Finding the Center of Mass of a Suspended Scaffold and Paint Cans System

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Homework Help Overview

The problem involves finding the center of mass of a system consisting of a horizontal scaffold and paint cans, suspended by two cables. The scaffold has a uniform mass and is subjected to forces from the paint cans and the tension in the cables.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of torque equations and the need for additional information, such as the tension in the left cable. There is a focus on the relationship between the forces and the center of mass location.

Discussion Status

Some participants have provided guidance on continuing with the current approach, while others have pointed out missing elements in the original poster's reasoning. Multiple interpretations of the problem setup are being explored, particularly regarding the distances involved.

Contextual Notes

There is a mention of the need to clarify the tension in the left cable and the importance of correctly identifying distances from the respective sides of the scaffold.

bravoman
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Homework Statement


A horizontal scaffold, of length 2.00 m and uniform mass 50.0 kg, is suspended from a building by two cables. The scaffold has dozens of paint cans stacked on it at various points. The total mass of the paint cans is 80.0 kg. The tension in the cable at the right is 830 N. How far horizontally from that cable is the center of mass of the system of paint cans?

12-32.gif


I can't seem to get the right answer, I believe I may be missing a force. I set the rotational axis to the left side

Homework Equations


Sigma F = 0

The Attempt at a Solution


I have tried to create a balance of torque equation:

(L/2)* MassScaffold * g + x*MassPaint*g - 830L = 0
 
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Hello Bravo, welcome to PF :)

Your relevant equation doesn't show up in your attempt at solution. What's the tension in the cable at the left ? When you know that, it's a piece of cake to establish where mg should be located (i.e. where the c.o.m. is) ! Then you still have to go to the c.o.m. of the paint cans only .
 
BvU said:
Your relevant equation doesn't show up in your attempt at solution.
I don't think you need it with the approach bravoman used. The attempt gives a linear equation which can be solved for x as everything else is known.
 
mfb is right and Bravo should continue on the path he (she?) started on. o:)
 
This should work. You didn't forget that x is the the distance from the left side, while the distance from the right side is needed for the answer?
 

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