Worked exercise — Archimedes' thrust on a human being in air

Problem

A fluid is anything that deforms under a force: even air is a fluid. So Archimedes’ thrust also acts on us as we walk through the atmosphere. Let’s calculate its magnitude for a person with $m = 70\;\text{kg}$, assuming the volume of the human body is about $V \approx 0{,}17\;\text{m}^3$ (that of a crate of similar size). The density of air is $\rho_\text{aria} \approx 1{,}2\;\text{kg/m}^3$.

Solution

Step 1 — Displaced mass of air. Archimedes’ thrust equals the weight of the displaced fluid; the mass of air that occupied the body’s volume is

$m_\text{aria} = \rho_\text{aria}\cdot V = 1{,}2\cdot 0{,}17 \approx 0{,}20\;\text{kg}$

Step 2 — Archimedes’ thrust.

$\abs{\vv F_A} = m_\text{aria}\cdot g \approx 0{,}20\cdot 9{,}81 \approx 2{,}0\;\text{N}$

$\ev{\abs{\vv F_A} \approx 2{,}0\;\text{N}}$

Step 3 — Comparison with weight.

$\abs{\vv P} = m g = 70\cdot 9{,}81 \approx 686{,}7\;\text{N}$

The thrust from the air is less than $0{,}3\%$ of the weight: negligible. But if the person were immersed in water ($\rho \approx 1000\;\text{kg/m}^3$) the thrust would rise to $\approx 1667\;\text{N}$, more than double the weight: and this is exactly why we float.

Linked atoms

Argomenti: Fluid statics and dynamics Concetti: Archimedes’ principle · Weight force