Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
61
Chapter 3: How to solve problems and spreadsheet analyses
Make sure the students are reporting the solution to the proper number of significant
figures. The solution set uses the ‘need-know-how-to-solve’ method. They should also be
encouraged to use spreadsheet analysis. Make sure they format their cells with the
appropriate number of significant figures; this will ensure they continually practice this
skill and not allow the calculator to decide on many non-significant figures and the
accompanying absurdities.
Often we use both ‘need-know-how-to-solve’ and spreadsheet methods, even when the
solutions are obvious and the arithmetic is equally obvious since practice is the key.
========================================================
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
62
3-1. Suppose the ranch in Example 1 was a circle instead of a square. Using the same
financial information ($320,000 available funds, $10,000 a mile for fence, and $100,000
per square mile land cost), what would be the diameter of the ranch?
Need: Diameter of circular ranch
Know: $320,000 available funds, $10,000 a mile for fence, and $100,000 per
square mile land cost
How: Use sketch to visualize
situation and then make appropriate
mathematical model.
Solve: Cost of fence = $10,000 × πD
Cost of area = $100,000 × πD2/4
Total cost = $10,000 × πD + $100,000 × πD2/4 = $320,000
Hence diameter given by 10π D2/4 + πD – 32 = 0
Solution using Excel, “Goal seek” equation solver
A
B
C
D
E
Use "goal seek" First guess at
Equation using Update using Updated
answer
in
miles
1st guess
goal seek
Equation
23
See Excel help
1.000
-21.004
1.828
-7.54E-05
24
A
B
C
Use "goal seek" First guess at
Equation using 1st guess
answer in miles
23
See Excel help
24
1
=10*PI()*B24^2/4 + PI()*B24-32
D
Update using goal
seek
E
Updated Equation
1.82838777060369
=10*PI()*D24^2/4 + PI()*D24-32
Diameter = 1.83 miles
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
63
3-2. Suppose the ranch in Example 1 was an equilateral triangle instead of a square.
Using the same financial information ($320,000 available funds, $10,000 a mile for
fence, and $100,000 per square mile land cost), what would be the length of one side of
the ranch?
Need: Diameter of equilateral ranch
Know: $320,000 available
funds, $10,000 a mile for
fence, and $100,000 per
square mile land cost
How: Use sketch to visualize
situation and then make
appropriate mathematical
model. Assume length of a side
is 2x
Solve: Cost of fence = $10,000 × 6x
Cost of area = $100,000 × √3 × x2
Total cost = $10,000 × 6x + $100,000 × √3 x2 = $320,000
Hence half side x given by 10 √3 x2+ 6x – 32 = 0
Solution using Excel, “Goal seek” equation solver
A
B
C
D
E
Use "goal seek" First guess at 1/2 Equation using Update using Updated
answer in miles 1st guess
goal seek
Equation
51
52 See Excel help
A
B
Use "goal seek" First guess at 1/2
answer in miles
51
52 See Excel help
1
1.000
-8.679
1.197
0.000
C
Equation using 1st guess
D
E
Update using Updated Equation
goal seek
=10*SQRT(3)*B52^2 + 6 * B52 - 32
1.19702198786=10*SQRT(3)*D52^2 + 6 * D52 - 32
Thus one side = 2x = 2.40 miles
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
64
3-3. Suppose the ranch in Example 1 was a rectangle with the long side twice as long as
the short side. Using the same financial information ($320,000 available funds, $10,000
a mile for fence, and $100,000 per square mile land cost), what would be the length of the
short side of the ranch?
Need: Short side of a 2:1 rectangular ranch
Know: $320,000 available funds, $10,000 a mile for fence, and $100,000 per
square mile land cost
How: Use sketch to visualize situation
and then make appropriate
mathematical model. Assume length
of short side is x
Solve: Cost of fence = $10,000 × 6x
Cost of area = $100,000 × 2 × x2
Total cost = $10,000 × 6x + $100,000
× 2 x2 = $320,000
Hence 20x2 + 6x -32 = 0
A
B
Use "goal seek" First guess at
answer x miles
81
82 See Excel help
A
Use "goal seek"
81
82 See Excel help
C
D
E
Equation using Update using Updated
1st guess
goal seek
Equation
1.000
-6.000
1.124
0.000
B
C
First guess at Equation using 1st guess
answer x miles
D
Update using goal
seek
E
Updated Equation
1
1.12377040092609
=20*D82^2 + 6 * D82 - 32
=20*B82^2 + 6 * B82 - 32
Thus the short side of the ranch is 1.12 miles.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
65
3-4. The great physicist Enrico Fermi used to test the problem solving ability of his
students at the University of Chicago by giving them the following problem: how many
piano tuners are there in the city of Chicago? Assume the population of Chicago is five
million people).
Need: ___________ Piano tuners in Chicago.
Know: Population of Chicago is 5. × 106 people.
How: Estimate 1 person in 100 Chicagoans owns a piano and gets it tuned
yearly. A standard working year is about 200 days. Assume tuner can tune
a piano in 2 hours and, with travel, he/she can tune 3 pianos per day. Use
the method of dimensional analysis to aid in the solution.
Solve: Except for unemployed piano tuners, the required tunings = the
number of tuners available.
(5.× 106/100.) × 1 [Chicagoans][piano/Chicagoan][tunings/piano yr] =
50,000 [piano tunings/year].
∴ 50,000/200 [piano tunings/year][year/working day] = 250 [piano
tunings/day]
# tuners = 250/3 [piano tunings/day] [tuners day/tuning] = 83.3 = 80
tuners.
If think the assumptions are poor just use the spreadsheet program blow
and change them to reasonable ones you prefer.
A
1 Population
2 Pianos
3 Freq. tuning
4 Tuned/yr
5 Tuned/day
6 Tuners
B
C
5000000
=B1*E1
1
per yr
=B2*B3 per yr
=B4/E2 per day
=B5/E3
D
Fraction pianos
Tuning days
Tuned/day/tuner
Copyright ©2010, Elsevier, Inc
E
=1/100
200
3
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
66
3-5. Of all the rectangles that have an area of one square meter, what are the dimensions
(length and width) of the one that has the smallest perimeter? Solve by graphing on a
spreadsheet.
Need: Minimum perimeter ____________ for 1 m2 enclosed rectangle.
Know: Perimeter, P = 2 × (L + W) where L is length, W is width and area
enclosed, A = L × W = 1 m2.
