# Gliding airplane

1. Feb 3, 2009

### moy13

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

When an airplane is in flight with the engine running, it is acted on by
four forces. These are:
(i) The weight of the airplane.
(ii) The thrust provided by the engine, which pulls the airplane forward in the direction it is moving.
(iii) The drag (air resistance), which pulls on the airplane opposite to the direction in which the airplane is
moving.
(iv) The lift provided by the wings, which pulls on the airplane in a direction perpendicular to the direction in
which the airplane is moving.

Fred has just earned his private pilot license. He is
flying at an altitude H in a Piper Warrior, a small
single-engine airplane of weight W. Suddenly,
without any warning, the engine stops. The airplane
begins to descend in a steady glide at a constant speed
v through the air, with the nose of the plane pointed
below the horizontal. In the glide, the magnitude of
the total drag on the airplane is D. What horizontal
distance will the plane glide before Fred has to
execute an emergency landing?
To test your general expression, evaluate for the case W = 10,000 N, D = 1500 N,
v=40 m/s, and H = 2000 m.

2. Relevant equations

components of velocity:
Vx = Vox * cos (a)
Vy = Voy * sin (a)

tan (a) = opposite / adjacent
sin (a) = opposite / hypotenuse
cos (a) = adjacent / hypotenuse

I'm not sure if this one is relevant to the problem:

Y = Yo + Vo*t

3. The attempt at a solution

I know I'm supposed to find the angle in which the plane is falling, but how do I do that?

I also know the forces remaining when the engine shuts off are the drag, the lift, and the weight, but these forces cancel since the velocity is constant. Because there is a constant velocity and because the forces canceled each other the plane is not in projectile motion.

The direction of the lift vector is perpendicular to the velocity and that makes lift perpendicular to drag as well. The weight is just straight down.

2. Feb 4, 2009

### Carid

The glide angle you need to find is the angle between the lift vector and the weight vector.
Once you've got the glide angle you can draw a triangle to find how far you can glide the aircraft.

Resolve the weight vector into one component opposing the lift and another operating in the forward aircraft direction. The component of the weight vector operating in the aircraft's forward direction is sufficient to cancel the drag. We know this because the speed of the aircraft is constant. So do the free body drawing and find the angle.

3. Feb 4, 2009

### moy13

so, could I use the law of cosines c^2 = a^2 + b^2 - 2ab cos(gamma),

where c = L - W, a = L, b = -W, and gamma is the angle between L and W?

4. Feb 5, 2009

### Carid

What is the force propelling him forward which exactly counters the 1500N drag?

5. Feb 5, 2009

### moy13

Would that be the thrust of the engine? But after the engine shuts off there is no more thrust right?

6. Feb 6, 2009

### Carid

It's the component of his weight in the forward direction. He's coming down a glidepath. Part of his weight exactly counteracts the lift of the wings, the other part balances the drag.