mikaelaq23 Posts: 13, Reputation: 1 New Member #1 Sep 20, 2009, 09:18 PM
How to find force needed for takeoff in a plane.
While waiting at the airport for your flight to leave, you observe some of the jets as they take off. With your watch you find that it takes about 34.9 seconds for a plane to go from rest to takeoff speed. In addition, you estimate that the distance required is about 1.6 km.

If the mass of a jet is 1.83 105 kg, what force is needed for takeoff
 Alty Posts: 28,318, Reputation: 5972 Pets Expert #2 Sep 20, 2009, 09:32 PM
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #3 Sep 21, 2009, 08:15 AM

I would say

Force > [1.83x10^5 x 9.81 =] 1795230 N

I may be wrong though.
I'm using the idea that the weight of the plane 1795230 kg, and that you need a greater force than that to overcome the gravitational force.
 ebaines Posts: 12,130, Reputation: 1307 Expert #4 Sep 21, 2009, 09:42 AM

I assume you're asking about the force from the jet's engines (thrust) required to accelerate to takeoff speed. From the data you provided you can use d= 1/2*a*t^2 to determine the plane's acceleration, then F=ma to determine the thrust from the jet engines to accelerate the plane at that rate.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #5 Sep 21, 2009, 10:01 AM

I would like to discuss it a little, if you don't mind ebaines.

From my point of view, the force that lifts the airplane up is the pressure of air that hits the wings of the plane. Since the wings are not completely horizontal but tilted a little, there are components which have to be considered. Only the one that pushes the plane up is 'used' to lift the plane (that's a fraction of the air which strikes the wing)

Now, if you take the acceleration of the plane, that would be in the horizontal direction, and so will the force, not overcoming the weight of the plane so that it can take off, right? :confused:
 ebaines Posts: 12,130, Reputation: 1307 Expert #6 Sep 21, 2009, 10:25 AM
Hello Unknown008. First let me say that I took a very simplistic, high school or college freshman physics class approach to this. Consequently I assumed constant acceleration down the runway (which I acknowledge the problem did not say to do), and that the runway is horizontal. Now, as the plane goes down the runway in the horizontal direction the engines accelerate the plane but do nothing to actually lift the plane directly (unless it's some kind of specialized craft like the Harrier "jump jet," which uses engine thrust directed downward to provide some - or all - of the lift). For a conventional airplane you can assume that all the lifting forces come from the wings, which act upon air flowing over them to create lower air pressure on the upper surface than the lower. It's this difference in air pressure that creates lift, and the difference in pressure is created by the velocity of the plane through the air. The air flowing over the wings create drag, which the engines must overcome. The magnitude of that drag increases approximately with the square of the plane's velocity through the air. So the thrust needed is a little more complicated than simply F=ma. A more realistic expression would be:

$
F = ma + Drag = ma + \frac 1 2 \rho A C_d v^2
$

where $\rho$ is the density of air, $A$ is the cross-sectional area of the plane into the oncoming wind, and $C_d$ is the coefficient of drag for the airplane and wing geometry. This can be a very difficult equation to solve, and hence usually in high school of college freshman physics they tell you to ignore air friction.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #7 Sep 22, 2009, 06:47 AM

Oh, OK, thanks. You're right, if I take all this it will make it more complex. I don't understand myself sometimes... In my first post, I took it way too simple, now, I'm making way too complicated. :(

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