Understanding Constraint Forces in Mechanics

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Constraint forces are those that restrict a particle's movement, such as the normal force between surfaces. Not all internal forces are constraint forces, but in rigid bodies, internal forces typically act as constraint forces. The principle of virtual work states that if a system is in equilibrium, the sum of the applied forces multiplied by their virtual displacements equals zero, which does not imply that all forces on each particle are zero. For systems with ideal constraint forces, if the sum of applied forces equals zero for any arbitrary displacement, it indicates the system is in equilibrium. Understanding these concepts is crucial for analyzing mechanical systems effectively.
pardesi
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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?
 
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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.
 
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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
 
pardesi said:
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

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