Maximum weight a bar and cables can hold before breaking

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Homework Help Overview

The problem involves determining the maximum weight W that can be supported by a bar and cables in a three-dimensional space, specifically at point A in the xy-plane. The setup includes a bar AO and cables AB, AC, and AD, with given strengths for each member. The original poster attempts to analyze the forces acting on the structure using free body diagrams and equilibrium equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of calculating loads in each structural member in terms of W using equilibrium equations. There is a focus on understanding how the applied load W affects the tension in the cables and compression in the bar. Some participants question the interpretation of forces and the application of static equilibrium equations.

Discussion Status

The discussion is ongoing, with participants providing guidance on using equilibrium equations to find the forces in the structure. There is a recognition that the original poster may be misinterpreting the forces and that further clarification is needed on how to relate the loads to the applied weight W. Multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the original poster has made a typographical error regarding the coordinates of point A, which may affect calculations. There is also mention of the need to consider the allowable strengths of the structural members as part of the analysis.

  • #31
Alison A. said:
So I am finding all the forces from the moments first then applying those to the force equations?
No, you use the one moment equation I indicated,

ΣMz = FAB(12) - FAC(18.46) = 0 to find FAB in terms of FAC

Then, you can substitute FAB back into the force equations to find the other forces in terms of W.
 
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  • #32
SteamKing said:
No, you use the one moment equation I indicated,

ΣMz = FAB(12) - FAC(18.46) = 0 to find FAB in terms of FAC

Then, you can substitute FAB back into the force equations to find the other forces in terms of W.
Alright I found FAB to be (4615/5539)W and FAC to be (3000/5539)W
 
  • #33
Alison A. said:
Alright I found FAB to be (4615/5539)W and FAC to be (3000/5539)W

And from there force equations I found FOA to be -0.6983W
 
  • #34
WOOOOOOOOOOOOOO:oldsurprised: I got the right answer. Wow that was a lot of work, I think I've gone through about 10 pages of paper. :bow:Thank you so much for sticking with me even when it seemed like I couldn't grasp the most simple concepts . Could you help me find the last part to my other problem you've been answering? That is my last problem... then I'm finally done.
 

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