calculate the upthrust acting on an object in a fluid using the equation F = ρgV (Archimedes’ principle)

Equilibrium of Forces – Upthrust (Archimedes’ Principle)

Learning Objectives (AO1 & AO2)

  • State the definition of upthrust (buoyant force) and the symbols ρ, g, V.

  • Write the hydrostatic‑pressure relation Δp = ρ g Δh and the derived formula p = p₀ + ρ g h.

  • Derive the upthrust formula Fup = ρ g V from the pressure‑difference argument.

  • Apply the formula to fully and partially submerged objects.

  • Construct a vector‑triangle for the forces acting on a body in a fluid and use it to find an unknown tension or normal reaction.

  • Predict whether an object will float, sink or remain neutrally buoyant by comparing its density with the fluid density.

Key Formulae

Hydrostatic pressure increaseΔp = ρ g Δh
Absolute pressure at depth hp = p₀ + ρ g h
Upthrust (buoyant force)Fup = ρfluid g Vdisplaced
Weight of an objectW = m g = ρobject Vobject g

Key Concepts

TermDefinition / Relevance
Upthrust (Buoyant Force)The net upward force exerted by a fluid on an immersed body.
Archimedes’ PrincipleThe upthrust equals the weight of the fluid displaced: Fup = ρfluid g Vdisplaced.
Hydrostatic PressurePressure varies linearly with depth according to Δp = ρ g Δh.
Equilibrium of ForcesFor a body at rest, the vector sum of all forces is zero: ∑ \vec F = 0.
Floating / Sinking Criterion

  • ρobject < ρfluid → Fup > W → floats (partly submerged).

  • ρobject = ρfluid → Fup = W → neutrally buoyant (fully submerged).

  • ρobject > ρfluid → Fup < W → sinks.

Derivation of the Upthrust Formula from Hydrostatic Pressure

  1. Take a thin horizontal slab of the submerged body at depth h with cross‑sectional area A and thickness dh.

  2. Pressure on the bottom face: pb = p₀ + ρ g h.

    Pressure on the top face: pt = p₀ + ρ g (h − dh).

  3. Resulting forces:

    Upward force on bottom Fb = pbA and Downward force on top Ft = ptA.

  4. Net upward force on the slab:

    \[

    \Delta F = F{b}-F{t}= \big[p{0}+ρgh\big]A-\big[p{0}+ρg(h-dh)\big]A = ρgA\,dh = ρg\,dV .

    \]

  5. Integrate over the whole displaced volume V:

    \[

    F_{\text{up}} = \int ρg\,dV = ρgV .

    \]

Vector‑Triangle Method for Equilibrium

When a body is at rest in a fluid the forces can be represented as a closed triangle drawn tip‑to‑tail.

  • Weight \(W = mg\) downwards.

  • Upthrust \(F{\text{up}} = ρgV{\text{disp}}\) upwards.

  • Supporting force Tension \(T\) or normal reaction \(N\) acting as required.

Mathematically: \(\vec W + \vec T + \vec F_{\text{up}} = \vec 0\).

Worked Vector‑Triangle Example

Problem: A wooden block (density 600 kg m⁻³, volume 2.5 × 10⁻³ m³) is attached to a light string and allowed to float in water. Find the tension in the string.

  1. Calculate the weight:

    \(W = ρ_{\text{block}} V g = (600)(2.5\times10^{-3})(9.81) = 14.7\ \text{N}\).

  2. Since the block floats, only a fraction \(f\) of its volume is submerged.

    \(F{\text{up}} = ρ{\text{water}} g fV = (1000)(9.81)f(2.5\times10^{-3}) = 24.5f\ \text{N}\).

  3. In equilibrium the vertical forces form a triangle: \(\vec W\) down, \(\vec F_{\text{up}}\) up, \(\vec T\) up (if the string pulls upward).

    Because the block is stationary, \(\displaystyle T = W - F_{\text{up}}\).

  4. For a floating block the displaced volume adjusts until \(F_{\text{up}} = W - T\).

    If the string is just enough to keep the block from sinking completely, assume the block is *just* fully submerged (maximum upthrust). Then \(F{\text{up,max}} = ρ{\text{water}} g V = (1000)(9.81)(2.5\times10^{-3}) = 24.5\ \text{N}\).

  5. Since \(F_{\text{up,max}} > W\), the string must act downward to keep the block from rising.

    Thus the required tension magnitude is

    \(T = F_{\text{up,max}} - W = 24.5 - 14.7 = 9.8\ \text{N}\) (directed upward on the block, downward on the string).

