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1. ENGINEERING ANALYSIS
Many physical problems of interest to engineers are modeled by mathematical analysis. In the following sections you will learn about a few such models. All of those models and the analysis methods used to construct them will share five key elements. One of them, numerical value, is familiar to you. Answering a numerical question requires coming up with the right number. But in engineering that is only one part of answering such a question. This section introduces other core elements of engineering analysis: variables, dimensions, units, and significant figures, as well as a fail-safe method of dealing with units and dimensions.
The essential idea to take away from this section is that arriving at the right numerical value in performing an analysis or solving a problem is only one step in the engineer's task. The result of an engineering calculation must involve the appropriate variables; it must be expressed in the appropriate units; it must express the numerical value (with the appropriate number of digits; or significant figures); and it must be accompanied by an explicit method so that others can understand and evaluate the merits and defects of your analysis or solution.
There is one variable introduced in this section that also has a strong claim to appear in another section that deals with energy and related subjects. That variable is force, and it is the scaffold on which much of modern engineering, as well as "classical" physics, relies. The definition of force and its associated units is crucial to what follows in much of this text. It is the strongest example of the notion of units and dimensions that appears in this text and thus has been placed in this section.
In addition to the preceding concepts, modern engineers have computerized tools at their fingertips that were unavailable just a generation ago. Because these tools pervasively enhance an engineer's productivity, it is necessary for the beginning engineer to learn them as soon as possible in his or her career. Today, all written reports and presentations are prepared on a computer. But there is another comprehensive computer tool that all engineers use: spreadsheets. This tool is another computer language that the engineer must master. We put its study in section 3 so you can soon get some practice in its use.
1.1 Variables
Engineers typically seek answers to such questions as, "How hot will this get?", "How heavy will it be?", "What's the voltage?" Each of these questions involves a variable, a precisely defined quantity describing an aspect of nature. What an engineering calculation does is different from what a pure mathematical calculation might do; the latter usually focuses on the final numerical answer as the end product of an analysis.
For example, p = 3.1415926 . . . is a legitimate answer to the question, "What is the value of p?" The question, "How hot?" is answered using the variable "temperature." The question "How heavy?" uses the variable "weight." "What voltage?" uses the variable "electric potential." For our purposes, variables will almost always be defined in terms of measurements made with familiar instruments such as thermometers, rulers, and clocks. Speed, for example, is defined as a ruler-measurement, distance, divided by a clock-measurement, time. This makes possible what a great engineer and scientist William Thomson, Lord Kelvin (1824-1907), described as the essence of scientific and engineering knowledge.
I often say that when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind: it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science, whatever the matter may be.
But expressing something in numbers is only the beginning of engineering knowledge. In addition to variables based on measurements and expressed as numbers, achieving Lord Kelvin's aspiration requires a second key element of engineering analysis: units.
1.2 Units
What if you are stopped by the highway patrol on a Canadian highway and get a ticket saying you were driving at "100"? You would probably guess that the variable involved is speed. But it would also be of interest to know if the claim was that you were traveling at 100 miles per hour (mph) or 100 kilometers
per hour (kph), knowing that 100 kph is only 62 mph. Units can and do make a difference! Although the fundamental laws of nature are independent of the system of units we use with them, in engineering and the sciences a calculated quantity always has two parts: the numerical value and its associated units, if any.1 Therefore, the result of any engineering calculation must always be correct in two separate categories: It must have the correct numerical value, and it must have the correct units.
Units are a way of quantifying the underlying concept of dimensions. Dimensions are the fundamental quantities we perceive such as mass, length, and time. Units provide us with a numerical scale whereby we can carry out a measurement of a quantity in some dimension. On the other hand, units are established quite arbitrarily and are codified by civil law or cultural custom. How the dimension of length ends up being measured in units of feet or meters has nothing to do with any physical law. It is solely dependent on the creativity and ingenuity of people. Therefore, the basic tenets of units systems often are grounded in the complex roots of past civilizations and cultures.
[1. Some engineering quantities legitimately have no associated units-for instance, a ratio of like quantities.]
