Pressure

Forces in Action - OCR A-Level Physics

Key Definition
Pressure
Force per unit area acting perpendicular to a surface. SI unit: pascal (Pa), where 1 Pa = 1 $N m^{-2}$.
$$p = \frac{F}{A}$$
$$p = \rho g h$$
  • PressureForce per unit area. Measured in pascals (Pa), where 1 Pa = 1 N m⁻². in a fluid acts equally in all directions at a given depth
  • The pressureForce per unit area. Measured in pascals (Pa), where 1 Pa = 1 N m⁻². due to a fluid column depends only on depth, densityMass per unit volume of a material. Measured in kg m⁻³. and g -- not on the shape of the container
  • Total pressureForce per unit area. Measured in pascals (Pa), where 1 Pa = 1 N m⁻². at a depth h = atmospheric pressure + $\rho$g h
  • Archimedes' principleThe upthrust on a body in a fluid equals the weight of fluid displaced.: the upthrust on an object submerged in a fluid equals the weight of fluid displaced
Worked Example [3 marks]
A swimming pool is 2.5 m deep and filled with water of densityMass per unit volume of a material. Measured in kg m⁻³. 1000 $kg m^{-3}$. Calculate the pressure due to the water at the bottom. Atmospheric pressure = 101 kPa. Use g = 9.81 $m s^{-2}$.
Show Solution
1
Pressure due to water

$p = \rho g h = 1000 \times 9.81 \times 2.5 = 24\,525 \text{ Pa} \approx 24.5 \text{ kPa}$

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2

Total pressure at the $bottom = atmospheric pressure + pressure due$ to water

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3

$Total = 101 + 24.5 = 125.5 \text{ kPa} \approx 126 \text{ kPa}$

[1]
Answer
Pressure due to $water = 24.5 kPa$; total pressure at the $bottom = 126 kPa$
Common Mistake MEDIUM
Students often: Watch out - it's tempting to skip add atmospheric pressure when asked for the total pressure at a depth in a fluid.
Instead: The equation $p = \rho g h gives$ the pressure due to the fluid column only. For total pressure, you must add atmospheric pressure on top.
Forces in Action Overview
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