Don't be fooled into thinking that only solids are 'heavy' and that we can approximate everything as a volume-less particle. Fluids, such as liquids, have a density, exert an upward buoyancy on anything that should displace them, have increasing pressure with depth and are used in hydraulics to exert huge forces.
Key Concepts
Density is a material property (like Resistivity), which means that it is a constant value for a given material irrespective of its dimensions. The density of copper wire is the same as the density of a copper block.
Density is defined as the ratio of the mass of an object to its volume:
\(\rho={m\over V}\)
- \(\rho\) is density (kg m-3)
- \(m\) is mass (kg)
- \(V\) is volume (m-3)
Solids (especially metals) have a high density due to their high concentration of particles; copper's density is approximately 9000 kg m-3. Liquids come next; water has a density of 1000 kg m-3. Gases have a low density; air has a density of approximately 1 kg m-3 at sea level.
At a surface
Pressure is the ratio of force acting to the surface area:
\(P={F\over A}\)
- \(P\) is pressure (Pa)
- \(F\) is force (N)
- \(A\) is surface area (m2)
At a depth
It can be helpful to think of our lives as taking place at the bottom of an ocean of atmosphere. The atmosphere presses down on us due to gravity. While we do not feel impeded by this, atmospheric pressure has a value of approximately 105 Pa.
Within a container of fluid, pressure increases with depth. This is because, the further down you go, the larger the force acting from the fluid above. Consider the force as emerging from the mass of the fluid, which in turn can be calculated using its density and volume. For a container of constant cross-sectional area:
\(P={F\over A}={m_f g\over A}={\rho_f V g\over A}\)
\(\Rightarrow P_f=\rho_f gd\)
- \(P_f\) is the pressure in a fluid (Pa)
- \(\rho_f\) is the density of the fluid (kg m-3)
- \(g\) is gravitational field strength (on earth, 9.81 N kg-1)
- \(d\) is depth (NB: downward from the surface) (m)
By Matt Cook - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=45705236
To find the total pressure acting at a depth in a fluid, we combine atomospheric pressure and the pressure due to the fluid:
\(P=P_0 +\rho_f gd\)
- \(P\) is total pressure (Pa)
- \(P_0\) is atmospheric pressure (on earth, 105 Pa)
- \(\rho_f\) is the density of the fluid (kg m-3)
- \(g\) is gravitational field strength (on earth, 9.81 N kg-1)
- \(d\) is depth (NB: downward from the surface) (m)
Archimedes' principle allows us to calculate the magnitude of fhe force of buoyancy. Buoyancy is equal to the weight of the fluid displaced, which we calculate using the fluid's density and volume (rather than mass):
\(B=\rho_f V_f g\)
- \(B\) is buoyancy (N)
- \(\rho_f\) is the density of the fluid (kg)
- \(V_f\) is the volume of fluid that is displaced by the object (m3)
- \(g\) is gravitational field strength (on earth, 9.81 N kg-1)
For an object that is fully submerged, the volume of the displaced fluid is equal to the volume of the object. For a floating object or an object that is partly submerged, only the volume of the object withn the fluid displaces fluid. Note that it is impossible for an object to float without displacing any fluid (however much polystyrene might appear to rest on top of a container of water!).
Pascal’s principle states that a change in pressure at any point in a confined fluid at rest is transmitted to all points in the fluid. For an incompressible fluid confined by two pistons, a change in pressure on the first piston causes an equal change in pressure on the second piston.
We can calculate these pressures using combinations of the following equation for the change in pressure:
\(\Delta P={\Delta F \over A}=\rho_f g \Delta d\)
- \(\Delta P\) is the change in pressure (Pa)
- \(\Delta F\) is the change in force (N)
- \(A\) is the cross-sectional area of a piston (m2)
- \(\rho_f\) is the density of the fluid (kg m-3)
- \(g\) is gravitational field strength (on earth, 9.81 N kg-1)
- \(\Delta d\) is the compression or retraction distance moved through by a piston (m)
One implication of Pascal's principle is the ability of a single human to raise the height of a much heavier object. Provided the area of the piston below the heavy object is much larger than the area of the piston where the human force is exerted, the human force can be much smaller:
\({\Delta F_1\over A_1}={\Delta F_2\over A_2}\)
This is an example of hydraulics, the use of fluids in engineering.
