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Tension in parachute cord on dragster

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-11-19_20-36-54.png

    2. Relevant equations
    Fn - T = ma

    3. The attempt at a solution
    Fn-T=ma
    810kg x 2.5m/s2 - T = 810 kg x -2.5m/s2
    T=0
     
  2. jcsd
  3. Nov 19, 2015 #2

    haruspex

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    What is this force "Fn"?
     
  4. Nov 19, 2015 #3
    sorry couldn't type subscript. It is the net force.
     
  5. Nov 19, 2015 #4

    haruspex

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    The net force is the sum of all the applied forces. What are the applied forces?
    The acceleration is what results from the net force: ##F_{net}=\Sigma F = ma##.
     
  6. Nov 19, 2015 #5
    applied forces are:
    -2025 kgm/s2 from the parachute
    I have a feeling we are missing other applied forces.
     
  7. Nov 19, 2015 #6

    haruspex

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    No, that is the only force of interest. (There is gravity and the normal force from the ground, but they cancel out.)
    So what is the net force? What equation does that give you?
     
  8. Nov 19, 2015 #7
    ok so - tension is the force transmitted through the cord on the object. Force of gravity and normal upward force are equal so cancel out but we are trying figure out horizontal force not vertical force anyways.
    We are lost.....
     
  9. Nov 19, 2015 #8

    SteamKing

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    You should ask yourself, if the tension in the parachute line is truly equal to zero, then why have the parachute attached to the dragster at all? What is the parachute doing?

    Isn't the correct form of this equation Fnet = ma ?

    By putting the deceleration force on both sides of the equation, how else can T be anything but zero?
     
  10. Nov 19, 2015 #9
    exactly what we thought - T cannot be zero as it wouldn't be doing any work.
    So then we thought T = net force on the object = ma = 810 kg x 2.5 m/s2 but that seems to be incorrect? Why?
     
  11. Nov 20, 2015 #10

    SteamKing

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    How do you know it's incorrect?
     
  12. Nov 20, 2015 #11
    haruspex said that was only the force of interest not the net force. Also 2025 N is not one of the options for an answer.
     
  13. Nov 20, 2015 #12

    haruspex

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    Does it not match any of the offered answers? The choices are only specified to two significant digits, so round your answer accordingly and then compare. What do you get?
     
  14. Nov 20, 2015 #13

    haruspex

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    I didn't say it was not the net force. If it's the only force of interest then it is the net force.
    And 2025 N is one of the answers (see my preceding post).
     
  15. Nov 20, 2015 #14
    ah ha it is becoming clearer. I appreciate you making me think.
     
  16. Nov 20, 2015 #15
    It has been many years since physics 12 - I am helping my daughter. She and I will laugh in the morning over thinking that 2025 N does not equal 2.0x10 to the third when rounded. Many thanks....
     
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