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Virtual displacements and Virtual Work

  1. Aug 3, 2011 #1
    Hi all,

    I am having trouble understanding Virtual displacements and related ideas and would appreciate your help!

    Let's say we have a simple pendulum and we take a snapshot of this system at some instant of time : 2wmm7iq.jpg

    A 'virtual displacement' is said to be :
    - consistent with the constraints on the system
    - occurs without the passage of time
    - is infinitesimal

    In the case above, the mass can undergo a virtual displacement along the pink dotted line, since any other displacement would cause the string/rod to elongate, thus violating the constraint. Also, the virtual displacement along the pink dotted line has to be infinitesimal as a large virtual displacement would cause a change in the length of the string/rod. The forces acting at this instant on the mass are the weight and the reaction force/tension. It would seem that the reaction force, which is the force maintaining the constraint is perpendicular to the virtual displacement and hence would do no 'virtual work'. But why is it important that the no time elapse during this displacement? Is it because the external and constraint forces may change directions and (possibly) magnitudes?

    Also, real displacements obey the system constraints (the mass moves in an arc). So apart from the idea that virtual displacements occur without time elapsing, what is the difference between the two?

    Finally, what I find really puzzling is that the principle of virtual work is an important step in deriving Lagrange's equation. How is it that some relations that are based on imaginary displacements give rise to equations relating physical quantities?

    Am I missing something here? Any help would be greatly welcomed. :)

  2. jcsd
  3. Aug 3, 2011 #2
    Good morning Boeing737 and welcome to Physics Forums.

    Virtual work must be the worst presented topic in most mechanics courses, yet is is one of the most useful techniques.

    Since I don't know what approach your course is taking to virtual work (there are several) I will be general.

    Firstly motivation

    Why do we do it?

    To make calculation easier! To replace a difficult (complicated) calculation with an easier (simple) one.

    You may have come across the idea, in statics, of taking moments about an unknown force to eliminate it from the equations, reducing the unknowns by one.

    Well virtual work does the same thing, although you may be able to eliminate more than one unknown.

    How does this happen?

    Well if we separate the forces acting on a system in equilibrium into

    1) Applied forces (also called external forces)

    2) Resisting forces or forces of constraint (also called internal forces)

    The principle of external work, which is really the principle of conservation of energy in disguise, states

    [tex]{\rm{\Sigma external work = \Sigma internal work}}[/tex]

    Note carefully neither the external nor internal work are necessarily zero, but if we can dream up a situation in which one is zero then the other is as well. This can then help our motivation of a simpler calculation.

    For single point masses we identify the constraint forces with

    The normal reaction against a surface, wire, hoop etc
    The tension (compression) in and inextensible string or rod and the reactions between two bodies connected by the same.
    The reactions between two bodies in rolling contact.

    With specific reference to your example. It can be solved by resolving forces (VW problems can always be solved in other ways).

    However we can identify the internal force with the tension in the string and if we dream up a (virtual) displacement at right angles to this tension then the work done by such a displacement is zero.

    Why must the displacement be small?

    Well in this case the weight is swinging on an arc you are approximating this arc by a straight line so you need a small displacement for this approximation to hold.

    However in general it is not necessary for the displacements to be small so long as they are 'compatible'.

    Your course must be unusual - virual work is not the normal ways of deriving Lagrange.

    go well
    Last edited: Aug 3, 2011
  4. Aug 3, 2011 #3


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    Hi boeing_737, Virtual work was a big mystery to me too, when I took mechanics. It was simpler to just skip the derivation and accept Lagrange's Equations as a given. But now I realize I'd been doing the virtual work thing anyway, just not calling it that.

    The "normal" way to solve a problem like the pendulum is to write out Newton's laws F - p· = 0.

    I'll have two equations, the x and y components. But in addition to the "external" force (gravity) they also involve the "internal" force T (string tension) which I don't care about. So I eliminate T algebraically by taking a linear combination. That linear combination happens to be the component in the "pink" direction, i.e. perpendicular to the string. So if I'd been thinking, I would have started writing down just that component in the first place and wouldn't ever have T to deal with.

    That's what the principle of virtual work (actually d'Alembert's principle) says to do. Let δx be a little vector pointing in the pink direction, and write down the component (F - p·)·δx = 0. T will not appear in this. Note that's because δx is perpendicular to T.

    Well, to generalize this idea, we choose a virtual displacement δx such that ∑Fint·δx = 0, and then the linear combinations of Newton's law, ∑(F - p·)·δx = 0 will not involve the internal forces Fint.

    It's after we do that, we suddenly realize, hey, ∑Fint·δx looks like a work! I can think of the condition as saying that the virtual work vanishes.
  5. Aug 4, 2011 #4
    Thanks Studiot and Bill for your answers. I think I am sort of getting it now. One thing that I don't get is this - At each instant, the position vector is perpendicular to the reaction force. If we consider an infinitesimal real displacement, then the real work done by the forces on the mass would be zero as long as the constraint force remains orthogonal to the displacement and doesn't change direction during the infinitesimal time interval. I suppose this is what we are trying to do with virtual displacements?

    Also, even though we say the virtual displacement is consistent with the imposed constraints, if we look at the pendulum example, any little displacement, even imaginary along the pink line, would result in the length of the string increasing and violating the constraints. In other words, the virtual displacement that you assume to be in agreement with the constraint leads the coordinates to take on values that are inaccessible during actual motion. (i.e., (x,y) can only take on certain values due to the length constraint, which are points on the arc travelled by the mass, but the virtual displacement 'delta x' along the pink line would enable (x,y) to have values outside the constraint, even though infinitesimally close to it) Does this not sort of contradict the very definition of virtual displacement?

    Thanks a lot for the help!
  6. Aug 4, 2011 #5
    Not really.

    That is why in this case you are limiting the virtual displacements to infinitesimal ones.

    As (if) you learn more you will see that sometimes the virual work is exactly zero and sometimes you can show it is 'second order' and therefore negligable compared to direct work.
    That would be the situation for a small displacement in this case.

    But as I remarked earlier the displacements need not be infinitesimal if you take the right ones. Such a situation would often be the case in structural engineering.
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