1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tension of a rope on a cone, fallacious solution

  1. Nov 26, 2014 #1

    ZetaOfThree

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    A rope of mass ##m## forming a circle is placed over a smooth round cone with half angle ##\theta##. Find the tension in the rope.
    2. Relevant equations
    ##\sum{F}=0##

    3. The attempt at a solution
    I know how to solve the problem, but I have another way that I think should work but it doesn't. I don't know why.
    The way that this problem is done is to consider the tension on a small element spanning an angle ##\alpha## of the rope as shown in the following image: 1.92_fig1.JPG . From there, we balance the forces on that small element as shown in the following image: 1.92_fig2.png . We find that $$T={m \over 2 \pi}g \cot{\theta}$$
    My question is why do we have to consider a small element of the rope? Why can't we consider half of the rope (that is, an element of the rope spanning ##\alpha = \pi##)? Therefore, the tension acts on the rope as in the following figure: Screen Shot 2014-11-26 at 12.39.54 PM.png , leading to a free-body diagram as in the following image: Screen Shot 2014-11-26 at 12.47.08 PM.png .
    This leads to the result $$T={m \over 4}g \cot{\theta}$$ What's going wrong?
     
  2. jcsd
  3. Nov 26, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    There is no tension at the end of the rope, and tension is not constant over the length.
     
  4. Nov 26, 2014 #3

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    In your alternate approach, I'll give you "2T" or "mg/2," but not both; "pi" ain't obvious to me yet.
     
  5. Nov 26, 2014 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    How is the normal force acting on different parts of your string?
    The rope is a full circle and has no end ;)
     
  6. Nov 26, 2014 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Oh, I misread how the rope is placed.
    In that case: the force is not acting in the same direction everywhere.
     
  7. Nov 27, 2014 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Just as a comment regarding alternative ways of approaching the problem: I would actually prefer to solve this one by using energy methods. The answer drops out quite naturally without having to consider individual parts of the rope.
     
  8. Nov 27, 2014 #7

    ZetaOfThree

    User Avatar
    Gold Member

    I know what's wrong with my alternative solution. The component of the normal force that opposes ##2T## is a sum of small normal forces acting in the direction opposite to ##2T##, which act on each element of the rope perpendicular to the surface of the cone. The force that opposes ##2T## can be found using integration and you get the correct answer.

    Yeah, virtual work does give the result in like two lines. I was trying to find as many solutions to this problem as a could. I should have been more careful with my alternative solution above. I assumed the the force from the cone would be in a direction at angle ##\alpha## from the horizontal. For half the rope, this is not the case.

    Thanks for the responses everyone!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Tension of a rope on a cone, fallacious solution
  1. Tension on rope (Replies: 1)

  2. Diffusion in Cone (Replies: 6)

Loading...