2. THE SI UNIT SYSTEM
The international standard of units is the SI system or, officially, the International System of Units (Le Systeme International d'Unites), which has been abbreviated to SI in many languages. It is the standard of modern science and technology and is based on MKS units (meter, kilogram, second). The fundamental units in the SI system are:
-- The meter (m), the fundamental unit of length
-- The second (s), the fundamental unit of time
-- The kilogram (kg), the fundamental unit of mass
-- The degree kelvin (K), the fundamental unit of temperature
-- The mole (mol), the fundamental unit of quantity of particles
-- The ampere (A), the fundamental unit of electric current Table 2.1 illustrates a variety of SI units that were all derived from proper names of scientists who made discoveries in each of the fields in which these units are used. Here are some rules regarding unit names.
-- All unit names are written without capitalization (unless they appear at the beginning of a sentence), regardless of whether they were derived from proper names.
-- When the unit is to be abbreviated, the abbreviation is capitalized if the unit was derived from a proper name.
-- Unit abbreviations use two letters only when necessary to prevent them from being confused with other established unit abbreviations (e.g., Wb for the magnetic field unit "weber" to distinguish it from the more common W, the watt unit of power), or to express prefixes (e.g., kW for kilowatt).
-- A unit abbreviation is never pluralized, whereas the unit's name may be pluralized. For example, kilograms are abbreviated as kg, and not kgs, newtons as N and not Ns, and the correct abbreviation of seconds is s, not sec. nor secs.
-- Unit name abbreviations are never written with a terminal period unless they appear at the end of a sentence.
-- All other units whose names were not derived from the names of historically important people are both written and abbreviated with lowercase letters-for example, meter (m), kilogram (kg), second (s), and so forth.
We assume you are familiar with mass, and you may indeed think you are, but it is not a trivial concept.
As you will soon learn, mass is not weight. In a gravitational field, mass certainly produces weight, but the mass is present even where there is no gravity such as in outer space. Mass is best considered as the quantity of matter; it is a property of the substance.
In examples that follow in this text, we introduce a failsafe method that will always allow you to develop the correct units. We will follow the numerical part of the question with square brackets [. . .] enclosing the unit conversions that are needed. For example, knowing there are 12 inches in one foot would produce the units conversion factor [12 inch/ft], so if we wanted to covert 12.7 ft^2 to in^2, we would write 12.7 [ft^2]
[12 in/ft]^2 = 12.7 x 144 [ft^2][in/ft]^2 = 1830 in^2.
Table 1 -- Some SI Units and Their Abbreviations
ampere (A) henry (H) pascal (Pa) becquerel (Bq) hertz (Hz) siemens (S) celsius (_C) joule (J) tesla (T) coulomb (C) kelvin (K) volt (V) farad (F) newton (N) watt (W) gray (Gy) ohm (O) weber (Wb)
[2. Non-SI unit systems do not generally follow this simple rule. For example, the English length unit, foot, could be abbreviated "f " rather than "ft." However, the latter abbreviation is well established within society, and changing it at this time would only cause confusion. ]
Although this may seem ponderous in this example, in examples that are more complicated it is essential to follow this methodology. We will attempt to be consistent in presenting solutions and follow this technique throughout this text.
There are many SI units pertaining to different quantities being measured and their multiples thereof (Tables 2 and 3, respectively). Some of the rationale for their fundamental units will become clearer as we proceed.
Table 2 has value beyond merely listing these units: It relates the unit's
name to the fundamental MKS units-that is, the fact that a frequency is expressed
in "hertz" may not be as useful as the fact that a hertz is nothing
but the name of an inverse second, s_1. In Table 2.3, multiples of these quantities
are arranged in factors of 1000 for convenience for very large and very small
multiples thereof.
Table 2 Some Derived SI Units
Table 3 SI Unit Prefixes
In recent years, a new subcategory of materials and technology known as Nanotechnology has arisen; it is so called because it deals with materials whose size is in the nanometer or 10^9 m range.
3. FORCE, WEIGHT, AND MASS
Central to any scientific set of units is the definition of force. You probably have had some prior introduction to this concept in your high school physics classes. For example, for a constant mass system, Newton's Second Law of Motion is correctly stated as follows:
Force on a mass is proportional to the acceleration it produces.
Newton's Second Law can be written in an equation form:
F / ma (Eqn. 1)
...where a is the acceleration of mass m. To convert Newton's force law proportionality into an equality, we need to introduce a constant of proportionality.
