AP Physics 1 & 2 Pressure & Pascal's Principle

Hi everyone. I am Dan Fullerton and I would like to welcome you back to Educator.com.0000

Today we are going to continue our study of fluids as we talk about pressure and Pascal's principle.0004

Our objectives are going to be to calculate pressure as the force a system exerts over an area, to explain the difference between gauge pressure and absolute pressure, and explain the operation of a hydraulic system as a function of equal pressure spread throughout a fluid.0009

Pressure -- pressure is the effect of a force acting upon a surface.0025

It is a scalar. It is a force per unit area and its units are Newtons per meter squared (N-m²), which are also known as Pascal's, which we typically abbreviate as (Pa).0030

If pressure if force per unit area, our formula for pressure is P = F/A.0040

Now it is important to know that the force is always perpendicular to the surface it is acting on.0047

Exerting pressure -- all states of matter can exert pressure.0063

You walk across an ice covered lake, you exert pressure on the ice equal to your weight, divided by the area, which contacts the ice.0067

That is why if you do not want to crack the ice, they teach you to spread your hands and feet out to spread out that force over a larger area, so you get less pressure.0074

If you walk on snow with snow shoes with large areas of contact, you increase the area; you reduce the pressure and you walk on top of the snow, that is why they are so big -- larger area.0085

Now, fluids exert outward pressure in all directions on the sides of any container holding the fluid.0096

Even Earth's atmosphere exerts pressure.0102

Atmospheric pressure is about 101,325 Pa, which will typically round you about 100,000 Pa or 10N/cm².0105

And you can even experience this by riding in an airplane as you change altitudes -- go up and down -- you may experience a popping sensation in your ears.0114

The pressure inside your ear, balances the pressure outside your ear in a transfer of air through some small tubes connecting your inner ear to your throat.0123

When that happens, you hear that or feel that popping sensation.0131

Let us take a look at our first example -- pressure on a keyboard.0138

Air pressure is approximately 100,000 Pa, what force is exerted on your keyboard when it is sitting flat on a desk if the area of the keyboard is 0.035 m².0141

Well if pressure is force divided by area, then that means force is pressure times area.0153

If our pressure is about 100,000 Pa × 0.035 m² (area), that will give us a force of about 3500N.0162

Let us take a look at an example of a sleepy fisherman.0183

A fisherman with a mass of 75 kg falls asleep on his four-legged chair of mass (5 kg).0186

If each leg of the chair has a surface area of 2.5 × 10^-4m² in contact with the ground, what is the average pressure exerted by the fisherman and chair on the ground?0192

Well, the force applied is going to be the force of gravity, so our pressure is going to be force/area, which is going to be mg/a.0205

Now our mass is 75 kg, the fisherman, plus 5 kg for the chair, times (g) about 10 m/s² all over the area...0217

...which is going to be the four corners of the chair, the four legs of the chair times the area of each leg, 2.5 × 10^-4m² or about 800,000 Pa.0226

Let us take a look at another one.0250

A scale which reads 0 in the vacuum of space is placed on the surface of Planet Physica.0252

On the planet's surface, the scale indicates a force of 10,000N.0258

Calculate the surface area of the scale given the atmospheric pressure on the surface of Physica is 80,000 Pa.0263

If pressure is force/area, then area is force/pressure.0271

That will be 10,000N/80,000 Pa (pressure), or about 0.125 m².0279

Let us see if we cannot rank some pressures.0297

Rank the following from highest pressure to lowest pressure upon the ground.0301

The atmosphere at sea level -- well that we know is right around 100 Pa.0305

A 7,000 kg elephant with total area of 0.5 m² in contact with the ground -- well pressure will be force/area or that will be 7,000 kg × g (10)/area of 0.5 or about 140,000 Pa.0313

A 65 kg lady in high heels with a total area of 0.005 m² in contact with the ground -- if pressure is force/area, that will be 65 kg × 10 m/s²/area (0.005) or about 130,000 Pa.0336

And finally a 1600 kg car with a total tire contact area of 0.2 m² -- the pressure equals force over area, so that will be 1600 kg × 10 m/s²/0.2 m² or about 80,000 Pa.0363

So, if I were to rank these from highest pressure to lowest pressure, I would start with the elephant (B), go to the lady in high heels, atmospheric pressure, and finally the car, so (B), (C), (A), (D).0384

Let us talk about pressure on a submerged object.0404

The pressure a fluid exerts on an object submerged in that fluid is determined by multiplying the density of the fluid by the depth submerged, all multiplied by the acceleration due to gravity.0407

We call that the gauge pressure, which is ρ (density) × (g) × (h).0417

If there is also atmosphere above the fluid, such as the situation here on Earth, you can determine the absolute or total pressure by adding in the atmospheric pressure which we will write as P0, which is about 100,000 Pa.0423

So absolute pressure is atmospheric pressure plus gauge pressure -- P0 + ρ, where ρ is the density of the fluid, (gh).0435

Let us take an example of gauge pressure.0448

Samantha spots buried treasure while scuba diving on her Caribbean vacation.0450

If she must ascend to a depth of 40 m to examine the treasure, what gauge pressure will she read on her scuba equipment?0454

The density of sea water is 1025 kg/m³.0461

Well, we want gauge pressure, so that is going to be ρ (fluid), (gh) or 1025 kg/m³ × g (10 m/s² × her depth of 40 m or about 410,000 Pa.0466

If we want to look at absolute pressure here, let us find the absolute pressure for Samantha in the same scenario.0493

Now we are looking for absolute pressure and that is P0 + gauge pressure and we are going to say atmospheric pressure is about 100,000 Pa, for simplicity...0500

