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Homework Help: Skydiver falling

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    I just had an exam in physics and there was this one easy question that I just couldn't wrap my head around.

    A skydiver (85kg) is accelerating at 3m/s^2. What is the force of the air resistance on him?

    2. Relevant equations
    f=ma and other mechanics

    3. The attempt at a solution
    I solved for the F of the skydiver which would equal 255 N. I then used Newton's third law to conclude that the air resistance must be 255 N up. But then I started thinking... If he's accelerating, there must be a Fnet... How, then, can he accelerate if the friction from the air equals the force of gravity?

    What should I have done?
  2. jcsd
  3. Sep 27, 2012 #2
    So you basically say that the air resistance is equal to and opposite to that of the force due to gravity.

    If the air resistance is *equal and opposite* to the force due to gravity, would he be accelerating?
  4. Sep 27, 2012 #3
    That's what I mean. That's what threw me off. I have no idea how to solve for the air friction... maybe something with 10m/s^2 and 3m/s^2?
  5. Sep 27, 2012 #4
    If they're equal and opposite, then

    [itex]F_{g} = -F_{res}[/itex]


    [itex]F_{net} = F_{g} + F_{res} = 0[/itex]

    can you accelerate if there is zero net force?
  6. Sep 27, 2012 #5


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  7. Sep 27, 2012 #6


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    Newtons third law: For every force acting, there is an equal and opposite force acting

    That 255N force is the net force - the vector sum of the acting forces.

    You don't apply Newtons Third Law to the net force.

    You should have been looking at the acting forces; which gave a net force of 255N
  8. Sep 28, 2012 #7
    Hmm... So... Would this be correct reasoning

    85kg 10m/s^2 =800N
    85kg 3m/s^2 = 255N

    800-255= 545

    The air friction is 545N

  9. Sep 28, 2012 #8


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    Numerically not bad - except that 85 * 10 is 850 not 800; meaning a final answer of 595N

    More appropriately.

    Acting force down = weight force which is mg. If you assume a value of g = 10, that means a weight force of 850N.
    Often we use the valu of 9.8 for g, which mans a weight force of 833N.

    The net Force is 255N down [as you already calculated].

    The other acting force is air resistance up.

    Weight of 850N down, in combination with an air resistance force of 595N up is how we get a net force of 255N down, resulting in an acceleration 3 m/s^2.
  10. Sep 28, 2012 #9
    Woops, you're right :D.
    Well, at least I figured it out. Thank you.
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