Video Transcript
Which of the following formulas
correctly shows the relationship between the pressure exerted by a fluid, the height
ℎ of the fluid above the point at which pressure is measured, the density 𝜌 of the
fluid, and the gravitational acceleration rate 𝑔? A) 𝑃 is equal to 𝑔 times ℎ
divided by 𝜌. B) 𝑃 is equal to 𝜌 times 𝑔
divided by ℎ. C) 𝑃 is equal to 𝑔 divided by 𝜌
times ℎ. D) 𝑃 is equal to 𝜌 times 𝑔 times
ℎ. E) 𝑃 is equal to ℎ divided by 𝜌
times 𝑔.
So in this exercise, we’re trying
to find the correct mathematical relationship between four quantities: pressure,
density, height, and gravity. We’re told that one of our five
answer options shows this correct relationship. To figure out which one it is,
there are two ways we can go about it. The first way — and certainly the
quickest way — is to recall off the top of our head the equation for the pressure
produced by a fluid. But let’s say that we’re not able
to recall that equation. Well, there is still a way we can
solve for the correct answer. The way we can do it is by looking
at the units on either side of each one of these equations.
Here’s what we mean by that. We can see that each one of these
five equations has four terms in it. Each one has pressure, it has
height, ℎ, it has density, 𝜌, and it has acceleration due to gravity, 𝑔. Now we can take each one of these
terms and we can look at the units that that term involves. For example, the units of pressure
are pascals, abbreviated Pa. And a pascal we can recall is equal
to a newton of force spread over a square meter of area. And we can further recall that
because a Newton is equal to a kilogram meter per second squared, then that means
the units of pressure are kilograms meters per second squared per meter squared,
which simplifies in the most basic of base units to a kilogram per meter second
squared.
The base unit of height on the
other hand is meters. The base unit of density is
kilograms per cubic meter. And we can recall that the units of
the acceleration due to gravity are meters per second squared. Now here’s the reason we’ve gone
through all this. If we look at the left-hand side of
each one of our five candidate equations, we see they all involve the pressure 𝑃 by
itself. We’ve seen from our analysis that
the units of pressure are kilograms per meter second squared. This means that these are the units
on the left-hand side of all of our candidate equations.
And for an equation to be true,
that is for it to be accurate, that means that the units on the right-hand side must
match these. They must also be kilograms per
meter second squared. If they’re not, then that means the
two sides of the equation can’t be equal. And that means that that particular
equation doesn’t correctly represent the relationship between these four values. So here is what we’ll do. We’ll go through these equations
one by one, starting with equation A. And we’ll analyse the units on the
right-hand side. If those units match up to the
units of pressure, then we’ll mark that equation as a possibly correct solution.
Starting out with equation A, the
right-hand side here has 𝑔 times ℎ divided by 𝜌. Looking over at the units of each
one of these terms, we see that the units of 𝑔 are meters per second squared, the
units of ℎ are meters, and the units of density are kilograms per cubic meter. In the numerator of this fraction,
these two factors of meters combine. And then if we multiply this
fraction by meters cubed divided by meters cubed, then that term meters cubed
cancels out in our denominator. And we have meters to the fifth
divided by second squared divided by kilograms or simply meters to the fifth divided
by second squared kilograms. When we compare these units to the
units we have for pressure, we see that there’s not a match, which means that option
A is not a reasonable candidate for the correct mathematical relationship.
Moving on to the right-hand side of
option B. Here, we have units of kilograms
per cubic meter for 𝜌 multiplied by meters per second squared for 𝑔 divided by
meters for ℎ. This numerator simplifies to
kilograms per meter squared second squared. And then if we divide both
numerator and denominator by meters, we have a final result of kilograms per cubic
meter second squared. This is close to the units for
pressure, but it’s not quite right. Therefore, option B isn’t our
answer either.
Moving to the right-hand side of
option C. Here we have meters per second
squared, the units of 𝑔, divided by kilograms per cubic meter, the units of 𝜌,
multiplied by meters, the units of height. A meter divided by meters cubed is
equal to one over meters squared. And then, if we multiply top and
bottom of this fraction by the inverse of the denominator, in other words multiply
it by meter squared per kilogram, then the meter squared per kilogram, the
denominator, cancels out with kilograms per meter squared. We’re left with a final result of
meters cubed per kilogram second squared, also not equal to the units of
pressure. So we cross option C off our
list.
Looking at the right side of
equation D. Here we have units of kilograms per
cubic meter multiplied by meters per second squared multiplied by meters. Combining all these factors of
meters, we get a result of kilograms per meter second squared. And we see that this is a match for
the units of pressure. So let’s put a star by option
D. And we’ll keep that in mind as a
possible answer choice.
Looking finally at the right-hand
side of option E. Here we have units of meters
divided by kilograms per cubic meter multiplied by meters per second squared. This denominator simplifies to
kilograms per meter squared second squared. And if we multiply both top and
bottom by meters squared second squared, this term cancels out in our denominator,
leaving us with meters cubed second squared per kilogram. We see this is not a match for the
units of pressure. And therefore, option E is off the
table. And this means that option D is our
answer. The pressure exerted by a fluid is
equal to the density of that fluid multiplied by 𝑔 multiplied by ℎ.
