Understanding Tension in a Hanging Chain on a Fixed Sphere - Homework Help

  • Thread starter Thread starter mooncrater
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the tension in a chain of mass per unit length k and length 1.5 m resting on a fixed smooth sphere with a radius of R = 2/π m. Participants identify that the tension must balance both the rightward force from the chain segment in contact with the sphere and the downward force from the hanging segment. A critical insight reveals that the tension varies along the chain and must be included in the force balance equations, particularly in the tangential direction, to achieve a correct solution.

PREREQUISITES
  • Understanding of basic mechanics, specifically force balance
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of tension in strings and chains
  • Concept of forces acting on objects in contact with curved surfaces
NEXT STEPS
  • Study the principles of tension in non-uniform chains
  • Learn about force balance in systems involving curved surfaces
  • Explore integration techniques for non-standard functions
  • Investigate the effects of varying tension along a chain
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics, as well as anyone involved in solving problems related to tension in chains and forces on curved surfaces.

mooncrater
Messages
215
Reaction score
18

Homework Statement

The question says
A chain of mass per unit length k and length 1.5 m rest on a fixed smooth sphere of radius R= 2/π m
. Tension in the thread is asked.The question figure is attached.

Homework Equations


N/a

The Attempt at a Solution


IMG_20150217_145935.jpg
IMG_20150217_145935.jpg

I think that the tension has to balance two types of forces :
1) rightward force of part AC(touching)
2)downward force of part CB(hanging)
Finding first type of force:
Using Fbd(attached) of small part taken in touching part
Then Ncos(theta)=(dm)g
→ N=(dm)g/cos(th.)...(1)
Length of that part =R(d(th.))
mass of that small part =Rk(d(th.))
Using Fbd rightward force on the small part =Nsin(th.)=(dm)g(tan(th.))
=Rkgtan(th.)(d(th.))
Now if I want to integrate it , i would have to use limits from 0 to π/2
BUT tan (th.) is not integrable b/w 0 to π/2 . So what's wrong with my answer ?
 

Attachments

  • IMG_20150217_145912.jpg
    IMG_20150217_145912.jpg
    30.7 KB · Views: 607
Physics news on Phys.org
I'm missing something in your force balance for the little chunk of chain. I see an N and an mg and they clearly don't balance. What force could possibly complete the picture ?
 
BvU said:
I'm missing something in your force balance for the little chunk of chain. I see an N and an mg and they clearly don't balance. What force could possibly complete the picture ?
It could have been friction BUT the sphere is smooth so , what can be the force (you are talking about ) left out that balances the tension of the thread ?
 
mooncrater said:
It could have been friction BUT the sphere is smooth so , what can be the force (you are talking about ) left out that balances the tension of the thread ?
But you don't have the tension in the chain in the equation. Indeed, that's what's missing. Remember, it will not be constant along the the chain.
 
You have to consider the tension on both sides of the element along the tangential direction!
 

Similar threads

Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K