Aeroplane follows circular trajectory-Tension? (geometry)

[Const@ntine]

Homework Statement

So, I have this problem here that's pretty basic, but the solution manual sets different axes, and I'm having a bit of trouble understanding the geometry part, meaning how he applies the given forces to the new axes.

A model airplane of mass 0.750 kg fl ies with a speed of 35.0 ms in a horizontal circle at the end of a 60.0 m control wire. The forces exerted on the airplane are shown in Figure P6.63: the tension in the control wire, the gravitational force, and aerodynamic lift that acts at θ = 20.0° inward from the vertical. Compute the tension in the wire, assuming it makes a constant angle of 20.0° with the horizontal.First of all, here's the main pic:

Homework Equations

It's the geometry part that gives me trouble.

The Attempt at a Solution

He finds the radius of the circular trajectory that the plane follows:

No trouble here, we just take the direction of T as the hypotenouse and going with cosθ = r/l finds the radius.

Then there's this:

English is not my native language, so I'm not sure whether he means that x and y are the crossed lines, or the lines that "fall on top" the directions of T & F respectively. From the test I'm getting the latter, but if I try to apply the usual cosx = adjacent/hypotenouse and sinx = opposite/hypotenouse, I can't figure out how he gets his results.

I'd appreciate some help!

PS: I know this isn't exactly a "problem" with statements and such, but it's only this part that troubles me, and I didn't know where else to post it.

 

[haruspex]

I'm not sure whether he means that x and y are the crossed lines, or the lines that "fall on top" the directions of T & F respectively.

The book defines x as at 20 degrees to the horizontal, along the line of the tether. Correspondingly, y is at 20 degrees to the vertical, along the line of the lift force.
Please clarify where you have difficulty with the resulting equations.

 

[Const@ntine]

The book defines x as at 20 degrees to the horizontal, along the line of the tether. Correspondingly, y is at 20 degrees to the vertical, along the line of the lift force.
Please clarify where you have difficulty with the resulting equations.

How he gets that Fgx = Fg*sin(20) & acx = ac*cos(20). I know that he uses the formulas of cosine and sine, but I can't figure out how he gets these particular results.

 

[haruspex]

How he gets that Fgx = Fg*sin(20) & acx = ac*cos(20). I know that he uses the formulas of cosine and sine, but I can't figure out how he gets these particular results.

What would you calculate for the component of Fg in the x direction?

 

[Const@ntine]

What would you calculate for the component of Fg in the x direction?

I worked at it again and realized I was looking at it wrong. The new axes confused me and I lost my train of thought. If I'm not mistaken, it should look something like this:

I go by the previous x & y axes, I take the directions the acceleration and Fg had on them, and then I create a net force based on the new axes. Correct? (there might be a misconception or two in my wording, but I'm still trying to get the hang of the english scientific terms).

 

[haruspex]

I worked at it again and realized I was looking at it wrong. The new axes confused me and I lost my train of thought. If I'm not mistaken, it should look something like this:

I go by the previous x & y axes, I take the directions the acceleration and Fg had on them, and then I create a net force based on the new axes. Correct? (there might be a misconception or two in my wording, but I'm still trying to get the hang of the english scientific terms).

Yes, that looks right. At least, I see you now get F_gx=F_gsin(θ). Is the rest ok now?

 

[Const@ntine]

Yes, that looks right. At least, I see you now get F[SUBgx][/SUB]=F_gsin(θ). Is the rest ok now?

Yeah, that was the only part I had trouble with, not with the formulas themselves. I've got the acceleration vectors on the left as well, so that's all.

Thanks for taking the time!