# Forces and state of equilibrium

1. Oct 3, 2011

### sdoi

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Trig

3. 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]

2. Oct 3, 2011

### grzz

Is the above the pulling force of the engine?

3. Oct 3, 2011

### sdoi

oh sorry, "If the forward thrust provided by the engine is..."

4. Oct 3, 2011

### grzz

OK
How many forces act on the plane?

5. Oct 3, 2011

### sdoi

Two forces, Negative force of gravity, and the thrust of the engine.

6. Oct 3, 2011

### grzz

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!

7. Oct 3, 2011

### sdoi

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.

8. Oct 3, 2011

### grzz

Correct. Here we better call the normal force by the name 'upward lift'.
So now you know the answer to(a) and (b).