Processing Math: Done
Solution 16.10a
From Mechanics
(Difference between revisions)
m (New page: <math>\begin{align} & m(4\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}+2\mathbf{j}) \\ & (4m+2\lambda m-3m)\mathbf{i}+(mV-\lambda mV-2\lambda m)\mat...) |
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Line 1: | Line 1: | ||
<math>\begin{align} | <math>\begin{align} | ||
- | & m( | + | & m(5\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}-2\mathbf{j}) \\ |
- | & ( | + | & (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} \\ |
\end{align}</math> | \end{align}</math> | ||
+ | |||
Consider the | Consider the | ||
<math>\mathbf{i}</math> | <math>\mathbf{i}</math> | ||
- | component. That is the <math>\mathbf{i}</math> component on the left hand side must be the same as the <math>\mathbf{i}</math> component on the right hand side. | + | component. That is the <math>\mathbf{i}</math> component on the left hand side must be the same as the <math>\mathbf{i}</math> component on the right hand side, that is zero. |
<math>\begin{align} | <math>\begin{align} | ||
- | & | + | & 5m+2\lambda m-3m-3\lambda m=0 \\ |
- | & 2\lambda | + | & 2-\lambda =0 \\ |
- | & \lambda = | + | & \lambda =2 \\ |
\end{align}</math> | \end{align}</math> |
Revision as of 16:23, 1 October 2010
m(2i−Vj)=(m+
m)(3i−2j)(5m+2
m−3m−3
m)i+(mV−
mV+2m+2
m)j=0i+0j
Consider the
m−3m−3
m=02−
=0
=2