What Is the Pressure on the Upper Wing Surface During Level Flight?

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The discussion centers on calculating the pressure on the upper wing surface of an airplane during level flight, given its mass and the pressure on the lower wing surface. A participant encounters a negative value for pressure, indicating a potential error in their calculations. The group identifies that the equations used were dimensionally inconsistent, particularly mixing pressures and forces. After correcting the equations, they arrive at a pressure value of approximately 6.6 x 10^4 Pa, suggesting that the discrepancy with the provided answer of 6.8 x 10^4 Pa may be due to a typo. The conversation emphasizes the importance of maintaining dimensional consistency in physics equations.
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Homework Statement


An airplane has a mass of 1.60x10^4 kg and each wing has an area of 40.0m^2. During level flight the pressure on the lower wing surface is 7.00x10^4 Pa. Determine the pressure on the upper wing surface.

http://img168.imageshack.us/img168/617/41426717vo3.th.jpg

Homework Equations



\sum F= F_L -F_u - F_{mg} = 0


The Attempt at a Solution



My basic problem is that I keep getting a negative number for the pressure which is odd

\sum F= F_L -F_u - F_{mg} = 0

F_u = F_{mg}- F_L

P= F/A

P_u= Mg - P_L A

((9.8m/s^2)*(1.60x10^4kg))-((40.0m^2)(7.00x10^4Pa))= P_u

-2,643,200 = P_u A

P_u= -66,080 Pa =>

my problem is why is it negative?? is it because it's force is downward??
Plus..the answer that was given (only the answer) was 6.80x10^4 Pa so I must have done something wrong. However I ccalculated it 3 times and I still get the same answer.

Thank you very much =D
 
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First of all, your rearranging of the equation has an error:

You start with

FL - Fu - Fmg = 0

Now add Fmg to both sides of the equation, and subtract FL from both sides:

- Fu = Fmg - FL

This is not what you arrived at. The discrepancy explains your errant minus sign
 
Your third equation also makes no sense. You have written:

Pu = Mg - PLA

But Pu is a pressure whereas Mg and PLA are forces. You can't equate a pressure to a force. Your equation is not dimensionally consistent. Does this help?
 
cepheid said:
First of all, your rearranging of the equation has an error:

You start with

FL - Fu - Fmg = 0

Now add Fmg to both sides of the equation, and subtract FL from both sides:

- Fu = Fmg - FL

This is not what you arrived at. The discrepancy explains your errant minus sign

Hm I still get the same answer...just without the negative. 6.60x10^4Pa

cepheid said:
Your third equation also makes no sense. You have written:

Pu = Mg - PLA

But Pu is a pressure whereas Mg and PLA are forces. You can't equate a pressure to a force. Your equation is not dimensionally consistent. Does this help?

I actually forgot to type that in. On paper I did add that in.
 
I get the same answer as you, too.
 
catkin said:
I get the same answer as you, too.

okay then, I'm suspecting it might be a typo.
 
I don't think you understand. This equation that you have:

Pu = Mg - PLA

is WRONG, because the thing on the left-hand side is a pressure, and the quantities on the right-hand side are forces. Therefore your equation is not dimensionally consistent: it is in error and you need to fix it.

Either have all pressures or all forces.
 
cepheid said:
I don't think you understand. This equation that you have:

Pu = Mg - PLA

is WRONG, because the thing on the left-hand side is a pressure, and the quantities on the right-hand side are forces. Therefore your equation is not dimensionally consistent: it is in error and you need to fix it.

Either have all pressures or all forces.

this is what I have:

P_u A= Mg- P_L A

(technically now all forces)
I used this in my calculations.

Is its still wrong?
 
Well, it should be

-P_u A= Mg- P_L A

because of what we talked about before with the sign error. But other than that, it looks ok.

When I plug in the numbers, I get

P_u = 6.6076 \times 10^4 \ \ \textrm{Pa}

So if the book says 6.8*10^4, then maybe it IS a typo.
 
  • #10
Sorry I misinterpreted what you were saying before about how you used the correct formula on paper, without the typo.
 
  • #11
cepheid said:
Well, it should be

-P_u A= Mg- P_L A

because of what we talked about before with the sign error. But other than that, it looks ok.

When I plug in the numbers, I get

P_u = 6.6076 \times 10^4 \ \ \textrm{Pa}

So if the book says 6.8*10^4, then maybe it IS a typo.

oh yep and I corrected that..

Okay, thanks :smile:
 
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