How: Set L = 1/W to guarantee 1 m2 area so that P = 2 × (1/W + W). Plot
P vs. W for minimum.
Solve: P minimum if L = W = 1.00 m.
B
1
2
3
5
6
7
8
9
D
B
C
Perimeter, m
4
C
10
11
12
13
14
15
16
D
W , m L , m P, m
L, m
P, m
1 W, m
0.50 2.00
5.00 2 =0.5
=1/B2
=2*(B2+C2)
4.9 1.67
0.60
4.53 3 =B2+0.1
=1/B3
=2*(B3+C3)
0.70
1.43
4.26
=B3+0.1
=1/B4
=2*(B4+C4)
4
4.7
0.80
4.10 5 =B4+0.1
=1/B5
=2*(B5+C5)
4.5 1.25
0.90 1.11
4.02 6 =B5+0.1
=1/B6
=2*(B6+C6)
4.3
1.00 1.00
4.00 7 =B6+0.1
=1/B7
=2*(B7+C7)
4.1
1.10 0.91
4.02 8 =B7+0.1
=1/B8
=2*(B8+C8)
1.20
4.07 9 =B8+0.1
=1/B9
=2*(B9+C9)
3.9 0.83
2
Area
=
1.00
m
1.30
0.77
4.14
=B9+0.1
=1/B10
=2*(B10+C10)
10
3.7
1.40 0.71
4.23 11 =B10+0.1 =1/B11
=2*(B11+C11)
3.5
1.50 0.67
4.33 12 =B11+0.1 =1/B12
=2*(B12+C12)
0.0
0.5
1.0
1.5
2.0
2.5
1.60 0.63
4.45 13 =B12+0.1 =1/B13
=2*(B13+C13)
1.70 0.59
4.58 14 =B13+0.1
=2*(B14+C14)
Length, m =1/B14
1.80 0.56
4.71 15 =B14+0.1 =1/B15
=2*(B15+C15)
1.90 0.53
4.85 16 =B15+0.1 =1/B16
=2*(B16+C16)
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
67
3-6. Suppose you want to make a cylindrical can to hold 0.01 m3 of soup. The sheet steel
for the can costs $0.01/m2. It costs $0.02/m to seal circular pieces to the top and bottom
of the can and along the long seam. What are the dimensions (radius and height) of the
cylindrical can that is least expensive to make?
Need: Minimum cost can = __________ $
Know: Volume of cylindrical can is V = π D2/4 × h where D is diameter and h is
height. Also seal length = h + 2π D. Total area of materials used = 2 × π D2/4 + π
D h. Multiply by respective cost factors.
How: Since V = 0.01 m3 is fixed, h = 4V/πD2
Solve: Multiply by cost factors and varying with the can’s diameter. Minimum
can cost = $0.03 or 3 cents per can. (This big can’s dimensions are 17 cm
diameter × 44 cm tall).
A
B
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
D
E
F
G
2
1 D, m
2
C
h, m
Area, m $ material Seal, m $ seal
Total $
0.01 127.32
4.00
0.04 127.39
2.55
2.59
0.03
0.05
0.07
0.09
0.11
0.13
0.15
0.17
0.19
0.21
0.23
0.25
0.27
0.29
0.31
0.33
0.35
14.15
5.09
2.60
1.57
1.05
0.75
0.57
0.44
0.35
0.29
0.24
0.20
0.17
0.15
0.13
0.12
0.10
1.33
0.80
0.58
0.46
0.38
0.33
0.30
0.28
0.27
0.26
0.26
0.26
0.26
0.27
0.28
0.29
0.31
0.01
0.01
0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
H
2 Constants
3
3 m can
4 $/m seal
2
5 $/m steel
14.34
5.41
3.04
2.14
1.74
1.57
1.51
1.51
1.55
1.61
1.69
1.77
1.87
1.97
2.08
2.19
2.30
I
0.01
0.02
0.01
Copyright ©2010, Elsevier, Inc
0.29
0.11
0.06
0.04
0.03
0.03
0.03
0.03
0.03
0.03
0.03
0.04
0.04
0.04
0.04
0.04
0.05
0.30
0.12
0.07
0.05
0.04
0.03
0.03
0.03
0.03
0.03
0.04
0.04
0.04
0.04
0.04
0.05
0.05
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
A
1 D, m
2 0.01
B
C
D
h, m
Area, m2
=4*$I$3/(PI()*A2^2) =0.5*PI()*A2^2+PI()*A2*B2
3 =0.02+A2 =4*$I$3/(PI()*A3^2)
4 =0.02+A3 =4*$I$3/(PI()*A4^2)
=0.5*PI()*A3^2+PI()*A3*B3
=0.5*PI()*A4^2+PI()*A4*B4
68
E
F
=C3*$I$5 =B3+2*PI()*A3 =E3*$I$4 =D3+F3
=C4*$I$5 =B4+2*PI()*A4 =E4*$I$4 =D4+F4
3
0.01 m can
0.07
$ total can cost
0.06
0.05
0.04
0.03
0.02
0.01
0.00
0
0.1
G
$ material Seal, m
$ seal
Total $
=C2*$I$5 =B2+2*PI()*A2 =E2*$I$4 =D2+F2
0.2
Diameter, m
Copyright ©2010, Elsevier, Inc
0.3
0.4
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
69
3-7. Suppose the mass used in Example 3.3 was increased from 10.0 kg to 20.0 kg, and
the wire stretched by twice as much. If the 20.0 kg mass was then used to stretch a 4.00 m
piece of the same steel wire, how much will it stretch?
(Example 3.3: A 2.00 m steel wire is suspended from a hook in the ceiling by with a mass
of 10.0 kg that is tied to its lower end; the wire stretches by 15.0 mm under this load. If
this same mass is used to stretch a 4.00 m piece of the same steel wire, how much will it
stretch?)
Need: Stretch = ____ mm for a 4.00 m piece of wire under a 20.0 kg load.
Know: Answer to example 3.3: A 4.00 m wire stretched 30.0 mm with a
10.0 kg load.
How: We are assuming a plausible relationship that a wire stretches
proportional to its length and proportional to its suspended weight.
Solve: Since the 4.00 m wire was stretched 30.0 mm with a 10.0 kg load,
our proportionality law suggests the same wire will stretch 60.0 mm under
a 20.0 kg load.
