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Newton's Second Law of Motion: Fnet=ma

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A 90 kg parachutist in free fall has an acceleration of 6.8 m/s^2.

    2. Relevant equations
    What is the frictional force provided by air resistance when she is accelerating at this rate?


    3. The attempt at a solution
    I tried this, assuming down is positive. I used the equation Ff=mg-ma, and I got the answer 270 N, but I thought frictional force was supposed to be in the opposite direction of the gravitational force. So shouldn't it be negative?
     
  2. jcsd
  3. Oct 2, 2011 #2
    It is however you took care of that by making the equation = mg-ma

    The negative takes care of it.

    The sum of forces is equal to ma therefore you get something like this

    ƩF = Fg + Ff which goes to

    ma = mg + Ff

    You actually have the right formula you just misinterpreted it.
     
  4. Oct 2, 2011 #3

    lewando

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    Gold Member

    You can say Ff is a vector consisting of magnitude 270N and direction "up". If you say "-270N" and "up" then the force would actually be downward.

    Your FBD should take care of this. If you have Fg pointing down and Ff pointing up then the sign information (for the purposes of writing down the correct equation) is contained in the direction of these vectors (even with your choice of "down" being a positive direction for force/acceleration).

    Fg [positive because it is pointing down] - Ff [negative because it is pointing up] = m*6.8m/s2 [positive 6.8 because it is accelerating downward]

    Note: A negative result for a force magnitude would indicate that the specified direction is off by 180°.
     
  5. Oct 2, 2011 #4
    Thanks!
     
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