Ian at 10:36, 1 April 2010

2010-04-01T10:36:38Z

←Older revision		Revision as of 10:36, 1 April 2010
Line 10:		Line 10:
	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

Ian: New page: In part a) the resultant $\mathbf{R}$ is seen to be $\begin{align} & \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\ \end{align}$ Also ac...

2010-04-01T10:34:27Z

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...

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 according to the theory

<math>\tan \alpha =\frac{V}{H}</math>
where <math>H</math> is the horizontal component of the resultant and

<math>V</math> is the verical component of the resultant.

Thus

<math>\tan \alpha =\frac{\text{3}\textrm{.}\text{93}}{\text{11}\textrm{.}\text{1}}=0\textrm{.}354</math>

giving

<math>\alpha =19\textrm{.}5{}^\circ</math>

Solution 4.9b - Revision history

Ian at 10:36, 1 April 2010

Ian: New page: In part a) the resultant \mathbf{R} is seen to be \begin{align} & \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\ \end{align} Also ac...

Ian: New page: In part a) the resultant $\mathbf{R}$ is seen to be $\begin{align} & \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\ \end{align}$ Also ac...