Suppose there exists a set of units for which the force F1 accelerates the mass m1 by a1. Then Newton's Second Law can be written as:
(Eqn. 2)
If we now eliminate the proportionality by dividing the general force-defining Equation (Eqn. 1) by Equation (Eqn. 2):
(Eqn. 3)
Clearly, the proportionality constant, now explicitly the ratio of a specific force to a specific mass and to a specific acceleration, is very important to the calculations made with this equation. We must choose both the magnitude and the dimensions of this ratio. Any consistent set of units will satisfy this equation. With this degree of flexibility, it is easy to see how a large number of different force systems have evolved.
In fact, the slug system is still preferred in the United States by some engineering specialties. Its obvious advantage is that the constant of proportionality for Newton's law is exactly unity so Newton's law can be written in the familiar form of F = ma, with the stipulation that mass m is in units of slugs. Its disadvantage is that you probably have little feel for the size of a slug.
The SI system is somewhat similar to the slug system of units but with the choice for the constant of proportionality being both easy and logical: Pick the ratio F1/m1a1 so that its numerical value is exactly 1, and pick it so that it effectively has no (i.e., [0]) dimensions.4 The unit of force thus identified is called the newton
when the unit of mass is the kilogram and the acceleration is exactly 1 m/s^2. It other words, the force F1 of 1 N is defined to be that which causes the acceleration a1 of exactly 1 m/s^2 when acting on a mass m1
of exactly 1 kg. Thus, in the SI system, F1/m1a1 _ 1 and Newton's Second Law is again written:
F = ma (Eqn. 4)
This simplification leads to the familiar form of Newton's Law of Motion that most of you have seen before. Since the conversion factor F1/m1a1,-that is, [N/(kg _ m/s2)]-is exactly unity, we have dropped it altogether.
Example 2.2
What is the force in newtons on a body of mass 102 g (0.102 kg) that is accelerated at 9.81 m/s2? Equation (Eqn. 4) is the principle we use:
F = ma = 0:102 _ 9:81 ½kg_½m=s2_ = 1:00 ½kg_m=s2_ = 1:00 N
The last two examples use an acceleration of special interest, which is that caused by Earth's gravity: g = 32.2 ft/s2 = 9.81 m/s2. Looking at Equation (Eqn. 4), the SI force acting on 1 kg mass due to gravity is not 1 N but is 9.81 N-a fact that causes distress to newcomers to the subject. Weight is thus just a special force-that due to gravity. In this sense, Equation (Eqn. 4) can be modified for the acceleration of gravity to yield that special force we call weight, W, by writing it as:
W = mg (Eqn. 5)
Example 2.3
What is the weight in newtons of a mass of 0.102 kg? Equation (Eqn. 5) is the underlying principle; the arithmetic is identical to that of Example 2.2-the force on 0.102 kg of mass (which is about 4 oz in common English units) is just 1.00 N; in other words, one newton is just about the weight of a small apple here on Earth! Perhaps this will help you mentally imagine the magnitude of a force stated in newtons.
In addition to SI units, a North American engineer must master at least one of the other systems that relates mass and force, one whose persistence in the United States is due more to custom than logic: it is the Engineering English system of units. In this system, the conversion factor between force and mass _ acceleration is not unity. Because of this we must carry an explicit proportionality constant every time we use this unit system. And because the proportionality factor is not unity, it requires the use of an explicit set of units as well.
4 Read [0] to mean dimensionless.
This system has also evolved into a rather unfortunate convention regarding both the pound unit and the definition of force. It was decided that the name "pound" would be used both for mass and weight (force).
Since mass and force are distinctly different quantities, a modifier had to be added to the pound unit to distinguish which (mass or weight) was being used. This was solved by simply using the phrase pound mass or
pound force with the associated abbreviations lbm and lbf, respectively, to distinguish between them.
In the English Engineering system it was decided that a pound mass should weigh a pound force at standard gravity. (Standard gravity accelerates a mass by 32.174 ft/s2.) This has the helpful convenience of allowing us the intuitive ability to understand immediately what is meant by, say, a force of 15 lbf. It would be the force you would experience if you picked up a rock of mass 15 lbm on the Earth's surface.
