Dynamics (mechanics) and virtual work

[luis20]

Hi everyone, I'm studying applied dynamics for college.

I have a simple question and then a more importante question:

F=dq/dt=ma
(torque)T=dL/dt

What do you call these equations, Newton-euler equations?
I use these equations all the time to relate forces and torques with motion, in certain mechanisms.

But suddenly, the principle of virtual work appears.
I think that virtual work is nothing more than a deduction of the previous equations, where we do F-ma=0 and T-dL/dt=0 and multiply by any displacement, since 0*displacement=0.
I heard that virtual work is used a lot! But I can't get it, what does virtual work bring new?

Another question of curiosity, virtual work was used at the beggining for static or dynamic?

Thanks a lot for the answers

 

[Bill_K]

Luis, I agree with you, the discussion given in most mechanics texts makes a mystery out of virtual work. It's usually skimmed over quickly, and it seems so obvious, why would anyone make a deal about it? Because in addition to being useful in its own right, it's a stepping stone on the path from Newton's laws of motion to Lagrange's equations.

The principle of virtual work is basically a restatement of static equilibrium. We know the condition is that the resultant force on each particle is zero: R_i = 0. So first we just add them all together and get this: δW = ∑ R_i·δr_i = 0 where δr_i are arbitrary virtual displacements. As you point out, this is trivial, adding up a bunch of zeroes to get zero.

Things get more interesting however if the system is subject to constraints, because then the displacements δr_i are not all independent. What to do in that case? Answer: we split the forces into applied forces F^a and constraint forces F^c: R_i = F^a_i + F^c_i. If we only allow virtual displacements that are compatible with the constraints, the constraining forces can't do any work: ∑ F^c_i·δr_i = 0. What we are left with is a relationship that involves just the applied forces, ∑ F^a_i·δr_i = 0. This is the Principle of Virtual Work.

Still looks almost the same, so why is it important? Because it's the way to state the equilibrium condition for a system with constraints. Since the δr_i are not independent, it is no longer true that the forces F^a_i are separately zero.

 

[luis20]

Thanks for the explanation!

I didn't understand your last paragraph. What you mean with equilibrium condition?

Anyway, everything we can conclude with virtual work, we can also conclude with Newton's laws right? So it just may be easier to use virtual work if, as you said, the virtual displacement is compatible with some of the constraint forces.

But friction, if it is a constraint force, does work in some situations...

 

[Bill_K]

everything we can conclude with virtual work, we can also conclude with Newton's laws right?

Doing it with Newton's Laws requires you to calculate all the forces including the constraint forces. Whereas doing it with virtual work you only need to deal with the external forces, so you've reduced the size of the problem.

But friction, if it is a constraint force, does work in some situations...

Yes, this method only works for rigid constraints.

I didn't understand your last paragraph. What you mean with equilibrium condition?

Virtual work is an expression of static equilibrium - nothing moves.

 

[luis20]

Doing it with Newton's Laws requires you to calculate all the forces including the constraint forces. Whereas doing it with virtual work you only need to deal with the external forces, so you've reduced the size of the problem.

Right. Yet, we can apply virtual displacements not compatible with some of constraint forces, so they will "do work" in that case right, even though they are rigid constraints?

Like a block attached to the floor and I apply a δy, so all goes up (in this case wouldn't be useful :)