Download A Primer on Dimensions and Units

A Primer on Dimensions and Units
Glen Thorncroft, Cal Poly State University
1 Dimensions vs. Units
Nearly every engineering problem you will encounter will involve dimensions: the length of a
beam, the mass of a concrete block, the time and velocity of an object’s fall, the force of the air resistance
on an airplane, and so forth. We express these dimensions using specific units: for example, length can be
expressed in feet, mass as kilograms, time as minutes, velocity as miles per hour, and force as newtons 1.
The goal of this section is to explain the use of dimensions and units in engineering calculations,
and to introduce a few of the standard systems of units that are used.
2 How Dimensions Relate to Each Other
Dimensions (as well as units) act just like algebraic symbols in engineering calculations. For
example, if an object travels 4 feet in 10 seconds, we can calculate its velocity. First, algebraically:
v=
d
,
t
where v is the symbol for velocity, d for distance, t for time. Plugging in the actual values (and units),
v=
( 4 ft )
ft
= 0.4
.
(10 s)
s
Thus we can see that velocity in this case has the units feet per second (ft/s). We can convert feet to
whatever we like: meters, miles, etc. We can also convert seconds to minutes, hours, days, etc. But the
dimensions are always the same:\
velocity =
[length]
.
[time]
There are two kinds of dimensions: (1) primary dimensions, like length and time, and (2) secondary
dimensions, like velocity, which are combinations of primary dimensions.
Because any given system of units we use has so many different measurements, standard units
have been developed to make communication (and science and commerce) easier. We will explore three
of these standard systems: the SI system, the British Gravitational system, and the English Engineering
System. There are more!
3 The SI system
The SI (Système International d’Unitès) system is the official name for the metric system. The
system is described as an “MLtT” system, because its primary dimensions are mass (M), length (L), time
(t), and temperature (T). The standard units are listed below.
Notice that newtons is not capitalized. It is standard not to capitalize the name of the unit, even though the unit
abbreviation is capitalized (i.e., N). It’s a confusing rule.
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1
Primary Dimension Standard Unit
mass (M)
length (L)
time (t)
temperature (T)
kilogram (kg)
meter (m)
second (s)
Kelvin (K)
Secondary units are derived from these primary units. For example, velocity has units of m/s, acceleration
is m/s2, and force has units of…?
How do we relate force to the primary units? Isaac Newton discovered that the force on an object
is proportional to its mass times its acceleration:
F ∝ ma .
If we plug dimensions into the above relation, we see that
Force ∝ [M]
Or, if we use primary SI units, we see that
Force
∝
[L]
.
[t] 2
kg ⋅ m
.
s2
In honor of Newton, it was decided to give this particular set of terms the name newton (N). It is defined
as
1N ≡ 1
kg ⋅ m
.
s2
(1.7)
So the unit of force in the SI system is the newton (N), defined as “the force required to accelerate a mass
of 1 kg to an acceleration of 1 m/s2 .” Why not 2 kg? Or 10 m/s2 ? Actually, the number is arbitrary, but
the number 1 is chosen for convenience.
Example 1.
An object has a mass of 80 kg. If the acceleration of gravity is 9.81 m/s2 , what is its weight?
Solution:
The weight of an object is the force of gravity on the object, which is given by
W = mg .
Plugging in values (and units) for m and g,
W = (80 kg)(9.81 m/s 2 ) .
(a)
As you can see, the result of the above calculation does give us the correct dimensions and units for
force. But for convenience, we know by definition that
1N = 1kg ⋅ m/s 2 .
2
get
Notice that we can manipulate the above equation slightly: If we divide both sides by 1 kg∙m/s2 , we
1N
=1 .
1 kg ⋅ m/s 2
Thus, if we multiply the right-hand-side of Equation (a) by the ratio above, we are merely multiplying
by one – and a unitless value of one – which doesn’t change anything:
 1N
W = (80 kg)(9.81 m/s 2 )
2
 1 kg ⋅ m/s

 .

Note that all the units cancel except for N, which yields
W = 784.8 N .
Comments:
1.
Note that we just used the definition of a newton as a kind of “conversion factor” to convert the answer
above into a more convenient form. To be honest, it’s not necessary to use newtons, and in fact some
engineers leave the units of force as kg∙m/s2 sometimes, because they know the units will cancel later.
But just remember that you want to express your final answer in as relatable units as possible, for your
audience’s understanding.
2.
Recall that we determined the gravitational force by the equation
W = mg .
Why didn’t we use Newton’s second law, F = ma , where a = g ? Isn’t that the same? Absolutely
not! GRAVITY IS NOT ACCELERATION. IT IS A FORCE (PER UNIT MASS). It only looks like
acceleration because it has units like that of acceleration (In fact, dimensionally, acceleration and force
per unit mass are the same). Think about this. What is the force of gravity acting on your body right
now? Are you in motion right now? If you are sitting still, you are not accelerating (relative to the
ground). Then a=0! So is the force on your body zero? No!
Remember that in stating Newton’s second law, F is the net force acting on the mass m. If the mass is
stationary, the net force is zero. That is, the force of gravity on your body is exactly balanced by the
force of the ground pushing up on you. You are in equilibrium, and therefore your acceleration is zero.
4 The British Gravitational System (“Slug” System)
The British Gravitational system of units is referred to as an “FLtT” system, because the primary
dimensions are force (F), length (L), time (t), and temperature (T). The standard units are:
Primary Dimension Standard Unit
Force (F)
length (L)
time (t)
temperature (T)
pound-force (lbf)
foot (ft)
second (s)
Rankine (R)
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If force is a primary dimension, how do we find the unit of mass? Mass is now a secondary
dimension; we have to derive it. Newton’s second law always holds:
F ∝ ma .
or, dimensionally,
[F] ∝ [mass]
If we use primary units, we see that
lb f
[L]
.
[t] 2
mass ⋅ ft
,
s2
∝
Rearranging the above,
lb f ⋅ s 2
.
ft
mass ∝
We need a name for the unit of mass. Let’s call it a slug! Then we’ll define it by
1 lb f ≡ 1
slug ⋅ ft
.
s2
(1.8)
We can interpret the above by saying, “one pound-force is the force required to accelerate 1 slug to an
acceleration of 1 ft/s2 .” Again, we could have defined the slug as 10 lbf ∙s2/ft, or 936.1 lbf ∙s2/ft, but for the
sake of simplicity, we choose 1 as the constant.
Example 2.
An object has a mass of 5.59 slugs. What is its weight in Earth’s gravity?
Solution:
As in Example 1, the weight of the object can be determined by
W = mg .
Substituting the mass and the value of standard Earth gravity, 32.174 ft/s2, into the above,
W = (5.59 slug)(32.174 ft/s 2 )
The units above are not useful as units of force. But we know by definition that 1 slug =1lbf –s2/ft, or
 1 lb f ⋅ s 2 /ft 


 1 slug  = 1 .


Multiplying the weight by the above gives
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 1 lb f ⋅ s 2 /ft 

W = (5.59 slug)(32.174 ft/s 2 )

 1 slug 
= 179.85 lb f .
We see that the units in the above relation cancel, leaving the more convenient units of force.
5 The English Engineering System (“Pound-Mass System”)
In the English Engineering system of units, the primary dimensions are are force (F), mass (M),
length (L), time (t), and temperature (T). Therefore this system is referred to as a “FMLtT” system. The
standard units are shown below:
Primary Dimension Standard Unit
Force (F)
mass (M)
length (L)
time (t)
temperature (T)
pound-force (lbf)
pound-mass (lbm)
foot (ft)
second (s)
Rankine (R)
In this system, force and mass are primary dimensions. They must still be related by Newton’s
second law:
F ∝ ma .
or, dimensionally,
[F] ∝ [mass]
[L]
.
[t] 2
If we use the primary English units, we see that
lb f
lb m ⋅ ft
∝
s2
,
We don’t need to define a new unit, but we need to determine a constant in order to make the above relation
exact. Let’s use 32.174! Then the relationship between pound-force and pound-mass is as follows:
1 lb f ≡ 32.174
lb m ⋅ ft
.
s2
(1.9)
So in words, “one pound-force is the force required to accelerate one pound-mass to 32.174 ft/s2.” Why
32.174? Because that just happens to be the value for the acceleration of gravity, g = 32.174 ft/s2 . This
value was chosen so that if an object has a mass of 10 lbm, its weight on the Earth will also be 10 lbf . This
“convenience” will become apparent later in one of the examples which follow. One final note: If we
compare Equation (3) with Equation (2), we see that slugs and pounds-mass are related by
1 slug = 32.174 lb m .
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(1.10)
Example 3.
An object has a mass of 180 lbm . What is its weight in Earth’s gravity?
Solution:
Again, the weight is given by
W = mg ,
which becomes
W = (180 lb m )(32.174 ft/s 2 ) .
To convert the units in the above equation into useful force units, we note that by definition, 1 lbf
=32.174 lbm-ft/s2 . Or,

1 lb f

 32.174 lb ⋅ ft/s 2
m


 =1 .


