Why Does a System Return to Its Original Position After Force is Removed?

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SUMMARY

The discussion centers on the behavior of a system returning to its original position after a force is applied and then removed. The user initially confuses momentum with moments, leading to a misunderstanding of equilibrium. The explanation clarifies that while the sum of moments may remain zero when a force is applied, real-world factors such as the non-uniformity of the bar and its rigidity affect the system's response. The distortion caused by compression and tension on either side of the bar disrupts the balance, resulting in oscillation before returning to equilibrium.

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There's something I struggle to understand.

In the picture I attached to this post , the first drawing is a system with the sum of momentum = 0 (equilibrium).

In the second drawing I increase temporarily the force on point A in order to get and angle of 20° compare to the last position and maintain the angle stable for a moment.

Why when I stop to force the angle of 20° does the system come back to the previous position (after some oscillation) since from my understading the sum of momentum is also at equilibrium?http://img199.imageshack.us/img199/7277/momentumg.th.jpg

For information I'm not a student

Thanks for your help
 
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A trivial point perhaps but I think you meant moments rather than momentum.Are you ignoring the weight of the bar itself or is it a non uniform bar where one of the marked forces is the weight?.Anyway, if you ignore the weight of the bar and assume it is perfectly rigid then when you push it down then the sum of the moments is still zero and it should, in theory, balance.However with a real, not perfectly rigid bar the distortion of both sides of the bar will be different to what it was previously, for example the force on the right side has a greater compression(squashing) component and the force on the left side left has a greater tension(stretching) component so these changes can upset the balance.
 
Last edited:
Dadface said:
A trivial point perhaps but I think you meant moments rather than momentum.Are you ignoring the weight of the bar itself or is it a non uniform bar where one of the marked forces is the weight?.Anyway, if you ignore the weight of the bar and assume it is perfectly rigid then when you push it down then the sum of the moments is still zero and it should, in theory, balance.However with a real, not perfectly rigid bar the distortion of both sides of the bar will be different to what it was previously, for example the force on the right side has a greater compression(squashing) component and the force on the left side left has a greater tension(stretching) component so these changes can upset the balance.

It is the moment and I was indeed ignoring the weight of the bar itself since it shouldn't have influence on the system equilibrium.

Thanks a lot for that explanation. it's very helpful.
 

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