Ian at 16:01, 2 April 2010

Ian — Fri, 02 Apr 2010 16:01:00 GMT

←Older revision		Revision as of 16:01, 2 April 2010
Line 1:		Line 1:
-	Resolving vertically	+	Resolving vertically upwards

Ian: New page: Resolving vertically $\begin{align} & {{T}_{1}}\cos {{50}^{\circ }}-mg=0 \\ & \\ & {{T}_{1}}=\frac{mg}{\cos {{50}^{\circ }}}=\frac{490}{0\textrm{.}643}=762\ \text{N} \\ \end{ali...$

Ian — Fri, 02 Apr 2010 15:54:59 GMT

New page: Resolving vertically <math>\begin{align} & {{T}_{1}}\cos {{50}^{\circ }}-mg=0 \\ & \\ & {{T}_{1}}=\frac{mg}{\cos {{50}^{\circ }}}=\frac{490}{0\textrm{.}643}=762\ \text{N} \\ \end{ali...

New page

Resolving vertically

<math>\begin{align}
& {{T}_{1}}\cos {{50}^{\circ }}-mg=0 \\
& \\
& {{T}_{1}}=\frac{mg}{\cos {{50}^{\circ }}}=\frac{490}{0\textrm{.}643}=762\ \text{N} \\
\end{align}</math>

Solution 5.2a - Revision history

Ian at 16:01, 2 April 2010

Ian: New page: Resolving vertically \begin{align} & {{T}_{1}}\cos {{50}^{\circ }}-mg=0 \\ & \\ & {{T}_{1}}=\frac{mg}{\cos {{50}^{\circ }}}=\frac{490}{0\textrm{.}643}=762\ \text{N} \\ \end{ali...

Ian: New page: Resolving vertically $\begin{align} & {{T}_{1}}\cos {{50}^{\circ }}-mg=0 \\ & \\ & {{T}_{1}}=\frac{mg}{\cos {{50}^{\circ }}}=\frac{490}{0\textrm{.}643}=762\ \text{N} \\ \end{ali...$