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Problem of the week #47 - February 18th, 2013

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Jameson

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Jan 26, 2012
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Suppose there are point masses located at the points in the plane described by the vectors $v_1=<1,0>$, $v_2=<-1,1>$ and $v_3=<-1,-1>$. Find masses $m_1,m_2,m_3$ such that the center of mass is the origin of the plane given that the sum of these masses is 1.

Recall that the center of mass vector can be found by the following equation: \(\displaystyle c = \frac{m_1}{m}v_1+\frac{m_2}{m}v_2+\frac{m_3}{m}v_3\), where $m$ is the sum of the masses.

Hint:
Since $m=1$ this is just solving a three variable system of equations. The fact that $m=1$ should be used twice.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) Sudharaka

Remark:
This problem is nothing more than solving a three-variable system of equations but using the definition of scalar multiplication of vectors and the bit of information about the sum of the three masses must both be applied to simply this into a form that is recognizable as a three-variable system.

1) For a vector with two real components, $<a,b>$ and a real number $c$ (not a vector, but a scalar) then $c<a,b>=<ca,cb>$. This principle should be familiar to most high school students who have taken math or physics.

2) Since $m_1+m_2+m_3=1$, then all of the terms that divide by $m$ can be written without the denominator. Dividing by 1 doesn't change a real number.

3) If the center, $c$ is desired to be the origin, that's the same as $<0,0>$ in vector form.

Using the above leads to $<0,0>=m_1<1,0>+m_2<-1,1>+m_3<-1,-1>$. Each vector has two components so we get a total of two equations from that, which isn't enough to have a unique solution if there are three variables but if we remember the extra information from the problem that $m_1+m_2+m_3=1$, we now have three equations and can solve for $m_1,m_2,m_3$.


Solution (from Sudharaka):
Substituting the relevant values for \(\displaystyle c = \frac{m_1}{m}v_1+\frac{m_2}{m}v_2+\frac{m_3}{m}v_3\) we get,

\[<0,0>=m_1<1,0>+m_2<-1,1>+m_3<-1,-1>\]

So we have the equations,

\[m_1-m_2-m_3=0~~~~~~~~~~~~~~(1)\]

\[m_2-m_3=0~~~~~~~~~~~(2)\]

\[m_1+m_2+m_3=1~~~~~~~~~~~~~(3)\]

Solving these equations we get,

\[m_1=\frac{1}{2},\,m_2=m_3=\frac{1}{4}\]
 
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