An apparatus for this measurement is called Searle’s apparatus: http://www.scool.co.uk/assets/learn_its/alevel/physics/stress-and-strain/stress-straingraphs/image4.jpg
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
70
3.8. Use the spreadsheet analysis in Example 3.4 to determine the total miles driven by
each driver and the average miles driven in each category for the following car renters:
Renter
City Miles
Geske
35
Pollack
27
Loth
14
Sommerfeld
12
Thunes
22
Lu
5.0
Suburban Miles
57
11
43
31
16
21
Highway Miles
93
275
159
305
132
417
Need: Average and total miles for drivers
Know: Matrix table of data as above
How: Spreadsheet analysis
Solve:
A
98
99
100
101
102
103
104
105
106
107
B
City Miles
Renter
Geske
Pollack
Loth
Sommerfeld
Thunes
Lu
35
27
14
12
22
5
57
11
43
31
16
21
93
275
159
305
132
417
Average miles
19
30
230
A
98
99
100
101
102
103
104
105
106
107
C
D
E
Suburban Miles Highway miles Total miles
B
185
313
216
348
170
443
Renter
City Miles
C
Suburban Miles
D
Highway miles
E
Total miles
Geske
Pollack
Loth
Sommerfeld
Thunes
Lu
35
27
14
12
22
5
57
11
43
31
16
21
93
275
159
305
132
417
=SUM(B100:D100)
=SUM(B101:D101)
=SUM(B102:D102)
=SUM(B103:D103)
=SUM(B104:D104)
=SUM(B105:D105)
Average miles
=AVERAGE(B100:B105)
=AVERAGE(C100:C105)
=AVERAGE(D100:D105)
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
71
3-9. Using the renter mileage information given in the previous exercise and the miles per
gallon information given in Example 3.5, determine the average gallons per journey
segment and average per driver for this set of drivers.
Need: Average gallons by segment and by driver
Know: Matrix table of data as above plus mpg data
How: Spreadsheet analysis
Solve:
A
117
118
119
120
121
122
123
124
125
126
127
128
129
130
B
City miles
Renter
E
Av gal/driver
Geske
Pollack
Loth
Sommerfeld
Thunes
Lu
35
27
14
12
22
5
57
11
43
31
16
21
93
275
159
305
132
417
Average gallons
1.6
1.7
8.9
City
Surburbs
mpg
12
A
118
119
120
121
122
123
124
125
126
C
D
Suburban Miles Highway Miles
B
9.7
13.4
9.7
14.5
7.8
17.6
Highway
18
C
26
D
Geske
Pollack
Loth
Sommerfeld
Thunes
Lu
35
27
14
12
22
5
57
11
43
31
16
21
93
275
159
305
132
417
Average gallons
=AVERAGE(B119:B124)/B$130
=AVERAGE(C119:C124)/C$130 =AVERAGE(D119:D124)/D$130
Copyright ©2010, Elsevier, Inc
E
=B119/B$130+C119/C$130+D119/D$130
=B120/B$130+C120/C$130+D120/D$130
=B121/B$130+C121/C$130+D121/D$130
=B122/B$130+C122/C$130+D122/D$130
=B123/B$130+C123/C$130+D123/D$130
=B124/B$130+C124/C$130+D124/D$130
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
72
3-10. Using the technique introduced in Example 8, create a spreadsheet graph of the
following data for the median annual salaries in dollars for engineers based on years of
experience, supervisory responsibility, and level of education 5.
Number of Years After BS Degree
5
10
15
25
0
Nonsupervisory
B.S. 55,341
63,649 73,162
M.S. —
79,875 86,868
Ph.D. —
—
91,352
Supervisory
B.S. —
—
72,632
M.S. —
99,367 109,450
Ph.D. —
—
—
80,207
85,116
90,134
97,463
98,053 108,747
80,739
110,360
110,877
92,029
113,916
132,800
35
92,748
110,289
122,886
107,844
117,146
147,517
Need: Graph of salaries corresponding to table
Know: To plot mathematical data, normally need “scatter” plot (with true
Cartesian x axis)
How: Use “Insert chart” in Excel and follow instructions and/or intuition
Solve:
Median salary by experience
160,000
140,000
Salary in $
120,000
Non Super, BS
Non super, MS
100,000
Non super, PhD
Super, BS
Super MS
80,000
60,000
Super PhD
40,000
20,000
0
0
10
20
30
40
Years since BS
5
These data are from the 2007 report of the Engineering Workforce Commission
of the American Association of Engineering Societies (AAES).
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
73
Problems 11- 13 involve the following situation:
Suppose that the weight of the gasoline in lbf in a car’s gas tank equaled the weight of the
car in lbf to the 2/3 power (If G = gasoline weight, then W = car weight, G = W2/3).
Assume further that gasoline weighs 8.0 lbf/gal, and gas mileage varies with weight
according to the empirical formula:
mpg = (84,500 lbf mi/gal) × (1/car’s weight in lbf) -3 miles/gal
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
3-11. What is the fuel usage of a 3.00 × 103 lbf car?
Need: mpg = ___________ for a 3.00 × 103 lbf car.
Know - How: Empirical formula for mpg:
Solve: mpg = (84,500 lbf mi/gal) × (1/car’s weight in lbf) -3. miles/gal
∴mpg = (84,500 lbf mi/gal) × (1/car’s weight in lbf) – 3. miles/gal =
84,500/3.00 × 103 [lbf mi/gal][1/lbf] – 3. [miles/gal] = 25 mpg.
Copyright ©2010, Elsevier, Inc
74
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
75
3-12. What is the heaviest car that can achieve a range of 600. miles?
Need: Heaviest car with range of 600. miles.
Know: Weight and fuel capacity related by G = W2/3 and a formula
relating mpg and vehicle weight. Thus you can have high mpg in a small
car, but it also has a small gas tank and vice versa.
How: Spreadsheet analysis and graphing range in miles vs. tank capacity
in US gallons.
Solve: The heaviest car with a 600. mile range weighs 3.69 × 103 lbf.
