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Newton's Third is Giving Me a Problem That is Absurd

  1. Mar 27, 2009 #1
    Suppose there's a rocket in space with a mas m, and an object also with mass m. Now suppose the rocket turns on and pushes the object with a constant force F. The object would then have an acceleration a. By Newton's Third Law the object pushes back with a force -F. So shouldn't the rocket have a negative acceleration (-a) and thus eventually slow to a stop? But what happens in reality is the rocket pushes the object faster and faster. I must be missing something fundamental here. Set me straight so I can stop thinking about this friggin' thing in the shower. Thanks.
     
  2. jcsd
  3. Mar 27, 2009 #2

    Doc Al

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    All Newton's 3rd law says is that if the rocket exerts a force on the object, then the object will exert an equal and opposite force on the rocket. Of course that's not the only force on the rocket--it's also blasting out hot fuel, which provides another force. (Otherwise, it would be more like a rock than a rocket. :wink:)
     
  4. Mar 27, 2009 #3
    But won't this blasting rocket fuel acting on the rocket be canceled out by the force the object exerts in the opposite direction on the rocket since Newton says the forces are equal in magnitude and opposite? Thus the rocket's net force is 0, and there will be no acceleration? I hear this is a problem a lot of beginning physics students make, but I don't see how saying the forces are acting on opposite objects solves the problem. I am confused. :(
     
  5. Mar 27, 2009 #4

    HallsofIvy

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    No. There are three forces involved here- The force the rocket blast applies to the rocket, the force the rocket applies to the mass, and the force the mass applies to the rocket. Yes, the last two cancel. Which means that the whole assembly accelerates at a= F/2m.
     
  6. Mar 27, 2009 #5
    But isn't the force of the rocket blast the force that acts on the object?
     
  7. Mar 27, 2009 #6

    Doc Al

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    Nope. You are mixing up the 3rd law force pairs. The force the exhaust fuel exerts on the rocket is equal and opposite to the force that the rocket exerts on the fuel. But that force has nothing to do with the force between the object and the rocket--that's a totally different force.
    The net force on the rocket is the sum of (1) the force of the object on the rocket and (2) the force of the exhaust fuel on the rocket. No reason to think that these cancel out. One hopes that (2) is much greater than (1).
    In my experience, the main confusion is thinking that the "equal and opposite" forces from Newton's 3rd law somehow cancel out. But they can't: They act on different bodies!

    Read this: https://www.physicsforums.com/showthread.php?t=90060
     
  8. Mar 27, 2009 #7

    Doc Al

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    Nope. The exhaust pushes on the rocket; the rocket pushes on the object. Two different forces.

    Imagine this similar situation. A frictionless table with two blocks sitting side by side. You push one block into the other. You push block A. Block A pushes block B. Ah... but Newton's 3rd law says that block B must push back on A with the same force. Yes! Does that mean it "cancels out" and nothing moves? No! You push on block A and block B pushes on block A--the net force determines how block A accelerates. (Hint: Your push is greater than block B's push.)
     
  9. Mar 27, 2009 #8
    Okay, I think I'm getting it. So say the object and rocket both have 10 kg masses, and the blaster exerts a force of 20 N on the rocket. How would you find the force the object and rocket exert on each other?
     
  10. Mar 27, 2009 #9
    Actually, I think I know how. Fnet = MsAs ---> 20 N = 20kg*a ----> a = 1 m/s. So the rocket moves at 1 m/s, then the sum of the forces on the rocket Frocket = Fblaster + Fobj ---> 1m/s*10kg = 20N + Fobj, so Fobj is -10 N?
     
  11. Mar 27, 2009 #10
    Good question!
    All you have to know is f=ma.

    The fuel is pushing the rocket, which is pushing the object. Since the object and the rocket are always moving in sync, we can pretend that, instead of two 10 kg masses, they are one 20 kg mass. We can do this to find the acceleration of the whole system. f=ma. 20N=20kg*a, so "a"=1 m/s^2

    Now, break it back into two objects. It's okay, we can do that. We're physicists.

    The force pushing the whole system is 20N, but the force that is pushing just the object doesn't come from the fuel. It comes from the rocket.
    Now, we know the mass and the acceleration of the object being pushed. f=ma, again. f=10kg*1m/s^2 so f=10N.

    AKA, the force of the rocket on the object is 10N.

    Newton's law says that that means that there is a force of 10N from the object back to the rocket.

    Did this make sense?
     
  12. Mar 27, 2009 #11
    Indeed, I believe I understand quite well now. Thanks for all the help guys.
     
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