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Deduction of thrust force

  1. Feb 9, 2010 #1
    Hello People actually i want to make sure of this explanation

    The external force acting on the rocket is zero
    Therefore : F = mdv/dt+vdm/dt=0

    Therefore mdv/dt=-vdm/dt=Fimp(Impulse force or thrust force)

    Well I have an explanation for this but I'm not sure about it

    The external force is zero according to newton's third law

    Mdv/dt is the reaction of the rocket on thrust force because the mass of the rocket is constant
    Vdm/dt is the force acting on the rocket and mass is variable bec. The mass of gasses varies from time to time.
    Since these forces are equal in magnitude and opposite in direction
    Therefore the sum of them is zero

    So is this explanation right?
     
  2. jcsd
  3. Feb 9, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hello Misr! Welcome to PF! :smile:
    Nooo :redface:

    Newton's third law says action is equal and opposite to reaction (and is true both for internal and external (pairs of) forces).

    It has nothing to do with any force being zero.


    No, the force from the gas is an external force on the rocket.

    Newton's second law (usually written F = ma) is really F = d(mv)/dt = mdv/dt + vdm/dt.

    F in this case will be the force from the gas.

    But a more direct method would be simply to use conservation of momentum. :wink:
     
  4. Feb 10, 2010 #3
    The external force acting on the rocket is zero
    Therefore : F = mdv/dt+vdm/dt=0

    No, the force from the gas is an external force on the rocket.
    I'm not the person who is saying that actually its a mysterious point in my book and i'm trying to explain it
    here it is an extract from the book

    [PLAIN]http://img693.imageshack.us/img693/3506/thrustforce.jpg [Broken]
    [/PLAIN]
    i can't understand this point
    i thought that the external force is the net force acting on the rocket

    why??

    Thanks
     
    Last edited by a moderator: May 4, 2017
  5. Feb 11, 2010 #4
    Hello there?
     
  6. Feb 11, 2010 #5

    tiny-tim

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    Hi Misr! :smile:

    I'm sorry, I must have deleted the email about your post yesterday without ever opening it. :redface:
    ah, your book means there is no external force on the rocket-and-gas together.

    So the momentum of the rocket-and-gas together doesn't change.

    What it calls Fi is the external force on the rocket alone (or on the gas alone). :smile:
    Because it is … that is Newton's second law …

    F = d(mv)/dt.

    (and F = mdv/dt + vdm/dt = ma + vdm/dt comes from the Chain Rule of calculus, and if m is constant, then it's just F = ma :wink:)
     
    Last edited by a moderator: May 4, 2017
  7. Feb 12, 2010 #6
    Yeah and i'm asking why?

    Thanks so much for helping me
     
  8. Feb 12, 2010 #7

    tiny-tim

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    Because of Newton's second law … applied force = rate of change of momentum …

    so if there's no external force, there's no change of momentum. :smile:
     
  9. Feb 16, 2010 #8
    i'll try to define my prob exactly

    First why is the external force = zero ?
    second : i still don't understand why F = mdv/dt+vdm/dt

    so can u explain this exactly
    Thanks so much
     
  10. Feb 16, 2010 #9

    tiny-tim

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    What else can it be (for the rocket-and-fuel combined)?

    The force of the rocket on the fuel, and vice versa, is purely internal, and there's no other forces.
    Newton's second law (usually written F = ma) is really F = d(mv)/dt

    F = ma is a short version, which only applies if the mass (m) is constant.

    Here, the mass is not constant, so you have to use the full version, F = d(mv)/dt.

    (and, using the Chain Rule, that's the same as mdv/dt + vdm/dt)
     
  11. Feb 17, 2010 #10
    Ok , thanks so much i think i got it but still have one more question

    ok if Fi= Fimp=-Vdm/dt=mdV/dt
    then which one is the force on the rocket and which one is the gas force
    Thanks again
     
  12. Feb 17, 2010 #11

    tiny-tim

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    let's see :rolleyes:

    assuming that Fimp is being measured in the same direction as V (eg both along the x-axis), and since dV/dt is positive (and dm/dt is negative), that means that Fimp is in the same direction as the rocket is going, so it must be the force on the rocket (and -Fimp must be the force on the gas). :wink:
     
  13. Feb 17, 2010 #12
    but mass and time are both scalar
    do u mean that Vdm/dt is negative and the velocity of the rocket upwards is positive?
    and we don't know exactly if -Vdm/dt or mdV/dt is the force on the rocket or the gas because both mass and velocity are changing ?

    thrust exerts a force on the rocket , the rocket answers by a force equal in magnitude and opposite in direction which makes the rocket accelerate upwards

    since there is no external force affecting the system (because the system is isolated)
    therefore the net force acting on the rocket is zero (the sum of thrust force and rocket's reaction on thrust = 0)

    therefore F=d(mv)/dt=0
    where Fis the net force acting on the system

    by using the product rule we get :

    F= mdV/d t+ Vdm/dt=0


    Therefore mdV/d =- Vdm/dt = Fimp
    Where the negative sign indicates the direction

    SO AM I RIGHT?
    If yes then write "right" if no say why?
    Thanks so much
     
    Last edited: Feb 17, 2010
  14. Feb 17, 2010 #13

    tiny-tim

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    m is the mass of the rocket-plus-remaining-fuel, so m is decreasing. :smile:
    You have a totally weird view of force :redface:

    force on the rocket makes the rocket accelerate,

    force by the rocket (on something else) has nothing to do with it.
    No, the net force acting on the rocket-plus-all-the-fuel (including the fuel that's gone) is zero.

    No, Fimp = d(mv)/dt, which is not zero.

    ah, now I see I took your word for it when you wrote …
    … no, that's wrong, Fimp = d(mv)/dt = vdm/dt + mdv/dt.
     
  15. Feb 18, 2010 #14
    Ok ,thanks so much for helping
     
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