How can lift be less than thrust that is less than weight? [duplicate]

Question

Physics in schools teaches two contradictory and mutually exclusive things:

1. That the upward lift force on an airplane in flight equal its weight (Lift = Weight = Mass x Gravity). This is based on applying Newtons 2nd law of motion (F = ma) to the airplane in flight.

2. However, modern commercial airplanes like the Boeing 747–400 can fly with thrust-to-weight ratio as low as 0.3. Here the engine thrust is only 0.3x the weight of the airplane, but this thrust is sufficient to push the airplane forward and generate enough lift to fly.

Therefore the upward force required for lift and flight must be a lot less than 0.3x the weight of the Boeing 747-400 (Lift < Thrust < Weight). Both statements cannot be true. Which is correct?

Answer 1

Your misunderstanding lies in your thought that lift is smaller than thrust, while in fact, lift is much larger than thrust.

The lift is provided by the wings. Their purpose is exactly to create a lift force (upwards force) while requiring relatively little thrust (forwards force). How well they do this is expressed by their lift-to-drag ratio (L/D ratio). A modern airplane can have an L/D of 20 or more. This means that for every newton of thrust, you can lift 20 newtons of weight!

You can think of wings the same way you think of an inclined plane: you need a lot less force to push a car up a gentle hill than up a steep hill. Indeed, if the hill is gentle, most people can push the car up the hill, while most people cannot single-handedly lift a car up. Wings work by pushing air down, and by pushing the air down at a relatively shallow angle (by going forwards real fast), you need less force to go forwards than to lift the plane.

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If you want to know more than this 'intuitive' explanation, perhaps I can recommend you this answer of mine. I will present a terse derivation below, with the aim to calculate the L/D (neglection parasitic drag) purely from first principles.

Lift is due to a certain mass flow $\dot{m}$ being given a certain velocity $v$ downwards: $$L=\dot{m}v.$$ This mass flow is due to the wings encountering a certain amount of air; as an approximation, you can think that the airplane only affects a circular 'tube' of air with the diameter equal to its wingspan. The energy per unit of time (power) necessary to impart this downwards momentum equals $$P=\frac{1}{2}\dot{m}v^2.$$

This power is provided by the combination of thrust $T$ and forwards velocity, $$P=Tu.$$ This gives us an expression for the thrust: $$T=\frac{P}{u} = \frac{1}{2}\dot{m}\frac{v^2}{u}.$$ Note that the thrust $T$ equals drag force $D$, so we can now calculate the L/D:

$$\frac{L}{D}=\frac{L}{T} = \frac{\dot{m}v}{\frac{1}{2}\dot{m}\frac{v^2}{u}} = 2\frac{u}{v}. $$

In other words, like I said before, the L/D ratio is exactly how 'shallow' the relative velocity vector of the affected airflow is. It follows that the forwards speed $u$ should be as large as possible to maximize the L/D. However, I only took into account the so-called "lift induced drag" which is purely the horizontal force it takes to create a vertical lift force. In reality, there is also parasitic drag, which scales with $u^2$ and will at some point dominate over the reduction in L/D.

Answer 2

Because a wing produces way more lift than drag.

Thrust must equal drag, not lift.

...thrust-to-weight ratio as low as 0.3.

That's a drag-to-lift ratio, actually. And that is why you find fixed wing aircraft everywhere, otherwise all aircraft would be VTOLs

Answer 3

For any airplane to fly level at a constant altitude, the amount of lift generated by its wings must be equal to its weight. If lift exceeds weight, the plane starts to climb. If weight exceeds lift, the plane starts to descend. In a constant rate climb/descent, lift equals weight; a simplification that ignores the propulsive force vector.

The engine & prop in a light plane like a 4-seat Cessna develops no more than a couple hundred pounds of thrust at takeoff which is much less than the weight of the plane. But that amount of thrust is enough to move the plane's wings through the air at a speed sufficient to make enough lift to get the plane off the ground.

Note that the propulsive thrust does not have to equal the plane's weight in order to make it fly; this is only true if the plane had no wings and you were trying to hoist it off the ground by pointing the plane straight up and "hanging" it on its propeller.

Answer 4

When you push your car on a flat area you need very little force to move it, much less than its weight

With the same force you are pushing the car, you notice that at the beginning it will get an accelerated motion and afterwards due to the drag mainly, the speed becomes steady even though keeping the same pushing effort

Therefore in order to make an aircraft accelerate you need just to overcome the drag

Acceleration means increasing the speed, thus the drag, but all you need is to overcome the drag not the weight.

To lift off you just need to get the necessary speed for the wings to provide, as a function of the angle of attack, that lift that is just above the aircraft weight