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

In summary, the problem is to find an expression for T(y), which is the tension at a point y along a vertical chain of length L with mass per unit length u. The chain is supported by a force 'f' to prevent it from falling. Using the equation T(y)-T(y+dy)=ug(dy), the tension at any point can be found by taking the integral of -ug dy, where dy is the change in length and u is the mass per unit length. This can be solved using integration techniques.
  • #1
jayjackson
2
0
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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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 

1. What is the purpose of solving T(y)-T(y+dy)=ug(dy) using an easier method?

The purpose of using an easier method to solve T(y)-T(y+dy)=ug(dy) is to simplify the process and make it more efficient. This can save time and resources, making it a more practical approach for many scientists.

2. How is this method different from other methods used to solve T(y)-T(y+dy)=ug(dy)?

This method is different from other methods because it involves breaking down the equation into smaller, more manageable parts. This allows for easier manipulation and calculation, resulting in a quicker and more straightforward solution.

3. What are the key steps involved in solving T(y)-T(y+dy)=ug(dy) using this method?

The key steps involved in solving T(y)-T(y+dy)=ug(dy) using this method include identifying the variables and constants, simplifying the equation, and applying any necessary mathematical operations to isolate the desired variable.

4. Are there any limitations to using this method for solving T(y)-T(y+dy)=ug(dy)?

Like any method, there are limitations to using this approach for solving T(y)-T(y+dy)=ug(dy). It may not be suitable for more complex equations or may not yield accurate results in certain scenarios. It is essential to consider the specific equation and its variables before deciding on the best method to use.

5. How can this method be applied to real-world scientific problems?

This method can be applied to real-world scientific problems by breaking down complex equations into smaller, more manageable parts. This allows for easier analysis and understanding of the problem, leading to more efficient and accurate solutions. It can be particularly useful in fields such as physics, engineering, and chemistry.

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