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

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A system returns to its original position after a force is removed due to the balance of moments, despite the temporary application of force. When the force is applied to angle the system, the distribution of forces changes, leading to different compression and tension on either side of the bar. This distortion affects the equilibrium, causing the system to oscillate before returning to its original state. Ignoring the weight of the bar simplifies the analysis, but in reality, the weight can influence the system's behavior. Understanding these dynamics clarifies why the system stabilizes after oscillation.
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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.
 
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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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