Force transferred across a fulcrum

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The discussion focuses on calculating force across a fulcrum, specifically when tension is released on one side. The formula N_1m_1=N_2m_2 is introduced, where N represents force and m represents distance from the fulcrum, but it applies only to equilibrium scenarios. When tension is removed, the force on the opposite side does not transfer but results in downward movement instead. The conversation shifts towards catapults, emphasizing the importance of energy, momentum, and angular momentum in such non-equilibrium situations. Understanding these concepts is crucial for ensuring stability and safety in designs involving fulcrums.
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Ok, I am wondering what the formula is for calculating force across a fulcrum. If you were to have tension on one side of a fulcrum, and release that tension, how much force is transferred to the other side of the fulcrum?

Thanks,

Mitch Guzman
 
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N_1m_1=N_2m_2 Where everything subscripted 1 is on one side and everything 2 is on the other, N is force, and m is distance away from the fulcrum. Rearranging it produces N_1\frac{m_1}{m_2}=N_2. This only useful for equilibrium problems, i.e. the arm is balanced and not moving. Removing the force on the one side, will only allow the force of the other side to move down, i.e. no force is tranfered.
Not sure what you mean by your wording and the fact that your example is a non-equilibrium problem. That sounds more like a catapult, in which case you would be more interested in energy, momentum, and angular momentum of the arm so you know how massive to make your base so it doesn't jump up and crush someone or something important.
 
I suppose it would be more like a catapult actually. I suppose I'll have to search up energy and angular momentum then.

Thanks :)
 
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