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Same string, two differents tensions. A concept clarification

  1. May 20, 2007 #1
    In problems which have a pulley with a certain mass and radius, and a massless string over it, we often consider the tension of the left part of the string different than the tension of the right part of the string.

    I thought that since the string is massless, the tensions should be the same in every element of the string, because of the newton law applied to such element leads to F=T1 - T2 = M.a =0.a=0 ==> T1=T2 right?
    Clearly i'm wrong, but Why??? Why is T1 different than T2 in problems with pulleys that can suffer rotation (mass and radius well-defined)?
     
  2. jcsd
  3. May 21, 2007 #2

    Doc Al

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    Staff: Mentor

    Your reasoning is fine for a continuous stretch of massless string--as long as there's nothing in the middle interacting with it. Realize that the pulley exerts forces on the string that are different in each direction. If the tension were the same on both sides of the pulley, how could the string accelerate? If the pulley--as well as the string--was massless, then your reasoning would work.
     
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