Equilibrium sum of forces question

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Homework Help Overview

The discussion revolves around an equilibrium problem involving forces, distances, and weights. The original poster seeks validation of their approach and expresses uncertainty about the simplicity of their solution.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the equilibrium by summing forces in both the X and Y directions and mentions a midpoint calculation. Some participants question the correctness of the original poster's approach, specifically regarding the tension in the rope and its effect on the forces acting on a sliding collar.

Discussion Status

Participants are actively engaging with the problem, with one providing a hint about considering the tension in the rope. The original poster acknowledges the feedback and indicates a willingness to revisit their calculations.

Contextual Notes

The original poster expresses a concern about overcomplicating the problem and mentions difficulties in finding a starting point for their solution. There is an indication of a need for clarification on the setup and assumptions related to the forces involved.

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Equilibrium question with distance and weight

Homework Statement


Would someone be able to tell me if my attempt at this is correct. It seems too easy. If it is not could you help?


Homework Equations


Sum of the forces in the X direction
Sum of the forces in the Y direction
Mid point = DF

The Attempt at a Solution


attached
 

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Last edited:
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I am just seeing if i am on the right path. A simple yes or no would be enough.
Thank you
 
No, it's not quite right. Hint: What's the tension in the rope? Then you can sum the forces on the sliding collar.
 
Thanks Doc Al
I will look at it again when i get home from work.
 
I was going about this the wrong way, making it too complicated.
1st what I did was get the angle where the rope attached to collar A and the rod, γ = cos-1 (40/50), γ=36.87 ; the to find x I just used tan36.87= 20/x  x = 20/tan36.87 so the result is x = 26.67in.
I find that I over think a problem too much but I guess that is because I find it hard to find a starting point.

Thanks
 

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