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Cambridge IGCSE0625

Pressure

Physics 0625 Chapter Notes

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Pressure
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In this chapter

  1. 1Pressure and the equation p = F/A
  2. 2Rearranging for force and area
  3. 3Force, area and everyday examples
  4. 4Pressure in a liquid
  5. 5Change in pressure with depth

1. Pressure and the equation p = F/A

Pressure is force per unit area.

To find it, use pressure = force / area. For an object standing on a surface, the force is the weight of the object, not its mass.

A force in N and an area in m² give a pressure in N / m², which is the Pa. An area in cm² gives N / cm².

p = F/A

Examiner insight

“A few candidates correctly recalled the equation P = F / A and then substituted a value for the mass of the bear instead of the weight of (force exerted by) the bear.” “The most common error was to multiply the force and the area to give 1536.”

Worked example 12 marks

Another student stands on the ground. The area in contact with the ground is 380 cm². The weight of the student is 530 N. Calculate the pressure exerted by the student on the ground. State the unit.

  1. 1

    p = F/A = 530/380 [1]

  2. 2

    = 1.4 N/cm² [1]

2. Rearranging for force and area

The same equation gives the force or the area. Force = pressure × area, and area = F / p.

Keep the units matched. With pressure in Pa, the area must be in m².

F = pA

A = F/p

Worked example 12 marks

The piston applies a force of 72 N to the brake pedal. The pressure exerted by the brake fluid on the piston is 8.4 × 10⁴ Pa. Calculate the cross-sectional area of the piston.

  1. 1

    A = F/p [1]

  2. 2

    = 72/(8.4 × 10⁴) = 8.6 × 10⁻⁴ m² [1]

3. Force, area and everyday examples

For the same force, a small area gives a high pressure. A large area gives a low pressure.

The tip of a drawing pin has a small area, so the pressure on the notice board is high.

The polar bear is about 7 times heavier than the person, yet the person exerts about 8 times the pressure. The person's contact area is 56 times smaller.

Examiner insight

“All the objects have the same mass but candidates did not realise that the maximum pressure will be exerted by the object with the smallest contact area with the surface.”

Worked example 13 marks

Table 3.1 contains information about the polar bear and the person.

(i) Calculate the weight of the polar bear.

(ii) Calculate the pressure that the polar bear exerts on the ice.

polar bearperson
mass / kg59084
area in contact with ice / m²1.120.020
  1. 1

    weight = 590 × 9.8 = 5800 N [1]

  2. 2

    p = F/A = 5800/1.12 [1]

  3. 3

    = 5200 Pa [1]

4. Pressure in a liquid

The pressure beneath the surface of a liquid increases with depth. It also increases with the density of the liquid.

The pressure due to a liquid depends on density and depth only, along with the gravitational field strength. It does not depend on the mass of an object in the liquid, or on the area of the liquid.

Examiner insight

“However a number of candidates thought that the less dense liquid would produce a greater pressure than the more dense liquid.”

Worked example 11 mark

Sea water is denser than fresh water. Where is the pressure on a diver greatest?

A 10 m below the surface of a fresh-water lake · B 10 m below the surface of the sea · C 20 m below the surface of a fresh-water lake · D 20 m below the surface of the sea

  1. 1

    D: 20 m below the surface of the sea. [1]

5. Change in pressure with depth

The change in pressure beneath the surface of a liquid depends on the density of the liquid, g, and the change in depth. Use g = 9.8 N / kg.

Between the surface and a depth h, the change in depth Δh is h. At the surface Δh is 0, so Δp is 0.

Δp is the change due to the liquid alone. Do not add or subtract air pressure.

Δ p = ρ g Δ h

Examiner insight

“A common misconception was to state that pressure change equals density times depth.” “A few candidates attempted to subtract a value for air pressure from their calculation of pressure change with depth of liquid.”

Worked example 12 marks

Under the ice is sea water with a density of 1030 kg / m³. Calculate the change in pressure Δp between the top of the sea water and a depth of 2.50 m.

  1. 1

    Δ p = ρ g Δ h = 1030 × 9.8 × 2.50 [1]

  2. 2

    = 25 235 ≈ 25 000 Pa [1]

Recap

  • Pressure = force per unit area: p = F / A.
  • Use the weight of an object as the force, not its mass.
  • A small area gives a high pressure for the same force.
  • Pressure in a liquid increases with depth and with density.
  • Extended: Δp = ρgΔh. Do not leave out g.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define pressure.1 mark
  2. A box weighing 240 N rests on a table. The area of its base is 800 cm². Calculate the pressure on the table in N / m².3 marks
  3. A crate of mass 45 kg stands on a floor. The area of the crate in contact with the floor is 0.30 m². Calculate the pressure on the floor.3 marks
  4. A skier wears wide skis. Explain why the skis stop her sinking into the snow.2 marks
  5. Oil is less dense than water. Two identical tanks are filled to the same depth, one with oil and one with water. State which tank has the greater pressure at its base. Explain your answer.2 marks
  6. A small pond and a large lake both contain fresh water. Each is 3 m deep. A student says the pressure at the bottom of the lake is greater because the lake has the larger area. State whether the student is correct. Explain your answer.2 marks
  7. A paving slab weighs 360 N. Its edges are 0.60 m, 0.40 m and 0.050 m. Calculate the greatest pressure the slab can exert on the ground.4 marks
  8. The density of fresh water is 1000 kg / m³. Calculate the change in pressure between the surface of a lake and a depth of 12 m.3 marks
  9. A tank of oil has a rubber plug in its base. The area of the plug in contact with the oil is 25 cm². The plug is pushed out when the force on it due to the oil reaches 49 N. The density of the oil is 800 kg / m³. Calculate the depth of oil at which the plug is pushed out.4 marks
  10. A tank holds a layer of oil 0.50 m deep floating on water 1.50 m deep. The density of the oil is 800 kg / m³ and the density of the water is 1000 kg / m³. The base of the tank is a rectangle 40 cm by 25 cm. Air pressure is 1.0 × 10⁵ Pa. Calculate the force on the base of the tank due to the two liquids.6 marks

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