Method to calculate work while moving two walls

[Mary2100]

Why would they consume any energy?

Work = force * displacement

When everything is stationary there is no displacement going on so no work is being done.

When things are moving the net force is zero (see above) so again no work is done.

I'm assuming there is no friction and the motors are "ideal".

Nothing move ? Look at that image:

At start, the position is 1, at final it is 2. The motor 1 take the black wall. The motor 2 take the pink wall. The motor 3 take the green wall. The black arm turns, no ? the green wall moves, no ? for me the motors 1 & 2 recover an energy. The motor 3 needs the energy recovered by the motors 1 & 2.

 

[sophiecentaur]

@Mary:
This argument seems to be along the same lines as the arguments people try to use to prove that they have invented a Perpetual Motion Machine. You have made a highly complicated experiment and done some maths which seems to produce a strange result.
1. No net work is done
2. Some net work is done
?
It's all lost in the fog of your over complicated model. Do you think that, somehow, a complicated enough model will produce a result that proves the gas laws are wrong? Personally, I would always look for a (possibly subtle) error in the model if it yields a result that goes against conventional Science.

 

[CWatters]

Nothing move ? Look at that image:

At start, the position is 1, at final it is 2. The motor 1 take the black wall. The motor 2 take the pink wall. The motor 3 take the green wall. The black arm turns, no ? the green wall moves, no ? for me the motors 1 & 2 recover an energy. The motor 3 needs the energy recovered by the motors 1 & 2.

Yawn. Go back and read my post again. I described two situations... one while it's stationary and one while it's moving.

I think I'm just about done with this thread. All the information you need to understand the problem has been well covered.

I'll just repeat.. In in order to work out the energy required to move any of the walls you need to consider all the forces acting on the wall. You haven't done that.

 

[berkeman]

This thread will be closed unless the next post by the OP is extremely lucid and helps us to clear up your confusion. Fair warning.