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A child is sitting at one end of a seesaw, at a distance of $2.12 \mathrm{~m}$ from the fulcrum. The fulcrum is at the middle of the seesaw. The weight of the child is $170 \mathrm{~N}$. His dad is sitting at the other end of the seesaw at a distance of $1.56 \mathrm{~m}$ from the fulcrum.
- Which force perpendicular to the seesaw should the dad exert to keep the seesaw in equilibrium?

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A child is sitting at one end of a seesaw, at a distance of 2,12 m from the fulcrum. The fulcrum is at the middle of the seesaw. The weight of the child is 170 N. His dad is sitting at the other end of the seesaw at a distance of 1,56 m from the fulcrum.

- Which force F perpendicular to the seesaw should  dad exert to keep the seesaw in equilibrium?

170*2,12 = F*1,56 

F = 170*2,12/156 = 231 N

 



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170·2.12 = Ρ·1.56

Ρ = 9010/39-----> Ρ = 231 N (circa)

(equilibrio alla rotazione attorno al fulcro)

@lucianop 👌👍👍



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