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Newton's law in non inertial frame.

  1. May 14, 2012 #1
    Please explain the concept of pseudo forces by considering an example of two cars say A and B, accelerating in the same direction. How will newton's law applied to the man in car B as noted by the man in car A differs from that of the man standing on the earth? And how will the man in car A apply Newton's law to himself?
     
  2. jcsd
  3. May 14, 2012 #2
    Fictitious forces have to do with the fact that a frame is accelerating compared to a non-accelerating frame. For each frame, Newton's law applies as usual. However, if people want to explain what is happening in the other frame, the must compensate for its acceleration. Let's consider the 2 cars in the two reference frames:

    As we know, F=ma. There is a force accelerating both cars. Let's now consider a third frame that is not accelerating. In respect to this frame, any object that is in the other two frames is accelerating, even if it is immobile in its own frame. Let's also consider that in this frame you have an observer C, who assumes that the objects he is watching accelerate (cars A,B) are in the same frame as he is. In this case, even if the cars are not accelerating in their own frames, he would have to assume that there is a fictitious force that is causing the acceleration of the cars, since he is watching them move (because their frames are moving).

    If the cars start accelerating in their own frame (due to a force), applying Newton's law requires observer C to add the new force to the fictitious force he had before, in order to get the acceleration. Each car of course would apply Newton's law to itself as usual.
     
  4. May 14, 2012 #3
    My problem is this: to the man C on the earth, both cars are accelerating(say is +ve direction of x). So, both are experiencing a force F=ma towards +ve x axis. But the man in car A thinks that the car B is not accelerating. So no force is acting on car B according to man in car A. Hence, to get the correct acceleration man A must add a fictitious force of magnitude F=ma IN THE DIRECTION OF HIS OWN ACCELERATION.(this is contrary to what we do in working out problems).
     
  5. May 14, 2012 #4

    Doc Al

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    Staff: Mentor

    Right. And that force is a 'real' interaction force that exists in every frame. (I will assume that C is an observer at rest in the inertial frame of the earth.)
    No net force.
    No, he must add a fictitious force opposite to the direction of his acceleration.
     
  6. May 14, 2012 #5
    But if he add a fictitious force in opposite direction he will write -F=ma i.e. a=(-F/m) getting acceleration in -x direction whereas to the observer in inertial frame it is in +x direction.
     
  7. May 14, 2012 #6
    No. To him the person B is at rest.

    He adds a fictitious force to the direction opposite to his acceleration, and that gives him

    [itex]F_B - ma = 0[/itex]

    From here, he finds the actual force that the person B is experiencing, which is in agreement with what the ground observer measures.
     
  8. May 14, 2012 #7
    Well Phyman, since it is a vector sum, it depends on the direction of the two acceleration vectors, the frame's and the car's so there are a number of combinations :biggrin:

    In any case, use your common sense in working out these problems and remember that you are describing nature. If, for instance, the car is on an accelerating frame and it's not moving to the eyes of the resting observer, then some force is countering the "fictitious force" due to the accelerating frame.
     
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