What is the tension in a hanging chain problem?

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The discussion centers on calculating the tension in a hanging chain with a weight of 2.0 N/m over a length of 20 m. Participants explore how to find the tension at various points along the chain, specifically three-quarters of the way up, 1 m from the top, and 1 m from the bottom. Key calculations involve understanding the total weight of the chain (40 N) and applying Newton's laws and free body diagrams. Confusion arises regarding the correct answers for the tension at specific points, with participants debating the signs and units involved. Ultimately, the tension is confirmed to be 40 N at the top and 0 N at the bottom, with 38 N at 1 m from the top.
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Homework Statement



A 20 m length of chain weighing 2.0 N/m is hung vertically from one end on a hook.
Answer in Newtons

1.What is the tension three quarters of the way up?

2.What is the tension 1 m from the top?

3.What is the tension 1 m from the bottom?

Homework Equations



F = ma

w = mg

The Attempt at a Solution



Base on my imagine, it has 2 forces total. One is the weight of the chain which is mg (going down); and another one is Newton's law (going up) which is F = ma.

But I look at the problem, it given mg = 2 N/m => m = 2 / 9.8 = 0.2 (what's units here??) I'm confused. Moreover, they give the length of the chain which is 20m. What can I do with it? The length? I really don't know how to start!
 
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huybinhs said:

Homework Statement



A 20 m length of chain weighing 2.0 N/m is hung vertically from one end on a hook.
Answer in Newtons

1.What is the tension three quarters of the way up?

2.What is the tension 1 m from the top?

3.What is the tension 1 m from the bottom?

Homework Equations



F = ma

w = mg

The Attempt at a Solution



Base on my imagine, it has 2 forces total. One is the weight of the chain which is mg (going down); and another one is Newton's law (going up) which is F = ma.

But I look at the problem, it given mg = 2 N/m => m = 2 / 9.8 = 0.2 (what's units here??) I'm confused. Moreover, they give the length of the chain which is 20m. What can I do with it? The length? I really don't know how to start!
The chain is not moving. What is its acceleration?? Then, Free Body Diagrams are a must. The basic weight unit is the Newton, and the basic mass unit is the kg. Are you at all familiar with free body diagrams?
 
PhanthomJay said:
The chain is not moving. What is its acceleration?? Then, Free Body Diagrams are a must. The basic weight unit is the Newton, and the basic mass unit is the kg. Are you at all familiar with free body diagrams?

The chain is not moving => the velocity is constant => you mean a = 0 m/s^2 ?

I'm new at Physics so I'm still weak on it. I draw my diagram and came up with 2 forces.

One is mg (pointing down) and one is N (pointing up). Then how can I do next?
 
I suddently think of this:

Given 2N/m, so we have length 20m => F = 2 N/m * 20 m = 40 N. Is this correct?
 
huybinhs said:
I suddently think of this:

Given 2N/m, so we have length 20m => F = 2 N/m * 20 m = 40 N. Is this correct?
That's the entire weight of the 20 m chain. If the chain weighs 40N, what force does the ceiling exert on the chain, in order that the sum of all forces acting on the chain are 0 (yes, it is not accelerating, or even moving at constant velocity; it just hanging there). Finding the ceiling force is always a good start before you examine the tensions at other points along the chain. You should note that if the chain weighs 2 N/m, then for example the first 10 feet of the chain weighs (2)(10) = 20 N, etc. Think Free Body Diagrams.
 
PhanthomJay said:
That's the entire weight of the 20 m chain. If the chain weighs 40N, what force does the ceiling exert on the chain, in order that the sum of all forces acting on the chain are 0 (yes, it is not accelerating, or even moving at constant velocity; it just hanging there). Finding the ceiling force is always a good start before you examine the tensions at other points along the chain. You should note that if the chain weighs 2 N/m, then for example the first 10 feet of the chain weighs (2)(10) = 20 N, etc. Think Free Body Diagrams.

If the chain weighs 40 N, the force is weight force = m * g
I thought a Free Body Diagrams as:

N - (m*g) = m*a (a = 0 ; m = 40N/9.8 = 4.08 kg)

=> 40N - (4.08kg * 9.8) = 0 N

right?
 
What is the tension three quarters of the way up?

It means (40 N * 75)/ 100 = 30 N = answer for question 1. Correct?
 
huybinhs said:
If the chain weighs 40 N, the force is weight force = m * g
I thought a Free Body Diagrams as:

N - (m*g) = m*a (a = 0 ; m = 40N/9.8 = 4.08 kg)

=> 40N - (4.08kg * 9.8) = 0 N

right?
What I think you are trying to say is that

N -mg = 0
N - 40 = 0
N =40 N, which is the ceiling force on the chain. This is a result of the application of Newton 1 to a free body diagram of the entire chain.
 
huybinhs said:
What is the tension three quarters of the way up?

It means (40 N * 75)/ 100 = 30 N = answer for question 1. Correct?
yes, correct!
 
