Processing Math: Done

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Revision as of 16:29, 1 October 2010 (edit) Ian (Talk | contribs) (New page: From part a) <math>(5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j}</math> Consider the <math>\mathbf{j}</math> component. That is...) ← Previous diff		Current revision (16:43, 27 March 2011) (edit) (undo) Ian (Talk | contribs)
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	& V=6 \\		& V=6 \\
	\end{align}</math>		\end{align}</math>
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-	Note that <math>V</math> is a dimensionless number.

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Current revision

From part a)

(5m+2m−3m−3m)i+(mV−mV+2m+2m)j=0i+0j

Consider the j component. That is the j component on the left hand side must be the same as the j component on the right hand side, that is zero.

mV−mV+2m+2m=0V−2V+2+4=0−V+6=0V=6