Ian at 16:10, 4 February 2011

2011-02-04T16:10:59Z

←Older revision		Revision as of 16:10, 4 February 2011
Line 21:		Line 21:


-	<math>F=8.6\ \text{N}</math>	+	<math>F=8\textrm{.}6\ \text{N}</math>

Ian: New page: Image:teori4.gif Using ${{F}^{2}}={{H}^{2}}+{{V}^{2}}$ and $\tan \alpha =\frac{V}{H}$ Here $V=-7\ \text{N}$ and $H=-5\ \text{N}$ ...

2010-03-23T12:33:01Z

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=-7\ \text{N}</math> and <math>H=-5\ \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=-7\ \text{N}</math> and
<math>H=-5\ \text{N}</math>.

Substituting the given values

<math>{{F}^{2}}={{\left( -5 \right)}^{2}}+{{\left( \grave{\ }-7 \right)}^{2}}=74</math>

giving

<math>F=8.6\ \text{N}</math>

Also

<math>\tan \alpha =\frac{-7}{-5}=1\textrm{.}4</math>
giving

<math>\alpha =-125\textrm{.}5{}^\circ</math>

Here the solution in the third quadrant has been chosen.

Solution 4.5d - Revision history

Ian at 16:10, 4 February 2011

Ian: New page: Image:teori4.gif Using {{F}^{2}}={{H}^{2}}+{{V}^{2}} and \tan \alpha =\frac{V}{H} Here V=-7\ \text{N} and H=-5\ \text{N}...

Ian: New page: Image:teori4.gif Using ${{F}^{2}}={{H}^{2}}+{{V}^{2}}$ and $\tan \alpha =\frac{V}{H}$ Here $V=-7\ \text{N}$ and $H=-5\ \text{N}$ ...