Solution 4.6c
From Mechanics
(Difference between revisions)
(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 =-...) |
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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 | ||
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+ | [[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