This convenience was accomplished by setting the ratio (F1/m1a1) = 1/32.174 [lbf_s2/lbm_ft]. In other words, 1 lbf is defined as the force that will accelerate exactly 1 lbm by exactly 32.174 ft/s2. The designers of the Engineering English system cleverly decided to define the inverse of the proportionality constant as
gc defined as:
gc _ 32:174 lbm _ ft lbf _ s2
(Eqn. 6)
The gc symbolism originally was chosen because the numerical value (but not the dimensions) of gc is the same as that of the acceleration in standard gravity in the English Engineering units system. However, this is awkward because it tends to make you think that gc is the same as (i.e., equal to) the acceleration due to local gravity, g, which it definitely is not. The constant gc is nothing more than a proportionality constant with dimensions of [mass _ length/(force _ time2)]. Because the use of gc is so widespread today in the United States and because it is important that you are able to recognize the meaning of gc when you see it elsewhere, it will be used in the relevant equations in this course except when we are using the much more convenient (and universal) SI units. For example, in the English Engineering unit system, we will henceforth write Newton's Second Law as:
F =
ma gc
(Eqn. 7)
The consequence of this choice for F1, m1, and a1 as expressed by gc is that you can easily calculate the force in lbf corresponding to an acceleration in ft/s2 and a mass in lbm.
Example 2.4
What is the force necessary to accelerate a mass of 65.0 lbm at a rate of 15.0 ft/s2? Since the problem is stated in English units, assume the answer is required in these units. Equation (Eqn. 7) is the principle used here:
F =
ma gc =
65:0 _ 15:0 32:174
½lbm_
ft s2 _ _ lbf _ s2
lbm _ ft _ _= 30:3 lbf
Notice how the units as well as the value of gc enter the problem; without gc our "force" would be in the nonsense units of lbm/ft/s2 and our calculated numerical value in those nonsense units would be 975.
There are other consequences, too. In a subsequent section of this book you will be introduced to the quantity kinetic energy. An engineer using the SI system would define kinetic energy as ½ mv2 (here the
v stands for speed). However, an engineer using the Engineering English system would define kinetic energy as ½ (m/gc)v2. The convenient mnemonic for all applications of the Engineering English system is:
In the Engineering English system, when you see a mass m, divide it by gc.
Of course, it is logically safer to argue the units using the [..] convention previously introduced. For example, if the grouping ½ mv2 were expressed in English units, it would be [lbm] [ft/s]^2 if gc were ignored. This is a meaningless collection of units, but using the gc conversion factor, the definition of kinetic energy now would read dimensionally as [lbm] [lbf_s^2/lbm_ft][ft/s]^2 = [ft_lbf], a legitimate unit of energy in the English system (see section 3).
Understand that the constant gc has the same value everywhere in the universe-that is, the value of 32.174 lbm_ft/lbf_s2, even if its domain of acceptance is confined to the United States! In this regard, as previously stated, it should not be confused with the physical quantity g, the acceleration due to gravity, which has different numerical values at different locations (as well as different dimensions from gc).
The concept of "weight" always has the notion of the local gravity associated with it. Weight thus varies with location-indeed only slightly over the face of the Earth but significantly on non-terrestrial bodies.
To restate what has been learned: the weight of a body of mass m on the surface of the Earth is the force on it due to the acceleration "g" due to gravity-that is,
In English Engineering units; W =
mg gc
(Eqn. 8a)
where g = 32.174 ft/s^2 and gc = 32.174 lbm_ft lbf _s^2
or in SI units,
W = mg (Eqn. 8b)
where g = 9.81 m/s2.
The weight of a body of mass m when the local gravity is g0 is mg0/gc in English units and mg0 in SI.
A person on the International Space Station experiences microgravity, a much smaller g than we do on Earth, and a person on the Moon experiences about 1/6 of g compared to a person on the surface of the Earth. However, if that person is an engineer using the Engineering English system of units, he or she must use the same numerical value for gc wherever in the universe he or she may be.
In summary, when you see Newton's Second Law written as F = ma (in physics books, for example), you must use a unit system in which the proportionality constant between force and mass _ acceleration is unity and is also effectively dimensionless, such as in the SI (MKS) system.5
Example 2.5
(a) What is the weight on Earth in Engineering English units6 of a 10.0 lbm mass?
(b) What is the weight on Earth in SI units of a 10 kg mass?
