Pressure

• Pressure is the force acting normally per unit area

• Pressure = Force

• Area

• P = F/A

F

P A

• SI Unit: Pascal (Pa) or Newton per square metre (N/m2)

Example 1

• The weight of a man is 600 N. Calculate the pressure he exerted on the floor if he is wearing a pair of track shoes and the area of contact of each shoe with the ground is 0.02 m2.

• Pressure = F/A

• = 600/(2x0.02) =15 000 Pa

Example 2

• A rectangular block of dimension 50cm by 30cm by 20 cm has a mass of 4 kg. Calculate the maximum and minimum pressure it can exert on the floor.

Example 2

• Weight of block

• = mg = 6 x 10 = 60 N

• Maximum pressure exerted

• = F/A(min)

• = 60/(0.2 x 0.3) = 1000 Pa

30 cm20 cm

50 cm

Example 2

• Minimum pressure exerted

• = F/A(max)

• = 60/ (0.5 x 0.3)

• = 400 Pa

30 cm

20 cm

50 cm

Liquid Pressure

• The pressure inside a volume of liquid depends on the depth below the surface of the liquid.

• The deeper it is, the greater the weight of the overlying liquid, and thus the greater the pressure.

Example 1: Liquid pressure increases with depth

Example 2: Liquid pressure increases with depth

• The thickness of the wall of dam increases downwards because the deeper it is, the greater the water pressure.

Damwaterland

Liquid Pressure: Formula

• The pressure at any point in a liquid at rest is given by:

• Pressure = hpg

• where h = height of liquid in metre

• p = density of liquid kg/m3

• g = gravitational acceleration in N/kg or m/s2

Proof

• Consider a cylindrical container of of area, A, filled with liquid of density, p, to a height, h

Area A

height h Density p

Proof

• Volume of liquid = Ah

• Weight of liquid

• = mg = Vpg = Ahpg

• Pressure on base

• = Force/ Area

• = Ahpg/ A = pgh

Area A

height h Density p

Liquid Pressure

• Pressure, P = hpg

• Pressure at any point in a liquid is independent of the cross-sectional area

Liquid Pressure

• A liquid always find its own level

• (Liquid will flow to equalise any pressure difference)

Liquid Pressure

• All points on the same level in a liquid have the same pressure

Liquid Pressure

• Pressure at any one depth in a liquid acts equally in all directions

liquid

Example 1

• The water level in a water tank is 12 m above the tap. What pressure forces water out from the tap? (Density of water = 1000 kg/m3).

Water tank

12 m

Example 1

• Pressure at the tap is due to the water in the pipe above it.

• Pressure = hpg

• = 12 x 1000 x 10 = 120 000 Pa

Water tank

12 m

Example 2

• A regular shaped object is immersed in water of density 1000 kg/m3.

• (a) Calculate the water pressure at the top and the bottom of the object.

• (b) What is the resultant pressure on the object?

0.2 m0.5 m object

Example 2

• (a) Pressure exerted by water at the top surface of the object

• = h1pg

• 0.2 x 1000 x 10 = 2000 Pa

0.2 m0.5 m object

Example 2

• Pressure exerted by water at the bottom surface of the object

• = h2pg

• = 0.5 x 1000 x 10

• = 5000 Pa

0.2 m0.5 m object

Example 2

• Resultant pressure on the object

• = 5000 - 2000

• = 3000 Pa (The object experiences an upward force)

0.2 m0.5 m object

Atmospheric Pressure

• The atmosphere is the layer of air surrounding the Earth. It extends up to 1000 km above the Earth surface.

• The weight of the air exerts a pressure on the surface of the Earth. This pressure is called the atmospheric pressure

Atmospheric Pressure

• The atmospheric pressure is about 1.03 x 105 Pa or 105 Pa

• or 10 m of water

• or 0.76 m of mercury (760 mmHg)

Simple Mercury Barometer

• The atmospheric pressure can be measured using a simple mercury barometer

760 mm

mercury

vacuum

Simple Mercury Barometer

• Height of mercury column for the mercury barometer:

• Pressure = hpg• 103 000 Pa = h x 13 600 kg/m3 x 10 N/kg

• h = 0.76 mHg

Water Barometer

• Water can be used in a barometer instead of mercury. However, the glass tube used need to be much longer because water has a much lower density than mercury.

