1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: For every action there is an equal and opposite reaction

  1. Jul 5, 2006 #1
    This isnt really a homework question but it is something that I have been learning about out of school and this thought just popped into my head:

    According to Newtons Third Law: "For evry action there is an equal and opposite reaction".

    Now if you take a Jet engine,
    It works on the fact that you are pushing force out of the back of the engine, so according to Newtons third law, there is a force pushing fowards.
    If this force is equal and opposite of that coming out of the back of the engine, shouldn't you just stay stationary?

    Thanks, Iceman123
  2. jcsd
  3. Jul 5, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Realize that the "action" and "reaction" forces in Newton's 3rd law always act on different bodies. Only if two equal and opposite forces act on the same body would they cancel.
  4. Jul 5, 2006 #3
    This law suggests that when body A exerts a force on body B , body B exerts the same magnitude of force , but at an opposite direction.
    When the forces are equal (equilibrium) , it means either 2 things :
    1. As you mentioned earlier in your post , the jet would be stationary .
    2. The jet maybe moving at a constant speed (no acceleration).This is because there's no external or resultant force(s) acting on the jet plane.

    I hope this will help you =)
  5. Jul 5, 2006 #4
    Thanks v much,
    Iceman 123
  6. Jul 5, 2006 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Just to be clear: The jet exerts a force on the exhaust gases; in turn, the exhaust gases exert an equal and opposite force on the jet. These two forces alone never produce equilibrium: one is a force on the gases; the other is a force on the jet.
  7. Jul 5, 2006 #6
    In jet engine there is not realy a force acting by the gas on the jet.

    One of the great consequences of Newton's third law is that in a system, the net momentum (mass time velocity) is conserve.

    And one consequence of the momentum conservation is that jet propulsion is possible.

    Before the time that the gases are emitted, the momentum of the system is zero.

    Suddently, gases are emitted let's say downward . So they have acquired a downward momentum.

    The jet must react because the net momentum must be zero, so the jet will acquire a momentum in the upward direction.
  8. Jul 5, 2006 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Of course, there is a force exerted by the gases on the jet, as Doc al correctly pointed out.
    How does the jet acquire this momentum? Why does it acquire this momentum? It's because the gas is exerting a force on the jet.

    If the momentum of an object changes it's because there is a net force acting on the object. That's simply Newton's second law
    [tex] \sum {\vec F} = {d {\vec p} \over dt} [/tex]
  9. Jul 5, 2006 #8
    Yeah, that's entirely how jets and rockets work is change in momentum. To make any jet or rocket analysis worthwhile you have to use calculus.
  10. Jul 8, 2006 #9
    Thanks guys, thats great!
  11. Jul 8, 2006 #10
    Just to clarify, so the engine exerts a force on the exhaust gasses and the gasses exerts a force on the engine which in turn make the plane move. Since the forces are of same magnitude, the gasses gain as much momentum as the plane but in the opposite direction but they move faster because they are much less massive?

    In reality, does it happen this way or is it the exhaust gasses that exerts the force on the surrounding air molecules and the air molecules exerts a force on the exahust gases making the plane move? That is, is the exhaust gas from the engine considered as a system with the plane?
  12. Jul 9, 2006 #11

    Doc Al

    User Avatar

    Staff: Mentor

    This is correct.

    While it's certainly true that the exhaust gases and surrounding air exert forces on each other, that doesn't help the jet move. Once the gases are expelled they cease to exert forces on the jet.

    Consider a rocket in empty space. The same principle applies: The rocket expels the exhaust gases, which in turn exert a thrust back on the rocket. No surrounding air to confuse the issue here.
  13. Mar 28, 2007 #12
    Need Translation

    OK. I've never taken physics nor have I ever learned calculus, so please bear with my elementary question. I have an ad I have to work on and found your equation which piqued my interest. What would the equation mean if I were to only use the first Greek symbol with the arrow above a word I would replace the F with. I want to send the subliminal message as apart of the ad "for every action there is a reaction" but I was curious what message I would be communicating if I stopped with just those two components of the equation.
  14. Mar 28, 2007 #13
    not much I'm afraid, the arrow hat only means that the quantity is a vector. A vector is a number or property with an "attitude". It implies that it operates in a specific direction. Thats it.
  15. Mar 28, 2007 #14
    Need Translation

    So does that whole equation represent "for every action there is a reaction"?

    Is there a physics/calculus symbol or image that would get that message across?
  16. Mar 28, 2007 #15
    Actually, the equation represents Newton's second law.
    whereas the statement "for every action there is an equal and opposite reaction" is Newton's third law.
  17. Mar 28, 2007 #16
    So how is Newton's third law written out as an equation?
  18. Mar 28, 2007 #17
    there is not an exact equation to express Third law.
    The statement "If a body A exerts force F on another body B, then body B exerts a force -F(minus F) on A, the two forces acting along the same line." is alone enough.(as far as i know)
  19. Mar 29, 2007 #18

    Doc Al

    User Avatar

    Staff: Mentor

    You could write it this way if you liked:
    [tex]\vec{F_{a/b}} = - \vec{F_{b/a}}[/tex]

    But Sourabh N's statement is more complete.
    Last edited: Mar 29, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook