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A Primer on Dimensions and Units
Glen Thorncroft
Mechanical Engineering Department
Cal Poly State University, San Luis Obispo
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.
The goal of this paper 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:
d
,
v=
t
where v is the symbol for velocity, d for distance, t for time. Plugging in the actual
values (and units),
( 4 ft )
ft
v=
= 0 .4
.
s
(10 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 easier. We will explore
three of these standard systems: the SI system, the British Gravitational system, and
the English Engineering System. There are more!

1

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.
Primary Dimension
mass (M)
length (L)
time (t)
temperature (T)

Standard Unit
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]

[L]
[t] 2

.

Or, if we use primary SI units, we see that
Force

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)

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.

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:
2

Primary Dimension
Force (F)
length (L)
time (t)
temperature (T)

Standard Unit
pound-force (lbf)
foot (ft)
second (s)
Rankine (R)

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]

[L]
.
[t] 2

If we use primary units, we see that
lb f

mass ⋅ ft
,
s2

Rearranging the above,
mass ∝

lb f ⋅ s 2
.
ft

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

(2)

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 one as the
constant.
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
Force (F)
mass (M)
length (L)
time (t)
temperature (T)

Standard Unit
pound-force (lbf)
pound-mass (lbm)
foot (ft)
second (s)
Rankine (R)

3

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

(3)

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 .

6. Examples
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 will not give us the correct units for force. But
we know by definition that

1N = 1kg ⋅ m/s 2 .

4

If we divide both sides by 1 kg·m/s2 , we get

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, 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 .
1. Note that we just used the definition of a newton as a “conversion factor” to convert the
answer above into a useful form.
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

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

5

⎛ 1 lb f ⋅ s 2 /ft ⎞
⎜⎜
⎟⎟ = 1 .
⎝ 1 slug ⎠
Multiplying the weight by the above gives

⎛ 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 appropriate units of force.

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
⎜⎜
2
⎝ 32.174 lb m ⋅ ft/s

⎟⎟ = 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 .

⎟⎟

Comments: 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.

7. The Proportionality Constant gc
As a final note, if you haven’t yet heard of gc (“g sub c”) in your studies, you will
soon enough. What is gc , and what do we do with it?
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
6

F=

ma
.
gc

Similarly, the gravitational force could be written as
W=

mg
.
gc

Comparing gc in the equation above with the “conversion factors” we used in the
examples, you can show that
gc = 1

kg ⋅ m/s 2
N

(SI system),

slug ⋅ ft/s 2
gc = 1
(“slug” system),
lb f
and

g c = 32.174

lb m ⋅ ft/s 2
(“pound-mass” system) .
lb f

You may choose not to use the gc approach in your calculations. 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.
8. Summary of Unit Systems
SI
System
(“Metric” system)
Primary Dim’s
MLtT
Mass
kg
Length
m
Force
N
Time
s
Temperature
K
Force-Mass
Relationship

gc

1N ≡ 1

gc = 1

kg ⋅ m
s2

kg ⋅ m/s 2
N

British Gravitational
English Engineering
(“slug” system)
(“pound-mass” system)
FLtT
FMLtT
slug
lbm
ft
ft
lbf
lbf
s
s
R
R
slug ⋅ ft
lb ⋅ ft
1 lb f ≡ 1
1 lb f ≡ 32.174 m2
2
s
s
1 slug = 32.174 lb m

gc = 1

slug ⋅ ft/s 2
lb f

g c = 32.174

lb m ⋅ ft/s 2
lb f

7