# Mechanics Goldstein doubt

1. May 2, 2008

### pardesi

in the "derivation" of 'principle of vitual work' Goldstein started with this
he wrote the force F acting on each particle as the sum of $$F_{a}$$ (Force applied) and $$F_{i}$$(Constraint force)
ok i have some three queries
1)What are constraint forces?
2)Are all internal forces constraint forces
3)He said the cond. of equilibriia of a system is equivalent to
$$\sum \vec{F_{a,i}} .d \vec{r_{i}} =0$$
that is the virtual work is $$0$$ i don't get how that happens(because virtual work being 0 doesn't imply force on each particle of the system is $$0$$ or is it just by definition?

Last edited: May 2, 2008
2. May 2, 2008

### siddharth

The constraint forces are the forces which restrict the particle to move in some way. One example is the normal force, when two surfaces touch. For a rigid body, if you look at the forces on a single constituent particle, then the internal forces are indeed constraint forces.

What Goldstein says is that, if $F_i$ is the total force on a particle, and for a virtual displacement $\delta r_i$, the product $F_i . \delta r_i$ is 0 since each particle is at equilibrium, and so the sum over all particles $\sum_i \bf{F}_i . \delta \bf{r}_i=0$.

If you write $F_i$ as the sum of an applied force $F_i^a$ and a constraint force $f_i$, and only consider systems where $\sum_i f_i.\delta r_i$ is assumed to be 0, the condition for equilibrium is now, $\sum F_i^a . \delta r_i=0$

This assumption holds for many cases. For example, in rigid bodies where the internal forces are constraint forces, then because of Newton's third law, the work due to the constraint forces is 0. Similarly, when the normal force from a surface is the constraint force, the work done is again 0, because the constraint force will always be perpendicular to the virtual displacement.

Last edited: May 2, 2008
3. May 2, 2008

### pardesi

i was clear about the rigid body part but does the notion of constarint force hold good anywhere else?
also about the principle of virtual work
it's clear that if the body is in equilibrium with constarint foprces doing no work then we have
$$\sum \vec{F_{i,a}}.\vec{dr_{i}}=0$$ but $$\sum \vec{F_{i,a}}.\vec{dr_{i}}=0$$ does not imply that the body has to be at equilibrium

4. May 2, 2008

### siddharth

For systems with ideal constraint forces, where $\sum f_i.\delta r_i$ is assumed to be 0, then if $\sum_i \bf{F}_i^a . \delta \bf{r}_i=0$ for any arbitrary displacement $\delta r_i$, it implies the system is in equilibrium. Just reverse the sequence of arguments.

Last edited: May 2, 2008
5. May 3, 2008

### pardesi

oh ok i think i got the point ....the thing is for any disp