Tipping Point of Stationary Barbell

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The discussion revolves around a physics problem involving a uniform bar supported at two points, with varying masses across its sections. The bar, measuring 2200mm in length and weighing 20kg, has sections A and C of equal mass and length (445mm each), with a load placed in the middle of section C. Participants emphasize the need for the original poster to show their calculations and understanding of the problem to receive effective help. Clarifications are requested regarding the measurements of sections A and C, as well as adherence to homework guidelines. A visual representation of the setup is suggested to aid in the analysis.
vedant_krish
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New user has been reminded to always show their work when posting schoolwork questions
TL;DR Summary: Please Help. I need an answer for an Investigation

if a uniform bar of length 2200mm is supported evenly on 2 points (1 and 2) 1190mm apart. The bar is comprised of 3 sections (A, B and C) of varying masses. Section A and C have equal mass and volume and is comprised of the parts of the bar hanging before Point 1 and extending past Point 2 respectively. The length of Section A and C are both 445mm. The total mass of the bar is 20kg. A Load of width 67mm and x kilogram mass is placed in the middle of Section C. For what value x does the bar start tipping? (Answer in terms of Mass of Section A or C)

Mentor note: Moved to homework forum, so no template.
 
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Welcome, @vedant_krish !

What calculations have you done to reach that needed answer?
What are you investigating?

Also, where the length of sections A and C (445 mm) is measured from?

2200 mm - 1190 mm ≠ 2 x 445 mm
 
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This is a homework thread. According to our rules, to receive help, you need to show some credible effort towards answering the question(s). How about telling us what you do know and why you think about this problem?

Please read, understand and follow our homework guidelines, especially item 4, here
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Also a figure that accompanies this problem would be helpful.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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