Water Barometer

• If water barometer is used, the height of the water column will be:

• Pressure = hpg• 100 000 Pa = h x 1000 kg/m3 x 10 N/kg

• h = 10 m

Mercury Barometer

• Determine the pressure at points A, B, C and D

A

B0.2 m

C

0.76 m

D0.3 m

Mercury Barometer

• Pressure at A = 0 mmHg

• Pressure at B = 0.2 mHg

• = hpg = 0.2 x 13600 x 10

• = 27200 Pa = 27.2 kPa

A

B0.2 m

C

0.76 m

D0.3 m

Mercury Barometer

• Pressure at C = 0.76 mHg

• Pressure at D = (0.76 + 0.3) mHg

• = 1.06 mHg

A

B0.2 m

C

0.76 m

D0.3 m

Pressure Difference• Manometer

• A manometer consists of a U-tube containing liquid and it is used to measure differences in gas or liquid pressure

X Y

Mercury

To gas supply

Manometer

• The height difference XY tells how much the gas pressure is different from atmospheric pressure

X Y

Mercury

To gas supply

Example 1

• There is no pressure difference between X and Z

• Pressure difference between XY

• = 12 cm Hg

X

Y

Mercury

To gas supply

Z

12 cm

Example 1

• If atmospheric pressure is 76 cm Hg

• Then Gas Pressure = pressure of X or Z

• = (76 + 12) cm Hg

• = 88 cm Hg

X

Y

Mercury

To gas supply

Z

12 cm

Example 2

• A U-tube with some mercury at the bottom is set up vertically and 12 cm of water is added into one arm of the tube. Methylated spirit is then added carefully into the other arm of the U-tube until the mercury level are the same in both arm.

15 cm12 cm

water

methylated spirit

mercury

AB

Example 2

• It is observed that the level of methylated spirit is higher. What is the density of methylated spirit if the methylated spirit column is 15 cm high? (Density of water = 1000 kg/cm3)

15 cm12 cm

water

methylated spirit

mercury

AB

Example 2• Let

• hw = height of water column from level A

• hm = height of methylated spirit column from level B

• PA = pressure at level A

• PB = pressure at level B

• Pa = atmospheric pressure

• pw =density of water

• pm = density of methylated spirit

15 cm12 cm

water

methylated spirit

mercury

AB

Example 2

• PA = Pa + hwpwg

• PB = Pa + hmpmg

• But PB = PA

• Pa + hmpmg = Pa + hwpwg

• 0.15 x pm x g = 0.12 x 1000 x g

• 0.15 pm = 0.12 x 1000

• pm = 800 kg/m3

15 cm12 cm

water

methylated spirit

mercury

AB

Hydraulic Systems

• Hydraulic system work by using liquid pressure. They make use of two properties of liquid

1. Liquids are incompressible.

2. If pressure is applied to a trapped liquid, the pressure is transmitted to all parts of the liquid.

Simple Hydraulic Systems

small area piston

Large area piston

Applied forceOutput force

Hydraulic Systems

• When a force of 20 N is applied to a small piston of 0.01 m2, the pressure exerted on the liquid is given by

• Pressure = Force/ Area

• = 20/ 0.01 = 2000 Pa

Hydraulic Systems

• This pressure is transmitted to the larger piston. If the larger piston has an area of 0.1 m2, the force on the large piston is

• Force = pressure x area

• = 2000 x 0.1

• = 200 N

Pressure - Volume Relationship of a gas

• Boyle’s Law states that

• for a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume.

• P α 1/V

• or P = k/V where k is a constant

P - V Relationship of a gas

• P = k/V• When a graph of Pressure against Volume

is plotted, the graph is as follows

P

V

P - V Relationship of a gas

• P = k/V

• When P is plotted against 1/V, a straight line is obtained as shown

P

1/V

P - V Relationship of a gas• Since P = k/V

• PV = k

• Thus, P1V1 = P2V2

P1V1

P2

V2

Example 1

• If a given mass of gas has a volume of 4.5 x 10-5 m3 at a pressure of 30 kPa, what will be the volume of the gas if the pressure is increased to 50 kPa while the temperature is kept constant?

Example 1

• By Boyle’s law, PV = constant

• Thus, P2V2 = P1V1

• 50 x V2 = 30 x (4.5 x 10-5)

• V2 = 30 x (4.5 x 10-5)/ 50

• V2 = 2.7 x 10-5 m3

Example 2

• A bicycle pump of length 45 cm as shown contains air with a pressure of 100 kPa. In order to force air into the tyre, it is necessary to move the piston 15 cm down the barrel. What is the pressure inside the tyre assuming the temperature is kept constant?

45 cm

15 cm

Example 2

• Let the cross-sectional area of bicycle pump be A cm2

• Thus, V1 = (45 x A) cm3

• V2 = (30 x A) cm3

45 cm

15 cm

Example 2

• Apply Boyle’s law

• P2V2 = P1V1

• P2 x 30A = 100 x 45A

• P2 = (100 x 45A)/30A

• P2 = 150 kPa

45 cm

15 cm

Example 3

• An air bubble at the bottom of a lake 40 m deep has a volume of 1.5 cm3. What si the volume of the air bubble when it rises to the surface of the lake? (Atmospheric pressure is about 10 m of water)

Example 3

• Pressure at lake surface

• = Atmospheric pressure

• = 10 m of water

• Pressure at bottom of lake

• = Atmospheric pressure + pressure exerted by 40 m of water

• = 10 + 40 = 50 m of water

Example 3

• Applying Boyle’s law,

• P1V1 = P2V2

• 10 x V1 = 50 x 1.5

• V1 = (50 x 1.5)/10

• V1 = 7.5 cm3