A
B
C
lbf/gal
2
D
E
8.0
3
4 wt car, lbf
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
A
4 wt car, lbf
5 1500
6 =A5+250
7 =A6+250
mpg
1500
1750
2000
2250
2500
2750
3000
3250
3500
3750
4000
4250
4500
4750
5000
5250
53
45
39
35
31
28
25
23
21
20
18
17
16
15
14
13
B
mpg
=84500/A5-2.9
=84500/A6-2.9
=84500/A7-2.9
lbf gas/car Gal/car Range, miles
131
16.4
875
145
18.2
824
159
19.8
781
172
21.5
744
184
23.0
711
196
24.5
683
208
26.0
657
219
27.4
634
231
28.8
612
241
30.2
592
252
31.5
574
262
32.8
557
273
34.1
541
283
35.3
526
292
36.6
512
302
37.8
498
C
lbf gas/car
=A5^(2/3)
=A6^(2/3)
=A7^(2/3)
D
Gal/car
=C5/$C$2
=C6/$C$2
=C7/$C$2
Copyright ©2010, Elsevier, Inc
E
Range, miles
=B5*D5
=B6*D6
=B7*D7
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
76
800
750
Range in miles
700
650
600
550
500
450
400
1500
2500
3500
4500
Wt of car in lbf
Copyright ©2010, Elsevier, Inc
5500
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
77
3-13) Suppose the formula for weight of the gas was G = Wb where b is can be varied in
the range 0.5 – 0.75. Graph the range of a 3.69 × 103 lbf car as a function of b.
Need: Range as function of 0.75> b >0.50
Know - How: Mpg formula, weight of a gallon of gas and that the weight
of a 3.69 × 103 lbm car is 3.69 × 103 lbf.
Solve: Notice in the graph just how sensitive the solution is to the
exponent. In effect, the car’s designer can control ‘b’ by appropriately
shaping and sizing the gas tank. Most cars have about a 400 mile range.
C
D
3 Wt of car
4
5 b
6
7
8
E
3.69E+03 lbf
G
Wt of gal gas
H
8.0 lbf
lbf gas/tank Gal/tank
mpg
Range, miles
0.50
6.07E+01 7.59E+00
2.00E+01
1.52E+02
0.51
6.59E+01 8.24E+00
2.00E+01
1.65E+02
0.52
7.16E+01 8.95E+00
2.00E+01
1.79E+02
C
3
F
D
E
F
G
Wt of car
3700
lbf
Wt of gal gas
=8
b
=0.5
=C6+0.01
=C7+0.01
=C8+0.01
lbf gas/tank
=$D$3^C6
=$D$3^C7
=$D$3^C8
=$D$3^C9
Gal/tank
=D6/$G$3
=D7/$G$3
=D8/$G$3
=D9/$G$3
mpg
=84500/$D$3-2.9
=84500/$D$3-2.9
=84500/$D$3-2.9
=84500/$D$3-2.9
Range, miles
=F6*E6
=F7*E7
=F8*E8
=F9*E9
4
5
6
7
8
9
Copyright ©2010, Elsevier, Inc
H
lbf
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
78
1.2E+03
Range in miles
1.0E+03
8.0E+02
6.0E+02
4.0E+02
2.0E+02
0.0E+00
0.50
0.55
0.60
0.65
Exponent b
Copyright ©2010, Elsevier, Inc
0.70
0.75
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
79
Problems 14–17: These problems are concerned with bungee jumping. At full stretch, the
elastic rope of original length L stretches to L + x. For a person whose weight is W lbf,
and a cord with a stiffness K lbf/ft, the extension x is given by this formula:
x=
W
W 2 2W × L
+
+
K
K
K2
and which can be written in spreadsheet script as
x = W K + sqrt W ∧ 2 K ∧ 2 + 2W * L K .
(
)
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
80
3-14. If the height of the cliff is 150. ft, K = 6.25 lbf/ft, L = 40.0ft and the person’s weight
is 150. lbf, will the person be able to bungee jump safely? Support your answer by giving
the final value for length = L + x.
Need: Total length of extended bungee cord in ft.
Know: Energy conservation equation: x =
W
W 2 2W × L
+
+
with W =
K
K
K2
150. lbf, K = 6.25 lbf/ft, and L = 40.0 ft.
How: Set up spreadsheet as alternative to hand solving.
Solve: Final stretched length = L + x = 114 ft < 150. ft and OK.
B
C
D
L, ft
5 K lbf/ft W, lbf
6
6.25
150
B
5 K lbf/ft
6 6.25
C
W, lbf
150
D
L, ft
40
E
x, ft
40
F
G
L + x, ft Max ht, ft
74
114
150
E
x, ft
=C6/B6+SQRT((C6/B6)^2+2*C6*D6/B6)
Copyright ©2010, Elsevier, Inc
F
L + x, ft
=D6+E6
G
Max ht, ft
150
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
81
3-15. Americans are getting heavier. What’s the jumper’s weight limit for a 40.0 ft
unstretched bungee with stiffness of K = 6.25 lbf/ft? Graph final length L + x vs. W for
weights from 100 lbf to 300 lbf in increments of 25 lbf. Print a warning if the jumper is
too heavy for a 150. ft initial height. (Hint: Look up the application of the ‘IF’ statement
in your spreadsheet program).
Need: Total length of extended bungee cord in ft.
W
W 2 2W × L
+
+
with
K
K
K2
300. > W > 100. and K = 6.25 lbf/ft, and L = 40.0 ft.
Know: Energy conservation equation: x =
How: Set up spreadsheet as alternative to hand solving.
Solve: Final stretched length = L + x < 150. ft. Only the most foolhardy
250 lbf person would risk it! Definitely not heavier!
B
C
2 K lbf/ft
6.25
6.25
6.25
6.25
6.25
6.25
6.25
6.25
6.25
3
4
5
6
7
8
9
10
11
B
2 K lbf/ft
3 6.25
4 6.25
5 6.25
C
W, lbf
100
=C3+25
=C4+25
D
L, ft
40
40
40
D
E
F
G
H
W, lbf
L, ft
x, ft
L + x, ft Max ht, ft Warning
100
40
55
95
150 alive
125
40
65
105
150 alive
150
40
74
114
150 alive
175
40
83
123
150 alive
200
40
92
132
150 alive
225
40
101
141
150 alive
250
40
109
149
150 alive
275
40
118
158
150 DEAD
300
40
126
166
150 DEAD
E
x, ft
=C3/B3+SQRT((C3/B3)^2+2*C3*D3/B3)
=C4/B4+SQRT((C4/B4)^2+2*C4*D4/B4)
=C5/B5+SQRT((C5/B5)^2+2*C5*D5/B5)
F
L + x, ft
=D3+E3
=D4+E4
=D5+E5
Copyright ©2010, Elsevier, Inc
G
Max ht, ft
150
150
150
H
Warning
=IF(F3 > G3, "DEAD", "alive")
=IF(F4 > G4, "DEAD", "alive")
=IF(F5 > G5, "DEAD", "alive")
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
82
350
Stupid!
