Engineering Practice Problem for a Pin-Connected Frame

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To solve the engineering practice problem for a pin-connected frame, it is recommended to treat the structure as two separate entities due to the central pin. Participants emphasize the importance of creating Free Body Diagrams (FBDs) to visualize forces acting on each section. The discussion highlights the need to apply 2D equilibrium equations, specifically focusing on force and moment calculations. Clarifications on assumptions regarding force equilibrium are encouraged to ensure accuracy in solving the problem. Properly documenting the work is essential for receiving further assistance.
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Homework Statement
Stuck on solving this practice problem for the pin connected frame
Relevant Equations
2D Equilibrium, Fx, Fy, M
1651165244951.png

Have identified what i need to find, but not sure how to proceed since there is a pin in the centre, do I treat it as two separate structures?
 
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engineerintraining said:
Homework Statement:: Stuck on solving this practice problem for the pin connected frame
Relevant Equations:: 2D Equilibrium, Fx, Fy, M

View attachment 300720
Have identified what i need to find, but not sure how to proceed since there is a pin in the centre, do I treat it as two separate structures?
Welcome to PF.

Per the PF Rules, you need to show your work before we can help. So yes, start by treating it as two separate structures and show the FBDs please. Thank you.
 
1651166148374.png

this is what I have, then based on this i was thinking that i could assume this:
1651166198165.png

Does that look right?
 
Sure. That is the force equilibrium for the right part. What else can you deduce?
 
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Thread 'Chain falling out of a horizontal tube onto a table'

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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