## Section 3.2 | Fluid Pressure

In this section, we are going to dive deeper (pun intended) into physics. It is to our benefit to recall a few basic concepts.

**MASS**

- Metric unit: gram, kilogram
- Imperial unit: none really (see text below)

Mass can be thought of as an object's ability to resist acceleration from a force. Sounds confusing? Well, you can get the hang of it, but there are some pitfalls to be aware of. First,

**mass is different than weight**. Weight is a force, mass is not!

In the metric system, mass is easy - it's just measured in grams (or kilograms) and it

**never**changes unless the structure of the object changes. An object with a mass of 70 kg on Earth will have a mass of 70 kg on the moon. When someone tells you that they "weigh" 70 kg, they are trying to tell you that, if they stood on your stomach, you would feel the same force that a 70-kg steel ball would exert. They actually don't "weigh" 70 kg. In fact, if you read ahead a little you will know that they should say that they weigh roughly 686 Newtons.

In the Imperial system, mass is messy. Most people think that the pound is the unit for mass, but this is technically not true. The pound is a unit of force. It depends upon the acceleration due to gravitational forces. A person who weighs 200 pounds on Earth will weigh less if they went to the moon. In fact, since weight (which is a force) is the product of mass and acceleration, and we can only change one of those quantities (without changing our distance from the center of the Earth drastically), most weight-loss programs should technically be called "mass-loss" programs. It literally means that if someone wants to lose weight (force), then they must become less massive.

**FORCE**

- Metric unit: Newton (\(1 N = 1kg \frac{m}{s^2}\))
- Imperial unit: Pound

Force is a push or a pull. It is also famously known as the product of mass and acceleration. That is, \[F = m a.\]

**DENSITY**

- Metric unit: none (\(\frac{kg}{m^3}\))
- Imperial unit: none (\(\frac{lb}{ft^3}\))

Density (a.k.a. volumetric mass density) is an object's mass per unit volume. That is, \[\rho = \frac{m}{V}.\]A way to think about this is take a unit cube of the object and measure the mass in there. That would be its density.

I should note that since density involves an object's mass, the Imperial concept of density is flawed (remember, in Imperial units the pound is used incorrectly for mass). Technically, the Imperial unit for density is measuring the

*specific weight*.

**PRESSURE**

- Metric unit: Pascal (\(1 Pa = 1 \frac{N}{m^2} = \frac{kg}{m \cdot s^2}\))
- Imperial unit: psi (\(\frac{lb}{in^2}\))

Pressure is perpendicular force per unit area. By "perpendicular force" I mean to say "the component of the force vector perpendicular to the surface". In notation, we have \[P = \frac{F}{A}.\]

In this section, we are only going to concern ourselves with hydrostatic force on thin plates submerged either wholly or partially into liquid. To prepare for the work we will be doing, it is necessary to bring together some of the formulas above. In particular, we will be using \[F = ma\] \[\rho = \frac{m}{V}\]and \[P = \frac{F}{A}.\]If we solve for \(m\) in the density formula, we get \[m = \rho V.\]Moreover, for a brief moment, let's just consider the volumes to be the chunk of water directly above a

*horizontally*submerged plate having surface area \(A\). Then the volume of water above this plate is \(V = A d\), where \(d\) is the depth of the plate. Putting this all together, we get the hydrostatic pressure on the plate to be \[P = \frac{F}{A} = \frac{m a}{A} = \frac{(\rho V) a}{A} = \frac{\rho A d a}{A} = \rho d a.\]What happens if the plate is*not*submerged horizontally? Well, the following theorem from physics allows us to say, "it doesn't matter."**THEOREM | UNIFORM HYDROSTATIC PRESSURE**

If a particle/person/plate is at a depth \(d\) in liquid, the pressure is the same in all directions.

This little fact allows us to now look at an object submerged in liquid vertically and only consider the "slices" of that object that occur at the same depth. That is, if we submerge a plate

*vertically*into water, the part of the plate that is 3 meters under the surface of the water experiences the same pressure. All we need to do is find the area of that slice of the plate to determine the hydrostatic*force*on that slice. Oh, that reminds me, we will be finding the hydrostatic force, so we will multiply everything in that last set of equalities by \(A\) to get \[F = \rho d a A.\]Hydrostatic Pressure from the Wolfram Demonstrations Project by Fernando Blanco Galán

**NECESSARY FACTS:**

In the metric system, \(\rho = 1000 \frac{kg}{m^3}\) and \(a \approx 9.8 \frac{m}{s^2}\). In the Imperial system, \(\rho a = \delta = 62.5 \frac{lb}{ft^3}\). Note the difference that the Imperial system needs the adjustment to include acceleration due to gravity.

**EXAMPLE 1 [online #1] | A standard aquarium**

A fish tank 4 feet long, 2 feet wide, and 2 feet deep is full of water.

- Find the hydrostatic
*pressure*on the bottom of the tank. - Find the hydrostatic
*force*on the bottom of the tank. - Find the hydrostatic
*force*on one end of the tank.

**EXAMPLE 2 [online #2] | Dipping plates**

A semicircular plate with radius 9 meters is submerged vertically in water so that the top is 1 meter above the surface. Find the hydrostatic force against one side of the plate.

**EXAMPLE 3 [online #3] | Going swimming**

A swimming pool is 10 feet wide and 20 feet long and its bottom is an inclined plane, the shallow end having a depth of 4 feet and the deep end, 7 feet. Assume the pool is full of water.

- Estimate the hydrostatic force on the shallow end.
- Estimate the hydrostatic force on the deep end.
- Estimate the hydrostatic force on one of the sides.
- Estimate the hydrostatic force on the bottom of the pool.*