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

[~christina~]

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

 

[cepheid]

First of all, your rearranging of the equation has an error:

You start with

F_L - F_u - F_mg = 0

Now add F_mg to both sides of the equation, and subtract F_L from both sides:

- F_u = F_mg - F_L

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

 

[cepheid]

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

P_u = Mg - P_LA

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

 

[~christina~]

First of all, your rearranging of the equation has an error:

You start with

F_L - F_u - F_mg = 0

Now add F_mg to both sides of the equation, and subtract F_L from both sides:

- F_u = F_mg - F_L

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

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

P_u = Mg - P_LA

But P_u is a pressure whereas Mg and P_LA 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.

 

[catkin]

I get the same answer as you, too.

 

[~christina~]

I get the same answer as you, too.

okay then, I'm suspecting it might be a typo.

 

[cepheid]

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.

 

[~christina~]

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?

 

[cepheid]

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.

 

[cepheid]

Sorry I misinterpreted what you were saying before about how you used the correct formula on paper, without the typo.

 

[~christina~]

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