### Video Transcript

In this video, we will learn how to
identify fundamental and derived quantities used in mechanics, such as length, time,
and velocity, and identify their units and unit conversions.

A unit is a defined magnitude of a
quantity. Any other magnitude of the same
quantity may be expressed as a multiple of the unit. For example, the kilogram is a unit
of mass. And any other mass, such as the
mass of a typical person, may be defined as a multiple of that unit. Specific units of measurement are
arbitrary and may be defined by convenience. For example, degrees and radians
were arbitrarily defined to be useful in certain situations.

Radians are typically more useful
in mathematical problems, such as trigonometry, since the base functions of sine and
cosine operate naturally with radians. Quantities in degrees, minutes, and
seconds must first be converted into radians before inputting them into
trigonometric functions. But they are a much more intuitive
way for humans to think about angles. Nevertheless, both of these units
describe the same quantity, also known as a dimension, in this case an angle.

Every system of units has a unit
for every physical quantity. And the most commonly used system
of units is the International System of Units, or SI. The International System of Units
consists of seven base units for the following base quantities: for mass, the
kilogram or kg; for time, the second, s; for length, the meter, m; for electric
current, the ampere, capital A; for amount of substance, the mole, mol; for luminous
intensity, the candela, cd; and for thermodynamic temperature, the Kelvin, capital
K. Each of these base quantities also
comes with a dimension symbol, which is used in dimensional analysis, which weβll
get to in a moment.

The International System of Units
forms a coherent system of quantities, meaning that any other physical quantity can
be expressed as a product of powers of any of the seven base units. So letβs look at an example for how
to derive the SI unit for a quantity that is not among the seven base units.

Velocity is equal to the distance
covered over a period of time. Which of the following is not a
unit of velocity? Is it (a) centimeters per second,
(b) meters per second, (c) meters per second squared, (d) meters per minute, or (e)
kilometers per hour?

Since velocity is equal to the
distance covered over a period of time, velocity π£ is given by the equation π£
equals π over π‘, where π is the distance and π‘ is the time elapsed. We can determine the dimension of
velocity by performing dimensional analysis on this equation. All this means is replacing every
variable on the right-hand side of the equation with its dimension. So, in this case, the distance π
has dimension length πΏ and the time elapsed π‘ of course has dimension time π. Both sides of the equation must
have the same dimension. Therefore, the dimension of π£ must
be the same as the dimension of π over π‘, πΏ over π.

We can now compare this with every
one of our answers to see which one does not have units representing dimension L
over π. So, for answer (a), we have the
centimeter, which is a multiple of the meter, the base unit for length. And we have second, which is the
base unit for time. So answer (a) is πΏ over π. So this is indeed a unit for
velocity.

For answer (b), we just have the
base unit for length over the unit for time. So this is also a unit for
velocity. For answer (c), we have the base
unit for length, the meter, divided by the base unit for second squared. So the dimension is πΏ over π
squared. Therefore, (c) is not a unit for
velocity. For completeness, for answer (d),
we have the base unit for length, again, divided by the minute, which is a multiple
of the second. So we have length over time once
again. And likewise, for (e), we have the
kilometer, a multiple of the meter, divided by the hour, a multiple of the
second. So we once again have πΏ over π. Therefore, our answer is (c). Meters per second squared is not a
unit of velocity.

Sometimes there may be more than
one step to deriving the SI unit for a quantity not among the seven base
quantities. Such quantities and units are often
given a special name. The International System of Units
contains a further 22 derived quantities. Here are a few examples. As you can see, many of these are
very fundamental quantities with units you may already be familiar with, such as the
newton for force. Some of them are even surprising,
such as the coulomb for electric charge. One might think that electric
charge is a more fundamental quantity than current. But it turns out that the electric
current is easier to quantify in practice.

