Forces and state of equilibrium

AI Thread Summary
The discussion revolves around calculating the upward lift force and air resistance acting on an aircraft in equilibrium. Given a gravitational force of 6.6x10^4N downward and an engine thrust of 1.3x10^4N westward, it is established that the sum of forces must equal zero due to constant velocity. The upward lift force is calculated to be 6.7x10^4N directed north of west, which balances the gravitational force. Additionally, it is clarified that air resistance must counteract the engine's thrust for the aircraft to maintain its flight. The conclusion emphasizes the necessity of both lift and air resistance forces for the aircraft's equilibrium.
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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).
 
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