### Video Transcript

In this video, weβre going to learn
about dimensional analysis, what it is, how to use it, and why itβs important. To begin, imagine that youβre in
the middle of a physics examination. Youβre working on a problem with
all sorts of different kinds of information provided. And the problem asks, ultimately,
for the acceleration, π, of the object described in the problem. You take some of the information
given in the problem statement and combine those values. Subtracting, adding, dividing, and
finally coming up with an answer on your calculator screen for the acceleration.

Of course, on an exam, we all want
to get the questions right as much as possible. And the question is, is there a way
that we could check our result for acceleration, π, to give us some confidence that
weβre on the right track? The answer is yes. And we would do it through a method
called dimensional analysis.

Dimensional analysis is analyzing
relationships between physical quantities using units. There are two main applications for
dimensional analysis. The first application is to verify
dimensional consistency. Every physically measurable
quantity has units that go along with it that describe that quality. For example, a measurement of force
has units of newtons, which can be simplified in the basic SI terms: kilograms,
meters, and seconds.

Thereβs a statement known as
Newtonβs second law, which says that the force on an object is equal to that
objectβs mass times its acceleration. This is a statement we could
evaluate for dimensional consistency. A measured value of force has units
of kilograms meters per second squared. Kilograms has a dimension of mass,
meters has a dimension of length, and inverse second squared has a dimension of π
to the negative two.

Evaluating the right side of our
equation, mass times acceleration, the dimensions of that product equals dimensions
of mass, π, multiplied by a length per time squared. So we see that, dimensionally,
Newtonβs second law is consistent. So thatβs one application of
dimensional analysis.

A second application is to convert
between units. Say youβre given a time in hours,
but you want to report that time in seconds. That would involve converting the
unit of time from hours to seconds. Or if you had a value of length in
meters and you wanted to express that length in units of inches. That also would involve unit
conversion, which would rely on dimensional analysis. Letβs get some practice trying
dimensional analysis out with a couple example problems.

A student is trying to remember
a formula from geometry. Assuming that π΄ corresponds to
area, π corresponds to volume, and all other variables are lengths. What missing dimension in the
formula π equals π΄ must the right-hand side of the equation be multiplied by
to make the formula dimensionally consistent?

Considering these two terms, π΄
corresponding to area and π to volume, if we write out their dimensions, the
dimensions of π΄ are length squared, while the dimensions of π are length
cubed. Considering the formula π
equals π΄ multiplied by something, weβre asked what dimension needs to fit in to
that blank spot. If we look at the equation from
a dimension perspective, we can see that, on the left-hand side, we have πΏ
cubed. And currently, on the
right-hand side, we have πΏ squared. So weβre missing a dimension of
πΏ.

This means we would want to
multiply the right-hand side of our equation by a length πΏ in order to make it
dimensionally consistent. And with this πΏ in place, our
equation is.

Now, letβs try a second practice
problem involving dimensional analysis.

Consider the physical
quantities π, π , π£, π, and π‘ with dimensions. Dimensions of π equals capital
π. Dimensions of π equals capital
πΏ. Dimensions of π£ equals πΏπ to
the negative one. Dimensions of π equals πΏ
times π to the negative two. And dimensions of π‘ equals
capital π. The equation π equals ππ
over π is dimensionally consistent. Find the dimension of the
quantity on the left-hand side of the equation.

We want to solve for the
dimension of the term capital π that appears on the left-hand side of the given
equation. Since weβre given an equation
that weβre told is dimensionally consistent, we can write that the dimensions of
π is equal to the dimension of π times the dimension of π divided by the
dimension of π. If we enter these dimensions
in. Where the dimension of π is
capital π. The dimension of π is capital
πΏ. And the dimension of π is πΏ
times π to the negative two. We see that, in this fraction,
the factors of length πΏ cancel out. And we can write it in a
simplified form as capital π times π squared. This is the net dimension of
the right-hand side of the equation. And since the equation is
dimensionally consistent, itβs also the dimension of capital π. So our result is capital π
times capital π squared.

In summary of dimensional analysis,
dimensional analysis involves analyzing relationships between physical quantities
using units. The symbol of brackets indicates
dimensions of some quantity. For example, the dimensions of
seconds, π , is equal to time, capital π. And finally, dimensional analysis
enables answer-checking.

Remember that hypothetical exam
question that asked us to solve for acceleration, π. We know the units of acceleration
are meters per second squared. That is, its dimension is length
per time squared. And we can check to see if the way
weβve combined given variables yields those units. If it does, we can be confident
that our numerical answer is likely on the right track. So dimensional analysis is very
helpful for letting us compare physical quantities.