...so 100,000 + 410,000 Pa (gauge pressure) = 510,000 Pa of pressure as the absolute pressure.0513

Let us talk about Pascal's principle.0530

When a force is applied to a contained incompressible fluid, the pressure increases equally in all directions throughout that fluid.0533

This is the foundation for hydraulic systems and things like barber shop chairs, construction equipment, and even car brakes.0540

In car brakes, in the movies, you will even sometimes see people cut the brake lines so the brakes do not work -- the fluid leaks out and the brakes no longer work because the fluid is no longer contained.0548

It must be a contained fluid or incompressible or nearly incompressible fluid.0558

Let us talk about force multiplication using Pascal's principle, also known as the basis of hydraulics.0566

To begin with, we have a force (F1) that we are applying to a piston of area (A1) and that is going to create a pressure of (P1), so (P1) is caused by force (F1) applied to a piston of area (A1) on a contained, incompressible fluid -- we have a closed container, incompressible fluid here.0573

Now over on the right hand side, the pressure on this piston, must be F2/A2.0594

Why? By Pascal's principle, you must have the same pressure anywhere throughout the fluid.0601

Well when you do that, let us take a look at the ramifications.0607

If P1 = P2, by Pascal's principle, then that means F1/A1 = F2/A2 or if I cross-multiply (F1)(A2) = (F2)(A1) or F2 = A2/A1 × F1.0611

What does this mean? You have increased the force -- if you apply a force (F1) and you have a different area on your two pistons, you can increase that force by the ratio of the areas.0637

If (A2) is five times larger than (A1), and you apply force (F1), you get five times that force on (F2).0650

You have effectively increased your applied force.0658

Now you do not really get anything for free here.0661

What you are going to end up having by conservation of energy is you are also going to have to push this piston five times further or you will get 1/5 the displacement that you would over here for the same displacement over there.0665

The total work done on each side has to be the same for conservation of energy, and that is going to be by the same ratio as the area multiplier, but you can multiply a force if the areas are a ratio of 100:1, you have increased your force by a factor of 100 and that is the principle behind hydraulic systems.0678

Let us take the example of a barber's chair when we apply this.0699

A barber raises his customer's chair by applying a force of 150N to a hydraulic piston of area 0.01 m².0702

If the chair is attached to a piston of area 0.1 m², how massive a customer can the chair raise?0711

Assume the chair itself has a mass of 5 kg.0717

Well, to solve this problem let us first determine the force applied to the larger piston.0721

We know (F2) must be equal to the ratio of the area, A2/A1 × F1, therefore, F2 = 0.1(A2)/0.01(A1) = 10 × F1 (150N) = 1500N.0728

This is the largest applied force you can have.0750

Now then, if we want to know how massive a customer the chair can raise, if our force is 1500N, that must be equal to the weight -- the maximum that we can handle -- therefore, the mass that you can handle is 1500N/g (10 m/s²) or 150 kg.0758

Now that 150 kg -- five of those kg have to be the chair, therefore, you could lift a customer of 145 kg, which is about 300 lbs.0782

Let us take a look at a hydraulic auto lift.0806

A hydraulic system is used to lift a 2,000 kg vehicle in an auto garage.0808

If the vehicle sits on a piston of area 0.5 m², and a force is applied to an area of 0.03 m², what is the minimum force that must be applied to lift the vehicle?0813

Well, starting with Pascal's principle, we know that P1 = P2, assuming we have a contained incompressible fluid, therefore F1/A1 = F2/A2.0825

We are looking for (F1), so that is going to be A1/A2 × F2, which is 0.03 m²/0.5 m² × (F2)...0838

...which is the weight of our vehicle, 2,000 kg × the acceleration due to gravity (10), therefore F1 must be 1200N.0855

We have to be able to apply 1200N in order to lift that 20,000N vehicle.0871

Let us take a look at the pressure on a penny.0881

A penny with a diameter of 19.05 mm sits on the bottom of the ocean, where we have a saltwater density of 1025 kg/m³ at a depth of 340 m.0883

What is the force on the penny?0894

Let us figure out the area of the penny first.0898

The area for the penny is πr² and the radius of the penny will be half the diameter or half of 19.05 mm...0900

...and that is 0.009525 m² or about 2.85 × 10^-4m².0913

Now, if we are looking for the force on the penny, let us start by finding the pressure.0928

The absolute pressure is the atmospheric pressure plus the gauge pressure (ρgh)...0935

...so that will be about 100,000 Pa plus our density of our fluid (1025), the acceleration due to gravity (10 m/s²) and our depth of 340 m or about 3,585,000 Pa.0943

Now to find the force if P = F/A, then that means F = (P)(A), or 3,585,000 Pa × 2.85 (area) × 10^-4m² = 1,022N (force).0965

That is quite a force on a little penny.0995

How about depth in freshwater -- a diver's pressure gauge reads 250,000 Pa in freshwater.1000

How deep is the diver?1007

Well, gauge pressure is (ρgh), therefore h = P/ρg or 250,000 Pa/1,000 kg/m³, our density of freshwater × g (10) or 25 m.1010

One more -- A pressure gauge reads 350,000 Pa, what is the absolute pressure?1040

Well, just to review, absolute pressure is atmospheric pressure plus (ρgh)...1049

...or 100,000 Pa (atmospheric pressure) + 350,000 Pa (gauge pressure) = 450,000 Pa (absolute pressure).1058

All right. Hopefully that gets you a great start with pressure and Pascal's principle1078

Thank you so much for your time and for watching us on Educator.com.1081

Looking forward to seeing you again. Make it a great day!1084