Will moving closer to the fulcrum impact the balance of a see-saw?

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Moving closer to the fulcrum on a see-saw affects balance due to changes in torque. When both an adult and a child move in 0.25m toward the fulcrum, the torque produced by the child increases. This increase in torque results in the child's end of the see-saw lowering. For balance to be maintained, the torques on both sides must be equal, and in this scenario, the child's end will drop. Thus, the movement toward the fulcrum does impact the balance of the see-saw.
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Suppose that an adult and a child are balanced on a see-saw. Each person now moves in toward the fulcrum a distance of 0.25m. What effect will this have on the balance of the see-saw?

Would it even have an effect?? Or would the childs end go to the ground?
 
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The best way to answer this question is to imagine there are some numbers. In what relation do the torques produced by the adult and the child have to be for the see-saw to be balanced?
 
The torque has to be equal on both sides to balance the see saw.
 
After applying some numbers, I came to the conclusion that the torque that the child applying would increase, therefore, causing the childs end to lower.
 
bengaltiger14 said:
After applying some numbers, I came to the conclusion that the torque that the child applying would increase, therefore, causing the childs end to lower.

You got it right.
 
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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