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Newton's Third Law and a rocket in space

  1. Feb 27, 2015 #1
    I am the kind of guy that always needs to return to the horse-and-carriage problem to hone in my understanding of Newton's Third Law. Here's my question. Assume a rocket in space is applying a force to another object of equal mass in space. Now, I understand the object in space is experiencing an unbalanced force, and so, it will accelerate in the opposite direction of the rocket.

    What always perturbs me is when I begin to think of the force acting on the rocket - it will experience a force in the opposite direction - what does this mean exactly? Does this force prevent it from moving forward? Here, I say no, because the system is moving the direction of the force as the object in space is moving forward due to the unbalanced force.

    Are we trying to say the opposite force experienced by the rocket is what prevents it from going through the object? I'm just trying to wrap my head around this opposite force, experienced by the rocket, that does not prevent it from moving forward (as this system, including the rocket, moves forward), yet is still felt by the rocket.
     
  2. jcsd
  3. Feb 27, 2015 #2

    A.T.

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    Acceleration is determined by the net force on the object, not a force.
     
  4. Feb 27, 2015 #3
    Understood. There is a net force in one direction, accelerating the system forward. But, from the rocket's perspective, it is experiencing an equal force in the opposite direction - what does this mean for the rocket? Does it mean the rocket will be unable to go through the object it is pushing again?
     
  5. Feb 27, 2015 #4

    A.T.

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    There is also a net force on the rocket accelerating it forward. The force from the exhaust is greater than the force from the pushed object.
     
  6. Feb 27, 2015 #5
    Yes, we've established the system is moving forward. Now, to my question, the rocket will experience a force in the opposite direction equal to the force it is applying to the object - this is the force pair that I'm interested in. Is this force responsible for the rocket not going through the object? For example, if the object was not strong enough to provide this force, it would collapse, and the rocket would go through the object.
     
  7. Feb 27, 2015 #6

    A.T.

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    You mean accelerating? Movement at constant speed doesn't require any forces.

    That is a weird way to parse it. The object will collapse due to the force acting on the object, not due to the force on the rocket. But since they are of equal magnitude it doesn't really matter how you put it..
     
  8. Feb 27, 2015 #7
    Yes accelerating forward. Is the rocket experiencing a force in the opposite direction because the object is intact; thus, able to provide that opposite force? Otherwise, if the object collapsed, the rocket would proceed through the object, with no opposite force.
     
  9. Feb 27, 2015 #8

    A.T.

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    Yes, if the object is destroyed by the push, the rocket can accelerate faster.
     
  10. Feb 27, 2015 #9

    anorlunda

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    What about a passenger in the rocket? The person is another object.

    The rocket exterts a force on the person, so the person accelerates with the rocket.

    The person exerts an equal and opposite force on the rocket so the rocket will accelerate slower than if there was no person on board.
     
  11. Mar 1, 2015 #10
    I just want to make sure I properly understand Newton's Third Law.

    Assume there's a horse with a mass 100kg and a cart with a mass 20kg. Assume the horse pushes off the ground with a force 1,000N, the ground,in turn, will push the horse forward with a force of 1,000N.

    If I look a the system as a whole, it will have a mass of 100 + 20 = 120. With a force of 1,000N, the system will accelerate 1,000/120 = 8.33 m/s^2

    So now, we can calculate the net force exerted on each object ( F = m * 8.33)

    Horse: F = 100 * 8.33 = 833.33N

    Cart: F = 20 * 8.33 = 166.667N A

    The following demonstrates Newton's Third Law

    The horse exerted a force of 1,000N, but experiences only a net force of 833.33N, a loss of 833N - 1000N = -166.667 B

    A + B = 0 <- Newton's Third Law
     
  12. Mar 1, 2015 #11

    A.T.

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    Yes, and those +/- 1000N between horse and ground are also a Newton 3rd pair.
     
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