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Virtual displacement and the D’Alembert principle

Classical mechanics

Virtual displacement and the D’Alembert principle

flickr.com/Tobi Mattingly

Virtual displacement and the D’Alembert principle

Objects moving with time frozen sounds like science fiction. But virtual displacement is a key tool in understanding equilibrium under constraint forces. And we try here to understand this concept as well as we can.

The concept of objects displacing in a certain time interval, is something that most of us began our studies in Physics with. The object changes its position from one point in space to another, and we say that the object is in motion. But what if we imagined the object move with time frozen? That is, an imaginary change in position of the object while the time stands still. This might seem very “unphysical”, after all we know that all motions cease when time stops. But we have stressed upon the use of the word imaginary here.

We call such a change in position in zero time (by an infinitesimal, teeny-weeny length), the virtual displacement of the object. But we must be careful: we here allow ourselves to imagine motion in zero time, but retain the limitations of the motion that exist in real time. That is, if an object’s motion is confined in some manner, for e.g., a book placed on a table top is free to move anywhere on or above the table, but not “into” the table; this restriction will be carried over to virtual displacement as well. And in a more general set up where the forces are constantly changing with time, we consider the forces only at the instant at which the virtual displacement happens.

Virtual work

Now to cause a change in the position of any object, in zero time or real time, requires a force (assuming the object was at rest initially). In fact, the very concept of force was in the first place introduced to account for a cause for displacements. Hence here, a force causing a virtual displacement will lead to a work done, a virtual work done. If \delta \mathbf{r}_i is the virtual displacement caused by a force \mathbf{F}_i, on the ith particle of a system, the virtual work will be

    \[\mathbf{F}_i.\delta \mathbf{r}_i\]

But if the net force \mathbf{F}_i on each particle is zero, then each particle is in mechanical equilibrium, and the virtual work done on each particle will be zero. The same argument holds for the entire system:

    \[\sum_i \mathbf{F}_i.\delta \mathbf{r}_i=0\]

Now, the force \mathbf{F}_i can be from two sources – it could be an externally applied force or it could be a force of constraint (the force that does not allow the book to move “into” the table). Therefore, we have,

    \[\mathbf{F}_i=\mathbf{F}_{i(ext)}+\mathbf{F}_{i(const)}\]

And the total virtual work done will be (for this equilibrium case),

    \[\sum_i (\mathbf{F}_{i(ext)}+\mathbf{F}_{i(const)}).\delta\mathbf{r}_i=0\]

Or,

    \[\sum_i \mathbf{F}_{i(ext)}.\delta\mathbf{r}_i+ \sum_i\mathbf{F}_{i(const)}.\delta\mathbf{r}_i=0\]

Avoiding the constraints

We very well know that work done is zero when the applied force is perpendicular to the direction of displacement. So, we could get rid of the second term above, by demanding that the forces of constraint be always perpendicular to the direction of the virtual displacement (i.e., we are dealing only with cases that satisfy this condition). So, forget frictional forces. Also, we could be dealing with constraints that are changing with time. Imagine an ant on a balloon into which air is being blown constantly, causing it to swell. The constraint for the motion of the ant here, is changing with time. But, when we started speaking of virtual displacements, we froze time, didn’t we? And so, we are speaking of the motion of the ant when the balloon is at a specific stage during its expansion. And in such a scenario, the force of constraint (the force that doesn’t allow the ant to simply diffuse into the balloon) will always be perpendicular to the (virtual) displacement of the ant.

With the contribution from the constraint forces gone, the total virtual work will be,

    \[\sum_i \mathbf{F}_{i(ext)}.\delta\mathbf{r}_i=0\]

Looking at this equation, it might seem like, \mathbf{F}_{i(ext)}=0. But no. From the physical perspective, \mathbf{F}_{i(ext)} is the external force experienced by the ith particle. When we began this discussion, we only demanded that the total force on each particle be zero, not that the external force (which is only a part of the total force) be zero. Even though we have arranged for the contribution of the constraint forces to the virtual work, to vanish, we still do have the constraint forces. And these in fact connect the coordinates \delta\mathbf{r}_i.

For e.g., if say our system had just two particles, the virtual work of the system would be (under equilibrium),

    \[\mathbf{F}_{1(ext)}.\delta\mathbf{r}_1+\mathbf{F}_{2(ext)}.\delta\mathbf{r}_2=0\]

But the coordinates \mathbf{r}_1 and \mathbf{r}_2 could be connected via the constraint equations. Say, we had the relation \mathbf{r}_1=2\mathbf{r}_2 as the constraint, then \delta\mathbf{r}_1=2\delta\mathbf{r}_2, and we would have

    \[2\mathbf{F}_{1(ext)}.\delta\mathbf{r}_2+\mathbf{F}_{2(ext)}.\delta\mathbf{r}_2=0\]

Or,

    \[(2\mathbf{F}_{1(ext)}+\mathbf{F}_{2(ext)}).\delta\mathbf{r}_2=0\]

Since \delta\mathbf{r}_2 was chosen arbitrarily, this would lead to the relation,

    \[2\mathbf{F}_{1(ext)}=-\mathbf{F}_{2(ext)}\]

As long as this relation holds, it is not necessary that \mathbf{F}_{1(ext)} and \mathbf{F}_{2(ext)} be zero. More generally, as we stated before, \mathbf{F}_{i(ext)}\neq 0, in general, because the \delta\mathbf{r}_i are not independent and are in fact connected via the constraint equations.

“Creating” equilibrium

Newton said, force is the rate of change of momentum. If you observe an object’s momentum change, there’s always a force behind it. (Note that we are now releasing ourselves from the clutches of equilibrium.) One of the most celebrated equations in Physics,

    \[\mathbf{F}=\mathbf{\dot{p}}\]

We could rephrase this statement as: if \mathbf{F} is the force that causes a change in momentum of the object, then we would require a force of magnitude \mathbf{\dot{p}} in the opposite direction, to keep the object in equilibrium. So, instead of imagining an object changing its momentum under the influence of a force, we could as well imagine the object at equilibrium, under the influence of two equal and opposite forces \mathbf{F} and \mathbf{\dot{p}}. So, our net force here is \mathbf{F}-\mathbf{\dot{p}}.

We can therefore write, for the virtual work done, under this balanced force,

    \[(\mathbf{F}-\mathbf{\dot{p}}).\delta \mathbf{r} =0\]

Or, speaking of a system of particles, the net virtual work done on the system will be,

    \[\sum_i(\mathbf{F}_i-\mathbf{\dot{p}}_i).\delta \mathbf{r}_i =0\]

Again, the old argument: the force here is \mathbf{F}_i=\mathbf{F}_{i(ext)}+\mathbf{F}_{i(const)}.
And so,

    \[\sum_i(\mathbf{F}_{i(ext)}-\mathbf{\dot{p}}_i).\delta \mathbf{r}_i +\sum_i \mathbf{F}_{i(const)}.\delta \mathbf{r}_i =0\]

Demanding again that the constraint forces be perpendicular to the virtual displacements, we are left with,

    \[\sum_i(\mathbf{F}_{i(ext)}-\mathbf{\dot{p}}_i).\delta \mathbf{r}_i =0\]

This is the famous D’Alembert’s principle. It gives us the condition under which a given system will remain at equilibrium.
In our next article, we will extend this principle into generalized coordinates and introduce concepts such the generalized force, which will ultimately help us understand motions of objects.

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Roshan Sawhil is a Physics postgraduate who rejoices both doing and explaining Physics. He also finds doing Philosophy as a leisure activity quite interesting. You can find and connect with him on Facebook and Twitter.

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