Higher Physics – Unit 1

1.5 – Density and Pressure

Density

The density of a substance is the mass per unit volume.

Vm

ρdensity(kg m-3) volume

(m3)

mass(kg)

The more mass in a given volume, the greater the density.

Density has the symbol, ρ (rho).

Example 1

A rectangular block of metal measures 20 cm x 15 cm x 10 cm and has a mass of 12 kg.

Calculate the density of the metal.

0.10.150.2V

kg 12m?ρ

Vm

ρ

0.00312

3m 0.003

3m kg 4,000ρ

Measuring Density of Air

Aim

To measure and compare the value for the density of air.

Method

Around bottomed flask is weighed with air inside it.

The air is evacuated using a vacuum pump and is reweighed.

The volume of the flask is measured by filling it with water and emptying into a measuring cylinder.

to vacuum pump

air

Results

g air flask of mass

g air no flask of mass

g extracted air of mass

kg

ml flask in air of volume

litres

3m

Vm

ρair

ρair

-3air m kg ρ

** Must Be Able To Describe How To Measure Density of Air **

** 1000 litres = 1 m3

**

Molecule Separation

The molecule separation for each of the three states of matter is considered.

solid liquid gas

The volume occupied by a substance in its solid form is approximately equal to the volume occupied when in liquid form.

The volume of a gas however is 1000 times greater than the volume of the same mass of the solid or liquid form of that substance.

As the mass is contained in a volume which is 1000 times greater, the density is 1000 times smaller.

d

10 d

10 d

10 d

The densities of gases are smaller than the densities of solids and liquids by a factor of 1000.

V 1000m

ρ

Summary

Separation

Volume Density

Solid

Liquid

Gas

1 1 1

1 1 1

10 1000 1/1000

Pressure

The pressure is defined as the force per unit area.

AF

Ppressure

(Pa) area(m2)

force(N)

The unit of pressure is the Pascal (Pa).

Alternative units for pressure are: N m-2.2m N 1Pa 1

Example 1

A 3 kg block has dimensions as shown.

0.3 m

0.2 m

0.05 m

Calculate the pressure the box exerts on the surface of a desk.

g mW

Weight of Box

9.8 3N 29.4W

Area of Base

blA 0.2 0.3

2m 0.06A

Pressure on Desk

AF

P

0.0629.4

Pa 490P

Worksheet – Density and Pressure

Q1 – Q20

Pressure in Fluids

Fluid is a general term which describes liquids and gases.

Any equations that apply to liquids at rest equally apply to gases at rest.

Atmospheric Pressure

At the Earth’s surface, the air exerts pressure in all directions.

This is caused by air molecules which move at high speeds in random directions. When they collide with surfaces they exert a force.

The force exerted by the atmosphere is 1x105 N per m2.

This force is known as the atmospheric pressure: Pa 101.01 5

At high altitudes there is less atmospheric pressure because there are fewer air molecules and less weight pushing down from above.

Pressure in Liquids

At the surface of a liquid the pressure is the same as atmospheric pressure.

Beneath the surface, the pressure in fluids varies directly with the depth and the density of the liquid.

h g ρP pressure(Pa)

density(kg m-3)

gravitational field strength(N kg-1)

depth(m)

Graph of Relationship

depth of liquid (m)

pressure (Pa)

density of liquid (kg m-3)

pressure (Pa)

Example 1

An object at 6 m is placed in a liquid with a density of 800 kg m-3.

Calculate the pressure due to the liquid.

m 6h-3m kg 800ρ

-1kg N 9.8g

h g ρP

?P

69.8800 47,040

Pa 104.7P 4

Example 2

A cube with sides 12 cm is submerged in water to a depth of 30 cm.

30 cm

(a) Calculate the pressure at the bottom surface of the cube due to the water.

cm 30h

-33 m kg 101ρ -1kg N 9.8g

h g ρP

?P

0.39.8101 3

m 0.3

Pa 2,940P

The density of fresh water is 1x103 kg m-3.

(b) Calculate the pressure at the top surface of the cube due to the water.

12- 30hcm 18

m 0.18-33 m kg 101ρ

-1kg N 9.8g

?P

h g ρP

0.189.8101 3 Pa 1,764P

(c) Calculate the force acting on the bottom surface of the cube.

A PF 0.01442940

N 42.3F

Pa 2,940P

0.120.12A 2m 0.0144

?F

Example 3 – (Atmospheric Pressure)

A mooring buoy is tethered to the bottom of a sea water loch by a vertical cable as shown.

(a) Calculate the total pressure on the top of the buoy.

1.5 m

0.30 mThe density of sea

water is 1.02x103 kg m-

3.

m 1.5h-33 m kg 101.02ρ

-1kg N 9.8g

h g ρP

?P

1.59.8101.02 3 Pa 14,994P

(b) Calculate the total pressure on the bottom of the buoy.

This pressure is only due to the sea water.

To find the total pressure, we need to add the atmospheric pressure from the air above.

5total 101.01 14,994P

Pa 115,994Pa 101.16 5

m 1.8h-33 m kg 101.02ρ

-1kg N 9.8g

h g ρP

?P

1.89.8101.02 3 Pa 17,992.8P

5total 101.01 17,992.8P

Pa 118,992.8Pa 101.19 5

Upthrust or Buoyancy

To hold an object in a stationary position a force, T, must be applied.

T must be equal in size but opposite in direction to the weight, W, of the object.

T

W

When the object is placed on the surface of a liquid the force, T, required to hold it stationary is less than before.

This is because the base of the object experiences an upward force from the liquid.

The upward force is called UPTHRUST (Fup).

Upthrust is caused by the pressure of the liquid on the base area of the object.On submerging the object further into the liquid, the force, T, required becomes even smaller.

This is because as depth increases the pressure increases, hence the upthrust increases. APFup

T

W Fup

APFup

T

W Fup

h

Maximum Upthrust

The MAXIMUM UPTHRUST is achieved just before the object becomes fully submerged.

After that the pressure on the top surface would cause a downward force cancelling out any further increase in Fup.

T

W Fup

h

Size of Upthrust (Submerged)

The size of the upthrust depends on the pressure difference between the top and bottom surface.

It does NOT depend on the depth .

T

W Fup

P2

P1T

W Fup

P3

P4

T

W Fup

P2

P1T

W Fup

P3

P4

Note

• P1 < P3

• P2 < P4

• However, the pressure difference in both situations is the same:

3412 PPPP

• So the upthrust on each object is equal.

Flotation

W Fup

Much Less Dense

When an object is much less dense than a liquid it will float on the surface.

This is because the liquid provides a large enough upthrust to balance the weight of the object, even at the surface.

Slightly Less Dense

When an object is only slightly less dense than a liquid it will become submerged to certain depth, h.

At this depth, the pressure has increased enough to produce an upthrust to balance the weight of the object.

W Fup

h

More Dense

When an object is more dense than a liquid it will sink because the weight is greater than the maximum upthrust (which is achieved just before submersion). W Fup

sinking

Buoyancy

When an object is held below the surface of a more dense liquid it remains there until the force holding it is removed.

This is because the UPTHRUST is greater than the weight at this depth due to increased pressure.

It will rise to a depth where the pressure is just enough to produce an upthrust equal to the weight of the object.

W

Fup

Worksheet – Pressure in Liquids and Buoyancy

Q1 – Q13