Solution 4.6c

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(New page: Resultant force <math>=80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}</math> Thus <math>\tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14</math> giving <math>\alpha =-...)
Current revision (16:13, 4 February 2011) (edit) (undo)
 
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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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As the vector is in the fourth quadrant
As the vector is in the fourth quadrant
 +
 +
[[Image:4.6c.gif]]

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

Image:4.6c.gif