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Newton's second law

  1. Dec 27, 2005 #1
    I'm reading through an example involving Newton's second law. The situation is that there are n particles surrounded by a sytem boundary. The picture consists of a bunch of circles (particles) enclosed by a closed loop.

    The forces acting on one of the particles of mass m_i consist of an external resultant force F_i and other external forces which is given by [tex]\sum\limits_{j = 1}^n {\mathop {F_{ij} }\limits^ \to } [/tex]. So Newton's second alw applied to the particle with mass m_i is:

    \mathop {F_i }\limits^ \to + \sum\limits_{j = 1}^n {\mathop {F_{ij} }\limits^ \to = \frac{d}{{dt}}\left( {m_i \mathop {V_i }\limits^ \to } \right)}

    F_ij ~ force of particle with mass j on particle with mass i.

    The example goes on to say that there are n such equations (presumably the one above) so to simplify they rewrite it as follows:

    \sum\limits_{i = 1}^n {\mathop {F_i }\limits^ \to } + \sum\limits_{i = 1}^n {\sum\limits_{j = 1}^n {\mathop {F_{ij} }\limits^ \to } } = \sum\limits_{i = 1}^n {\frac{d}{{dt}}\left( {m_i \mathop {V_i }\limits^ \to } \right)}

    I don't know how the double summation works. For a single summation for example [tex]\sum\limits_{k = 1}^n k [/tex] I just take k, replace it with one and repeat for all integers from 1 to n. The sum would then be S = 1 + 2 + 3...+ n = (n/2)(n+1). I'm not sure if the double summation is similar. Can someone explain? Thanks.
  2. jcsd
  3. Dec 27, 2005 #2
    First do the summation over j, then you have to do j summations over i :

    [tex] \sum\limits_{i = 1}^n {\sum\limits_{j = 1}^n (F_{ij}) = \sum\limits_{i=1}^n (F_{i1}+F_{i2}+...+F_{in}) = \sum\limits_{i=1}^n (F_{i1})+\sum\limits_{i=1}^n (F_{i2})+...+ \sum\limits_{i=1}^n (F_{in}) [/tex]

    Last edited: Dec 27, 2005
  4. Dec 28, 2005 #3
    Thanks for your explanation Marlon.
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