(c) What is the mass of a 10.0 lbm object on the Moon (local g = 1/6.00 that of Earth)?
(d) What is the weight of that 10.0 lbm object on the Moon?
5Or in the cgs (centimeter-gram-second) system, the now outdated predecessor to SI.
6Generally, you should give your answers in the natural set of units that is suggested by the problem.
Solution
(a) W = mg/gc in English units.
Thus, W = 10.0 _ 32.2/32.2 [lbm _ ft/s^2]/[lbm_ft/lbf_s^2] = 10.0 lbf.
(b) W = mg in SI units. Thus, W = 10.0 _ 9.81 [kg _ m/s^2 ] = 98.1 [kg m/s^2] = 98.1 N. (See Table 2.2 for the definition of newtons in terms of MKS fundamental units.)
(c) Mass is a property of the material. Thus, the object still has a mass of 10.0 lbm on Earth, on the Moon, or anywhere in the cosmos.
(d) W = mg/gc in English units. On the Moon, g = 32.2/6.00 = 5.37 ft/s^2.
Thus, on the Moon, W = 10.0 _ (5.37)/ 32.2 [lbm _ ft/s^2]/[lbm_ft/lbf_s^2] = 1.67 lbf.
Until the mid-twentieth century, most English-speaking countries used one or more forms of the Engineering English units system. But because of world trade pressures and the worldwide acceptance of the SI system, many engineering textbooks today present examples and homework problems in both the Engineering English and the SI unit systems. The United States is slowly converting to common use of the SI system.
However, it appears likely that this conversion will take at least a significant fraction of your lifetime. So to succeed as an engineer in the United States, you must learn the Engineering English system. Doing so will help you avoid future repetition of such disasters as NASA's embarrassing loss in 1999 of an expensive and scientifically important Mars Lander due to an improper conversion between Engineering English and SI units7 (see Exercise 30 at the end of this section).
4 SIGNIFICANT FIGURES
Having defined your variable and specified its units, you now calculate its value. Your calculator will obediently spew out that value to as many digits as its display will hold. But how many of those digits really matter? How many of those digits actually contribute toward achieving the purpose of engineering, which is to design useful objects and systems and to understand, predict, and control their function in useful ways? This question introduces into engineering analysis a concept you have possibly seen in your high school science or mathematics courses: the concept of significant figures.
Even the greatest scientists have made the same kind of howling errors by quoting more significant figures than were justified. Newton wrote that "the mass of matter in the Moon will be to the mass of matter in the Earth as 1 to 39.788" (Principia, Book 3, proposition 37, problem 18). Since the ratio of the mass of the Earth to the mass of the Moon is actually Me/Mm = 81.300588, it is clear that Newton had gone wrong somewhere.
His value of Me/Mm to five significant figures was completely unjustified.
The use of the proper number of significant figures in experimental work is an important part of the experimentation process. Reporting a measurement of, say, 10 meters, or as 10. meters (notice the period after the zero), or as 10.0 meters, or as 10.00 meters, implies something about how accurately the measurement was made. The implication of 10 meters as written is that the accuracy of our measuring rule is of the order of _10 meters. However, 10. meters implies the measurement was good to _1 meter. Likewise, 10.0 meters implies accuracy to 0.1 meter and 10.00 meters to 0.01 meters and so on, a convention we will try to maintain in this book. Unless they are integers, numbers such as 1, 30, and 100 all have only one significant figure.
The concept of significant figures arises since arithmetic alone will not increase the accuracy of a measured quantity.9 If arithmetic is applied indiscriminately it might actually decrease the accuracy of the result. The use of significant figures is a method to avoid such blunders as 10/6 = 1.666666667 (as easily obtained on many electronic calculators), whereas the strict answer is 2! (since 6 and 10 are apparently known only to 1 significant figure here, and, if they represent real physical measurements, they apparently have not been measured to the implied accuracy of the arithmetical operation that produced 1.666666667.) In this sense, physical numbers differ from pure mathematical numbers.
"Exact" numbers (numbers such as in 1 foot = 12 inches, or numbers that come from counting, or in definitions such as diameter = 2 _ radius) have no uncertainty and can be assumed to have an infinite number of significant figures. Thus they do not limit the number of significant figures in a calculation.
Definition
A significant figure10 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.
