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Force is ill-defined? contradictory definitions?

  1. Feb 18, 2012 #1
    Correct me if I am wrong but as far as I know, force is generally defined in three ways ways:
    1) [tex] F = \frac{d p}{d t} [/tex]
    2) [tex] F = m\dot v [/tex]
    3) [tex] F = ma [/tex]

    This is all well in good usually...until the case arises when mass is variable.
    Then two contradictory cases arise:
    If we take definition 1...we get
    1) [tex] F + u \frac{dm}{dt} = m\frac{dv}{dt} [/tex]
    If we take definition 2 we defined the quantity [tex] u \frac{dm}{dt} [/tex] as thrust and part of force and thus
    we get [tex] F= u \frac{dm}{dt} +\cdots =ma =m \frac{dv}{dt} [/tex]

    Am I missing something here?
    Thanks.
     
    Last edited: Feb 18, 2012
  2. jcsd
  3. Feb 18, 2012 #2

    Pengwuino

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    There's a few errors and things I don't get. For one, the force is given by [itex]F = {{dp}\over{dt}}[/itex], not by partials. Two, did you get your = and + mixed up in the 4th equation? Finally, if the mass is variable, the 2nd equation no longer applies.
     
  4. Feb 18, 2012 #3

    Ken G

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    Yes, we can say that none of those three equations are reliable in general, though do work in some situations. The first one isn't generally right because it cites an explicit derivative with respect to the symbol t, but forces produce a real rate of change of momentum, not just an explicit dependence on time (the partial derivative is often used for a situation where every place and time has a momentum associated with it, and you are only interested in the dependence on time, not place, whereas the full time derivative is a derivative along a trajectory, so the momentum can be changing explicitly both with location and time as we track that trajectory through the values of momentum in space and time). The second equation is wrong because it refers to a coordinate derivative, the rate of change of velocity, but in accelerating or curvilinear coordinates the coordinate rate of change of velocity will not equal the forces (unless one also includes fictitious forces to cancel out the coordinate accelerations). The third equation is wrong even if a is the true acceleration and F is the real net force, because it doesn't work when the mass is variable. So the one correct equation in general is F = dp/dt, as noted above.
     
  5. Feb 18, 2012 #4
    Thanks for the replies and explanations. yes I meant to put total derivative for the first equation but I do not believe I made any errors to equation 4.

    Ken, or anyone else, would you mind explaining what you meant by two? If I changed [tex] \frac{dv}{dt}[/tex] to [tex] \ddot x [/tex] would this change the error?

    I guess my main confusion is that in http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Variable-mass_systems They give the fourth equation but later on mention how
    "Under some conventions, the quantity (u dm/dt) on the left-hand side, known as the thrust, is defined as a force (the force exerted on the body by the changing mass, such as rocket exhaust) and is included in the quantity F."
    I am somehow misunderstanding what they are saying here?
    Thanks!
     
  6. Feb 19, 2012 #5

    Ken G

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    Yes I think equation 4 is fine, it wasn't clear what u is (it seems we'd need u=-v).
    It's sort of a technical point about the difference between the actual acceleration a (read by an accelerometer) and a coordinate acceleration like dv/dt (or x double-dot too). The former is physically real, being a measurable, but the latter is just a coordinate computation, and can be different in different coordinates or in accelerating frames. You can fix that by making sure that you mean v and x are vectors, so are coordinate-independent abstract notions, rather than a coordinate velocity or a spatial component. In curvilinear coordinates, in coordinate notation the dv/dt term will spawn additional terms that are just a matter of the coordinates chosen, and accelerating coordinates will spawn fictitious force terms. Those are all handled automatically if the v in dv/dt is interpreted as a vector and not a coordinate velocity, so your expressions are fine if that technical point is made clear.
    I don't know that you are misunderstanding anything, that all sounds fine. It's all consistent with F = dp/dt (where F and p are coordinate-independent vectors).
     
  7. Feb 19, 2012 #6
    Ah I see what you are saying.

    Also what I mean is that my 4th and 6th equations are not consistent (I believe). The 4th assume F = dp/dt , but I don't think the 6th does? I interpret 6 as defining the LHS of 4th as force.
     
  8. Feb 19, 2012 #7

    Ken G

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    The two are consistent in that they only differ in the semantics of what we call a "force." The same issue crops up with noninertial or coordinate forces-- shall we call the centrifugal force a force and put it in with the list of forces on the LHS, or shall we call it a correction to the ma term and put it on the RHS? Note that which side of the equation only matters insofar as you have to reverse the sign if you put it on the other side, but that's essentially the difference between a "centrifugal" force and a "centripetal" force. The "correct" way is to count the fictitious force as a correction to the coordinate acceleration to get it to be a real acceleration, and put it all on the RHS with the dp/dt (where if p is interpreted as a vector then this is all done automatically when the coordinates are chosen). However, it is a standard convenience to use the language of the "centrifugal force", and change its sign and put it in with the other forces on the LHS. Neither is considered wrong, it's just a question of being clear.
     
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