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- Thread starter courtrigrad
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Thanks

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BobG

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[tex]cos \phi = \frac{F \cdot dl}{F * dl}[/tex]

If you substitute that for [tex]cos \phi[/tex] in the first, the problem simplifies to the last.

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Work is defined as the line integral over some path L of the scalar product of r and F. Now a scalar product of two vectors r and F is equal toplugpoint said:

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[tex]\vec{r} \cdot \vec{F} = ||\vec{r}||*||\vec {F}||* cos( \theta)[/tex]

The product of F with the cosine of the angle theta between r and F denotes the component of F parallel with vector r.

marlon

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