The number 234 has three significant figures, and the number 7305 has four significant figures since the zero within the number is a legitimate significant digit. But leading zeroes before a decimal point are not significant.
Thus, the number 0.000452 has three significant figures (4, 5, and 2), the leading zeroes (including the first one before the decimal point) being place markers rather than significant figures.
How about trailing zeroes? For example, the number 12,300 is indeed twelve thousand, three hundred, but we can't tell without additional information whether the trailing zeroes represent the precision11 of the number or merely its magnitude. If the number 12,300 was precise only to _100, it has just three significant figures. If it were truly precise to _1, then all five figures are significant. In order to convey unequivocally which ending zeroes of a number are significant, it should be written as 1.2 _ 104 if it has only two significant figures, as 1.23 _ 104 if it has three, as 1.230 _ 104 if it has four, and as 1.2300 _ 104 (or 12300.-note the decimal point) if it has five.
The identification of the number of significant figures associated with a measurement comes only through knowledge of how the measurement was carried out. For example, if we measured the diameter of a circular shaft with a ruler, the result might be 3.5 inches (two significant figures), but if it were measured with a digital micrometer (Figure 2.1), it might be 3.512 inches (four significant figures).
Engineering calculations often deal with numbers having unequal numbers of significant figures. A number of logically defensible rules have been developed for various computations. These rules are actually the result of strict mathematical understanding of the propagation of errors due to arithmetical operations such as addition, subtraction, multiplication, and division. Rule 1 is for addition and subtraction.
Rule 1
The sum or difference of two values should contain no significant figures farther to the right of the decimal place than occurs in the least precise number in the operation.
10There are many good web sites on the Internet that deal with this concept. The following site has a self-test that you can use to check yourself: http://science.widener.edu/svb/tutorial/sigfigures.html
[The word "precision" has a specific meaning: it refers to how many times using independent measurements we can reproduce the number. If we throw many darts at a dartboard and they all cluster in the double three ring, they are precise but, if we were aiming at the bull's eye, they were not accurate! ]
For example, 113.2 + 1.43 = 114.63, which must now be rounded to 114.6. The less precise number in this operation is 113.2 (having only one place to the right of the decimal point), so the final result can have no more than one place to the right of the decimal point. Similarly, 113.2 - 1.43 = 111.77 must now be rounded to 111.8.
This is vitally important when subtracting two numbers of similar magnitude since their difference may be much less significant than the two numbers that were subtracted. For example, 113.212 _ 113.0 = 0.2 has only one significant figure even though the "measured" numbers each had four or more significant figures.
There is another rule for multiplication and division, as follows.
Rule 2
The rule for multiplication and division of figures is:
The product or quotient should contain no more significant figures than are contained by the term with the least number of significant figures used in the operation.
For example, (113.2) x (1.43) = 161.876, which must now be rounded to 162, and 113.2/1.43 = 79.16, which must now be rounded to 79.2 because 1.43 contains the least number of significant figures (i.e., three) in each case.
Finally a rule for "rounding" is as follows.
Rule 3
The rule for rounding numbers up or down is:
When the discarded part of the number is 0, 1, 2, 3, or 4, the next remaining digit should not be changed. When the discarded part of the number is 5, 6, 7, 8, or 9, then the next remaining digit should be increased by one.
FIGURE 1--Digital Micrometer
[There is another round-off rule corresponding to Rule 3: The so-called Bankers' Rule was used before computers to check long columns of numbers. When the discarded part of the number is exactly 5 followed only by zeros (or nothing), then the previous digit should be rounded up if it is an odd number, but it remains unchanged if it is an even number. It was meant to average out any rounding bias in adding the columns. ]
For example, if we were to round 113.2 to three significant figures, it would be 113. If we were to round it further to two significant figures, it would be 110, and if we were to round it to one significant figure, it would be 100 with the trailing zeroes representing placeholders only. As another example, 116.876 rounded to five significant figures is 116.88, which further rounded to four significant figures is 116.9, which further rounded to three significant figures is 117. As another example, 1.55 rounds to 1.6, but 1.54 rounds to 1.5.
SUMMARY
To summarize this section, the results of an engineering analysis must be correct in four ways. It must involve the appropriate variables. It must be expressed in the appropriate units. It must express the correct numerical value with the appropriate number of significant figures.
