I've only just begun reading through Goldstein's Classical Mechanics. This section is on D'Alembert's Principle and Lagrange's Equations. The following quote is from the beginning of the section.

Note that if a particle is constrained to a surface that is itself moving in time, the force of constraint is instantaneously perpendicular to the surface and the work during a virtual displacement is still zero, even though the work during an actual displacement in the time $dt$ does not necessarily vanish.

What does this mean? I've tried to think of it in terms of a bead moving on a wire, or a ball on a moving plane—but it isn't obvious to me why the "virtual work" done by the surface of constraint is zero, and the "real work"... isn't?

Consider a a ball on a plane that is moving up. The plane is surely doing work on the ball, fighting against gravity to accelerate it upwards. In what situation is the work done on the ball by the plane zero (outside of the situation where time is frozen, in which case there's nothing interesting).


1 Answer 1


First of all if the body is moving up the plane. I would consider the ball to be small . Then if there is no sliding friction which is one of the primary conditions of d alemberts principle!! We can say the constraint force to be only the one normal to the surface exerted by the surface on the ball. Which is perpendicular to every position of the ball. So virtual work as well as real work is 0.and for gravity! Don't feel upset it is a conservative force and will spend the energy to reduce the velocity as you go up. The virtual work would always be zero. (by the plane). If someone else is doing work that is his fancy ND that will be taken into account while noting the actual work done on the system . I hope it helps !! Thank you very much.


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