Use flashcards to practise your recall.
Use quizzes to practise application of theory.
START QUIZ!
1000 kg of water is contained in a 1 m3 rigid cube. The cube and contents are falling freely close to the Earth. Atmospheric pressure is 100 kPa.
The pressure exerted by the fluid on the bottom of the box is:
Free-fall is equivalent to zero gravity so there is no pressure due to depth.
The container is rigid and contains no air so no there is pressure due to the atmosphere.
1000 kg of water is contained in a 1 m3 rigid cube by a frictionless piston. The cube and contents are falling freely close to the Earth. Atmospheric pressure is 100 kPa.
The pressure exerted by the fluid on the bottom of the box is:
Free-fall is equivalent to zero gravity so there is no pressure due to depth.
Atmospheric pressure acts on the piston. Since the piston is frictionless, the entirety of this atmospheric pressure acts on the fluid and is transmitted everywhere in the fluid.
1000 kg of water is contained in a 1 m3 rigid cube by a frictionless piston of mass 100 kg. The cube and contents are resting on the Earth. Atmospheric pressure is 100 kPa.
The pressure exerted by the fluid on the bottom of the box is:
The total pressure is the combination of atmosopheric pressure, the pressure due to depth and the pressure due to the piston:
\(P = P_0 + P_f + P_\text{piston} = 10^5 + 1000\times 10\times 1+{100\times 10\over 1}\)
A diver measures the pressure in a lake at 50 m and 100 m. The ratio \(P_{50}\over P_{100}\) is:
The trick is to remember to add atmospheric pressure. Atmospheric pressure is equivalent to the pressure at a depth of 10 m.
Since pressure at depth is proportional to depth for constant density and gravitational field strength:\({P_{50}\over P_{100}}={50+10\over 100+10}={6\over 11}\)
A 10 m deep swimming pool is constructed on the Moon. What is the ratio \(\text{pressure at the bottom of the Moon pool}\over \text{pressure at the bottom of an identical pool on the Earth}\)?
The Moon has no atmoshere and a gravitational field strength that is one sixth of the Earth.
Since pressure is proportional to gravitational field strength, the pressure on the Moon is one sixth that on Earth. This is doubled since 10 m depth of water on Earth is equivalent to atmospheric pressure.
This image is from a Geogebra simulation.
Calculate \(F_2\).
\({F_1\over A_1}={F_2\over A_2}\Rightarrow F_2=F_1{A_2\over A_1}\)
Since \({r_2\over r_1}=2\), \({A_2\over A_1}=4\Rightarrow F_2=3\times 4\)
This image is from a Geogebra simulation.
Calculate \(\Delta h\).
The pressure acting, \(P = {20\over 10\times10^{-4}} = 20 \text{ kPa}\)
This pressure is equal to \(\rho_f g\Delta h\):
\(\Delta h={20 000\over 1000\times 10}=2 \text{ m}\)
A rectangular block with square base of side 4 m and height 5 m is immersed in water in a room with no air.
If the bottom of the block is at a depth of 10 m, the pressure on the bottom surface of the block is:
Since no air is present, there is no atmospheric pressure. The depth of the bottom surface is 10 m.
\(P= ρ_fgd = 1000\times10\times10 = 100\text{ kPa}\)
A rectangular block with square base of side 4 m and height 5 m is immersed in water in a room with no air.
If the bottom of the block is at a depth of 10 m, the pressure on the top surface of the block is:
Since no air is present, there is no atmospheric pressure. The depth of the top surface is 5 m.
\(P= ρ_fgd = 1000\times10\times5 = 50\text{ kPa}\)
A rectangular block with square base of side 4 m and height 5 m is immersed in water in a room with no air.
If the bottom of the block is at a depth of 10 m, the buoyant force is:
\(B=\rho_fV_fg\)
The volume of fluid displaced is equal to the volume of block \(= 4\times4\times5 = 80\text{ m}^3\)
\(B = 1000\times 80\times 10 = 800\text{ kN}\)
NB: The depth of the block is irrelevant once it becomes fully submerged.
How much of AHL Static fluids have you understood?
Twitter 
Facebook 
LinkedIn