- #1
Benny
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I'm reading through an example involving Newton's second law. The situation is that there are n particles surrounded by a system 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 \sum\limits_{j = 1}^n {\mathop {F_{ij} }\limits^ \to }. So Newton's second alw applied to the particle with mass m_i is:
<br /> \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)} <br />
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:
<br /> \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)} <br />
I don't know how the double summation works. For a single summation for example \sum\limits_{k = 1}^n k 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.
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 \sum\limits_{j = 1}^n {\mathop {F_{ij} }\limits^ \to }. So Newton's second alw applied to the particle with mass m_i is:
<br /> \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)} <br />
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:
<br /> \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)} <br />
I don't know how the double summation works. For a single summation for example \sum\limits_{k = 1}^n k 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.
