Three bys trying to balance a seesaw

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SUMMARY

To balance a seesaw with three boys, the mass of each boy and their distances from the fulcrum must be considered. The seesaw is 3.52 meters long, with one boy weighing 53.0 kg and positioned at one end, and another weighing 32.8 kg at the opposite end. The third boy, weighing 23.0 kg, must position himself at a calculated distance from the center to achieve equilibrium. The balancing equation involves the moments created by each boy's weight and distance from the fulcrum.

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Three boys are trying to balance on a seesaw, which consists of a fulcrum rock, acting as a pivot at the center, and a very light board L = 3.52 m long, see figure below.

Two boys are already on either end. One has a mass of m1 = 53.0 kg, and the other a mass of m2 = 32.8 kg. How far from the center should the third boy, whose mass is m3 = 23.0 kg, place himself so as to balance the seesaw?

i know how to do it but with just to boys, but with three boys idont have the minimum idea... can anyone helps me?...
 
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Naldo6 said:
Three boys are trying to balance on a seesaw, which consists of a fulcrum rock, acting as a pivot at the center, and a very light board L = 3.52 m long, see figure below.

Two boys are already on either end. One has a mass of m1 = 53.0 kg, and the other a mass of m2 = 32.8 kg. How far from the center should the third boy, whose mass is m3 = 23.0 kg, place himself so as to balance the seesaw?

i know how to do it but with just to boys, but with three boys idont have the minimum idea... can anyone helps me?...

Solving the problem with two boys and three boys is done exactly the same way--there is just one more term in your equation. So try doing it the same way, and if you get stuck, please post your work so that we can help you.
 

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