How Much Lift Force Do Airliner Wings Generate at Cruising Speed?

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Homework Help Overview

The discussion revolves around calculating the lift force generated by the wings of a commercial airliner cruising at a specific speed, considering the dimensions of the wings and the density of air at cruising altitude. The problem involves applying principles from fluid dynamics, particularly Bernoulli's equation, to understand the relationship between airspeed and pressure differences above and below the wings.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore Bernoulli's equation and its application to the problem, questioning how to relate the forces exerted by air on the wings. There are attempts to clarify the role of pressure differences and how they contribute to lift. Some participants express uncertainty about the calculations and the relevance of specific equations.

Discussion Status

The discussion is ongoing, with participants sharing thoughts and calculations related to the pressure differential and lift force. Some guidance has been offered regarding the application of Bernoulli's principle, but there is still a lack of consensus on the exact calculations and methods to be used.

Contextual Notes

Participants note the constraints of the problem, including the need to adhere to introductory physics principles and the implications of using specific equations. There is also mention of homework guidelines regarding the posting of certain types of questions.

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Homework Statement



A commercial airliner cruises at a speed of 284 meters per second. Each wing is 28.8 meters long and has an average width of 1.73 meters. The density of air at cruising altitude is 0.47 kilograms per cubic meter. How much lift force is generated by the wings if the air below the wings travels at the plane's cruising speed and the air above the wings travels 9.8 percent faster? Hint: Planes have two wings.


Homework Equations



Bernoulli's?

The Attempt at a Solution



No real clear idea where to start...
 
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What is Bernoulli's equation?
 
p +1/2(rho)v^2 + (rho)gh = p + 1/2(rho)v^2 + (rho)gh

so the pressures should cancel since it atmospheric...

and so if the
air above: 311.832 m/s
air below: 284 m/s

1/2(.47)311.832 + I am not really sure what to do beyond this...
so somehow i should relate the force exerted by the air on wing above to the force exerted by the air on the wing below

so F/A = F/A?

im full of a lot of random thoughts
 
There's not a lot of pressure difference due to gravity on the top and the bottom of the wing, so we can leave out the rho*gh part.

p +1/2(rho)v^2 = constant

What happens to p when the the velocity, v is increased over the top of the wing?
 
I'm not completely sure about this, but i read something about NASA scientists deducing an equation for finding the force of lift on a wing:

Flift=1/2*density of air*(velocity of air in opposite direction)2*area
of wing*wing coeficcient

I barely think this is going to work, unless you know how to find out the wing coeficcient, which I don't know. I just read this on the net. Just trying to help.

BTW...you didn't get this equation from an engineering book, did you, 'cause if you did, I think you're supposed to have posted it in the engineering section, not in 'introductory physics!
 
This is an AP Physics B question... so it should be just involving Bournelli's... and i still can't quite get it...

so I have P + 1/2(rho)v^2 + (rho)gh = P + 1/2(rho)v^2 + (rho)gh

so I simplified to P + 1/2(rho)v^2 = P + 1/2(rho)v^2

so I know (rho),v,and I can get P because P=F/A and I know A and atmospheric pressure is at cruising altitude...

so then I just compare the two and the difference will be the force?
 
Last edited:
bleedblue1234 said:
p +1/2(rho)v^2 + (rho)gh = p + 1/2(rho)v^2 + (rho)gh

so the pressures should cancel since it atmospheric...

This is the flaw in your thinking. What comes out of Bernoulli's equation is the result that in order for the equality to be satisfied, the air pressures above and below the wings should be slightly different due to the different flow rates. That is why there *is* a lift force in the first place. So, obviously you should be getting the answer that p below the wings is slightly higher than p above them.
 
cepheid said:
This is the flaw in your thinking. What comes out of Bernoulli's equation is the result that in order for the equality to be satisfied, the air pressures above and below the wings should be slightly different due to the different flow rates. That is why there *is* a lift force in the first place. So, obviously you should be getting the answer that p below the wings is slightly higher than p above them.

yes i figured that... so how can i calculate the pressure differential per each wing?

alright so i got it down to F/A = F/A + 3897.05 somethings...
 
bleedblue1234 said:
yes i figured that... so how can i calculate the pressure differential per each wing?

Er...not really a question that makes any sense. The pressure below the wings has a certain value (period, no matter where exactly you are or what direction you look in). Same with the pressure above the wings. This pressure difference leads to a net force on the wings. The wing geometry comes into play because of the relationship between force and pressure. Pressure is the force per unit area. Does that help?
 
  • #10
cepheid said:
Er...not really a question that makes any sense. The pressure below the wings has a certain value (period, no matter where exactly you are or what direction you look in). Same with the pressure above the wings. This pressure difference leads to a net force on the wings. The wing geometry comes into play because of the relationship between force and pressure. Pressure is the force per unit area. Does that help?

alright so i calculated my 1/2(rho)v^2 values...
then subtracted over..

so i got F/A = F/A + 3897.05 somethings...
and i can calculate the area under and above the wing (they will be equal) so
F = F/A + 3897.05 * A?
 
  • #11
hrm still having trouble
 
  • #12
I got it thank you...
 

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