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Air Resistance Problem

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data
    An 84 kg person is parachuting and experiencing a downward acceleration of 3.0 m/s2. The mass of the parachute is 5.5 kg.

    a) What upward force is exerted on the open parachute by the air?
    (b) What downward force is exerted by the person on the parachute?

    2. Relevant equations
    Fnet=mA
    Fg=mg
    W=mg

    3. The attempt at a solution
    How many forces are acting on him? Parachute force, air resistance, weight (mg), and acceleration? Is that the only forces? Would the equation look like Fnet = mg-Fp-Fair=mg ???? I cannot seem to come up with the correct answer. Would a) and b) be equal?
     
  2. jcsd
  3. Feb 15, 2010 #2

    PhanthomJay

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    The parachute force is the force of air resistance (ignoring any small air resistance on the person), so don't count it twice. Also, acceleration is NOT a force. So there are just 2 forces acting on the person, the net sum of which must be in accord with Newton 2. You say 'a' and 'b' are equal....reasoning?
     
  4. Feb 15, 2010 #3
    Sometimes they put more applicational problems instead of actual computing it was just a total guess. But newtons law says Fnet=mg-Fp=ma correct? Is that the equation for this type of problem?
     
  5. Feb 15, 2010 #4

    PhanthomJay

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    Yes, but be careful to identify to which object or system of objects you are applying Newton's laws.
     
  6. Feb 15, 2010 #5
    I'm confused, if these numbers are correct it seems like the answer would turn out exactly the same. Fp=mg(823.2)-ma(252)=571 Yes?
     
  7. Feb 15, 2010 #6
    Anybody?
     
  8. Feb 15, 2010 #7
    for part a you seem to be quite close
    but I think what you might need is
    [tex](m_{1}+m_{2})a=-m_{1}g-m_{2}g+F_{p}[/tex]
    where
    [tex]m_{1}[/tex] is the mass of the person
    [tex]m_{2}[/tex] is the mass of the parachute
    g=9.8
    a=3.0
     
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