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...)
Current revision (10:36, 1 April 2010) (edit) (undo)
 
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where <math>H</math> is the horizontal component of the resultant and
where <math>H</math> is the horizontal component of the resultant and
-
<math>V</math> is the verical component of the resultant.
+
<math>V</math> is the vertical component of the resultant.
Thus
Thus

Current revision

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 vertical 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