Newton's Second Law of Motion presents additional challenges in dimensional analysis, since so many possibilities are open to define its proportionality constant. In the MKS system, the proportionality constant is unity, so we can write it as F = ma; in the English Engineering system, the proportionality constant gc is chosen to have a value of 32.174 lbm_ft/lbf_s^2 and Newton's law is written as F = ma/gc. Dependent quantities such as kinetic energy are likewise modified in the English Engineering system of units by dividing quantities that contain lbm by gc.
EXERCISES
To help get you in the habit of applying these elements, in the following exercises you will be graded on use of all the elements we have discussed in this section. In these problems, where necessary, assume that on the surface of the Earth, g = 9.81 m/s^2 = 32.2 ft/s^2 (i.e., each to three significant figures). Make sure you are reporting the solution to the proper number of significant figures.
1. If a US gallon has a volume of 0.134 ft^3 and a human mouth has a volume of 0.900 in^3, then how many mouthfuls of water are required to fill a 5.00 US gallon can? (A: 1.29 _ 10^3 mouthfuls)
2. Identify whether you would perform the following unit conversions by definition, by conversion factors, by geometry, or by scientific law.
a. How many square miles in a square kilometer?
b. How many microfarads in a farad?
c. What is the weight on Earth in N of an object with a mass of 10.0 kg?
d. How many square miles on the surface of the Earth?
3. The height of horses from the ground to their shoulder is still measured in the old unit of hands. There are 16 hands in a fathom and 6.0 feet in a fathom. How many feet high is a horse that is 13 hands tall? (A: 4.9 ft)
4. An acre originally was defined as the amount of land that an oxen team could plow in a day. Suppose a team could plow 0.4 hectare per day, where a hectare is 10^4 m^2. There are 1609 meters in a mile. How many acres are there in a square mile?
[ Distinguish US gallons from the old English measure of Imperial gallons; 1.00 Imp. gallon = 1.20 US gallons. ]
[ The furlong or "furrow-long" was the distance of 220 yards that the oxen could plow in a day times a width of one "chain" or 22 yards. Multiplied together they defined the area of an acre.]
5. There are 39 inches in a meter. What is the area in the SI system of the skin of a spherical orange that is 4.0 inches in diameter? (A: 3.3 _ 10_2 m^2)
6. There are 39 inches in a meter. What is the volume in the Engineering English system of a spherical apple that is 10. cm (note the decimal point here) in diameter?
7. If the pressure in the tire on your car is 32.0 lbf/in^2 (or psi), what is its pressure in SI units?
8. Suppose the mass in Example 1 was 50.0 slugs. What would be its weight in lbf (pounds force)?
9. What would the 5.00 slug mass in Example 1 weigh on the Moon where the acceleration of gravity is only 1/6 of that on Earth?
10. What would be the force on the body in Example 2 if its mass were 856 grams?
11. What would be the weight of the body in Example 3 on the Moon where the acceleration of gravity is just gMoon = 1.64 m/s^2?
12. What force would be necessary in Example 4 if the mass were 735 lbm?
13. What is the value and units of gc in the Engineering English system on the Moon?
14. Acceleration is sometimes measured in g's, where 1.0 g = 9.8 m/s^2. How many g's correspond to the steady acceleration of a car doing "zero to sixty" in 10.0 seconds? (A: 0.27 g)
15. What is your mass in kilograms divided by your weight in pounds? Do you have to step onto a scale to answer this question? How did you answer the question?
16. If power (measured in W, or watts) is defined as work (measured in J, or joules) performed per unit time (measured in s), and work is defined as force (measured in N or newtons) _ distance (measured in m) and speed is defined as distance per unit time (measured in m/s), what is the power being exerted by a force of 1000. N on a car traveling at 30. m/s? (Assume force and speed are in the same direction, and treat all numbers as positive.) (A: 3.0 _ 104 W)
17. A rocket sled exerts 3.00 x 10^4 N of thrust and has a mass of 2.00 x 10^3 kg. In how much time does it do "zero to sixty"? How many g's (see Exercise 14) does it achieve?
