Mass on a Spring semi-submerged in a liquid

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The discussion centers on the calculations involving a mass on a spring submerged in liquid, with a focus on the definitions of variables related to density and height. The participant expresses confusion over differing interpretations of the variable definitions but concludes that their approaches yield equivalent results. They suggest eliminating gravitational force from the equations to simplify the problem. Ultimately, the participant asserts that their final result depends solely on the spring constant, height, density, gravitational acceleration, and volume. The conversation highlights the importance of consistent variable definitions in physics problems.
MatinSAR
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Homework Statement
According to the figure, the metal cylinder with mass M and volume V is connected to a fixed point from above by a spring with spring constant k and it floats inside a liquid of density p so that half of its height is inside the liquid. What weight should we put on it so that two-thirds of the height is inside the liquid?
Relevant Equations
Fluid mechanics.
Hi , Can someone tell me wether my answer is corrrect or false ?
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I'm getting something different from you. The difference is I define ##d'## in terms of ##d## and ##h## differently. Actually, they work out equivalently. If you just eliminate ##Mg## via your first equation, and put ##L## in terms of ##d## and ##h## , I think we will get the same result.

My end result is purely a function of ##k,h,\rho,g,V##
 
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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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