Ian at 15:15, 6 April 2010

Ian — Tue, 06 Apr 2010 15:15:23 GMT

←Older revision		Revision as of 15:15, 6 April 2010
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		+
		+	Resolving horisontally to the right
		+
		+	<math>\begin{align}
		+	& T2\cos {{40}^{\circ }}-T1\cos {{50}^{\circ }}=0 \\
		+	& \\
		+	\end{align}</math>
		+
		+	giving
		+
		+	<math>T2=\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1</math>
		+
		+	Resolving vertically upwards.
		+
		+	<math>T1\cos {{40}^{\circ }}+T2\cos {{50}^{\circ }}-mg=0</math>
		+
		+	Substituting for
		+	<math>T2</math>
		+	in this equation gives an equation only containing
		+	<math>T1</math>.
		+
		+	<math>\begin{align}
		+	& T1\cos {{40}^{\circ }}+\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1\cos {{50}^{\circ }}-mg=0 \\
		+	& \\
		+	& T1\left( \cos {{40}^{\circ }}+\frac{{{\cos }^{2}}{{50}^{\circ }}}{\cos {{40}^{\circ }}} \right)=mg \\
		+	& \\
		+	& T1\left( 0\textrm{.}766+0\textrm{.}539 \right)=50\times 9\textrm{.}8 \\
		+	& \\
		+	& T1=\frac{490}{1\textrm{.}3}=375\ \text{N} \\
		+	\end{align}</math>
		+
		+
		+	Also using
		+	<math>T2=\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1</math>
		+
		+	<math>T2=\frac{0\textrm{.}642}{0\textrm{.}766}\times 375=315\ \text{N}</math>

Ian: New page: Image:5.8.gif

Ian — Tue, 06 Apr 2010 14:43:49 GMT

New page

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