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Say a beam, 3m long from end to end, has a pivot placed at its centre of mass, at half its length i.e. 1.5m. Objects with weight 200N each are placed on both ends of the beam. The system is in equilibrium and thus no turning effect.

Now a force of 30N is applied on one of the ends until the beam turns 15deg to the horizontal. When this force is removed, the moments on both sides are balanced. But, the beam turns, until it becomes horizontal?! If resultant moment is zero, how can the object turn?

Is the moment really not zero, or is there some potential energy transferred to the beam which converts when the force is released?

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  • $\begingroup$ How are the objects "placed" on the beam? The center of mass of the beam plus the objects is not necessarily at the pivot point of the beam. $\endgroup$
    – alephzero
    May 13, 2021 at 18:53
  • $\begingroup$ You need to include the speed of the 30N force is applied, from which, you can get the impact force. During an impact, an object's energy is converted into work (kinetic energy, KE = 0.5 × m × v^2). If the 30N force is applied slowly, the beam will not be tilted farther than it was depressed to with the load, and will regain its balance mainly through the difference in potential energy and friction. of the pivot $\endgroup$
    – r13
    May 13, 2021 at 18:53
  • $\begingroup$ Even if objects are placed perfectly on centerline, any bending of the beam under weight will move CM downward, when supported at the point that was CM when it was on the ground. CM moving down stabilizes it (causing it to return to horizontal). $\endgroup$
    – Pete W
    May 13, 2021 at 19:18
  • $\begingroup$ to me what you are describing (assuming you are using an beam with finite thickness, supported at the middle from the bottom side, and using a weight of finite dimension placed on top of the beam) does not hold true. I mean I would not expect the beam to return upon removing the 30N weight. I would expect it to continue rotating until it reached the vertical position, both because of the angular momentum and due to the shift of the CM's of all masses involved in the problem. So maybe, you need to provide a sketch with more clarification, and also describe with more precision the setup. $\endgroup$
    – NMech ♦
    May 13, 2021 at 19:53
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    $\begingroup$ @r13 IF the OPs question says the force = 30N, you can not also control the speed (or displacement) of the point where the force is applied. Newton's 2nd law says that either the force depends on the velocity or the velocity depends on the force - they can't be independent of each other. $\endgroup$
    – alephzero
    May 13, 2021 at 22:50

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