Buoyant Force — Hot-Air Balloon Lift (Worked Example)

Medium

Problem

A hot-air balloon envelope holds 2000 m³ of hot air (ρ = 0.90 kg/m³) surrounded by cool air (ρ = 1.20 kg/m³). What is the net buoyant lift? (g = 9.8 m/s²)

Given

V = 2000 m³
ρ_out = 1.20 kg/m³
ρ_in = 0.90 kg/m³
g = 9.8 m/s²

Find

The net upward force.

Principle

Buoyancy lifts with the outside air's weight; the hot air's own weight pulls down. Net lift = (ρ_out − ρ_in)Vg.

Formula

F_{net} = (\rho_{out} - \rho_{in})Vg

F_net = (ρ_out − ρ_in)Vg

Solution

Substitute: $F_{net} = (1.20 - 0.90)(2000)(9.8)$
Difference: $0.30\ \mathrm{kg/m^3}$
Multiply: $F_{net} = 0.30 \times 2000 \times 9.8 = 5880\ \mathrm{N}$

5880 N upward

Unit check

kg/m³ × m³ × m/s² = N — correct.

What this means physically

5880 N lifts about 600 kg — envelope, basket, burner and a few passengers. Heating the air more (lower ρ_in) increases lift; cooling it lands the balloon.

Common mistake

Using only ρ_out·Vg as 'the answer'. That is the gross buoyancy — subtract the hot air's weight for the net lift that actually flies.