300
250
W, lbf
Alive
Dead
200
150
100
50
0
75
95
115
135
155
L + x, final extended length, ft
Copyright ©2010, Elsevier, Inc
175
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
83
3-16. If the height of the parapet is 200. ft, and the weight of the person is 150. lbf, and
the unstretched length L = 45.0 ft, find a value of K that enables this person to stop
exactly five feet above the ground.
Need: Value of K to ensure a stop at L + x = 195 ft.
Know: W = 150. lbf, L = 45.0 ft and x =
W
W 2 2W × L
+
+
K
K
K2
How: One way would be to write the extension equation explicitly in the
form K = function of (W, L, x). But that would take work. Another way is
to use an equation solver built into Excel called “Goal Seek”. However a
simple way is to just use our previous spreadsheet and vary K. From the
previous examples, it appears that K < 6.25 lbf/ft. Use range 1.0 to 6.25 in
increments of 0.25 lbf/ft as a guess range and graph the result.
Solve: By interpolation, the value for K is 2.60 lbf/ft. Presumably the 5 ft
clearance is to allow for his or her body length so shorter people win this
round.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
B
2 K lbf/ft
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
3
4
5
6
7
8
9
10
11
12
13
14
C
W, lbf
150
150
150
D
D
L, ft
45
45
45
E
F
G
H
W, lbf
L, ft
x, ft
L + x, ft Max ht, ft
150
45
340
385
195
150
45
279
324
195
150
45
238
283
195
150
45
208
253
195
150
45
186
231
195
150
45
169
214
195
150
45
155
200
195
150
45
143
188
195
150
45
134
179
195
150
45
125
170
195
150
45
118
163
195
150
45
112
157
195
E
F
x, ft
=C3/B3+SQRT((C3/B3)^2+2*C3*D24/B3)
=C4/B4+SQRT((C4/B4)^2+2*C4*D4/B4)
=C5/B5+SQRT((C5/B5)^2+2*C5*D5/B5)
L + x, ft
=D3+E3
=D4+E4
=D5+E5
Warning
DEAD
DEAD
DEAD
DEAD
DEAD
DEAD
DEAD
alive
alive
alive
alive
alive
G
H
Max ht, ft
195
195
195
Warning
=IF(F3 > G3, "DEAD", "alive")
=IF(F4 > G4, "DEAD", "alive")
=IF(F5 > G5, "DEAD", "alive")
450
400
L + x, Total drop in ft
B
K lbf/ft
3 1
4 =B3+0.25
5 =B4+0.25
2
C
84
350
300
250
200
150
100
50
0
0.00
1.00
2.00
3.00
4.00
K, Stiffness, lbf/ft
Copyright ©2010, Elsevier, Inc
5.00
6.00
7.00
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
85
3-17. By copying and pasting your spreadsheet from problem 15, find and plot the values
of L needed (in ft) vs. W, weight of jumper (in lbf) for successful bungee jumps (coming
to a stop 5 ft above the ground) for K = 6.25 lbf/ft and from a cliff of height 150. ft above
the ground. The graph should cover weights from 100. lbf to 300. lbf in increments of 25
lbf. (Hint: The function ‘Goal seek’ under ‘Tools’ is one way to solve this exercise.)
Need: Value of L to ensure a stop at L + x = 145 ft.
W
W 2 2W × L
+
+
K
K
K2
How: One way would be to write the extension equation explicitly in the
form L = function of (W, K, x). But that would take work.
Know: K = 6.25 lbf/ft, 100 < W < 300. lbf, and x =
Solve: First way: Just use our previous spreadsheet and manually vary L
until we achieve L + x = 145 ft.
Second way: Use “Goal seek” (under “Tools”). Guess all the initial
unstretched length in column D is 50 ft, or 75 ft etc. – just something
reasonable). In Goal Seek, set cell = F3 (Excel will translate this as $F$3);
set “To value” as 145 (ft). Finally set “By changing cell” D3 (Excel will
translate this as $D$3) and hit “OK”. Cell D3 will then show the solution
for that case. Repeat for the rest of the table.
B
2 K lbf/ft
3
4
5
6
7
8
9
10
11
6.25
6.25
6.25
6.25
6.25
6.25
6.25
6.25
6.25
C
D
E
F
G
W, lbf
L, ft
x, ft
L + x, ft Max ht, ft
100
77.1
68
145
150
125
69.0
76
145
150
150
61.5
83
145
150
175
55.0
90
145
150
200
49.0
96
145
150
225
43.0
102
145
150
250
37.0
108
145
150
275
32.0
113
145
150
300
27.0
118
145
150
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
86
80
Locus of bungee length
vs. wt. for 145 ft plunge.
Unstretched bungee, ft
70
60
50
40
30
20
100
150
200
Person's wt., lbf
Copyright ©2010, Elsevier, Inc
250
300
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
87
Problems 18–20: The fixed costs per mile traveled of operating an automobile is
approximately 20% of the initial price of the car. Thus the operating cost/ mile = 0.20/yr
× (purchase price of automobile)/(miles driven per year) + (price of gasoline/gallon) ×
(gallons used per mile). In the problems that follow, assume that the automobile is driven
2.00 × 104 miles per year. Assume gasoline costs $5.00/gallon.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
88
3-18. Estimate the operating cost per mile of an automobile with a price of $15,000 that
gets 30. miles per gallon.
Need: Cost per mile to own a car.
Know: Annual mileage = 2.00 × 104. Cost of gas is $5.00/gallon and mpg
= 30.
How: Cost of ownership is in $/mile = 0.20 × (purchase price of
automobile)/(miles driven per year) + (price of gasoline/gallon) × (gallons
used per mile)
Solve: Cost of ownership per mile = 0.20 × $15,000/(2.00 × 104)
[$/yr][yr/mile] + $5.00 × 30. [$/gallon][gallons/mile]= $0.32/mile.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
89
3-19. If one were to double the price of the automobile in problem 18, what would its gas
mileage have to be in order to cost the same to operate per mile as the automobile in
problem 18?
Need: Gas mileage if total operating costs = $0.32/mile.
Know: Annual mileage = 2.00 × 104 and initial cost of car is $30,000.
How: Cost of ownership is in $/mile = 0.20 × (purchase price of
automobile)/(miles driven per year) + (price of gasoline/gallon) × (gallons
used per mile)
Solve: 0.32 = 0.20 × $30,000/(2.00 × 104) [$/yr][yr/mile]+ $5.00/mpg
[$/gallon][gallon/mile] = 0.30 + $5.00/mpg [$/mile]
Î ∴$5.00/mpg = $0.02; hence mpg = 250 mpg!
