Fluid Pressure

Key Stage 4

Meaning

Fluid Pressure is the force per unit area on a surface due to particles colliding with that surface.

About Fluid Pressure

The force responsible for pressure in a fluid is the weight of all the particles above the object.

The pressure in a fluid decreases with height and increases with depth. The deeper into a fluid the greater the pressure.

Since pressure increases with depth then the bottom of an object will experience more pressure than the top of the object. This causes a force called upthrust.

Equation

NB: You do not need to remember this equation.

Pressure = (Density of the Fluid) x (Gravitational Field Strength) x (Depth in the fluid)

\( P = \rho g h\)

Where

\( P\) = Pressure at that depth.

\( \rho\) = Density of the fluid (This symbol is a 'rho' not a 'p')

\( g \) = Gravitational Field Strength

\(h\) = Depth in the fluid

Calculating Pressure in a Fluid

A whale dives to a depth of 1300m below sea level. Given the density of water is 1000kg/m3, calculate the pressure on the whale due to the water correct to two significant figures.	Normal atmospheric pressure at sea level is 101kPa. A miner works in a mine shaft 550m below the surface of the Earth. Given the density of the atmosphere at this level is 1.2kg/m3, calculate the pressure on the miner correct to two significant figures.	A piece of Lithium is submersed 4.3cm deep in oil prevent it from oxidising. The oil has a density of 932kg/m3, calculate the pressure on the Lithium due to the oil correct to two significant figures.
1. State the known quantities Density = 1000kg/m3 Gravitational Field Strength = 9.8N/kg Depth = 1300	1. State the known quantities Density = 1.2kg/m3 Gravitational Field Strength = 9.8N/kg Depth = 550m	1. State the known quantities Density = 932kg/m3 Gravitational Field Strength = 9.8N/kg Depth = 4.3cm = 0.043m
2. Substitute the numbers into the equation and solve. \( P = \rho g h\) \( P = 1000 \times 9.8 \times 1300\) \( P = 12740000Pa\) \( P \approx 13000000Pa\)	2. Substitute the numbers into the equation and solve. \( P = \rho g h\) \( P = 1.2 \times 9.8 \times 550\) \( P = 6468Pa\) Since the miner also has normal atmospheric pressure 101kPa is added. \( P_{Total} = 6468 + 101000\) \( P_{Total} = 107468Pa\) \( P_{Total} \approx 110000Pa\)	2. Substitute the numbers into the equation and solve. \( P = \rho g h\) \( P = 932 \times 9.8 \times 0.043\) \( P = 392.7448Pa\) \( P \approx 390Pa\)

References

AQA

Fluid pressure, pages 169-171, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA

Fluid pressure, pages 58, 59, GCSE Physics; The Revision Guide, CGP, AQA

Edexcel

Fluid pressure, pages 318, 319, GCSE Physics, CGP, Edexcel

Fluids and pressure, pages 202-203, GCSE Physics, Pearson Edexcel