler:
F  ma provided m in kg and a in m/ s
2
Example: How many N to accelerate 3.51 kg
by 2.25 m/s2?
• Ans: F = 3.51 x 2.25 [kg][m/s2] = 7.88 N
Significant figures
 Arithmetic cannot improve the accuracy of a result
 10 meters, 10. meters, 10.0 meters and 10.00 meters are not
identical
 10 meters implies you have used a 10 meters scale; 10. meters
implies you have used a 1 meter scale; 10.0 meters implies you
have used a 0.1 meter scale and 10.00 meters implies you have
used a 0.01 meter scale
Tutorial on the Use of Significant
Figures
http://www.chem.sc.edu/faculty/morgan/
resources/sigfigs/index.html
Rules for deciding the number of significant
figures in a measured quantity:
 (1) All nonzero digits are significant:
 1.234 g has 4 significant figures,
 1.2 g has 2 significant figures.
 (2) Zeroes between nonzero digits are significant:
 1002 kg has 4 significant figures,
 3.07 mL has 3 significant figures
 http://www.chem.sc.edu/faculty/morgan/resources/sigfig
s/sigfigs3.html
Rules for deciding the number of significant
figures in a measured quantity:
 (3) Leading zeros to the left of the first nonzero digits are not
significant; such zeroes merely indicate the position of the
decimal point:
 0.001 oC has only 1 significant figure,
 0.012 g has 2 significant figures.
 (4) Trailing zeroes that are also to the right of a decimal point
in a number are significant:
 0.0230 mL has 3 significant figures,
 0.20 g has 2 significant figures.
 http://www.chem.sc.edu/faculty/morgan/resources/sigfig
s/sigfigs3.html
Rules for deciding the number of significant
figures in a measured quantity:
 (5) When a number ends in zeroes that are not to the right of a
decimal point, the zeroes are not necessarily significant:
 190 miles may be 2 or 3 significant figures,
 50,600 calories may be 3, 4, or 5 significant figures.
 The potential ambiguity in the last rule can be avoided by the use
of standard exponential, or "scientific," notation. For example,
depending on whether the number of significant figures is 3, 4, or
5, we would write 50,600 calories as:
 5.06 × 104 calories (3 significant figures)
 5.060 × 104 calories (4 significant figures), or
 5.0600 × 104 calories (5 significant figures).
 http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigf
igs3.html
Significant figures
 Thus 10/6 = 2 and not 1.66666667 etc. as displayed in your
calculator
 A significant figure is any one of the digits 1, 2, 3, 4, 5,
6, 7, 8, 9, and 0. Note that zero is a significant figure
except when it is used simply to fix the decimal point
or to fill the places of unknown or discarded digits.
Significant figures
 1.23 has 3 sig. figs.
 4567 has 4 sig. figs.
 0.0123 has three sig. figs.
 12,300 has three sig.figs. (The trailing zeroes are place holders
only)
 1.23 x 103, 1.230 x 103, 1.2300 x 103 have 3, 4, and 5 sig.
figs. respectively
Rules for mathematical operations
 In carrying out calculations, the general rule is that the accuracy of a
calculated result is limited by the least accurate measurement involved in
the calculation.
 (1) In addition and subtraction, the result is rounded off to the last
common digit occurring furthest to the right in all components.
Another way to state this rule is as follows: in addition and subtraction,
the result is rounded off so that it has the same number of decimal places
as the measurement having the fewest decimal places (or digits to the
right). For example,
 100 (assume 3 significant figures) + 23.643 (5 significant figures) =
123.643,
 which should be rounded to 124 (3 significant figures). Note, however, that
it is possible two numbers have no common digits (significant figures in the
same digit column).
 http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.h
tml
Rules for mathematical operations
 (2) In multiplication and division, the result should be
rounded off so as to have the same number of significant
figures as in the component with the least number of
significant figures. For example,
 3.0 (2 significant figures ) × 12.60 (4 significant figures) =
37.8000
 which should be rounded to 38 (2 significant figures).

http://www.chem.sc.edu/faculty/morgan/resources/sigfig
s/sigfigs3.html
Rules for rounding off numbers
 1) If the digit to be dropped is greater than 5, the last retained
digit is increased by one. For example,
 12.6 is rounded to 13.
 (2) If the digit to be dropped is less than 5, the last remaining digit
is left as it is. For example,
 12.4 is rounded to 12.
 (3) If the digit to be dropped is 5, and if any digit following it is
not zero, the last remaining digit is increased by one. For example,
 12.51 is rounded to 13.
 http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigf
igs3.html
Rules for rounding off numbers
 (4) If the digit to be dropped is 5 and is followed only by
zeroes, the last remaining digit is increased by one if it is odd,
but left as it is if even. For example,
 11.5 is rounded to 12,
 12.5 is rounded to 12.
 This rule means that if the digit to be dropped is 5 followed
only by zeroes, the result is always rounded to the even digit.
The rationale for this rule is to avoid bias in rounding: half of
the time we round up, half the time we round down.
 http://www.chem.sc.edu/faculty/morgan/resources/sigfig
s/sigfigs3.html
Rules for rounding off numbers
 When using a calculator, if you work the entirety of a long calculation without writing
down any intermediate results, you may not be able to tell if an error is made. Further,
even if you realize that one has occurred, you may not be able to tell where the error is.
 In a long calculation involving mixed operations, carry as many digits as possible through
the entire set of calculations and then round the final result appropriately. For example,
 (5.00 / 1.235) + 3.000 + (6.35 / 4.0)=4.04858... + 3.000 + 1.5875=8.630829...
 The first division should result in 3 significant figures. The last division should result in 2
significant figures. The three numbers added together should result in a number that is
rounded off to the last common significant digit occurring furthest to the right; in this
case, the final result should be rounded with 1 digit after the decimal. Thus, the correct
rounded final result should be 8.6. This final result has been limited by the accuracy in
the last division.
 http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html
Significant figures – example
 Round off 123.456 − 123.0
 123.456 has 6 sig. figs.
 123.0 has 4 sig. figs.
 But 123.0 is the least precise of these numbers with just 1
figure to right of decimal place
 Thus 123.456 − 123.0 = 0.456 = 0.46 = 0.5
 The moral: In this course you will be graded on
significant figures – read your text for all the relevant
rules of round-off!
Homework
 Book Problems
 18, 19, 22, 25, 26, 27, 28
18
19
22
Summary
Engineering problems need precise mathematics
 But not more precise than can be justified (see text, Chapter 1)
 Units must be consistent

[…] method is very helpful in maintaining correct units
 Newton’s 2nd law defines force and gives rise to different sets of
units
In SI, force = ma and wt = mg
In English units, force = ma/gc and wt = mg/gc
gc is a universal constant that defines force in lbf and g is merely the
acceleration due to gravity on Earth
Significant Figures are important in engineering
calculations.

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