Solution 4.9b
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(New page: In part a) the resultant <math>\mathbf{R}</math> is seen to be <math>\begin{align} & \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\ \end{align}</math> Also ac...)
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Revision as of 10:34, 1 April 2010
In part a) the resultant \displaystyle \mathbf{R} is seen to be
\displaystyle \begin{align} & \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\ \end{align}
Also according to the theory
\displaystyle \tan \alpha =\frac{V}{H} where \displaystyle H is the horizontal component of the resultant and
\displaystyle V is the verical component of the resultant.
Thus
\displaystyle \tan \alpha =\frac{\text{3}\textrm{.}\text{93}}{\text{11}\textrm{.}\text{1}}=0\textrm{.}354
giving
\displaystyle \alpha =19\textrm{.}5{}^\circ
