# Plane in flight - Dynamics Question

1. Apr 17, 2017

### Ryansf98

1. The problem statement, all variables and given/known data
"Figure 5 depicts a 30 000 kg aircraft climbing at an angle θ = 15˚ when the thrust T = 180 kN. The aircraft’s speed is 300 km/hr and its acceleration is 2m/s2. If the radius of curvature of the path is 20 km (i.e., θ is decreasing), compute the lift and drag forces on the aircraft."

a = 2m/s^2
T = 150000N
Theta = 15 degrees
Radius of Curvature (K) = 20000m
v = 300km/hr => v = 83.3 m/s^1

2. Relevant equations
W=mg, F=ma, a = v^2/K

3. The attempt at a solution
I was going to treat this similar to an object moving up an inclined plane. Although, I don't know if this would work.

I determined W = mg, W= 294,300N.
I then thought using a = 83.3^2/20000 = 0.35 m/2^1.
Therefore, F = ma, F = 30,000 * 0.35 = 10,500N.

Therefore F + F(f) (i.e. Wsin(Theta)) = Drag.

Upon calculating, I got close to the answer for drag but I believe only by coincidence. As I considered you would have to work out normal and tangential components individually, therefore a= v^2/K would be in the normal direction, not the tangential like I used it for.

I also am unsure on how to calculate Lift if I cannot treat this an incline plane problem.

2. Apr 17, 2017

### kuruman

I am not sure I understand what you are doing here. Which F is this and in what direction does it point?

3. Apr 17, 2017

### haruspex

That 0.35 is using only the centripetal acceleration. You are given an acceleration of 2ms-2,though it is not clear whether that is the overall acceleration or just the increase in speed. Is this F just the net force up the slope?
I can't make sense of that. Are you saying F(f) stands for W sin(θ)? Doesn't that add to drag? Where is T in that equation?

4. Apr 18, 2017

### Ryansf98

This is where I was confused, as I was using a incorrectly.

I tried to use a = v^2/K, but as you both noted I wasn't using it corretly but I don't know how to use it.

So this a would be a(centripetal)? The 2m/s^2 is the overall accleration as far as I'm aware.

[QUOTE="I can't make sense of that. Are you saying F(f) stands for W sin(θ)? Doesn't that add to drag? Where is T in that equation?[/QUOTE]

Yeah that's what I was working there. I was going to determine each component then sum them into a net force at the end.

5. Apr 18, 2017

### Ryansf98

I ended up redoing some of it. I calculated R = 284,272N, then used a(centripetal) = v^2/K, => a = 0.35 m/s^2.

I then used F(centripetal) = ma(centripetal), = 10,408N.

Then I done R - F(centripetal) to calculate lift. I got the answer on the sheet, but is that the correct process?

I also still can't figure out drag.

6. Apr 18, 2017

### kuruman

You are given the magnitude of the acceleration vector a = 2 m/s2. You have calculated the centripetal component to be 0.35 m/s2. Can you figure out from these two numbers the acceleration in the direction of motion? Hint: The direction of motion is perpendicular to the centripetal direction.

7. Apr 18, 2017

### haruspex

Yes. I don't see where you defined R, but it looks like it is W cos(θ). Centripetal is the radial component of the resultant, so =R-lift.