Forces involved in principle of virtual work

[phantomvommand]

Homework Statement

See picture below

Relevant Equations

Work done = Fd

The answer is as such: There’s only one way for the system to move: the rectangle can deform into a parallelogramso that the left horizontal arm moves up, and the right horizontal arm moves down by thesame amount. Then the total virtual work done on the scale by the weights is zero, so thesystem can be in equilibrium no matter where on the arms the weights are placed.

While I can understand this, this assumes that there is 0 work done by reaction forces at the joints, so the net work done on the system is entirely due to gravity. How fair is this assumption?

 

[TSny]

The forces in the joints are internal forces of the system. At a joint where two bars meet, the forces that the two bars exert on each other are action-reaction forces.

 

[phantomvommand]

The forces in the joints are internal forces of the system. At a joint where two bars meet, the forces that the two bars exert on each other are action-reaction forces.

Oh right, forgot about that! Thanks!