Lift

Lift is the force of a fluid on a moving object perpendicular to the direction of motion. A plane stays up by pushing air down.

Why (momentum model, MacKay):

• Newton 3: wings push air down ↔ air pushes wings up (= lift). Level flight (constant height): lift = $Mg$
• Newton 2: force = rate of change of momentum, $F = \frac{\Delta(mv)}{\Delta t} = \dot m\, v$ → $Mg = \dot m_{air}\, v_{lift}$ (like throwing 2 kg of balls down every second at 10 m/s → 20 N recoil upward)
• Air deflected per second: the plane sweeps the area $A_{lift}$ at forward speed $v$ → $\dot m_{air} = \rho A_{lift} v$ $v_{lift} = \frac{Mg}{\rho A_{lift} v}$
• Power = kinetic energy given to that air per second: $P_{lift} = \frac12 \dot m_{air} v_{lift}^2 = \frac12 \frac{(Mg)^2}{A_{lift}\,\rho\, v}$

Why $\propto 1/v$: faster → more air per second → a smaller downward speed is enough → less energy ($\propto v_{lift}^2$).

Optimal speed: per distance, drag rises ($\propto v^2$) and lift cost falls ($\propto 1/v^2$) → there is a minimum: $\frac{E}{d} = \frac12 c_{drag} A_{drag} \rho v^2 + \frac12 \frac{(Mg)^2}{A_{lift}\rho v^2}$ $\frac{d}{dv} = 0$ → at $v_{opt}$ both terms are equal, and $\rho v_{opt}^2 = \frac{Mg}{\sqrt{c_{drag} A_{drag} A_{lift}}}$. Thinner air (high altitude) → higher $v_{opt}$, same minimum energy per distance.

Related: Drag, Buoyancy. Archimedes’ principle, Momentum and Force