Calculating Work Done on a Hanging Chain Using Mass Density and Length

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MidgetDwarf
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A 3.0m long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that 2.0 m of the chain remains on the top level and 1.0m hangs down vertically. At this point ( the 1.0m segment that is hanging) is sufficient to pull the entire change down. How much work is preformed on the chain by the force of gravity as the chain falls from the point where 2.0m remains on the scaffold to the point where the entire chain has left the floor?

Assume that the chain has linear weight density 18N/m.

My attempt.

I'm not sure if my method was correct but i got the answer.

Since i know the chain is hanging 1m from the vertical axis and at this point, the force drops it. I can ignore the 1 meter and just think of the 2m left on the scaffold as falling vertically.

I know that chain is stretched, so I can use Force of Spring equation.

Fs=kX, where k=18N/m. I assume K is the linear weight density because it is in the units of the K in the formula.

This step i am unsure of, I know I can do Centroid of Mass and go from there, however the Centroid of Mass is explained further in the book. I am trying to do this problem with solely the information in the Introduction to Work Problem.

I take the integral from 0 to 2, of kX. When i integrate i get W= 0.5k*4.
Therefore W=72J which is the answer.

Is the thinking correct? Or do I have to use mgh?
 
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MidgetDwarf said:
A 3.0m long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that 2.0 m of the chain remains on the top level and 1.0m hangs down vertically. At this point ( the 1.0m segment that is hanging) is sufficient to pull the entire change down. How much work is preformed on the chain by the force of gravity as the chain falls from the point where 2.0m remains on the scaffold to the point where the entire chain has left the floor?

Assume that the chain has linear weight density 18N/m.

My attempt.

I'm not sure if my method was correct but i got the answer.

Since i know the chain is hanging 1m from the vertical axis and at this point, the force drops it. I can ignore the 1 meter and just think of the 2m left on the scaffold as falling vertically.

I know that chain is stretched, so I can use Force of Spring equation.

Fs=kX, where k=18N/m. I assume K is the linear weight density because it is in the units of the K in the formula.

This step i am unsure of, I know I can do Centroid of Mass and go from there, however the Centroid of Mass is explained further in the book. I am trying to do this problem with solely the information in the Introduction to Work Problem.

I take the integral from 0 to 2, of kX. When i integrate i get W= 0.5k*4.
Therefore W=72J which is the answer.

Is the thinking correct? Or do I have to use mgh?

Not sure,why you are treating the chain like a spring .

You need to work with the Center of Mass .
 
Vibhor said:
Not sure,why you are treating the chain like a spring .

You need to work with the Center of Mass .

I am trying to avoid the center of mass, since the center of mass is shown 3 chapter later an I'm trying to solve this problem without using it, only stuff in the introduction to Work.

I'm thinking that letting Z=weight density. So the force of gravity will be denoted by F=zY. Then integrating from [0,2] ( I can do 1 to 3 but its the same).

I think this is now viable.
 
MidgetDwarf said:
I am trying to avoid the center of mass, since the center of mass is shown 3 chapter later an I'm trying to solve this problem without using it, only stuff in the introduction to Work.

Ok.

Consider the hanging part of the chain and an element 'dx' at a distance 'x' from the top .Find the work done on this elemental part by force of gravity . Integrate under proper limits .You will get the answer :smile:.

Please work with symbols so that your work is easy to understand. You can treat λ as mass density and 'L' as length of the chain.