What this problem says the car must be very efficient to offset the upfront cost of buying
the vehicle if the cost of running it is to be held to 32 cents/mile. Not too likely!
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
90
3-20. Suppose that the purchase price of automobiles varies with weight according to the
formula that cost in $ = weight in lbf × $8.00, and gas mileage varies according to mpg =
(84,500 mile-lbf/gal)/W - 3. miles/gal. Graph the cost per mile of operating a car as a
function of the car’s weight, in increments of 500. lbf from 2000. lbf to 5000. lbf.
Need: Graph of operating cost for owing a car.
Know: Annual mileage = 20,000 and annualized cost of car is 20% of
purchase price. Gas costs $5.00 per gallon.
How: Cost of ownership is in $/mile = 0.20 × (purchase price of
automobile)/(miles driven per year) + (price of gasoline/gallon) × (gallons
used per mile)
Solve:
G
1
2
3
4
A
C
Cost $
1
2
3
4
5
6
7
8
2,000
2,500
3,000
3,500
4,000
4,500
5,000
A
Wt, lbf
16000
20000
24000
28000
32000
36000
40000
B
Cost $
1
2 2000
3 =A2+500
4 =A3+500
5 =A4+500
6 =A5+500
7 =A6+500
8 =A7+500
=A2*$I$3
=A3*$I$3
=A4*$I$3
=A5*$I$3
=A6*$I$3
=A7*$I$3
=A8*$I$3
I
Annual
miles
Cost of gas, $/gal
Cost of car, per lbf
Fixed cost ratio
B
Wt, lbf
H
2.00E+04
$5.00
$8.00
0.20
D
Annual mpg
operating
$0.16
$0.20
$0.24
$0.28
$0.32
$0.36
$0.40
E
39
31
25
21
18
16
14
C
Annual operating
cost/mile
=I$4*B2/$I$1
=I$4*B3/$I$1
=I$4*B4/$I$1
=I$4*B5/$I$1
=I$4*B6/$I$1
=I$4*B7/$I$1
=I$4*B8/$I$1
F
Fuel
Total
cost/mile cost/mile
$0.13
$0.29
$0.16
$0.36
$0.20
$0.44
$0.24
$0.52
$0.27
$0.59
$0.31
$0.67
$0.36
$0.76
D
E
F
mpg
Fuel cost/mile
Total cost/mile
=84500/A2-2.9
=84500/A3-2.9
=84500/A4-2.9
=84500/A5-2.9
=84500/A6-2.9
=84500/A7-2.9
=84500/A8-2.9
=I$2/D2
=I$2/D3
=I$2/D4
=I$2/D5
=I$2/D6
=I$2/D7
=I$2/D8
=E2+C2
=E3+C3
=E4+C4
=E5+C5
=E6+C6
=E7+C7
=E8+C8
Copyright ©2010, Elsevier, Inc
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91
$0.80
$0.70
$/mile
$0.60
$0.50
$0.40
$0.30
$0.20
2,000
2,500
3,000
3,500
4,000
Vehicle wt, lbf
Copyright ©2010, Elsevier, Inc
4,500
5,000
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
92
Problem 21 - 23: In visiting stores, one finds the following prices for various things.
Broccoli crowns cost $2.89 per pound. Soft drinks cost $2.00 per two-liter bottle (a liter
is 0.001 m3). A new automobile weighs 2.50 × 103 lbf and costs $1.50 × 104. A dozen
oranges, each of which is 0.06 m in diameter, costs $2.05. A 1.5 lb package of chicken
thighs costs $5.35. A dictionary weighs 5.00 pounds and costs $20. A refrigerator weighs
200. lbf and costs $900. Assume that one cubic meter of any solid object or liquid weighs
1.00 × 104 N.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
93
3-21. For the objects listed above, make a table and graph of the cost of objects in dollars
as a function of their weight in newtons. It is suggested this graph uses a line graph using
line with markers displayed at each data value (get rid of the unwanted line at using the
format series function). The value of the line graph is that everything plotted is at the
same horizontal displacement and not dependent on its value.
Need: Table and graph of costs/N of various items
Know: Prices of items, density = 1.00 × 104 N/m3
How: Express weight for all units in newtons.
Use conversion factor for 1.00 lbf Î N is 9.81/2.205 [m/s2] [kg]/[lbf] =
4.45 [N]/[lbf].
Also volume of sphere (orange) is πD3/6
Solve:
A
Object
1
2 Cost, $
3 Unit wt, lbf
3
4 m of liquid
Wt in N
5
6 $/N
E
Dozen oranges
1
2 2.05
3 --4
B
Broccoli
C
Drinks
D
Automobile
2.89
1
2
--0.002
15000
2500
=(B3/2.205)*9.81 =C4*10000 =(D3/2.205)*9.81
=B2/B5
=C2/C5
=D2/D5
F
Chicken
G
Dictionary
H
Refrigerator
5.35
1.5
20
5
900
200
5 =12*PI()*(0.06^3)*10000/6 =(F3/2.205)*9.81 =(G3/2.205)*9.81 =(H3/2.205)*9.81
Copyright ©2010, Elsevier, Inc
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94
$1.6
$1.4
$1.2
$0.8
$0.6
$0.4
Copyright ©2010, Elsevier, Inc
Refrigerator
Dictionary
Chicken
Dozen oranges
Automobile
$0.0
Drinks
$0.2
Broccoli
$/N
$1.0
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
95
3-22. What (perhaps surprising) simple generalization about the cost of things might one
make based on the table and graph of problem 21?
Need: Generalization from the solution to problem 21.
Know: Initial data set, table and graph of problem 21.
How: Look the data and deduce the information contained therein.
Solve: Look at the diversity of input data: cars, refrigerators, oranges, soft
drinks etc. Look at the graph: all of these items cost within less than one
order of magnitude. By eye, they all cost in order of magnitude around a
buck a newton of weight! More accurately the average is about 70
cents/N. A remarkable factoid!
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
96
3-23. Name a product or group of products that does not fit the generalization you made
in problem 22, and add and label the point on the graph in problem 21. To get a better
perspective use a log scale for the y-axis, $/N.
Need: Product with high cost to weight ratio.
Know: Hand made items, consumer electronics, drugs (licit and illicit!),
camera.
