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"virtual" displacements?

  1. May 23, 2015 #1
    1. Why is the D'Alembert's Principle

      ∑i(Fi−miai)⋅δri=0
      stated in terms of "virtual" displacements instead of actual displacements?

      what does the 'virtual displacement' really means and what is its physical significance??
     
  2. jcsd
  3. May 23, 2015 #2

    Doc Al

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    Staff: Mentor

  4. May 25, 2015 #3
    The concept of virtual displacement is not immediate so you have to study it well in suitable texts/documents.
    Here I want to give just a very simple example, which doesn't want to be comprehensive at all, but is useful (I think) to have at least some idea of the concept.
    There is a point mass m which can move frictionless along an horizontal bar, in the direction x. The bar, which is the constraint for the point mass, can move vertically along direction y. So, the actual displacement of the point mass is along x and along y: dr = (dx,dy).
    In this case the virtual displacement, instead, is the one only referred to the constraint: δr = (dx,0).

    --
    lightarrow
     
  5. May 26, 2015 #4
    Because in this way only active forces can be considered. Active forces are those forces which doesn't arise in order to make the particle follow the constraints (these last are constraint forces). In the simple example I made in my previous post, active forces are those which make the particle move along the bar (horizontal forces) while constraint forces are those directed vertically, which prevent the mass point to escape the bar. Usually the constraint is still, or it moves in a specific, given way, so the problem is much simpler to solve if it's focalized in finding how the points move in virtual displacement only. But there are also theoretical reasons to focalize on those displacements only, which will be more clear whe you will study lagrangian mechanics.
    --
    lightarrow
     
  6. May 26, 2015 #5
    thnx bro...
     
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