All of these derived units may be
written as a product of powers of the SI base units. So, for example, the newton may be
defined as one kilogram raised to the power of one times one meter raised to the
power of one divided by one second squared. Some quantities, such as plane
angle, are in fact what we call dimensionless or dimension one. This is because they are products
of the SI base units all raised to the power of zero.

Taking the plane angle as an
example, we can think of an angle as a kind of proportion. For example, the angle π radians
is equal to one-half of the circle, or one-half of the circleβs circumference,
divided by its full circumference. Both of these quantities are
lengths. So the radian, in fact, has SI base
units of meters over meters or meters to the zero. And of course anything raised to
the power zero is just one. Some of these units, for example,
the volt for electric potential difference, can get fairly complicated in terms of
the SI base units. Letβs look at an example of how to
derive these more complex units from the seven base units.

The kinetic energy, which is
measured in joules, is given by the rule π equals one-half ππ£ squared. Which of the following units is
equal to the joule? (a) Kilogram per meter per second
squared. (b) Kilogram times meter per
second. (c) Kilogram times meter squared
per second squared. (d) Kilogram per meter squared per
second squared. Or (e) kilogram times meter per
second squared.

Consider the equation for the
kinetic energy: π equals one-half ππ£ squared, where π is the kinetic energy, π
is the mass, and π£ is the velocity. In an equation representing
physical quantities, the dimensions of the quantities on both sides must be the
same. Therefore, the dimension of the
kinetic energy π must be equal to the dimension of one-half times the dimension of
π times the dimension of π£ squared. One-half is just a number. Therefore, it is dimensionless, or
dimension one. π is just mass, so it has
dimension mass. Velocity squared is a little more
complicated since it is not a base SI dimension.

To find velocity squaredβs
dimension in terms of the SI base dimensions, we will need to consider the equation
for velocity as well. Recall that the velocity is given
by the displacement π divided by the time elapsed π‘. We now have velocity expressed
purely in terms of base SI quantities, length and time. The displacement has dimension
length πΏ and the time elapsed has dimension time π, not to be confused with the π
for kinetic energy. Therefore, π£ squared equal to π
squared over π‘ squared has dimension πΏ squared over π squared. Putting all these together in the
original equation for kinetic energy, we get the dimension of kinetic energy π is
equal to one times π times πΏ squared over π squared.

We can now substitute in the SI
base unit for each of these quantities. Dimension one has no specific
unit. For mass π, we have the
kilogram. For length πΏ, we have the meter, in
this case squared. And for time π, we have the second,
in this case also squared. Putting all these together, we get
the unit for kinetic energy in terms of the SI base units: kilogram meter squared
per second squared. Comparing this with our possible
answers, we can see this matches with (c) kilogram times meter squared per second
squared.

Sometimes it might not be
convenient or helpful to express the unit of a quantity in terms of the SI base
units. For example, the torque π
represents the turning effect of a force and is given by the equation π equals
π
π, where π
is the force and π is the perpendicular distance from the pivot
point. If we were to perform dimensional
analysis on this equation to find the SI unit for torque, we would start by noting
that π
has dimension force and π has dimension length, the latter being an SI base
dimension. To find the dimension for force, we
would start with Newtonβs second law, force equals mass times acceleration, then
convert this to mass times velocity over time, and convert this to mass times
distance over time over time.

Simplifying the right-hand side, we
would get ππ over π‘ squared, which is entirely in SI base quantities and has
dimension ππΏ over π squared. Therefore, the dimension of force
in terms of SI base dimensions is ππΏ over π squared. So force has an SI base unit of
kilogram meter per second squared.

Substituting the dimension of force
into the equation for torque, we get the dimension for torque equal to ππΏ squared
over π squared. Therefore, the unit for torque in
terms of the SI base units is kilogram meter squared per second squared. This is exactly the same unit as
the unit for energy, even though the two quantities are entirely different. This shouldnβt be too surprising
though, since in a different context force times distance also gives the work done,
or energy. This is partly why we choose to
give derived quantities, such as energy, special names for their units, since it
contextualizes the unit.

