Fluids: pressure, upthrust and viscosity

What the exam asks

• Calculating pressure from $p = F/A$ and from $\Delta p = \rho g\Delta h$, including total pressure = atmospheric pressure + $\rho gh$.
• Explaining upthrust as a pressure difference, and using upthrust = weight of fluid displaced ($F = \rho_{\text{fluid}}gV$).
• Flotation (upthrust = weight) and apparent weight (weight − upthrust).
• Laminar and turbulent flow, viscosity, and Stokes’ law with its conditions (Edexcel and IAL).
• The terminal-velocity force balance $W = U + F_D$, and the falling-ball practical with its uncertainties (Edexcel and IAL).

Core ideas

Pressure in a fluid

Pressure is force per unit area, $p = \dfrac{F}{A}$, measured in pascals ($1\ \text{Pa} = 1\ \text{N m}^{-2}$).

Derivation of $\Delta p = \rho g\Delta h$. Take a column of fluid of height $\Delta h$ and base area $A$. Its mass is $m = \rho V = \rho A\Delta h$, so its weight is $\rho A\Delta h\,g$. Dividing by the area gives the extra pressure at the bottom: $\Delta p = \dfrac{\rho A\Delta h\,g}{A} = \rho g\Delta h$.

• The pressure at a given depth acts equally in all directions, and depends only on depth and density, not on the shape of the container.
• Total (absolute) pressure at depth $h$ is $p = p_{\text{atm}} + \rho gh$, with $p_{\text{atm}} \approx 1.0 \times 10^5\ \text{Pa}$. Read the question: “pressure due to the water” means $\rho gh$ only.

Upthrust and Archimedes’ principle

A submerged object has more fluid pressure on its lower surface than on its upper surface, because the lower surface is deeper. For a block of area $A$ and height $\Delta h$:

This is Archimedes’ principle: the upthrust equals the weight of fluid displaced.

• A fully submerged object’s upthrust does not depend on depth, for an incompressible liquid.
• Floating: upthrust = weight, so $\rho_f gV_{\text{sub}} = \rho_o gV$. The fraction submerged is $\dfrac{V_{\text{sub}}}{V} = \dfrac{\rho_o}{\rho_f}$.
• Apparent weight (for example, a newton-meter reading in water) = weight − upthrust.

Viscosity and flow (Edexcel and IAL)

• Laminar (streamline) flow: the fluid moves in layers that do not mix, and the velocity at any point is constant. Adjacent layers can move at different speeds.
• Turbulent flow: eddies and vortices form, the layers mix, and the velocity at a point changes abruptly. It happens at high speed and around non-streamlined shapes.
• Viscosity $\eta$ (unit Pa s) measures a fluid’s resistance to flow. For liquids it falls sharply as the temperature rises, which is why warm syrup pours more easily.

Stokes’ law gives the viscous drag on a small sphere moving slowly through a fluid with laminar flow:

Here $r$ is the radius. The law fails for large or fast objects, where the flow becomes turbulent.

Terminal velocity of a falling sphere

Three forces act on the sphere: its weight $W$ (down, constant), the upthrust $U$ (up, constant), and the drag $F_D$ (up, increasing with speed). At release, $F_D = 0$ and the acceleration is greatest. As the speed rises the drag rises, so the resultant force and the acceleration fall. Terminal velocity is reached when:

A graph of $v$ against $r^2$ is therefore a straight line through the origin, with gradient $\dfrac{2g(\rho_s - \rho_f)}{9\eta}$.

The falling-ball practical (Edexcel Core Practical 4)

• Use a tall tube of liquid with marker bands or rubber bands. Start the first band well below the surface so the ball is already at terminal velocity.
• To check terminal velocity, time the ball over two equal consecutive distances. The times should agree.
• Measure the ball’s diameter with a micrometer. Find its density from its mass and volume, and the liquid’s density with a balance and measuring cylinder.
• Keep the temperature constant (record it with a thermometer), because $\eta$ depends strongly on it.
• Improve the timing with light gates or a video with a timer in frame, which reduces reaction-time and parallax errors. Use a wide tube, because walls close to the ball increase the drag.

Worked examples

Exam technique

• Draw the free-body diagram first. For any object in a fluid, list the weight, the upthrust and (if it is moving) the drag. Most method marks come from writing the correct balance equation.
• “Explain why the ball reaches terminal velocity” needs a four-link chain: drag increases with speed → resultant force decreases → acceleration decreases → when $W = U + F_D$ the resultant force is zero, so the velocity is constant.
• Unit conversions: g cm⁻³ × 1000 gives kg m⁻³; mPa s ÷ 1000 gives Pa s; convert diameters in mm to radii in m.
• Stokes’ law conditions are a common 1–2 mark item: a small sphere, low speed, laminar flow (no turbulence).
• Gradient questions: if the axes are $v$ against $r^2$, the gradient is $\dfrac{2g(\rho_s - \rho_f)}{9\eta}$. Rearrange for $\eta$ symbolically before putting numbers in.

Common mistakes

Quick recap

• $p = F/A$; $\Delta p = \rho g\Delta h$; total pressure = $p_{\text{atm}} + \rho gh$.
• Upthrust comes from the pressure difference between the bottom and top surfaces, and equals the weight of fluid displaced, $\rho_f gV$.
• Floating: upthrust = weight; the fraction submerged is $\rho_o/\rho_f$.
• Laminar flow has non-mixing layers; turbulent flow has eddies. Liquid viscosity falls as the temperature rises.
• Stokes’ law $F = 6\pi\eta rv$ applies to small, slow spheres in laminar flow.
• Terminal velocity: $W = U + F_D$, so $v = 2r^2g(\rho_s - \rho_f)/(9\eta)$ and .