  • #10
2.What is the tension 1 m from the top?
It means 20m - 1m = 19m => (19m * 40 N)/20m = 38 N. Correct?
 
  • #11
huybinhs said:
2.What is the tension 1 m from the top?
It means 20m - 1m = 19m => (19m * 40 N)/20m = 38 N. Correct?
Yes! I wonder what it is at that the very bottom (bonus question worth 10 points).
 
  • #12
PhanthomJay said:
Yes! I wonder what it is at that the very bottom (bonus question worth 10 points).

I answers it but it's INCORRECT. I mean 38 N => incorrect somehow? What did I do wrong?

The last question I did correct. It is 2 N/m * 1 m = 2 N.
 
  • #13
btw what do u mean "bonus question worth 10 points" ?
 
  • #14
huybinhs said:
btw what do u mean "bonus question worth 10 points" ?
Oh since you were on a roll with now all answers to the 3 questions corrrect, i just thought i'd challenge you with a 4th question I made up on my own, which if you answered correctly, and if I was your teacher, i'd give you extra credit. So my question to you is, if you choose to accept the challenge and answer it, is
what is the tension at the very bottom of the chain?
 
  • #15
PhanthomJay said:
Oh since you were on a roll with now all answers to the 3 questions corrrect, i just thought i'd challenge you with a 4th question I made up on my own, which if you answered correctly, and if I was your teacher, i'd give you extra credit. So my question to you is, if you choose to accept the challenge and answer it, is

I think 40 N. Am I correct? LOL
 
  • #17
huybinhs said:
I think 40 N. Am I correct? LOL
Sorry, no bonus. It's 40 N at the top, but 0 at the bottom. BTW, your answer of T= 38N at 1 m from the top is correct, i don't know why you were marked wrong;I'll give you a bonus anyway for being smarter than the book...:wink:
 
  • #18
PhanthomJay said:
Sorry, no bonus. It's 40 N at the top, but 0 at the bottom. BTW, your answer of T= 38N at 1 m from the top is correct, i don't know why you were marked wrong;I'll give you a bonus anyway for being smarter than the book...:wink:

Got it. T = -38 N is more correctly. Agree?
 
  • #19
huybinhs said:
Got it. T = -38 N is more correctly. Agree?
Oh, here we go again with the plus and minus sign...tension forces always pull away from the objects on which they act...typically, tension Is considered a positive number...The tension at 1 m from the top is 38 n down acting on the top 1 m of chain; by Newton's 3rd Law, the tension is 38 N acting up on the lower 19 m of chain at that same point. Just say T =38 N.
 
  • #20
PhanthomJay said:
Oh, here we go again with the plus and minus sign...tension forces always pull away from the objects on which they act...typically, tension Is considered a positive number...The tension at 1 m from the top is 38 n down acting on the top 1 m of chain; by Newton's 3rd Law, the tension is 38 N acting up on the lower 19 m of chain at that same point. Just say T =38 N.

But when I submit 38 N. It's INCORRECT! Why? I think because the last question is + 2 N. Agree?
 
  • #21
huybinhs said:
Could u help me this problem:

https://www.physicsforums.com/showthread.php?t=381058

I appreciate that!
Now I know you've not been getting very good help with that question, but nonetheless, you shouldn't double post. I'll give it a look in the morning, under the original post, unless someone beats me to it...
 
  • #23
huybinhs said:
But when I submit 38 N. It's INCORRECT! Why? I think because the last question is + 2 N. Agree?
Oh, maybe, but the plus and minus sign really lacks meaning unless you identify the part of the chain you are looking at...give it a shot at neg 38 N...I'm maybe not as smart as a 5th grader :rolleyes:
 
  • #24
huybinhs said:
My homework is due at 8AM. Please look at it now!

https://www.physicsforums.com/showthread.php?t=381134
Gee, you're wearing me down...I've got to get up at 5...give it your best shot...19.6N is the max static friction available before it starts to slide..so what's F?? Good night , good morning, whatever...
 
  • #25
PhanthomJay said:
Oh, maybe, but the plus and minus sign really lacks meaning unless you identify the part of the chain you are looking at...give it a shot at neg 38 N...I'm maybe not as smart as a 5th grader :rolleyes:

Hang on... I just submited -38, and INCORRECT. What's wrong? I'm confused...
 
  • #26
Anyone help me with the second question?

It's not 38 nor -38 nor 2 :(
 
  • #27
huybinhs said:
Anyone help me with the second question?

It's not 38 nor -38 nor 2 :(
What have you done differetly for question 2 than you've done for the other 2 parts? Nothing. Sometimes, just sometimes, the 'book' is wrong...don't forget your units .....
 
  • #28
PhanthomJay said:
What have you done differetly for question 2 than you've done for the other 2 parts? Nothing. Sometimes, just sometimes, the 'book' is wrong...don't forget your units .....

You r correct, man! The system went wrong. too bad!
 

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