Torque does not in fact have its
own special unit. And the most common convention is
to use the newton-meter, one newton times one meter. Even though these are technically
the same units, whenever we see joules, we are usually talking about energy or work
done. And whenever we see newton-meters,
we are usually talking about moments or torque.

Another common convention is to use
units with different magnitudes from the SI base units. For example, the SI base unit for
speed is meters per second. But in different contexts, for
example, the speed of a car, a different unit might be used for convenience, like
kilometers per hour. However, itβs often necessary to
work with quantities in different units. In such cases, we usually convert
all quantities into SI base units first. Letβs look at an example of how to
convert units of different magnitudes into SI base units.

What is 2.87 times 10 to the seven
dynes in newtons?

The dyne is defined as one gram
times centimeter per second squared and is particularly useful when dealing with
very small forces, such as surface tension. The newton is a derived SI unit and
is given in terms of the SI base units as one kilogram times meter per second
squared. We can now proceed by converting
all of the units on the right-hand side of the equation for the dyne into SI base
units.

So to start with, one gram is one
kilogram divided by 1000, which is equal to 10 to the negative three kilograms. And similarly, one centimeter is
equal to one meter divided by 100, which is equal to 10 to the negative two
meters. We already have the last term,
second squared, in terms of the base SI units. So we donβt need to do
anything. Substituting these conversions into
the equation for the dyne, we get one dyne equals one times 10 to the negative three
kilograms times 10 to the negative two meters over second squared. Collecting powers of 10, we get 10
to the negative five kilogram meter squared per second squared. So we have one dyne equals 10 to
the negative five newtons.

If we now multiply both sides of
the equation by 2.87 times 10 to the seven, we get the number of dynes in terms of
newtons. Simplifying the right-hand side, we
get our final answer. 2.87 times 10 to the seven dynes
equals 287 newtons.

Sometimes converting into SI base
units may not be as straightforward as multiplying by a power of 10. In the final example, letβs look at
how to convert from units with nonstandard magnitudes to SI base units and vice
versa.

Which of the following is
false? (a) 72 kilometers per hour equals
20 meters per second. (b) Three kilometers per minute
equals 50 centimeters per second. (c) 15 meters per second equals 54
kilometers per hour. Or (d) 42 meters per minute equals
70 centimeters per second.

To verify each equation, we need to
convert the units on one side of the equation into the units on the other side of
the equation. Letβs convert the units from the
left-hand side of each equation into the units on the right-hand side of each
equation and then compare the coefficients to see if they are true or false.

So, starting with (a), we need to
convert kilometers into meters and hours into seconds. One kilometer is equal to 1000
meters, and one hour is equal to 3600 seconds. We therefore have 72 kilometers per
hour equals 72 times 1000 meters per 3600 seconds. It can be helpful to write these as
fractions so we can see the multiplication and division more clearly. So we have 72000 meters divided by
3600 seconds. Simplifying this, we obtain 20
meters divided by one second, or 20 meters per second. Therefore, equation (a) is
true.

Next, for equation (b), we need to
convert kilometers into centimeters and minutes into seconds. One kilometer equals 1000
meters. So in centimeters, one kilometer is
100000 centimeters, or 10 to the five centimeters. And similarly, one minute equals 60
seconds. Therefore, we have three kilometers
per minute equals three times 10 to the five centimeters per 60 seconds. Simplifying this fraction, we get
three kilometers per minute equals 5000 centimeters per second. Therefore, the false equation is
(b) three kilometers per minute equals 50 centimeters per second.

Letβs finish this video by
recapping some key points. The International System of Units,
or SI, consists of seven base units for seven base physical quantities. Every other physical quantity has a
unit that may be expressed as a product of powers of these seven base units. The unit for any physical quantity
in terms of the SI base units may be found from an equation for that quantity using
dimensional analysis. And finally, units of nonstandard
magnitudes may be converted to SI units by multiplying by the appropriate
factor. For example, one hour equals 3600
times one second.