From Mechanics

Revision as of 16:02, 23 March 2010 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (16:13, 4 February 2011) (edit) (undo) Ian (Talk | contribs)
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	Thus		Thus
-	<math>\tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14</math>	+	<math>\tan \alpha =\frac{-11\textrm{.}3}{80\textrm{.}9}=-0\textrm{.}14</math>
	giving		giving

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Current revision

Resultant force \displaystyle =80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}

Thus

\displaystyle \tan \alpha =\frac{-11\textrm{.}3}{80\textrm{.}9}=-0\textrm{.}14

giving

\displaystyle \alpha =-8\textrm{.}0{}^\circ

As the vector is in the fourth quadrant