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Tension equation

  1. Mar 30, 2009 #1
    I have a question about this equation (link below). I am working on a tension problem and I don't understand how you go from Sum of F= Ft-mg to
    v= [square root of] (Ft-mg)r /m

    The progression from one to the other is confusing me. I don't see how can use (mg) in one part of the equation and in the next part it is (v)squared and the back to (mg) at the end.


    http://img3.imageshack.us/img3/4830/equationi.jpg [Broken]


    I hope my question was not written too unclear and any help would be amazing. Thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 31, 2009 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi msudawgs267,

    I would not say the expression on the left is progression from one step to another. The starting point is Newton's law in the form:

    [tex]
    \sum F = m a
    [/tex]

    and then they are saying that

    [tex]
    \sum F \to F_T - mg
    [/tex]

    and

    [tex]
    m a \to m\frac{v^2}{r} \ \ \ \mbox{(for centripetal acceleration)}
    [/tex]

    Setting these two new expressions equal to each other and solving for v gives the answer.
     
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