How to Balance a Seesaw with Three Children?

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Homework Help Overview

The problem involves balancing a seesaw with three children of different weights: 150N, 250N, and 180N. The seesaw is 4.0m long and pivoted at its center, prompting a discussion on how to position the third child to achieve balance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the concept of torque and its application to balance the seesaw, questioning how to assign clockwise and counterclockwise torques. There are attempts to set up equations based on torque calculations and considerations of distances from the pivot point.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning their understanding of torque directionality. Some guidance has been provided regarding the need for consistent sign assignment in torque calculations, and there is a recognition of the need to visualize the problem through diagrams.

Contextual Notes

There is confusion regarding the assignment of torque signs and the interpretation of units used in the calculations. Participants express uncertainty about the placement of the third child and its implications for balancing the seesaw.

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A 150N child and a 250N child sit at opposite ends of a 4.0m seesaw pivoted at its center. Where should the third child whose weight is 180N sit in order to balance the seesaw?

Do you need to solve this problem or answer it where the 3rd child will be put
 
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Please show what you've tried.
What equation(s) do you think you're supposed to use?
 
TORQUE = LENGTH * FWboy
is it okay to assign clockwise and cclockwise to every number?
clockwise
TORQUE = 2.0m * 150N
TORQUEboy1 = 300 N * m
counterclockwise
TORQUEboy2 = 2.0m * 250N
TORQUEboy2 = 500N
waaaa I am so dumb
what now
 
Total torque must equal 0 to keep the seesaw from rotating.
You are using the correct formula, can you figure out what the "lenght"(/distance) should be for the 3rd boy for the torque to equal 0?
 
oh

Torqueboy1 + Torqueboy2 + Torqueboy3 = 0
300N.m + 500 N * m + 180 N * M<-(unknown) = 0

i know how but I've got a problem with Clockwise and C.Clockwise where can i assign them...

plss helpp...
thx
 
Last edited:
You should draw a diagram, and think where the seesaw will rotate if only one boy is placed on eg. the left.
You can choose the sign (+/-) however you like, but all torque causing clockwise rotation must have the same sign and it must be opposite to the ones causing counter-clockwise rotation.

I'm extremely confused about your markings, what are N.m, N * M, <-(unknown) ?
 
oh its the Length of the third boy * Its Normal Force..
weighing 180N to balance the seesaw..

so i was just wondering if the third boy can be assigned at the CC or C rotation...?.. icant assign it as positive or negative
helpp..

thx again
 
Which way would the seesaw rotate if the 3rd boy wasnt placed there? To balance this rotation out, where should the 3rd boy be placed?
 
hmm......
3rd boy should be with the boy weighing 150N because it's lighter
 
Last edited:
  • #10
Yes, correct, sort of. It's not because of the lighter weight, but because of the smaller torque.
 
Last edited:
  • #11
yup i get it now thanks :)
 

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