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The meaning of the D'Alembert's Principle

  1. Sep 2, 2012 #1
    The D'Alembert's Principle states that:

    [itex] \sum_s [\underline{ F_s^{applied}} - \frac{d}{dt} (\underline{p_s}) ] \cdot \underline{δr_s} = 0 [/itex]
    s - labels particles

    That is when [itex] F_s [/itex] doesn't include the constraint forces, and the virtual displacement is reversible, and compatible with the constraints.

    My question is - doesn't it just say that:

    - if there are no constraints, Newton's laws are obeyed, with force being [itex] F_s [/itex] - (the parenthesis is zero)
    - if there are holonomic constraints, we can only displace the object perpendicular to the constraint forces - (the dot product is zero)

    Does this principle also say something about non-holonomic constraints? And if so, can anyone give an example?

    And what exactly is the difference between reversible and irreversible virtual displacement? If a displacement is virtual, and if displacing by dx is possible, then also displacing back by -dx should be possible. So how can we have an irreversible displacement at all?
    Last edited: Sep 2, 2012
  2. jcsd
  3. Sep 2, 2012 #2
    Whenever you use virtual displacements you introduce another condition into the system.

    You introduce (geometric) compatibility.
    This condition limits what virtual displacements you can employ.

    D'Alembert's principle allows you to add imaginary forces(not virtual forces they are different) to a non equilibrium system to employ the equations of equilibrium.
  4. Sep 3, 2012 #3
    I understand the bit about virtual displacements.

    Let's think of a ball rolling off a solid sphere.

    https://dl.dropbox.com/u/94695102/fizyka/kula.jpg [Broken]

    One way of looking at it is that there is a reaction force [itex]F_R[/itex], etc.

    But we can also think that the ball is constrained to be outside of the sphere, and that puts a limitation on the possible virtual displacements.
    In this picture [itex]F_R[/itex] is a constraint force, and [itex]F_G[/itex] an applied force. Also: [itex]\frac{d}{dt}p = F_{net}[/itex] And in my book there's this formula that I've given in the previous post.

    When the ball doesn't touch the sphere, [itex] F_{net} = F_G [/itex] so clearly the parenthesis is zero so it works.

    When the ball still rolls on the sphere, the formula apparently works only is we displace it along its surface - this would be the only possibility if the constraints were holonomic - then the displacement is perpendicular to the parenthesis which equals [itex]F_R[/itex].

    But if we raise it outwards from the surface, which is compatible with the constraints, the dot product isn't zero, and neither is the parenthesis. Where's my mistake?

    And before I try to digest your second point - is "imaginary force" the same thing as "fictitious force"?
    Last edited by a moderator: May 6, 2017
  5. Sep 3, 2012 #4
    That's a quick yes.

    For instance one way to tackle a body, mass m, travelling in a circle is to introduce an imaginary or fictitious centrifugal force F = mass x central acceleration = mrω2
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