18. A person pushes a crate on a frictionless surface with a force of 100. lbf. The crate accelerates at a rate of 3.0 feet per second. What is the mass of the crate in lbm? (A: 1.07 _ 10^3 lbm)
19. The force of gravity on the Moon is one-sixth (i.e., 1/6.0) as strong as the force of gravity on Earth.
An apple weighs 1.0 N on Earth. (a) What is the mass of the apple on the Moon, in lbm? (b) What is the weight of the apple on the Moon, in lbf? (Conversion factor: 1.00 kg = 2.20 lbm)
20. How many lbf does it take for a 4.0 _ 10^3 lbm car to achieve 0 to 60 mph in 10. seconds? (A: 1.1 _ 10^3 lbf)
21. Suppose a planet exerted a gravitational force at its surface that was 0.6 the gravitational force exerted by Earth. What is gc on that planet?
22. Suppose you were going to accelerate a 2000. kg car by the Rube Goldberg contraption shown in the following figure. The fan (A) blows apples (C) off the tree (B) into the funnel and thus into the bag (D). The bag is pulled downward by the force of gravity (equal to the weight of the apples in the bag), and that force is transmitted via the pulley (E) to accelerate the car (F). About how many apples each weighing 1.00 N would have to fall into the bag in order to achieve 0 to 60.0 mph in 7.00 seconds? Assume the filled bag applies a constant force to the car, equal to the weight of the apples in the bag. (A: 7.66 _ 10^3 apples)
23. Calculate with the correct significant figures: (a) 100/(Eqn. 0 _ 10^2), (b) (1.0 _ 10^2)/(Eqn. 0 _ 10^2). (A: 0.5, 0.50)
24. Calculate with the correct significant figures: (a) 10/6, (b) 10.0/6, (c) 10/6.0, (d) 10./6.0, (e) 10.0/6.00.
25. What is 2.68 x 10^8 minus 2.33 x 10^3 to the correct significant figures? (A: 2.68 x 10^8)
26. A machinist has a sophisticated micrometer that can measure the diameter of a drill bit to 1/10,000 of an inch. What is the maximum number of significant figures that should be reported if the approximate diameter of the drill bit is: (a) 0.0001 inches, (b) 0.1 inches, (c) 1 inch?
27. Round off to three significant places: 1.53, 15.345, 16.67, 102.04, _124.7, and 0.00123456.
28. Suppose you were going to design a front door and doorway to fit snugly enough to keep out the drafts, yet to be easy to open. (You are not showing off precision carpentry here, but merely designing a convenient ordinary door by standard methods.) The dimensions are to be given in inches. To how many significant figures would you specify the length and width of the door and doorway? [Assume a standard door is 30.000 by 81.000. ]
29. You are browsing the Internet and find some units conversion software that may be useful in this course.
You would like to download the software on your PC at school and use it in this course. What do you do?
a. Check with the Internet site to make sure this software is freeware for your use in this course.
b. Just download the software and use it because no one will know.
c. Download the software at home and bring it to school.
d. Never use software found on the Internet.
[The subject of tolerancing is important in mass manufacturing to ensure a proper fit with one part and another; since each part will not be exact their combined tolerance will determine how well, if at all, they fit together. It is the subject of significant statistical analysis. Applying the methods rigorously allowed the Japanese automotive industry to eclipse those of the rest of the world in terms of quality (see section 15, Manufacturing Engineering).]
Suggested method: Apply the Fundamental Canons and fill in an Engineering Ethics Matrix: 30. On December 11, 1998, the Mars Climate Orbiter was launched on a 760 million mile journey to the Red Planet. On September 23, 1999, a final rocket firing was to put the spacecraft into orbit, but it disappeared.
An investigation board concluded that NASA engineers failed to convert the rocket's thrust from pounds force to newtons (the unit used in the guidance software), causing the spacecraft to miss its intended 140-150 km altitude above Mars during orbit insertion, instead entering the Martian atmosphere at about 57 km. The spacecraft was then destroyed by atmospheric stresses and friction at this low altitude. As chief NASA engineer on this mission, how do you react to the national outcry for such a foolish mistake?
a. Take all the blame yourself and resign.
b. Find the person responsible, and fire, demote, or penalize that person.
c. Make sure it doesn't happen again by conducting a software audit for specification compliance on all data transferred between working groups.
d. Verify the consistent use of units throughout the spacecraft design and operations.
Suggested method: Apply the Fundamental Canons and fill in an Engineering Ethics Matrix
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