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If I'm flying a regular Cessna 172 and its perfectly trimmed and flying in circles holding the same altitude, is the aircraft in equilibrium? Or steady flight?

(The definition of equilibrium flight condition occurs when the sum of all forces and moments about the center of gravity is zero)

I would appreciate the clarification.

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Equilibrium means all forces sum to zero. Per Newton's law $\sum F=ma$ this means that the total acceleration is also zero.

In a steady turn, the flightpath is constantly changing from a straight line. There is a centripetal acceleration that is nonzero: a continuous acceleration perpendicular to the direction of movement.

Working through Newton's law the other way round, this means that the sum of forces cannot be zero. Therefore, an aircraft in a steady-state turn is not experiencing a force equilibrium.

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  • $\begingroup$ thank you! what if it would be a steady level flight would that be in equilibrium or not nesseceraly? $\endgroup$
    – YamchaAviator
    Jun 10, 2021 at 23:44
  • $\begingroup$ lastly, I think this would clarify any doubts I might have.... is a straight and level flight moving with a constant velocity in equilibrium ??? (not necessarily? could not be in equilibrium if the speed is constant? should I assume that if its moving with constant speed at level straight flight the sum of Lift, Weight, Drag and Thrust should be zero?) $\endgroup$
    – YamchaAviator
    Jun 10, 2021 at 23:52
  • $\begingroup$ Unaccelerated flight is always due to equilibrium of forces. That means level flight but also a steady climb or descent. Only if the flight path direction changes, or if velocity changes, you will have forces not in equilibrium. $\endgroup$
    – Sanchises
    Jun 11, 2021 at 5:40
  • $\begingroup$ @Sanchises i think that steady climb should not be considered an equilibrium, since, by definition, the lift is larger that the weight of the aircraft $\endgroup$
    – Grigoris L.
    Jun 11, 2021 at 9:03
  • $\begingroup$ @Grigoris No no no no no. If that were the case you would keep accelerating upwards at an increasing velocity. In fact, lift is smaller than weight in a climb due to thrust carrying part of the weight. See aviation.stackexchange.com/q/40921/4108 Regardless of the exact magnitude and definition of lift, weight, drag and thrust, they must sum to zero if we are not to keep accelerating. $\endgroup$
    – Sanchises
    Jun 11, 2021 at 9:14

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