# Homework Help: Mass of Airplane

1. Jan 20, 2014

### zodiacbrave

1. The problem statement, all variables and given/known data

The wings have a total surface area of 250 m^2. The air above the wing is 300 m/s and the air below the wing is 250 m/s. The ambient air pressure is around 0.5 atm. The airplane is descending at 20 m/s, therefore there is an upward drag on the wings.

What is the mass of the airplane?

2. Relevant equations

Pt + (1/2)pVt^2 = Pb + (1/2)pVb^2

3. The attempt at a solution

Fw = m * 9.8 m/s^2. Pair = 5.05 x 10^4 Pa

Density of air at height of plane, pair = (po / Po)* Pair = 0.645 kg/m^3

Plugging into Pt + (1/2)pVt^2 = Pb + (1/2)pVb^2

Pb - Pt = 16.125 Pa

Fw = msg - Flift.

I'm not sure how to deal with the force pressing down caused by 20 m/s

Thank you

2. Jan 20, 2014

### BvU

What is Fw if the speed of descending is constant ?

3. Jan 20, 2014

### kyaranger

9.8 m/s^2 + 20 m /s * the mass of the airplane?

4. Jan 20, 2014

### rcgldr

The problem statment is somewhat confusing, it gives the speeds of the air above and below the wings and also states that the aircraft is descending, but that descent rate also affects the speed of the affected air. The problem doesn't seem to provide sufficient information to determine the forward speed of the aircraft, or it's effective slope angle, or the amount of drag involved. For a conventional wing, the speed of the aircraft would be closer to the speed of the air below the wing than the speed of the air above the wing.

Last edited: Jan 20, 2014
5. Jan 21, 2014

### BvU

Couldn't it be as simple as Fw = (Pb - Pt) * Area ?

6. Jan 21, 2014

### rcgldr

Since the aircraft is descending, it's angled downwards a bit, so lift, which is apparently to be approximated as (Pb - Pt) * area, is angled forwards a bit. The upward forces are the vertical components of lift and drag, and the downwards components are gravity (weight) and the vertical component of thrust (which is unknown). The speed of the aircraft would be somewhere between 250 m/s and 300 m/s, so you have a range for speed, but not an exact speed.

7. Jan 21, 2014

### haruspex

Is it angled down, or is it merely that the lift is not sufficient to maintain height (and the vertical speed is effectively a terminal velocity)?

8. Jan 21, 2014

### rcgldr

Assuming the pilot isn't forcing the situation with up elevator input, the tail will weathervane upwards (relative to the aircraft) in response to a descent, so typically the aircraft will be angled downwards. If the pilot is forcing the situation, then it's possible that the aircraft could even be pitched upwards a bit during descent, like an aircraft on final approach. I'm not sure how to combine Bernoulli and drag effects under these circumstances.

Last edited: Jan 22, 2014
9. Jan 22, 2014

### Staff: Mentor

My guess is that the question is intended to deal with by using some heavily simplified model and quite simple physics.

Not that I have any idea about what the model is intended to be. My first thought was similar to that signaled by haruspex, that the vertical speed is to be understood as the terminal velocity, but I don't see how to use this information.

10. Jan 22, 2014

### BvU

Don't see the logic: at touchdown it's the rear wheels that hit the ground first. The thud betrays the fact that is descends, yet the nosewheel lands latest (fortunately).

And forget the thrust. It's definitely not a glider, but from the problem formulation it might just as well be!

Last edited: Jan 22, 2014
11. Jan 22, 2014

### rcgldr

That's because the pilot feeds in up elevator to keep the nose of the aircraft up. The problem becomes more complicated if the aircraft is not angled downwards into the direction of flight, or at least drag becomes a bigger factor.

Since the aircraft is in a steady descent, the problem could assume the aircraft is a glider.

Last edited: Jan 22, 2014
12. Jan 22, 2014

### BvU

Big glider with 250 m2 wing area!

Is Zodiacbrave still in the picture ? How do you end up with Pb - Pt = 16.125 Pa if 1/2 * (3002 - 2502) = 13750 and ρ≈0.65 ?

I would ignore a few tiny details and estimate m = 13750 * 0.65 * 250 / 9.8 = 230 ton, about half a 747. They fly around with speeds around 1000 km/h. Gliders don't.

But then, I'm only a physicist, non-frequent flyer and definitely not an aeronautics expert.

I would ignore the 20 m/s if wing speeds are already given. Might be quite wrong, so please correct me. Titlting is also negligible: cosine effect and the whole thing wouldn't be much different if it descended without tilt.

I like this problem!

13. Jan 22, 2014

### rcgldr

No posts other than the original. Perhaps the problem statement was adjusted or he solved it using some assumption.

Since the aircraft is descending at a constant rate, part of the force opposing gravity is the vertical component of drag, so the sum of total thrust and drag must result in a net drag force.

Except that the problem states that part of the drag force (the vertical component) is opposing drag.

Ignoring the effects of the aircraft's orientation, assume it's path is θ radians below horizonal. Using a calculated pressure differential, lift = pressure differential x wing area, an absolute value independent of the mass. The vertical component of lift = lift x cos(θ). The weight of the plane = m g. The vertical component of drag = m g - lift x cos(θ) = drag x sin(θ). Assuming the aircraft isn't accelerating, the net force is zero:

0 = m g - lift x cos(θ) - drag x sin(θ)

m g = lift x cos(θ) + drag x sin(θ)

The angle of descent is somewhere within a range depending on the aircrafts total speed versus descent speed:

arctan(20 / 250) <= θ <= arctan(20 / 300)

Even if the angle of descent is known, there are still two unknowns in the equation above, mass and drag. If the problem included the information required to determine the angle of the aircraft's path (or its total speed), and a lift / drag ratio for the given conditions, then the problem could be solved using the apparent assumptions implied by the problem statement.