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I am having trouble understanding the kinds of virtual displacements which are permitted for a given constrained system. I have a specific example in mind:

A block of wood resting on a table parallel to the ground.

In this case, the block is at rest (forces balanced). From all examples I've seen, the virtual displacements of the block are taken to be along the surface of the table. Since the normal reaction is normal to any displacement on the surface of the table, the work done by the constraint force (normal reaction of the table) is zero.

But why don't we take virtual displacements to be such that the block leaves the surface of the table. This is in accordance to the constraint, as the constraint is simply that the block shouldn't be lower in height than the surface of the table.

And if we do so, the condition that the constraint forces doing no work would be violated.

UPDATE: Uldreth has pointed out the error in my reasoning by pointing out that the force of constraint would be zero as the block leaves the surface. The original question has been answered, and now I have a follow-up question:

Even though the block is in equilibrium, and the normal force (constraint force) is zero if the block takes off the table, the work done on the block if the virtual displacement is off the table is non-zero. This stands in contrast to d'Alembert's reasoning that the work done on an object in equilibrium over a virtual displacement is zero!!

What is the reason that the virtual displacements wouldn't include any displacement where the block leaves the table's surface?

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Usually in this case, the constrait is assumed to be the surface of the table itself - eg. the block is not allowed to leave the surface of the table anyhow. Or more mathematically speaking, the configuration space is restricted to the submanifold that represents the table's surface.

Of course this is an idealization (even if we don't account for the block leaving upwards, it could break the table).

If the situation requires the constrait to be only as such that the block should not be lower than the table, then I am not sure, but I assume that in this case, if you displace the block upwards, the constrait forces will do no work. This is because if $\mathbf{F}=\alpha\mathbf{n}$ is the constrait force, then if you take a curve $\gamma:\mathbb{R}\rightarrow\mathbb{R}^3$ that is not within the surface, but $\gamma(0)$ is on the surface, then for $$W=\int_0^{t_1}\langle\mathbf{F}\circ\gamma(t),d\mathbf{l}\rangle$$ $\mathbf{F}$ would only be nonzero at $t=0$, since the constrait force is zero off the surface - which means the integral will be zero, because the force is only nonzero on a set of measure zero.

Otherwise, in usual circumstances, eg. when the block is totally constraited to a lower dimensional submanifold, you can think of virtual displacements being only different from real displacements, if the constrait also has a time-evolution. In that case the set of all virtual displacement is the set of all displacements your block can make if you "freeze" the time evolution of the constrait. This also applies though, in the case when your block is allowed to leave the table, but then, as I said, the constrait force is zero off the table.

EDIT:

A follow-up question: I understand that the constraint force of the table always does zero work, but what about the applied force of gravity? If we move off the surface, the force of gravity does non-zero work on the block. Thus, despite all the forces being balanced, non-zero work will be done on the object if the virtual displacements are anything off the table. This is in contradiction to D'Alembert's principle.

Remember that this works for constrait forces and virtual displacements. If you move the block off the table then then constrait forces are once again zero, and they still don't do work.

This says nothing against nonconstrait forces. If, instead of a table parallel with the ground, you take an incline with angle of inclination $\varphi$ with the ground, then the downward pointing gravitational force, $\mathbf{F}_G$, will admit decomposition to normal and tangential parts. The normal parts will be cancelled by the constrait force, but the tangential part will not be cancelled, and the object will accelerate on the constrait surface. However in this case, the constrait force will be orthogonal to the trajectory, so it still does no work. But the force of gravity does work.

Since you seem to struggle with the concept (which is absolutely no problem, this is a quite abstract concept!), let me illustrate this whole principle of virtual work by several examples.

Difference between virtual and real displacements arise for example, if your incline is accelerating. Let us take a snapshot of the system with the accelerating incline at $t=0$, but let us also draw the total trajectory of the particle. You will see that at $t=0$, but with the trajectory drawn, the curve will not be tangential to the incline, since the incline accelerates. A real displacement is along the trajectory. A virtual displacement is along the incline, as the incline is frozen in time. The real displacement is not tangential to the incline, so the contrait force actually does positive work, but no virtual work is being done.

Here is an ultra crappy gimp drawing I made in 2 mins:

enter image description here

The red line is the trajectory, $dr$ is the real displacement, $\delta r$ is the virtual displacement.

Also, the principle of virtual work does not say that the constrait forces cannot do work for virtual displacements, but that they cannot do positive work. If you use a nongeometrical constrait, for example a stationary incline with friction added, then friction is also a constrait, however it is velocity-dependent. In this case the normal force and the friction together will point backwards, and will impede the acceleration of the object down the incline. Since the incline is stationary, all displacements are virtual, yet the constrait force will do negative work on the object. This is fine, as they only cannot do positive work.

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  • $\begingroup$ Oh I see. I was assuming that when you take virtual displacements, you freeze time. But I was also freezing the forces on the object, instead of just freezing the force field, which in this case would be zero off the surface of the table. $\endgroup$ – Sidd Sep 2 '15 at 19:58
  • $\begingroup$ A follow-up question: I understand that the constraint force of the table always does zero work, but what about the applied force of gravity? If we move off the surface, the force of gravity does non-zero work on the block. Thus, despite all the forces being balanced, non-zero work will be done on the object if the virtual displacements are anything off the table. This is in contradiction to D'Alembert's principle. $\endgroup$ – Sidd Sep 2 '15 at 23:09
  • $\begingroup$ @Sidd Edit added! $\endgroup$ – Bence Racskó Sep 3 '15 at 6:12
  • $\begingroup$ The distinction between virtual and real displacements was very well made. But d'alembert's principle states that virtual work of non-constraint force is zero for an object at rest. As for the incline, the object has imbalanced forces on it, hence the virtual work done by non-constraint forces (gravity) would be non-zero. And thus my problem still persists. For the block on the table, the virtual work of gravity should be zero. But that essentially limits virtual displacements to only the plane of the table. $\endgroup$ – Sidd Sep 3 '15 at 18:47

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