# Solving systems of equations with 5 unknowns, 4 equations

When solving systems of equations for first-year physics, I am told that to be able to find the solution for a system of $n$ unknowns (if one exists), I need at least $n$ equations.

In this system, I am looking for $a$. The known quantities are $\theta_0$ and $\theta_1$.

$$T_0 \cos \theta_0 = mg$$ $$T_0 \sin \theta_0 = \frac{m v ^{2}}{\ell \sin \theta_0}$$ $$T_1 \cos \theta_1 - mg = ma$$ $$T_1 \sin \theta_1 = \frac{m v ^{2}}{\ell \sin \theta_1}$$

In this system of equations, I had 5 unknown quantities: $T_0$, $T_1$, $m$, $\ell$, $a$. But I was able to successfully solve for $a$ with four equations. Where is the logical error in my understanding?

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By solve, you mean you set some variable to zero and then proceeded for other four? – hjpotter92 Feb 26 '13 at 18:52
If you set $\frac{mv^{2}}{\ell}$ equal to some variable $x$ and then use that in your equations. You are now down to 4 variables (the exact values of $m$, $v$ and $\ell$ aren't required to find $a$) – Shaktal Feb 26 '13 at 18:55
@BackinaFlash I isolated $a$ from (3). Substituted for $T_1$ from (4). Divided (2) by (1) and isolated and plugged $\ell$ from that expression into what I had. – dal102 Feb 26 '13 at 18:55
@Shaktal Are you saying I should solve for $m$, $v$, and $\ell$ in terms of $x$ in all the other equations too? – dal102 Feb 26 '13 at 18:57
You can just give a name to $\frac {v^2}\ell$. Maybe call it $y$. Because $v$ and $\ell$ only appear in that combination, you really only have four variables: $T_0, T_1, m, y$ and your four equations can solve for them. You will not be able to get separate values for $v$ and $\ell$ without another equation. – Ross Millikan Feb 26 '13 at 19:29

You can solve for $\ell$ from the start by plugging in the first equation to the second equation, so I don't think you made a mistake. Ie. from the first equation get an expression for m, plug into second equation, eliminate T0 and then just compute $\ell$.

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The command you're looking for is \ell, in between dollar signs. – Cameron Buie Feb 26 '13 at 19:06
Yes I'm a noob, thank you for your help. \ell , and I still can't get it right, sigh. – Michael Summerville Feb 26 '13 at 19:15
You need the dollar signs. You can right click on any $\LaTeX$ and select Show Math As ->TeX commands to see how it is done. As Cameron Buie said, I put \ell between dollar signs to get $\ell$ – Ross Millikan Feb 26 '13 at 19:24
Thx, also thx for correcting the post. $\ell$ huzzah! – Michael Summerville Feb 26 '13 at 21:23

Divide all the equations by $m$. You get a system of four equations in the four unknowns $T_0/m$, $T_1/m$, $\ell$ and $a$.

It's been a while since I did any even semi-serious mechanics, so take the following with a grain of salt. The physical reason why the above works is that the system scales with the mass $m$. If you double the mass, you will simply double the forces (tensions?) $T_0$ and $T_1$. Any other scaling factor works the same way. The object will have the same acceleration irrespective of its mass, when it is undergoing the same motion. Only the forces involved will scale accordingly.

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A better way of putting is that if you multiply the mass with a constant, and all the forces with the same constant (keeping speed et cetera fixed), you should get another physically valid solution with acceleration retaining the same value. – Jyrki Lahtonen Feb 27 '13 at 5:39

There in no law forbidding a system from being solvable when there are more unknowns then equations. $0x=y$ for example. This usually implies that an unknown has no or limited effect on the others. (Of course, if mass is $0$, which I hear occurs with photons, that would have an effect.)

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