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Parachutist's fall is because of balanced force or unbalanced force?

  1. Apr 5, 2014 #1
    A parachutist's fall is due to gravity of the earth which pulls down the weight of the parachutist towards it. Initially there is no air resistance. The air resistance increases and acts against gravity. There is a phase the velocity is constant. When parachute opens, air resistance increases significantly and the parachutist accelerates upwards and velocity decreases until they balance. My question is that if the forces were balanced, the Air Resistance= Gravity and net force on the parachutist is 0 N, how come does the Parachutist move?

    http://astarmathsandphysics.com/gcse-physics-notes/gcse-physics-notes-the-motion-of-a-parachutist.html
     
  2. jcsd
  3. Apr 5, 2014 #2

    Bandersnatch

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    Try applying Newton's 1st law of motion to the parachutist at the moment when the two opposing forces equalise.

    And welcome to PF, by the way!
     
  4. Apr 5, 2014 #3
    :cool: Thank you Bandersnatch but here do you mean to say that the object moves at a constant velocity of 9.8 m s-1 irrespective of the acceleration, which is 0.
     
  5. Apr 5, 2014 #4

    Bandersnatch

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    As the 1st law says, the object will move at whatever velocity it has, unless an unbalanced force acts on it.
    If the net force is 0, the acceleration will be 0.
    So, an object moves at a constant velocity not "irrespective of acceleration" but because the acceleration is 0.

    What is the velocity a falling object will have when the air drag force becomes equal to the gravitational force?(clue: it's mentioned on the graph in your link)
     
  6. Apr 5, 2014 #5
    do you know about viscosity?

    force experienced by a body with radius 'r' travelling through a fluid of viscosity 'η' with speed 'v' is
    F=6∏ηrv
    body moves with a constant velocity called terminal velocity 'V' when external force (weight in this case) equals viscous force

    so, 6∏ηrV=mg
    V=mg/6∏ηr
    air is also a fluid and has viscosity
    viscous force is directly proportional to radius and velocity
    if radius is large velocity will be small and vise versa
    so all bodies will fall with a specific velocity unless its density is smaller than air like hydrogen
     
  7. Apr 5, 2014 #6
    Bandersnatch Before the parachute opens its 50 m s-1 and after it open its 5 m s-1 according to the graph.
     
  8. Apr 5, 2014 #7
    before the parachute opens radius of cross section is small (of man).so velocity is large .
    after it opens radius increases,so velocity decreases
     
  9. Apr 5, 2014 #8
    Thank You very much... I understood it finally, @basheer uddin and @bandersnatch... can you please telp me how to tag users in a reply?
     
  10. Apr 5, 2014 #9

    Bandersnatch

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    Yes, it's called terminal velocity. It's different before and after, because the drag force is different - which is basically what basheer uddin is saying.
    At those two velocities the acceleration is 0 and the Newton's 1st law applies.

    edit: do you mean how to quote posts? There's a "quote" button in the lower-right corner of each post. The "M" button next to it lets you quote multiple posts.
     
    Last edited: Apr 5, 2014
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