Mechanics problem involving trig and forces.

1. Aug 1, 2010

chrisrock

A fully loaded Cessna-182 airplane of mass 1260 kg has an engine failure when flying with an airspeed of 126 km/h at an altitude of 2870 m on a calm day. It then glides at a constant glide angle (which is the direction of flight below the horizontal) towards a safe landing at this constant speed of 126 km/h experiencing a drag force of 1280 N that opposes the direction in which the plane is moving.
Please use: g = 9.81 m s-2

[PLAIN]http://img805.imageshack.us/img805/5625 [Broken]

I would like to know what equations and values i should be using to find the glide angle of the plane.

I tried using the forces given, weight and drag, with trig but as I know that the drag force is not a component force of weight I didn't have high hopes. So of course i got the wrong answer.

Will upload the working once I draw the diagram I drew in paint.

I know the answer is 5.94 degrees.

Thanking you in advance. Been banging my head around this problem I just don't see what to use to work out the glide angle.

Last edited by a moderator: May 4, 2017
2. Aug 2, 2010

Lok

If something is falling, what equations do we use?

If you have a Force (be it Drag) exerted over a distance what can you get out of it?

3. Aug 2, 2010

chrisrock

hey lok i presume we are talking about the kinematics equations. I will have a look back and report to you in a min but i dont see how i can get the angle using those equations.

actually its late in NZ so imma c this tomorrow but i think i got it, just need to manipulate the equations properly.

Last edited: Aug 2, 2010
4. Aug 2, 2010

Lok

Well you know the speed it moves forward (or at the unknown angle) and all you have to find is the speed it is falling considering it is dropping at a constant speed attracted by gravity and slowed down to a constant speed by friction.

And then there is a bit of Trig...

Last edited: Aug 2, 2010
5. Aug 2, 2010

chrisrock

Hey lok sorry for the late reply i did work it out last night. just been busy and havent been able to put it up i will in an hour or 2 though thanks for the help.

6. Aug 3, 2010

Np. Anytime.

7. Aug 3, 2010

chrisrock

I used the drag force divided by the weight force, and then took the inverse sin of the answer.

Θ = sin-1 ((drag force)/(mg))

Unfortunately i cant explain y i used this as that will require me to do a diagram which i find time consuming, and as i'm short on time i cant help it SORRY !