Thrust,drag and lift on airplane Question

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SUMMARY

The discussion centers on calculating atmospheric drag (F(D)) and wing lift (F(L)) for an airplane given engine thrust (F(T) = 130 kN) and weight (W = 180 kN). Participants concluded that in static equilibrium, drag equals thrust, thus F(D) is 130 kN, and lift equals weight, making F(L) 180 kN. To find the distance (s) to the line of action of the drag force, moments can be taken about the point where the weight acts.

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Homework Statement



If the engine thrust is F(T)=130kN and the plane's weight is W=180kN,determine the atmospheric drag F(D) and the wing lift F(L).
What is the distance s to the line of action of the drag force?


The Attempt at a Solution



Well I have trouble with this question because I don't know where to start.In my statics book there is nothing on airplanes or about drag,thrust and lift.

I was thinking like the resultant force = F(T)- F(D)

I don't know the relationship between the drag,thrust and lift and couldn't find anything in my statics book.
So any hint or advice how to start it would be greatly appreciated or is there any equation I need to use.
Thanks.

http://img685.imageshack.us/img685/1364/11212009113824am.jpg
 
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its flying at constant velocity... so all accelerations must be zero, ie. static equilibrium...
 
so then the drag is 130 and the lift is 180
and to determine s I can take moments at point where W is pointing
is this right?
 

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