Ian: New page: Image:teori4.gif Using ${{F}^{2}}={{H}^{2}}+{{V}^{2}}$ and $\tan \alpha =\frac{V}{H}$ Here $V=8\ \text{N}$ and $H=-2\ \text{N}$ ....

Ian — Tue, 23 Mar 2010 11:22:21 GMT

New page: Image:teori4.gif Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> and <math>\tan \alpha =\frac{V}{H} </math> Here <math>V=8\ \text{N}</math> and <math>H=-2\ \text{N}</math>....

New page

[[Image:teori4.gif]]

Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math>

and

<math>\tan \alpha =\frac{V}{H} </math>

Here

<math>V=8\ \text{N}</math> and
<math>H=-2\ \text{N}</math>.

Substituting the given values

giving

Also

<math>\tan \alpha =\frac{8}{-2}=-4</math>

giving

<math>\alpha =-76\textrm{.}0{}^\circ</math>

However the force is in the second quadrant so this angle is not correct.

Another possible angle which gives the same tan is

<math>\alpha =180-76\textrm{.}0=104\textrm{.}0{}^\circ</math>

This is in the second quadrant and thus is the correct answer.

Solution 4.5c - Revision history

Ian: New page: Image:teori4.gif Using {{F}^{2}}={{H}^{2}}+{{V}^{2}} and \tan \alpha =\frac{V}{H} Here V=8\ \text{N} and H=-2\ \text{N}....

Ian: New page: Image:teori4.gif Using ${{F}^{2}}={{H}^{2}}+{{V}^{2}}$ and $\tan \alpha =\frac{V}{H}$ Here $V=8\ \text{N}$ and $H=-2\ \text{N}$ ....