- #1
Kashmir
- 466
- 74
- Homework Statement
- Found it on internet, couldn't understand why it works
- Relevant Equations
- $Sigma T. x=0$
Why does ##Sigma T. x=0## hold for tension
- Thread starter Kashmir
- Start date
-
- Tags
- Equilibrium Physics Tension
- #2
- 42,753
- 10,475
Consider the work done on the (ideal) pulleys.
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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