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Ratio of Net Thrust to Weight during a Climbing Flight

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runstreet

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I came up with two solutions as shown in the picture.
I knew the one on the left is wrong, but I couldn't figure out why?

Can someone please clarify?? Thanks in advance.
 
I'm no expert in dynamics, but the jet is climbing, so it is overcoming gravitational acceleration. Your equations ignore the vertical acceleration of the plane.

BA
 
Do you mean the lift of the aircraft??
 
Hi runstreet,

Yes, I mean lift. The plane weighs W. Your equation sinθ(T - R) = W says that the sum of vertical forces is zero. That is true in statics. An object weighing W resting on the ground has a reaction of W. What force is required to give it an upward velocity of Vsinθ where V is the velocity of the plane measured along the slope theta?

BA
 
I would like to modify my earlier comments as follows:

In order to arrive at a constant velocity, the jet plane must accelerate and there would be forces associated with that acceleration. However, once a constant velocity has been reached, there is no acceleration, which means the system is in equilibrium.

I now believe that the correct answer to the question is 1/sinθ, as you have shown on the solution at the left.
EDIT: But wait a minute...that can't be correct, because as θ approaches zero, i.e. level flight, W/sinθ increases without limit. If the jet plane is in equilibrium after it reaches constant velocity, then T - R = 0 and the answer to question 4. is 0.

Don't hang your hat on this analysis. I think we need the help of an aeronautical engineer to solve this problem. I believe I flunked the exam. Help!!!

BA
 
Thrust is still required to climb at constant velocity -- the gain in potential energy comes from the work produced by the thrust; with zero thrust, the plane becomes the classic ballistic projectile in a parabolic trajectory.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
Well, I am not sure how you would express that in a way that would solve the problem, but I will mull it over in my poor old brain for a little longer. Maybe, while I am doing that, someone will come up with the answer to the problem.

BA
 
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