Question about lift in steady climb

In a climb you are pitched up and therefore there is a vertical component of the thrust. The lift in the vertical direction will reduce by the same amount to keep you in equilibrium, else you would accelerate.

If you draw an aircraft with the four main forces on it, the rotate it by 20 degrees say, and via vectors bring the force components into the horizontal and vertical that might illustrate it better.

(Although that is a horrendous description. I’ll.try to draw what I mean later)

It’s a nasty question.

In a climb you are pitched up and therefore there is a vertical component of the thrust.

Exactly, I agree, and probably the question writer was banking on that, but it doesn’t have to be the case! – it depends on what angle the wings are screwed on, flaps, etc.

If the prop axis (the thrust axis) is horizontal but the wings are making lift at that airframe angle, the plane will still climb.

For example a TB20 will full flaps will be close to horizontal, draggy as hell, but it will climb (at full power )

And the wings are supporting 100% of the aircraft weight!

If a plane is climbing (no matter how it does it) there is excess thrust which is being converted into potential energy. If the airspeed is constant, no kinetic energy is being traded (it remains constant) so the potential energy increase is coming wholly from the excess thrust.

Disregarding vertical component of thrust, The lift would be slighly greater than the weight, just to overcome the resistance of the air to the vertical motion of the plane (which shouldn’t be very significant). If there wasn’t that resistance component, the lift would be greater than weight on start of climb (while vertical speed increasing) then it would be equal in the steady climb, then less when levelling off and losing vertical speed. At level off, it would be back again to equal.

The question is actually rather meaningless without a concise drawing showing vectors. An aircraft can climb at a constant speed, and the engine can be pointed slightly down, slightly up, or east or whatever. The resultant total lifting force is L_wing + L_engine in y direction. The aircraft can climb or descend with the lift produced by the wing being both larger or smaller than the weight of the aircraft.

Yes – I think to achieve a climb you need the sum of

• the vertical part of the wing lift, and
• the vertical part of the prop thrust (if there is any)

to exceed the aircraft weight.

So, contrary to what I said above, if the prop thrust is horizontal, the wing lift must exceed the aircraft weight.

And if there is zero wing lift, the prop thrust alone must exceed the aircraft weight – an obvious case with some military aircraft which can go straight up.

Peter wrote:

So, contrary to what I said above, if the prop thrust is horizontal, the wing lift must exceed the aircraft weight.

That’s to start the climb. If you disregarded the resistance of the air to the climbing motion of the plane, then the a lift > weight would produce a constantly accelerated (= vertical speed always increasing)

Guys, it’s from the PPL question bank, let’s not over think it.

The wings will be attached horizontally, the propeller not inclined and the aircraft will fly level at 0° pitch. I.e. the aeroplanes we used to draw when we were three years old

Let’s not confuse the OP with real world examples, and adding complexities like flaps. That will come later in more advanced theory.

italianjon wrote:

The wings will be attached horizontally, the propeller not inclined and the aircraft will fly level at 0° pitch. I.e. the aeroplanes we used to draw when we were three years old

In that case the lift is greater than weight by 1/2*rho*S*Cz*vertical_speed^2, to overcome the air resistance on the vertical axis, but that would be negligible so wouldn’t be surprised if answer was “equal”

But you do not need to pitch up in order to climb. Assuming you are in straight and level cruise with 0° pitch, you just need to increase power and maintain 0° pitch to initiate a climb. In that case you have 0 vertical component of thrust.

EDIT: I guess this was already covered…