Interactive Lecture
Pressure, Buoyancy, and Forces on Submerged Surfaces
by Dr Arnaldo Delli Carri
Fluid statics, also known as hydrostatics, is the branch of fluid mechanics that studies fluids at rest. The fundamental principle is that pressure in a stationary fluid varies only with depth and is independent of the shape of the container.
Pressure at any point in a static fluid depends only on the vertical depth below the free surface. That's why dams are thicker at the bottom and why submarines must withstand enormous pressure at great depths.
Pressure is defined as the normal force per unit area exerted by a fluid on a surface:
\[p = \frac{F}{A}\]
Where:
In a fluid at rest, pressure increases linearly with depth due to the weight of the fluid above:
\[p = p_0 + \rho g h\]
Where:
There are two common ways to express pressure:
\[p_{\text{absolute}} = p_{\text{gauge}} + p_{\text{atm}}\]
\[p_{\text{gauge}} = \rho g h\]
Explore how pressure increases linearly with depth. Adjust fluid density and tank depth to see the pressure gradient visualised through colour and pressure arrows.
A pressure change at any point in a confined incompressible fluid is transmitted equally to all points in the fluid:
A pressure change transmits equally to every point in a confined fluid, so a small force on a small piston can raise a large force on a bigger one. That's the basis for hydraulic jacks, brakes, and presses.
\[\frac{F_1}{A_1} = \frac{F_2}{A_2}\]
Therefore: \[F_2 = F_1 \times \frac{A_2}{A_1}\]
See how a small force on a small piston creates a large force on a larger piston. This is the principle behind hydraulic lifts, brakes, and presses.
The total force on a submerged planar surface is the product of the pressure at the centroid and the surface area:
\[F = p_c \times A = \rho g h_c \times A\]
Where:
When an object is immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced:
\[F_b = \rho_{\text{fluid}} g V_{\text{displaced}}\]
Where:
Observe how objects float or sink based on their density relative to the fluid. The buoyant force equals the weight of displaced fluid.
This simulation demonstrates a hinged gate submerged in water with a horizontal force P applied. Adjust the parameters to see how the equilibrium changes and what minimum force P is required to prevent liquid leakage.
| Concept | Equation | Description |
|---|---|---|
| Pressure Definition | \(p = \frac{F}{A}\) | Force per unit area |
| Hydrostatic Pressure | \(p = p_0 + \rho g h\) | Pressure at depth h |
| Gauge Pressure | \(p_{\text{gauge}} = \rho g h\) | Pressure above atmospheric |
| Pascal's Law | \(\frac{F_1}{A_1} = \frac{F_2}{A_2}\) | Pressure transmission in fluids |
| Buoyant Force | \(F_b = \rho_f g V_d\) | Archimedes' Principle |
| Force on Surface | \(F = \rho g h_c A\) | Total force on submerged surface |
A water tank is filled to a depth of 8 metres. Calculate the gauge pressure and absolute pressure at the bottom of the tank. Use \(\rho_{\text{water}} = 1000\) kg/m³, \(g = 9.81\) m/s², and \(p_{\text{atm}} = 101.325\) kPa.
Show Solution
Solution:
Given:
Calculate Gauge Pressure:
\[p_{\text{gauge}} = \rho g h = 1000 \times 9.81 \times 8\]
\[p_{\text{gauge}} = 78,480 \text{ Pa} = 78.48 \text{ kPa}\]
Calculate Absolute Pressure:
\[p_{\text{absolute}} = p_{\text{gauge}} + p_{\text{atm}}\]
\[p_{\text{absolute}} = 78.48 + 101.325 = 179.805 \text{ kPa}\]
Answer: Gauge pressure = 78.48 kPa, Absolute pressure = 179.81 kPa
A hydraulic lift has a small piston with diameter 5 cm and a large piston with diameter 25 cm. If a force of 200 N is applied to the small piston, what force can be lifted on the large piston?
Show Solution
Solution:
Given:
Calculate Areas:
\[A_1 = \frac{\pi d_1^2}{4} = \frac{\pi (0.05)^2}{4} = 1.963 \times 10^{-3} \text{ m}^2\]
\[A_2 = \frac{\pi d_2^2}{4} = \frac{\pi (0.25)^2}{4} = 4.909 \times 10^{-2} \text{ m}^2\]
Apply Pascal's Law:
\[F_2 = F_1 \times \frac{A_2}{A_1} = 200 \times \frac{4.909 \times 10^{-2}}{1.963 \times 10^{-3}}\]
\[F_2 = 200 \times 25 = 5000 \text{ N} = 5 \text{ kN}\]
Answer: The hydraulic lift can lift a force of 5000 N (5 kN), which is 25 times the input force.
A solid cube with side length 0.5 m and density 800 kg/m³ is fully submerged in water (density 1000 kg/m³). Calculate the buoyant force and determine if the cube will float or sink.
Show Solution
Solution:
Given:
Calculate Volume:
\[V = a^3 = (0.5)^3 = 0.125 \text{ m}^3\]
Calculate Weight of Cube:
\[W = \rho_{\text{cube}} g V = 800 \times 9.81 \times 0.125 = 981 \text{ N}\]
Calculate Buoyant Force:
\[F_b = \rho_{\text{water}} g V = 1000 \times 9.81 \times 0.125 = 1226.25 \text{ N}\]
Compare Forces:
Since \(F_b = 1226.25 \text{ N} > W = 981 \text{ N}\), the buoyant force exceeds the weight.
Answer: The buoyant force is 1226.25 N. Since this is greater than the weight, the cube will float.