OCR · GCSE · Physics
Course version: J249 · Version from Sept 2018
Use the relationship between the force, the pressure and the area in…
Full specification wording
Forces in action: use the relationship between the force, the pressure and the area in contact (to include: an understanding of how simple hydraulic systems work) (content found only in the physics separate science qualification)
Pressure is force spread over the area in contact, so p = F ÷ A, with p in pascals (Pa), F in newtons (N) and A in square metres (m²).
Checked against the OCR J249 specification (opens in a new tab) on .
- Point 21 of 21
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Pressure is force spread over the area in contact, so p = F ÷ A, with p in pascals (Pa), F in newtons (N) and A in square metres (m²). Rearranged, F = p × A and A = F ÷ p. A small area gives a large pressure for the same force, which is why a sharp pin pierces easily. A simple hydraulic system uses a liquid, which is virtually incompressible, to transmit pressure. Push a small piston of area A₁ with force F₁ and the pressure p = F₁ ÷ A₁ passes through the liquid to a larger piston of area A₂, giving F₂ = p × A₂. Because A₂ is larger, F₂ is larger than F₁, so a hydraulic system multiplies force, though the smaller piston must move further. This content appears only in the physics separate science qualification.
What earns the marks
- Uses p = F ÷ A with pressure in pascals, force in newtons and area in square metres.
- Rearranges the relationship correctly to F = p × A or A = F ÷ p before substituting values.
- Explains that a smaller area in contact produces a greater pressure for the same force.
- Applies the same pressure throughout a simple hydraulic liquid to show that a larger piston area gives a larger output force.
- Converts an area given in cm² to m² by multiplying by 1 × 10⁻⁴ before calculating.
Common mistakes and how to put them right
Uses area in square centimetres without converting to square metres.
Put it right: Multiply the value in cm² by 1 × 10⁻⁴ to obtain m².
Divides area by force instead of force by area.
Put it right: Write p = F ÷ A and check the units give pascals.
Assumes a hydraulic system multiplies energy as well as force.
Put it right: State that force is multiplied while the smaller piston moves a greater distance.
Treats the liquid as compressible.
Put it right: Explain that the liquid transmits pressure because it is virtually incompressible.
Exam tips
- Write the equation, then rearrange it symbolically before inserting any numbers.
- Check that every quantity is in SI units, converting cm² to m² and N to kN where needed.
- For hydraulic questions, calculate the transmitted pressure first, then use it to find the force on the second piston.
Try the lesson
The start of the Forces in action: use the relationship between the force, the pressure and the area in contact (to include: an understanding of how simple hydraulic systems work) (content found only in the physics separate science qualification) lesson, made from checked teaching notes for this specification point.
Have a go first
A force of 60 N acts on a piston with an area of 0.20 m². Using pressure = force ÷ area, what pressure does this produce?
Listen
Pressure is force spread over the area in contact, so p = F ÷ A, with p in pascals (Pa), F in newtons (N) and A in square metres (m²). Rearranged, F = p × A and A = F ÷ p. A small area gives a large pressure for the same force, which is why a sharp pin pierces easily. A simple hydraulic system uses a liquid, which is virtually incompressible, to transmit pressure. Push a small piston of area A₁ with force F₁ and the pressure p = F₁ ÷ A₁ passes through the liquid to a larger piston of area A₂, giving F₂ = p × A₂. Because A₂ is larger, F₂ is larger than F₁, so a hydraulic system multiplies force, though the smaller piston must move further. This content appears only in the physics separate science qualification. A strong answer earns marks for these points. Uses p = F ÷ A with pressure in pascals, force in newtons and area in square metres. Rearranges the relationship correctly to F = p × A or A = F ÷ p before substituting values. Explains that a smaller area in contact produces a greater pressure for the same force. Applies the same pressure throughout a simple hydraulic liquid to show that a larger piston area gives a larger output force. Exam tip: Write the equation, then rearrange it symbolically before inserting any numbers.
Key words
Say what each word means before you turn the card.
Pressure
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Simple hydraulic system
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Area conversion
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Force multiplication
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See every definition
- Pressure
- Force spread over the area in contact, measured in pascals (Pa), with force in newtons and area in square metres.
- Simple hydraulic system
- Uses a liquid, which is virtually incompressible, to transmit pressure, so a larger piston area gives a larger output force.
- Area conversion
- An area given in cm² is converted to m² by multiplying by 1 × 10⁻⁴ before calculating.
- Force multiplication
- Because the larger piston area is larger, the output force is larger than the input force, though the smaller piston must move further.
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