How is Translation Equilibrium Achieved in a Torque Experiment?

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Translation equilibrium in a torque experiment is achieved when the net force acting on the system equals zero, which occurs alongside rotational equilibrium. In the lab, different masses are attached at varying distances from the pivot point on a meter stick, creating a balance of forces. Although the forces differ due to varying masses and gravitational acceleration, their moments about the pivot point counteract each other. This balance ensures that the system remains in a state of equilibrium. Understanding this concept involves recognizing that the distribution of forces and their respective distances from the pivot contribute to achieving both translational and rotational equilibrium.
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Homework Statement



We recently did a lab where we calculated various torque values for various systems. Basically, we attached mass by a hanger at different areas on a balanced/pivoted meter stick.

At the end of the lab, he asks the question "How is the condition for translation equilibrium Fnet = 0 satisfied in this experiment?

I think I understand why, but I can't really explain it in a comprehensible manner.

The Attempt at a Solution



I believe it is because we're establishing rotational equilibrium and at equilibrium any net force acting on the system is 0 (by definition). But.. how do I explain this? The forces are all different (considering Mass * Acc Due To Gravity), but they even out because they're placed at different distances from the pivot point to put the system into rotational equilibrium.
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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