Forces and state of equilibrium

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sdoi
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Homework Statement


Assuming an aircraft has a force of gravity of 6.6x10^4N [down] acting on it as it travels at a constant velocity of 6.4x10^2 km/h [w]. If the engine is 1.3x10^4N [W] determine:
(a) The upward lift force on the plane
(b) The force due to air resistance on the plane


Homework Equations


Trig


The Attempt at a Solution


(a) I drew out a vector diagram which looks very basic with Fg and Fe( force of the engine)
Because the aircraft is moving at a constant velocity we know that ƩF=0
I went about it in a very basic was, so I'm not sure of it is correct:
ƩFx=1.3x10^4N ƩFy= -6.6x10^4N
ƩF=√(1.3x10^4N)^2 + (-6.6x10^4N)^2
F=6.7x10^4N
therefore, the upward life force on the plane is 6.7x10^4N[North of west]
 
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sdoi said:
If the engine is 1.3x10^4N [W]
Is the above the pulling force of the engine?
 
oh sorry, "If the forward thrust provided by the engine is..."
 
OK
How many forces act on the plane?
 
Two forces, Negative force of gravity, and the thrust of the engine.
 
If those are the only two forces on the plane then the plane will fall down along the resultant of these two forces.
THINK again!
 
In order for the plane no remain airborne there needs to be a normal force as well to cancel out the force of gravity and air resistance to counter the forward thrust of the engine.
 
Correct. Here we better call the normal force by the name 'upward lift'.
So now you know the answer to(a) and (b).