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Parachute- acceleration due to gravity

  1. Feb 29, 2012 #1
    Hey, just started learning about forces and having trouble getting my head around forces acting on a parachutist and acceleration due to gravity. I understand that at terminal velocity Fg = Fair friction, my problem is when he opens his chute and starts to float down is he still accelerating to earth at g? His mass doesn't change so should g not change? Is he accelerating at g to a smaller velocity? just can't see how he has the same acceleration due to gravity when he is in free fall and with the chute open. Thanks.
     
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  3. Feb 29, 2012 #2

    tiny-tim

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    Hey ZB08! :smile:
    You're confusing acceleration and force. :redface:

    Weight is a force, not an acceleration.

    He doesn't have an "acceleration due to gravity", he has a weight (a force).

    He has an acceleration due to all the forces acting on him.​

    (And his acceleration if he's "floating" down is probably zero. :wink:)
     
  4. Feb 29, 2012 #3
    Well, think about it. When someone pulls a parachute during a fall from a plane, do they keep accelerating like a rock?

    Think about how the forces add together. Remember that the net force acting on a body is equal to the sum of the forces acting on it. And also think about in which direction these forces are pointed.
     
  5. Feb 29, 2012 #4
    Thanks for the replies. What i gathered today in class was that W = mg, and so I thought that just because he opened his chute, he didn't change his mass and so he should still have same weight and accelerate at the same rate. Am I right from what ye are both saying is that W = mg is just one part of it. When he opens his chute he deccelerates, and because of the force due to air resistance is opposite to W then due the net force between these two he has a new acceleration. Fnet = ma. Ya?
     
  6. Feb 29, 2012 #5
    right, net force is the sum of the forces acting on a body. In this case he has two forces, the force of gravity and the force due to air resistance.

    just remember that each force is a vector, and remember that they equal each other in terms of magnitude, and also remember which direction they are pointed
     
  7. Mar 1, 2012 #6

    tiny-tim

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    Yup! :biggrin:
     
  8. Mar 1, 2012 #7
    Thanks! I was reading another thread on this and it stated that the F air before would equal the F air after, as while the parachute has a bigger surface area the air molecules are hitting at a slower velocity. So again the F air after is also equal to W. So kinda confused on the different velocities. Am I right in saying that the difference in the "terminal velocity" in free fall and the final velocity when the parachute opens is because of that momentary imbalance in forces when he opens the chute and decelerates to the new velocity. Thanks again.
     
  9. Mar 1, 2012 #8

    tiny-tim

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    No.

    Terminal velocity means that the acceleration is zero, so yes by definition W = Fair.

    If the parachutist is in "free-fall" terminal velocity, and then opens his parachute, Fair suddenly becomes greater, and so W - Fair is negative, and his speed gets less.

    Fair depends on speed, so that means Fair also gets less, until it equals W. :wink:
     
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