In a vector‑triangle diagram the three arrows (W, Fup, T) close to form a triangle, confirming equilibrium.

Worked Numerical Examples

Example 1 – Fully Submerged Cube in Water

StepCalculation
Volume displaced\(V = (0.10\ \text{m})^{3}=1.0\times10^{-3}\ \text{m}^{3}\)
Fluid density\(\rho_{\text{water}} = 1000\ \text{kg m}^{-3}\)
Upthrust\(F_{\text{up}} = \rho g V = (1000)(9.81)(1.0\times10^{-3}) = 9.81\ \text{N}\)
Weight of cube (ρ = 800 kg m⁻³)\(W = \rho_{\text{cube}} V g = (800)(1.0\times10^{-3})(9.81)=7.85\ \text{N}\)
Equilibrium check\(F{\text{up}} > W\) → the cube would rise unless a string of tension \(T = F{\text{up}}-W = 1.96\ \text{N}\) holds it.

Example 2 – Partially Submerged Sphere in Oil

A solid sphere of radius \(r = 0.05\ \text{m}\) is placed in oil (\(\rho_{\text{oil}} = 800\ \text{kg m}^{-3}\)). Only 60 % of its volume is submerged.

StepCalculation
Total volume of sphere\(V_{\text{sphere}} = \frac{4}{3}\pi r^{3}=5.24\times10^{-4}\ \text{m}^{3}\)
Displaced volume\(V{\text{disp}} = 0.60\,V{\text{sphere}} = 3.14\times10^{-4}\ \text{m}^{3}\)
Upthrust\(F{\text{up}} = \rho{\text{oil}} g V_{\text{disp}} = (800)(9.81)(3.14\times10^{-4}) = 2.47\ \text{N}\)
Weight (ρ = 900 kg m⁻³)\(W = \rho{\text{sphere}} V{\text{sphere}} g = (900)(5.24\times10^{-4})(9.81)=4.62\ \text{N}\)
Result\(F_{\text{up}} < W\) → the sphere sinks until a larger fraction is submerged or it contacts the container bottom.

Common Mistakes to Avoid

  • Using the wrong volume: For partially submerged bodies use the *displaced* volume, not the total geometric volume.

  • Neglecting Δp = ρgΔh: The buoyant force originates from the pressure difference between the bottom and top surfaces; omitting this step can cause sign errors.

  • Confusing density with mass: Density must be in kg m⁻³; mass is in kg. Convert when only one is given.

  • Forgetting supporting forces: When a string or surface is present, include its tension or normal reaction in the vector‑triangle; otherwise you may incorrectly predict floating.

  • Assuming full submersion for objects that actually float partially – always check the floating‑criterion first.

Practice Questions

  1. A solid sphere of radius \(0.05\ \text{m}\) is suspended in oil (\(\rho_{\text{oil}} = 800\ \text{kg m}^{-3}\)). Find the upthrust.

  2. A wooden block of density \(600\ \text{kg m}^{-3}\) and volume \(2.5\times10^{-3}\ \text{m}^{3}\) is placed in water. Determine the magnitude of the upthrust and state whether the block will float or sink.

  3. A rectangular metal bar (density \(7800\ \text{kg m}^{-3}\), volume \(1.0\times10^{-4}\ \text{m}^{3}\)) is attached to a light string and lowered into water. Calculate the tension in the string when the bar is fully submerged.

  4. Explain, using the formula \(F_{\text{up}} = \rho g V\), why a submarine can dive by taking in seawater and surface by expelling it. Include a short sketch of the forces.

  5. Using the vector‑triangle method, a floating cube (density 500 kg m⁻³, side = 0.2 m) is attached to a string that pulls upward. The string tension is 3 N. Find the fraction of the cube’s volume that is submerged.

Summary

  • Hydrostatic pressure increases with depth: Δp = ρ g Δh, p = p₀ + ρ g h.

  • The pressure difference between the bottom and top of an immersed body gives a net upward force: Fup = ρ g Vdisplaced.

  • Equilibrium of forces is expressed by the vector‑triangle \(\vec W + \vec F_{\text{up}} + \vec R = \vec 0\).

  • Floating, sinking or neutral buoyancy are decided by comparing object density with fluid density.

  • Once the displaced volume and fluid density are known, the buoyant force follows instantly from Fup = ρ g V.

Suggested diagram: (a) Fully submerged cube showing weight \(mg\) downwards, upthrust \(F_{\text{up}}\) upwards, and a supporting tension \(T\) if present. (b) Vector‑triangle of the three forces closing to zero.