Stress around Fulcrum - How to Justify Net Momentum?

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'roidbreaker
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"Stress" around fulcrum

helo,

I have a couple of forces described in f''(t), and the f'(t) yields a momentum vector plane like pic:

YTb1kO7h.png


in this picture, we have a solid square plate of some hard material, and an axis where the white rod is.

Since the sum of the vectors on each side equal 0, the square is not rotating. What justification can I give to add -f'(t) to the system and demonstrate that f''(t)+(-f''(t))=0, and from that that the forces in f''(t) neutralize themselves and have no effect as long as the material does not transition into deformation?

Can I just say that the net momentum around the axis is 0 (by adding it all up), and cite conservation of energy?

nMvQlAxh.png


thanks!
 
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Confusion: Is your square static or rotating ?
You are using the term; “Momentum” as applied to a moving mass; but do you mean “Moment” about an axis.
Force is not Energy unless something moves. Without movement here, conservation of energy is not applicable.
 
Baluncore said:
Confusion: Is your square static or rotating ?
You are using the term; “Momentum” as applied to a moving mass; but do you mean “Moment” about an axis.
Force is not Energy unless something moves. Without movement here, conservation of energy is not applicable.

discrete parts of the square would have various different "momentums" as indicated by the arrows, but the square as a whole isn't rotating, because the net momentum on both sides of the axis is zero.

actually forget the axis, the axis isn't doing anything and probably just confusing things.