How: Look at your computer!
Solve: A 25 N laptop computer costs $1500, or about $60/N. Plot this
point on your graph. Point to the y axis and click; go to Format, Select
Axis, Scale, Logarithmic scale. Change the origin to $0.1/N. By stretching
the ordinate with a log scale, all of the values can be distinguished and not
all crowded along y ~ $0/N as they would display with an arithmetic
ordinate scale.
$100.0
$10.0
$/N
Copyright ©2010, Elsevier, Inc
Computer
Refrigerator
Dictionary
Chicken
Dozen oranges
Automobile
Drinks
$0.1
Broccoli
$1.0
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
97
3-24. An unnamed country has the following population of passenger cars on its roads as
determined by 250 kg mass differences. You have to make these data clear to the
undersecretary to that country’s transport minister. Plot these data by two methods: 1) as
a “pie chart” and 2) as a histogram to show the distribution in an effective manner.
Need: Information display for non-technical audience showing
distribution of the weight of cars (for this audience, do not distinguish the
concepts of ‘weight’ and ‘mass’!)
Know: Spreadsheets have multiply display types. Use pie charts and
histograms.
1
2
3
4
5
6
7
8
9
10
B
C
Upper limit, kg % all vehicles
1000
12.1
1250
13.1
1500
15.4
1750
18.6
2000
14.8
2250
9.2
2500
7.5
2750
6.3
4000
3.0
1
2
3
4
5
6
7
8
9
10
B
Upper limit, kg
1000
=B2+250
=B3+250
=B4+250
=B5+250
=B6+250
=B7+250
=B8+250
4000
C
% all vehicles
12.1
13.1
15.4
18.6
14.8
9.2
7.5
6.3
=100-SUM(C2:C9)
How: Pie Chart: Insert, Charts, Pie, Next, Series, Values: 'Prob
17'!$C$2:$C$10, Category labels: 'Prob 17'!$B$2:$B$10, Next: Category
labels. Finally toggle off Legend and Finish. You can display as colors
(default) but in the black-and-white of this text we will use patterns (see
under Format, Cells)
Histogram: Insert, Charts, Column, Next, Series1: Values: = 'Prob
17'!$C$2:$C$10, Category x axis labels: = 'Prob 17'!$B$2:$B$10, Finally
toggle off Legend and Finish.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
Cars by kg category
2750
4000
98
Wt distribution of cars by kg
1000
20
2500
18
16
1250
2250
14
12
Problems 25 - 27 deal with “Hubbert’s Peak 6”.10This is a model of supply and demand for
oil. It looks at the amount of available oil and its
rate of consumption to draw
8
conclusions about continuing the current course6of our oil-based economy.
1500
2000
4
2
0
1000 1250 1500 1750 2000 2250 2500 2750 4000
kg
1750
6. M. King Hubbert was a geologist with Shell Oil who, in the 1950’s, pointed out that the US supply of oil
was going to fall short of demand by the 1970’s. His methods have since been applied to world oil
production and, based on demand exceeding production, predicts an on-going oil supply crisis.
Copyright ©2010, Elsevier, Inc
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99
3-25. Suppose the world originally had three trillion (3. × 1012) barrels (abbreviation is
‘BBL’) of oil and its exploration began in 1850. Suppose 10.% of the remaining
undiscovered oil has been found in every quarter century since 1850. Call the discovered,
but not yet consumed oil, "reserves". Suppose oil consumption was 1. × 108 barrels in
1850 and further suppose oil consumption has grown by a factor of 5 in every quarter
century since 1850.
When will the oil start to run out? (That is, when will the reserves become negative?).
Give your answer to the nearest 25 years and provide a spreadsheet showing reserves and
consumption as a factor.
Need: Model of oil reserves
Know: Total available oil = 3. × 1012 BBL; initial consumption rate in
1850 = 1. × 108 BBL/yr; Oil consumption growth rate = factor of 5 every
25 years.
How: Discovered oil grows at 10% of undiscovered oil and so forms a
progression; reserves increase by discoveries and are reduced by
consumption.
Solve: Reserves turn negative by about 2025.
1850
=B7+25
=B8+25
=B9+25
=B10+25
=B11+25
=B12+25
=B13+25
=B14+25
5
0
=C7+25
=C8+25
=C9+25
=C10+25
=C11+25
=C12+25
=C13+25
=C14+25
3000000000000
=D7*(1-$C$4)
=D8*(1-$C$4)
=D9*(1-$C$4)
=D10*(1-$C$4)
=D11*(1-$C$4)
=D12*(1-$C$4)
=D13*(1-$C$4)
=D14*(1-$C$4)
100000000
=E7*$D$4
=E8*$D$4
=E9*$D$4
=E10*$D$4
=E11*$D$4
=E12*$D$4
=E13*$D$4
=E14*$D$4
Copyright ©2010, Elsevier, Inc
F
Oil Reserves
0.1
E
Consumption
D
Consumption
growth rate (per
quarter century)
Undiscovere
d oil
6
7
8
9
10
11
12
13
14
15
Year
3
4
5
C
Discovery
ability
25 yr
increment
B
0
=F7+$C$4*D7-E7
=F8+$C$4*D8-E8
=F9+$C$4*D9-E9
=F10+$C$4*D10-E10
=F11+$C$4*D11-E11
=F12+$C$4*D12-E12
=F13+$C$4*D13-E13
=F14+$C$4*D14-E14
BBL of oil by category
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
9.E+12
7.E+12
Oil reserves
Oil consumption
Undiscovered oil
5.E+12
3.E+12
1.E+12
-1.E+12
-3.E+12
-5.E+12
1850 1900 1950 2000 2050
Year
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100
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
101
3-26. Suppose the world originally had 10 trillion (10. × 1012) barrels of oil. Use the data
of problem 25 to again predict when the oil will start to run out.
Need: Model of oil reserves
Know: Total available oil = 10. × 1012 BBL; initial consumption rate in
1850 = 1. × 108 BBL/yr; Oil consumption growth rate = factor of 5 every
25 years.
How: Discovered oil grows at 10% of undiscovered oil and so form a
progression; reserves increase by discoveries and are reduced by
consumption.
BBL of oil by category
Solve: Surprisingly the oil reserves turn negative by about 2040, only 15
years later than the previous case. The extra 7 trillion BBLs of oil did not
much help the longevity of the reserves. Much of it was simply consumed
more quickly than the previous case! The consequences of exponential
growth in resource demand were explored in an influential study 7 in the
early 1970’s but its predictions were not universally accepted.
