Force needed to raise a backpack

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The discussion highlights a misunderstanding regarding the diagram depicting a hiker pulling a backpack with a force at an angle to the vertical, which is labeled incorrectly as a vertical force. It questions the relationship between the force exerted by the hiker on the rope and the tension in the lateral section of the rope. Additionally, it points out the oversight of not considering the contribution from the right half of the rope in the analysis. Clarifying these points is essential for accurately understanding the mechanics involved in raising the backpack. Overall, the conversation emphasizes the importance of correctly interpreting forces and tensions in such scenarios.
syllll_213
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The question is asking how the force needed to raise the pack change, but I am unsure if my working and reasoning is right. Particularly, I am unsure if I can assume both the angles (theta) I labelled to be equivalent because I sort of assumed the force pulled by the guy to be equal to the force pulling on the backpack. Some comment on my working would be appreciated thank you TT.
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F=ma
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The diagram provided seems to have misled you a little. It shows the hiker pulling down at some angle to the vertical but labels a vertical force ##\vec F##.
How does the force with which the hiker pulls on the rope relate to the tension in the lateral section of the rope?

Also, you seem to have forgotten the contribution from the right half of the rope.
 
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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