Solution 4.6c - Revision history

Ian at 16:13, 4 February 2011

Ian — Fri, 04 Feb 2011 16:13:54 GMT

←Older revision		Revision as of 16:13, 4 February 2011
Line 4:		Line 4:
	Thus		Thus

-	<math>\tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14</math>	+	<math>\tan \alpha =\frac{-11\textrm{.}3}{80\textrm{.}9}=-0\textrm{.}14</math>

	giving		giving

Ian at 16:02, 23 March 2010

Ian — Tue, 23 Mar 2010 16:02:29 GMT

←Older revision		Revision as of 16:02, 23 March 2010
Line 11:		Line 11:

	As the vector is in the fourth quadrant		As the vector is in the fourth quadrant
		+
		+	[[Image:4.6c.gif]]

Ian: New page: Resultant force $=80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}$ Thus $\tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14$ giving $\alpha =-...$

Ian — Tue, 23 Mar 2010 15:58:20 GMT

New page: Resultant force <math>=80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}</math> Thus <math>\tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14</math> giving <math>\alpha =-...

New page

Resultant force
<math>=80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}</math>

Thus

<math>\tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14</math>

giving

<math>\alpha =-8\textrm{.}0{}^\circ</math>

As the vector is in the fourth quadrant

Solution 4.6c - Revision history

Ian at 16:13, 4 February 2011

Ian at 16:02, 23 March 2010

Ian: New page: Resultant force =80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N} Thus \tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14 giving \alpha =-...

Ian: New page: Resultant force $=80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}$ Thus $\tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14$ giving $\alpha =-...$