Solve T(y)-T(y+dy)=ug(dy) | Easier Method?

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    Chain Tension
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Homework Help Overview

The discussion revolves around finding an expression for the tension T(y) at a point y along a vertically hanging chain, given the relationship T(y) - T(y+dy) = ug(dy), where u is the mass per unit length and g is the acceleration due to gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the mass of the chain below a given point and the implications of the tension difference equation. Some express uncertainty about how to proceed with solving the equation, while others suggest considering limits or integration as potential methods.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning how to approach the solution. Some guidance has been offered regarding the relationship between the change in tension and the mass of the chain, but no consensus has been reached on a specific method to solve the equation.

Contextual Notes

Participants note that the full problem statement is not provided, which may affect the clarity of the discussion. There is also mention of a force 'f' that supports the chain, though its role is not fully explored.

jayjackson
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Homework Statement
Find an expression for T(y) which is the tension at the point y along the chain.

Chain is length L, take the top of the chain hanging vertical to be 0, and the length from top to bottom be L. T(y) tension at point y down the chain. T(y+dy) tension at the point y+dy, which is below y. It has mass per unit length u. The section from T(y) down to T(y+dy) supports the weight ug(dy), which has mass u(dy) as g is just gravity. There is also a force 'f' supporting the chain from falling.

Using this information, find an expression for T(y).
Relevant Equations
T(y)-T(y+dy)=ug(dy)
T(y)-T(y+dy)=ug(dy) is what I have got. How would I solve this? Or is there a simpler method.
 
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jayjackson said:
Homework Statement:: Find an expression for T(y) which is the tension at the point y along the chain.

Chain is length L, take the top of the chain hanging vertical to be 0, and the length from top to bottom be L. T(y) tension at point y down the chain. T(y+dy) tension at the point y+dy, which is below y. It has mass per unit length u. The section from T(y) down to T(y+dy) supports the weight ug(dy), which has mass u(dy) as g is just gravity. There is also a force 'f' supporting the chain from falling.

Using this information, find an expression for T(y).
Relevant Equations:: T(y)-T(y+dy)=ug(dy)

T(y)-T(y+dy)=ug(dy) is what I have got. How would I solve this? Or is there a simpler method.
You haven’t provided the full problem statement, but I gather this just a chain hanging vertically.
The simplest way would be just to consider the mass of the chain below a given point. But from the equation you got you can easily take limits to get an integral.
 
haruspex said:
You haven’t provided the full problem statement, but I gather this just a chain hanging vertically.
The simplest way would be just to consider the mass of the chain below a given point. But from the equation you got you can easily take limits to get an integral.
im unsure on how to solve the T(y)-T(y+dy) equation. How do i go about it?
 
jayjackson said:
im unsure on how to solve the T(y)-T(y+dy) equation. How do i go about it?
T(y+dy)-T(y) is the change in T, which is written dT. So your equation is dT=-ug dy.
Do you know how to integrate?
 

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