Module 2

Lesson 1.8
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Lesson 1.5
Lesson 1.6
Lesson 1.7
Lab 1.5
Lab 1.6
Lab 1.7
Lab 1.8

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LESSON 1.8 - PRESSURE

Objectives
On completion of this lesson, you should be able to:
Define: Pressure, Atmospheric Pressure, The Pascal, Compressibility.
Explain: The effect of depth and density on pressure exerted by liquids and gases under the influence of gravity. Explain how the compressibility of the atmosphere affects the pressure it exerts at various altitudes.
Calculate: Pressure exerted by a liquid at a particular depth. The force exerted on a particular area by a liquid or gas under pressure.

MINI LAB

CHOICE OF ACTIVITIES

1. Can with holes: Drill three holes at various distances from the bottom on the side of a large tin can. Fill the can with water and observe the effect of depth on the speed of the water leaving through the holes.
2. Hydraulic jack: Use a hydraulic jack to show how a small force on a small area can be converted to a larger force on a larger area.
   

Pressure: The force acting on a certain area.

Fluid: A material that can flow: usually a liquid or a gas. Some materials flow extremely slowly.

Density: The mass per unit volume of a substance.

g: The gravitational field vector. The gravitational force on an object in a vertical direction towards the earth. At the earth’s surface this is equivalent to 9.81 N/kg.

Pascal: The SI unit of pressure. 1Pa. is equivalent to 1 N acting on 1 m².
Some authors prefer to use N/m² for reporting pressures in the SI system of units.

Compressibility: The ability of a material to reduce its volume when exposed to an increase in pressure. Gases are compressible. Liquids are generally regarded as incompressible.

Pressure
Pressure is the force that is exerted on a particular area. A moderate force exerted on a large area creates a small pressure. A moderate force on a small area creates a large pressure. The unit of pressure is the Pascal. This is equivalent to a force of one Newton acting on 1 square meter.
It’s more convenient to report pressures in terms of kilo-Pascals or even mega-Pascals.

Pressure in fluids
Because liquids and gases can easily change shape, pressure at any point in a liquid or gas exerts a force in all directions.

Pressure in fluids depends on gravity, depth and density
Liquid pressures are often due to gravity. The deeper the liquid, the greater the pressure.
Scuba divers are well aware of this because they use more air at higher pressures and have limitations on how long they can stay down at higher pressures.
For example: At a depth of 10 meters in water, the pressure is roughly twice that of the atmosphere
The pressure at any point in a liquid depends on the force of gravity, the density of the liquid and the height of liquid above the point.

Atmospheric pressure: Normal atmospheric pressure is roughly 101300 Pa. This is due to the weight of air above the earth’s surface. Nearly the earth’s entire atmosphere is located within 30 kilometers of the earth’s surface.
The mass of air directly above a 1 meter-square on the ground is 10 326 kg. The weight of this air creates a pressure at ground level of 101300 Pa.
Because air is compressible, there is more air in a cubic meter close to the earth’s surface than there is further up. This is why aircrafts that fly at high altitudes need to be pressurized and oxygen masks are needed if the pressure is lost. Because air is compressible, its density changes as its pressure changes and the formula for calculating air pressure is not as simple as that for liquids.

Hydraulic equipment
Using pistons with different sizes can cause hydraulic equipment to create large forces. A hydraulic jack uses a small piston to create a high pressure in the liquid connected to a larger piston. This pressure on the larger piston creates a much larger force that enables the jack to lift a heavy object. Hydraulic brakes in motor vehicles and hydraulic pistons in heavy equipment work on the same principle.

Pressure = Force / Area

In fluids: Pressure = r gh

Where: r = density (kg/m²)

g = gravitational field vector (9.81 N/kg)

h = height or depth of fluid

Example 1.8.1 Pressure on floor
A man with a mass of 80 kg rests his weight on one of the heels of his shoes. The area of the heel in contact with the floor is 16 cm² or 0.0016 m². Calculate the pressure exerted by his heel on the floor.

Solution
The downward force is 80 x 9.81 which is 784.8 Newtons.
The heel exerts a pressure of 784.8 / 0.0016 Newtons per square meter on the floor.

Pressure = 490,500 Pascals or 490.5 kilopascals.

Some stiletto heels have an area of about 1 square centimeter.
If a person weighing 50 kg rests on this, the pressure is roughly 4900 kPa or 4.9 mega-Pascals

Example 1.8.2 Pressure in liquids
If a tank that is open to the atmosphere contains water with a density of 1000 kg per cubic meter, the pressure above atmospheric pressure at a depth of 10 meters is:

1000 x 9.81 x 10 = 981,000 Pascals

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Review Questions

1. What is the pressure exerted on the floor by a person who weighs 50 kg and rests all of her weight on a stiletto heel with an area of 1 square centimeter?
2. Why is it a good idea to let the air out of the tires if a car gets stuck in soft sand?
3. If a probe is used to measure the pressure at a certain depth in a water tank, would a more accurate reading be obtained if the probe was pointing upwards, sideways or downwards (at the same depth)?
4. Why is it easier to compress gases than liquids?
5. What is the pressure (above atmospheric pressure) at a depth of 30 meters in salt water with a density of 1100 kilograms per cubic meter?
6. What is the pressure above atmospheric pressure at the bottom of a beaker containing a liquid with a density of 950 kg/m³ if the depth of the liquid is 8 cm?
7. If nearly the earth’s entire atmosphere is located in a region that is roughly 30 kilometers deep, why is the pressure at 15 km above the earth’s surface not roughly half of that at the earth’s surface?
8. A hydraulic jack has a small piston with a diameter of 1cm and a larger piston with a diameter of 3cm. How much pressure is exerted on the circular surface of the larger piston if the smaller piston exerts a pressure of 300 kPa on the oil connecting the circular surfaces of the two pistons? (Assume that the surfaces of the pistons are at the same depth in the oil)
9. A hydraulic jack has a small piston with a diameter of 1cm and a larger piston with a diameter of 4cm. How much force is exerted on the circular surface of the larger piston if the smaller piston exerts a force of 300 N on the oil connecting the circular surfaces of the two pistons? (Assume that the surfaces of the pistons are at the same depth in the oil)

Lesson 1.8 Pressure

1. 50 kg has a weight of 50 x 9.81 N = 490.5 N
   1 square cm has an area of 1/(100 x 100) = 0.0001 m².
   The pressure is thus 490.5 / 0.0001 = 4,905,000 Pa.
2. The weight of the car can be spread over a larger area and the pressure on the sand is reduced considerably.
3. The pressure is the same in all directions. There would be no difference between the probe readings.
4. I suppose that the easiest answer is that gases are compressible and liquids are usually regarded as incompressible. The reason for this is that the gas molecules are not held together with the same forces that hold liquid particles together and they can move about more freely and thus move closer to each other or further apart.
5. 1100 x 9.81 x 30 = 323,730 Pa.
6. 950kg/m³ x 9.81N/kg x 0.08m = 745.6 Pa.
7. The density of the air drops with pressure. It varies as the air pressure varies.
8. The pressure is the same: 300 kPa. (300,000 Pa.)
9. Area of small piston in contact with the oil = p x (1/2)² = 0.7854 cm²
   Area of larger piston in contact with the oil = p x (4/2)² = 12.57 cm²
   The force on the larger surface will be multiplied by the ratio of the areas:
   Force = (12.57/0.7854) x 300N = 4800 N