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Two Hookean Springs Connected in Series

  1. Feb 24, 2007 #1
    I was just going through the derivation for the equivalent spring constant of two springs (of different stiffness) connected in series, and I realized that there's one thing I don't understand. In the process of this derivation, it is necessary to write down the net force acting on the point connecting the two springs, and then set this equal to zero. This seems to imply that the two forces are the action-reaction pair dictated by Newton's Third Law. My question is: why is this the case? Given that the connecting point between two springs does accelerate, why do we set the net force on this point equal to zero?
     
  2. jcsd
  3. Feb 24, 2007 #2
    The mass of the point is (effectively) zero. When something has zero mass, the forces on it must be balanced because the alternative would (by N2) imply infinite acceleration.
     
  4. Feb 24, 2007 #3
    Thanks, that makes sense. But do the forces of the two springs on this point constitute an action-reaction pair?
     
  5. Feb 25, 2007 #4
    Not as such. An action/reaction pair are the two forces exerted by two objects on each other when they interact. (For example, a book sitting on a table exerts a normal force downward on the table, while the table exerts a normal force upward on the book.) So, if we're going to talk about the forces on the connecting point due to the two springs, these would be paired with the forces the connecting point exerts back on the springs.

    That said, since there's no massive object sitting at the connecting point, it's just as valid to think of the two springs directly exerting forces on each other. And, in that picture, the two forces are an action/reaction pair.
     
  6. Feb 25, 2007 #5
    Thank you. This might sound like a naive question, but is it always valid to say that the net force on any individual point in a rigid body is zero (even when the body experiences a net acceleration)?
     
  7. Feb 25, 2007 #6
    No. What makes this special is that, in talking about the point at the interface between the springs, you're basically defining that you're not including any part of either spring. In general, it's really only valid to talk about vanishingly small bits of an object, not individual points, because we need a well-defined way to add them back together to get the whole object. So, then, we'll have infinitesimally small bits of the object with infinitesimal masses, dm. These, then, must feel a net force, [tex]dF = adm[/tex]
     
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