Three people are balanced on a uniform seesaw - Torque

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Homework Help Overview

The problem involves three individuals balanced on a uniform seesaw, with a specific weight provided for the seesaw itself. The objective is to determine the mass of one individual and the normal force at the fulcrum, utilizing the principle of torque.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply a torque equation based on a previous example, questioning whether the weight of the fulcrum should be included in their calculations. Participants discuss the interpretation of the seesaw's weight and its effect on the balance.

Discussion Status

Participants are actively engaging in clarifying the setup of the problem, with some suggesting adjustments to the distances used in the torque calculations. There is acknowledgment of the seesaw's weight as a factor in the balance, and guidance has been provided regarding the correct interpretation of distances relative to the fulcrum.

Contextual Notes

There is a mention of a potential misunderstanding regarding the mass of the fulcrum, with clarification that it does not have mass and should be treated differently in the calculations. The discussion reflects a need for precise definitions of distances in relation to the fulcrum.

Gewitter_05
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Three people are balanced on a uniform seesaw -- Torque

Homework Statement


Three people are balanced on a uniform seesaw which has a weight of 150 N as shown in the figure.
Find the mass of person B and the normal force acting on the fulcrum.


Homework Equations


Tcw=Tccw


The Attempt at a Solution


There was a problem that involved mass and a fulcrum in my notes, it doesn't have a force or weight of the fulcrum, but I tried to set it up like the example to find the mass of person B

(mC)(5.4)=(mA)(0)+(mB)(1.2). I then solved for mass B. I am not sure if I have to also put the weight of the fulcrum in there or not.
 

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There is no mass of a fulcrum. The 150 N refer to the weight of the horizontal seesaw. You can just treat it like an additional person, sitting at the center of mass of the seesaw. The sketch shows this.
(mC)(5.4)=(mA)(0)+(mB)(1.2)
Those are not the distances to the fulcrum.
 
So would I change the distances by subtracting the number from the fulcrum. Instead of mB being 1.2, it would be 0.8, mC 3.4 and mA 2?
 
Right. And don't forget the weight of the seesaw.
 
Cool. Thanks!
 

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