Multiplying this constant with the weight gives

1 lb f
W = (180 lb m )(32.174 ft/s 2 )
2
 32.174 lb m ⋅ ft/s
= 180 lb f .
Comment:



Note that in Earth’s gravity, and the “pound-mass system,” the values of mass and weight are the same!
In fact, that’s how the relationship between lbf and lbm was defined. Remember, though, that the units
represent different dimensions: lbf represents force, while lbm represents mass. So it is NEVER
acceptable to write “1 lbf = 1 lbm.” This is not dimensionally correct; it is like saying that “1 apple
= 1 orange.”
6 The Proportionality Constant gc
As a final note, if you haven’t yet heard of gc (“g sub c”) in your studies, you might. It’s sometimes
referred to as the gravitational constant, and it is a less-common (some may say it’s obsolete, or oldfashioned) way to deal with the force-mass units relationship. So if you run across this term, how does it
work?
Did you notice that, in every example above, we had to multiply the weight we calculated by a
“conversion factor” to make the units come out right? Well, what some people do is just employ a factor,
called gc , directly in the equation they are using. For example, Newton’s second law could be written as
F=
ma
.
gc
Similarly, the gravitational force could be written as
W=
mg
.
gc
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Comparing gc in the equation above with the “conversion factors” we used in the examples, you can show
that
gc = 1
gc = 1
and
kg ⋅ m/s 2
N
(SI system),
slug ⋅ ft/s 2
(“slug” system),
lb f
lb m ⋅ ft/s 2
(“pound-mass” system) .
g c = 32.174
lb f
I advise you not to use the gc approach in your calculations. This technique can be confusing because you
have to remember when you have to include gc in your general equation. But, as you can see from the
example problems, we ignored gc entirely; as long as you ALWAYS keep track of ALL your units, you will
know when you need to perform unit conversions in order to cancel certain units. Think of the definitions
(1.7), (1.8), and (1.9) as a wild card that you can insert into a calculation when you need to simplify the
units.
A summary of the basic unit systems is presented in Table 1.1. Memorize the force-mass
relationships, and always use them explicitly in your calculations.
I can’t over-emphasize this point: NEVER DO UNIT CONVERSIONS IN YOUR HEAD. Always
show them, no matter how trivial. Incorrect units are a leading cause of mistakes in calculations, sometimes
leading to tragic results. Being explicit with your unit calculations will help you catch your own mistakes,
and help your audience understand your calculations (and convince them that you know what you are
doing).
System
Primary Dim’s
Mass
Length
Force
Time
Temperature
Force-Mass
Relationship
gc
Table 1.1. Summary of Unit Systems
SI
British Gravitational
(“Metric” system)
(“slug” system)
MLtT
FLtT
kg
slug
m
ft
N
lbf
s
s
K
R
1 lb f ≡ 1
kg ⋅ m
1N ≡ 1 2
s
gc = 1
kg ⋅ m/s 2
N
gc = 1
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English Engineering
(“pound-mass” system)
FMLtT
lbm
ft
lbf
s
R
lb ⋅ ft
slug ⋅ ft
1 lb f ≡ 32.174 m2
2
s
s
1 slug = 32.174 lb m
slug ⋅ ft/s 2
lb f
g c = 32.174
lb m ⋅ ft/s 2
lb f
7 Review Exercises (my solutions follow)
Problem 1. The pressure acting on a 1.25 in2 test specimen equals 15 MPa. What is the force (in N)
acting on the specimen?
Answer
Problem 2 The weight of a large steel cylinder is to be computed from measurements of its diameter and
length. Let its length L be equal to 3.32 m and its diameter d equal to 0.3605 m. Suppose that the density
of the steel equals 7835 kg/m3. Calculate the weight of the cylinder (in N) and report your result in a clear
and unambiguous form.
Answer
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8 Solutions to Review Exercises
Problem 1. The pressure acting on a 1.25 in2 test specimen equals 15 MPa. What is the force (in N)
acting on the specimen?
Problem 2 The weight of a large steel cylinder is to be computed from measurements of its diameter and
length. Let its length L be equal to 3.32 m and its diameter d equal to 0.3605 m. Suppose that the density
of the steel equals 7835 kg/m3. Calculate the weight of the cylinder (in N) and report your result in a clear
and unambiguous form.
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