Fluid Pressure
Contents
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
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