Solution 4.5a - Revision history

Ian at 10:42, 23 March 2010

Ian — Tue, 23 Mar 2010 10:42:14 GMT

←Older revision		Revision as of 10:42, 23 March 2010
Line 9:		Line 9:
	Here		Here

-	<math>V=A</math> and	+	<math>V=3\ \text{N}</math> and
-	<math>H=B</math>.	+	<math>H=4\ \text{N}</math>.

	Substituting the given values		Substituting the given values
-

	<math>\begin{align}		<math>\begin{align}
-	& {{F}^{2}}={{4}^{2}}+{{3}^{2}}=25 \\	+	{{F}^{2}}={{4}^{2}}+{{3}^{2}}=25 \\
-	~~& \text{giving} \\~~	+
-	~~& F=5\ \text{N}~~ \\	+
	\end{align}</math>		\end{align}</math>

		+	giving
		+
		+	<math>\begin{align}
		+	F=5\ \text{N} \\
		+	\end{align}</math>

	Also		Also

Ian at 10:35, 23 March 2010

Ian — Tue, 23 Mar 2010 10:35:30 GMT

←Older revision		Revision as of 10:35, 23 March 2010
Line 2:		Line 2:

	Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math>		Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math>
		+
		+	and
		+
		+	<math>\tan \alpha =\frac{V}{H} </math>
		+
		+	Here
		+
		+	<math>V=A</math> and
		+	<math>H=B</math>.

	Substituting the given values		Substituting the given values
		+

	<math>\begin{align}		<math>\begin{align}
-	& {{F}^{2}}={{4}^{2}}+{	+	& {{F}^{2}}={{4}^{2}}+{{3}^{2}}=25 \\
-	{3}^{2}}	+	& \text{giving} \\
-	=25 \\	+	& F=5\ \text{N} \\
-	& F=5 \text{ N} \\	+
	\end{align}</math>		\end{align}</math>

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

-	~~Here~~	+	Also
-	~~<math>V=3\ \text{N}</math>~~	+
-	~~and~~	+
-	~~<math>H=4\ \text{N}</math>.~~	+
-		+
-	~~Thus~~	+

	<math>\tan \alpha =\frac{3}{4}=0\textrm{.}75</math>		<math>\tan \alpha =\frac{3}{4}=0\textrm{.}75</math>
		+
	giving		giving
		+
	<math>\alpha =36\textrm{.}9{}^\circ</math>		<math>\alpha =36\textrm{.}9{}^\circ</math>

Ian: New page: Image:teori4.gif Using ${{F}^{2}}={{H}^{2}}+{{V}^{2}}$ Substituting the given values $\begin{align} & {{F}^{2}}={{4}^{2}}+{ {3}^{2}} =25 \\ & F=5 \text{ N} \\ \end...$

Ian — Tue, 23 Mar 2010 09:58:33 GMT

New page: Image:teori4.gif Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> Substituting the given values <math>\begin{align} & {{F}^{2}}={{4}^{2}}+{ {3}^{2}} =25 \\ & F=5 \text{ N} \\ \end...

New page

[[Image:teori4.gif]]

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

Substituting the given values

<math>\begin{align}
& {{F}^{2}}={{4}^{2}}+{
{3}^{2}}
=25 \\
& F=5 \text{ N} \\
\end{align}</math>

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

Here
<math>V=3\ \text{N}</math>
and
<math>H=4\ \text{N}</math>.

Thus

<math>\tan \alpha =\frac{3}{4}=0\textrm{.}75</math>
giving
<math>\alpha =36\textrm{.}9{}^\circ</math>

Solution 4.5a - Revision history

Ian at 10:42, 23 March 2010

Ian at 10:35, 23 March 2010

Ian: New page: Image:teori4.gif Using {{F}^{2}}={{H}^{2}}+{{V}^{2}} Substituting the given values \begin{align} & {{F}^{2}}={{4}^{2}}+{ {3}^{2}} =25 \\ & F=5 \text{ N} \\ \end...

Ian: New page: Image:teori4.gif Using ${{F}^{2}}={{H}^{2}}+{{V}^{2}}$ Substituting the given values $\begin{align} & {{F}^{2}}={{4}^{2}}+{ {3}^{2}} =25 \\ & F=5 \text{ N} \\ \end...$