# Archimedes' Principle ## Upthrust on an Object in a Fluid - Pressure in a fluid increases with depth due to the weight of the fluid layers above - The deeper you go, the greater the mass of fluid pressing down from above, resulting in higher pressure - This pressure variation with depth can be quantified using the **hydrostatic pressure equation**: $$\Delta p = \rho g \Delta h$$

$\Delta p = \rho g \Delta h$ where $\Delta p$ is change in pressure (Pa), $\rho$ is fluid density (kg m$^{-3}$), $g$ is gravitational field strength (N kg$^{-1}$), and $\Delta h$ is change in height (m)

### Deriving the Hydrostatic Pressure Equation - Consider a horizontal area $A$ submerged at a certain depth in a fluid - The pressure on this area is caused by the weight $W$ of the fluid column directly above it - If the fluid at rest, all points within it are in equilibrium - Pressure therefore acts equally in all directions at each point | Step | Description | Equation | |------|-------------|----------| | 1 | Weight of fluid column | $W = mg$ | | 2 | Express mass using density | $m = \rho V = \rho A h$ | | 3 | Substitute into weight equation | $W = \rho A h g$ | | 4 | Calculate pressure on area $A$ | $p = \frac{F}{A} = \frac{W}{A} = \frac{\rho A h g}{A} = \rho h g$ | - Within any volume of fluid, there is a pressure difference between the top and bottom surfaces - The pressure at the bottom is greater because more fluid sits above it, creating additional weight - This pressure difference depends on the vertical separation (change in height) between the two surfaces - This leads to the general equation: $\Delta p = \rho g \Delta h$ ### Visualizing Pressure vs Depth The graph below shows how total pressure increases linearly with depth in seawater (density 1025 kg m$^{-3}$), starting from atmospheric pressure (100 kPa) at the surface:

- The linear relationship confirms that pressure increases uniformly with depth - Every 10 m of depth adds approximately 100 kPa of pressure - At 20 m depth, the pressure has roughly tripled compared to the surface ## Archimedes' Principle

An object submerged in a fluid at rest experiences an upward buoyancy force (upthrust) equal to the weight of the fluid displaced by the object

- When an object is immersed in a fluid, it displaces a volume of that fluid - The displaced fluid would have had its own weight - The object experiences an upward force equal to this weight of displaced fluid - This upward buoyancy force is called **upthrust** ### Calculating Upthrust - The magnitude of upthrust can be calculated using: $$F = \rho g V$$ - Where: - $F$ = upthrust (buoyancy force) in newtons (N) - $\rho$ = density of the fluid in kg m$^{-3}$ - $g$ = gravitational field strength in N kg$^{-1}$ (typically 9.81 N kg$^{-1}$) - $V$ = volume of fluid displaced in m$^3$ - Since $m = \rho V$, the upthrust equals $F = mg$, which is the weight of the displaced fluid - Note that it's the density of the **fluid**, not the object, that matters for calculating upthrust ### Floating and Sinking - An object continues to sink until the weight of fluid it displaces equals its own weight - An object **floats** when the magnitude of upthrust equals the weight of the object **Conditions for floating:** - Upthrust $=$ Weight of object - $\rho_{\text{fluid}} g V_{\text{displaced}} = m_{\text{object}} g$ - The denser the fluid, the less volume needs to be displaced for an object to float **Why submarines can rise and sink:** - Submarines control their buoyancy by adjusting their average density - Taking water into ballast tanks increases density (submarine sinks) - Expelling water from ballast tanks decreases density (submarine rises) - When average density equals water density, the submarine maintains constant depth

Objects float when their average density is less than the fluid's density; they sink when their average density is greater

**Example: Hot air balloons** - Hot air is less dense than cool air - The balloon displaces a large volume of cool air - Weight of displaced cool air exceeds weight of hot air inside balloon plus basket - Net upward force causes balloon to rise - Balloon reaches equilibrium altitude when upthrust equals total weight ## Worked Example: Pressure at Depth A submarine descends into the ocean where seawater has a density of 1025 kg m$^{-3}$. Atmospheric pressure at the surface is 100 kPa. Calculate the depth at which the total pressure reaches 300 kPa. **Answer:** **Step 1:** Identify the change in pressure due to depth alone - Total pressure = Hydrostatic pressure + Atmospheric pressure - $300 \times 10^3 = \rho g h + 100 \times 10^3$ - Change in pressure: $\Delta p = 300 \times 10^3 - 100 \times 10^3 = 200 \times 10^3$ Pa **Step 2:** Use hydrostatic pressure equation $$\Delta p = \rho g \Delta h$$ **Step 3:** Rearrange for depth $$h = \frac{\Delta p}{\rho g} = \frac{200 \times 10^3}{1025 \times 9.81} = 19.9 \text{ m} \approx 20 \text{ m}$$

Always add atmospheric pressure when calculating total pressure; hydrostatic pressure alone gives only the pressure change due to depth

## Worked Example: Floating Objects A wooden log has a density of 650 kg m$^{-3}$ and a total volume of $V_{\text{log}}$. It floats in a freshwater lake where water density is 1000 kg m$^{-3}$. What fraction of the log's volume is submerged? **Answer:** **Step 1:** Apply Archimedes' Principle - For a floating object: Upthrust = Weight of object - Weight of log: $W_{\text{log}} = m_{\text{log}} g$ - Upthrust from displaced water: $W_{\text{water}} = m_{\text{water}} g$ **Step 2:** Since the log floats in equilibrium $$W_{\text{log}} = W_{\text{water}}$$ $$m_{\text{log}} g = m_{\text{water}} g$$ $$m_{\text{log}} = m_{\text{water}}$$ **Step 3:** Express masses using density equation - $m = \rho V$ - $\rho_{\text{log}} V_{\text{log}} = \rho_{\text{water}} V_{\text{submerged}}$ **Step 4:** Find the fraction submerged $$\frac{V_{\text{submerged}}}{V_{\text{log}}} = \frac{\rho_{\text{log}}}{\rho_{\text{water}}}$$ **Step 5:** Calculate the fraction $$\frac{V_{\text{submerged}}}{V_{\text{log}}} = \frac{650}{1000} = 0.65$$ - Therefore, 65% of the log is submerged beneath the water surface - The remaining 35% (or $0.35 V_{\text{log}}$) floats above the waterline **Key insight:** The ratio of densities directly gives the fraction submerged for any floating object in equilibrium