2.E+13
Oil reserves
2.E+13
Oil consumption
Undiscovered oil
1.E+13
5.E+12
0.E+00
-5.E+12
-1.E+13
1850
1950
2050
Year
7. Donella H. Meadows et al., The Limits to Growth: A Report for the Club of Rome's Project on the
Predicament of Mankind, (Potomac Associates, 1972)
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102
3-27. Repeat problem 25 but instead of assuming the exponential growth in
consumption continuing unabated by a factor of 5 in every quarter century since
1850, curtail growth since 2000 and assume consumption has stayed constant since
then. Again predict when the oil will start to run out.
Need: Model of oil reserves
Know: Total available oil = 10. × 1012 BBL; initial consumption rate in
1850 = 1. × 108 BBL/yr; Oil consumption growth rate = factor of 5 every
25 years until the year 2000 and then it will remain constant.
How: Discovered oil grows at 10% of undiscovered oil and so form a
progression; reserves increase by discoveries and are reduced by
consumption.
Solve: By curtailing the exponential growth in consumption, the reserves
do not turn negative until about the year 2100
BBL of oil by category
1.E+13
1.E+13
8.E+12
Oil reserves
Oil consumption
Undiscovered oil
6.E+12
4.E+12
2.E+12
0.E+00
-2.E+12
-4.E+12
1850
1950
2050
Year
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103
3-28. Your friend tells you that the “Know, Need, How, Solve” problem solving method
seems overly complicated. He/she just wants to find the answer to the problem in the
quickest possible way, say by finding some formula in the text and plugging numbers
into it. What do you tell him/her?
a)
b)
c)
d)
Go ahead and do whatever you want, then you’ll flunk and I’ll survive.
Talk to the instructor and have him/her explain why this methodology works.
Find someone who has used this method and ask to copy their homework.
Explain why this technique will leads to a fail-safe method of getting the correct
answer.
Options a) Go ahead
Canons
Hold
paramount the
safety, health
and welfare of
the public.
Perform
services only
in the area of
your
competence
Issue public
statements
only in an
objective and
truthful
manner
Act for each
employer or
client as
faithful agents
or trustees
Avoid
deceptive acts
Conduct
themselves
honorably …
b) Talk to
instructor
c) Copy
homework
d) Explain
method
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
No
Does not
apply
Yes
No
Does not
apply
Yes
No
Solution: This is not an ethical issue. A teacher should not require you to use a certain
method. The most a teacher can do is say, “if you don’t use the method, you may get a
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
104
F.” If you don’t use the method and you do get a F, then maybe you do have a problem.
But it is not an ethical problem.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
105
3-29. You Email a classmate in this course for some information about a spreadsheet
homework problem. In addition to answering your question, your classmate also attaches
a spreadsheet solution to the homework. What do you do?
a) Delete the spreadsheet without looking at it.
b) Look at their spreadsheet to make sure he/she did it correctly.
c) Copy the spreadsheet into your homework and change the formatting so that it
doesn’t look like the original.
d) Email the spreadsheet to all your friends so that they can have the solution too.
Apply the Fundamental Canons: Engineers, in the fulfillment of their
professional duties, shall:
1) Hold paramount the safety, health and welfare of the public. Does not
apply
2) Perform services only in areas of their competence. Does not apply
3) Issue public statements only in an objective and truthful manner.
Turning in homework implies a public statement that the work
turned in is wholly your own. The only way this canon can be met
here is not looking at the spreadsheet. Do a).
4) Act for each employer or client as faithful agents or trustees. The
teacher has an employer/client role here; being a faithful agent
requires you to do a).
5) Avoid deceptive acts. Looking at the spreadsheet and then
submitting any homework solution is a deceptive act. Do a).
6) Conduct themselves honorably, responsibly, ethically, and lawfully so
as to enhance the honor, reputation, and usefulness of the profession.
Again, do a).
In Engineering Ethics Matrix format:
Options a) Delete
Canons
Hold
paramount the
safety, health
and welfare of
the public.
Perform
services only
in the area of
your
competence
Issue public
statements
b) Look at and c) Copy
check
d) E-mail to
friends
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
only in an
objective and
truthful
manner
Act for each
employer or
client as
faithful agents
or trustees
Avoid
deceptive acts
Conduct
themselves
honorably …
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Does not
apply
Yes
Does not
apply
No
No
Does not
apply
No
No
Solution: No tension here. Do a).
Copyright ©2010, Elsevier, Inc
106
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
107
3-30. Stephanie knew Adam, the Environmental Manager, would not be pleased with her
report on the chemical spill. The data clearly indicated that the spill was large enough that
regulations required it to be reported to the state. When Stephanie presented her report to
Adam, he lost his temper. "A few gallons over the limit isn't worth the time it's going to
take to fill out those damned forms. Go back to your desk and rework those numbers until
it comes out right. What should Stephanie do? 8
a. Tell Adam that she will not knowingly violate state law and threaten to quit.
b. Comply with Adam’s request since he is in charge and will suffer any
consequences.
c. Send an anonymous report to the state documenting the violation.
d. Go over Adam’s head and speak to his supervisor about the problem.
Options
Canons
Hold
paramount
the safety,
health and
welfare of
the public.
Perform
services
only in the
area of your
competence
Issue public
statements
only in an
objective
and truthful
manner
Act for each
employer or
client as
faithful
agents or
trustees
Avoid
deceptive
acts
a. Tell
Adam
b. Comply
Yes
No. Would
knowingly
put public at
risk
Yes
Yes
c. Send
anonymous
report
Yes
Yes
d. Go over
Adam’s head
Yes
Yes
Does not
apply
Maybe - no
statement
may be
viewed as
approval
Does not
applyaction is not
public
No - as an
agent, you
are expected
to alert
management
to potential
problems
Yes – is not
deceptive
Maybedepends on
the basis for
Adam’s
request
No. Going
behind
employer’s
back is not
faithful
service
Yes
No- act is
deceptive
Nodeceiving
employer
Yes
8
Does not
apply- action
is not public
Abstracted from Engineering Ethics: Concepts and Cases at
http://wadsworth.com/philosophy_d/templates/student_resources/0534605796_harris/cases/Cases.htm.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
Conduct
themselves
honorably
…
Yes
No
No
Yes
You probably have to go above Adam’s head (even though that could be very
uncomfortable).
Copyright ©2010, Elsevier, Inc
108