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Homework Help: Please Help Grd 11 student

  1. Dec 13, 2004 #1
    hi ive been doing rocets recently and i have to a report and analysis of my rocket

    everything is going fine, i just got stuck on one question
    i have to calculate the force exerted on the rocket from the parachute

    so when the rocket reached its maximum, my parachute edjected....as it is coming down...wut is the force that the parachute is exerting on the rocket

    ive thought aalot of about it but i just cant figure it out
    i calculated my Acceleration Net to be around 15m/s

    Fnet was around 1.27N
    maximum height reached was 90.0m
    mass of rocket was 66.01g <<<< this is without engine btw

    i dunno if that helps :uhh:
    i just need some direction/suggestion

    Last edited: Dec 13, 2004
  2. jcsd
  3. Dec 13, 2004 #2
    shouldnt the force be mg - because it should be in terminal velocity which is constant speed??
  4. Dec 13, 2004 #3
    If you wait sufficiently long and the velocity of the descending rocket comes into equlibrium, according to Newton's Second and Third laws, there is no net acceleration of the parachute / rocket system. When that is the case all forces internal to the system must yield zero. The parachute pulls up on the cords that tie it to the rocket, while the cords pull the parachute down. Simultaneously (sp?) the cords pull up on the rocket while the rocket pulls down on the cords.

    [tex]T-mg=0 \therefore [/tex]
    [tex]F_{p}-mg=0 \rightarrow F_{p}=mg [/tex]

    I hope that helps.
    Last edited: Dec 14, 2004
  5. Dec 13, 2004 #4
    Yupp - I agree with bobp718. Welcome to physicsforums btw ;)
  6. Dec 14, 2004 #5
    what does Fp and T represent in that solution?

    sry...im new to physics...its my year taking it so im unfamiliar with these terms

    thnx for the reply
  7. Dec 14, 2004 #6
    [tex]F_{p}[/tex] was intended to be understood as the force that the parachute applied to the system.
    [tex]T[/tex] was intended to be understood as the tension in the cord uniting the rocket and the parachute.
  8. Dec 14, 2004 #7
    How about the air resistance?>
  9. Dec 14, 2004 #8


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    In this case, because of the function of the parachute, the air resistance would be the equivalent of the Force exerted by the parachute. So you do not need to calculate air resistance, once you find Fp you will already have it.
  10. Dec 14, 2004 #9
    thank you soo much
    it makes sense now
    i really appreciate the help
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