Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have an exam in two days for first year university physics. Often for dynamics problems, I am required to solve algebraic systems of equations by hand, and this can be very daunting.

When I see the solutions, however, the steps that the solver took to seem very clean and almost obvious. Are there some rules of thumb that physicists use to solve small systems of equations, either by elimination or substitution?

Here is an example. Find $\frac{m_1}{m_2}$ in terms of only $\theta$ where the known quantities are $\theta$, $m_1$, $m_2$.

$$F_T \sin \theta = m_1 g$$ $$F_T \cos \theta = m_1 a$$ $$F_T \sin \theta + F_N \cos \theta = m_2 g$$ $$F_N \sin \theta - F_T \cos \theta = m_2 a$$

share|cite|improve this question
up vote 1 down vote accepted

A rule of thumb would be to get rid of unwanted variables. For example, since we're only interested in $\frac{m_1}{m_2}$ and $\theta$, we can get rid of $F_T$.

$$F_T \sin \theta = m_1 g$$ $$F_T \cos \theta = m_1 a$$

Rearrange to get

$$\frac{m_1 g}{\sin \theta}=\frac{ m_1 a}{\cos \theta} \hspace{20mm}\text{ ...Eq 5}$$

I don't know whether getting this 'Equation 5' is part of the solution, but my point is get rid of unwanted variables.

share|cite|improve this answer

It is sometimes useful to add and/or subtract equations. Eq. 3 minus Eq. 1 gives you

$$F_N \cos \theta = (m_2 - m_1) g$$

and eq. 4 + eq. 2 gives you

$$F_N \sin \theta = (m_1 + m_2) a.$$ Divide these two guys, and you then obtain

$$(m_2 - m_1)\frac{g}{a} \tan \theta = m_2 + m_1.$$ This should be easy to solve for $m_1/m_2.$

share|cite|improve this answer
This expression alone can be solved for $m_2/m_1$? Would you be able to go a little bit further, because I don't see it. – dal102 Feb 25 '13 at 20:05
The one step you're missing: by expanding the LHS, you'll be able to rewrite it as $X m_1 = Y m_2$ where $X$ and $Y$ depend on $g,a,\theta.$ (So do this!) From there on, it should be easy. – Vibert Feb 25 '13 at 23:09

One "trick" I use often is to resolve the equations along coordinate systems that decouple my forces as much as possible. For example with your equations you can write them in a rotated coordinate system as

$$\begin{matrix} F_T\sin\theta = m_1\,g \\ F_N \cos\theta = m_1\,a \\ F_N = m_2 ( g\,\cos\theta+ a\,\sin\theta) \\ -F_T = m_2 ( a\,\cos\theta - g\,\sin\theta ) \end{matrix} $$

So now you can add/subtract equations to eliminate variables easier.

For example adding #1 to $\sin\theta$ times #4 creates

$$ F_T\sin\theta + \sin\theta (-F_T) = m_1\,g + \sin\theta (m_2 ( a\,\cos\theta - g\,\sin\theta ) ) $$ $$ a\, m_2 \sin\theta \cos\theta = g (m_2 \sin^2\theta - m_1 ) $$ $$ a = \frac{g (m_2 \sin^2\theta - m_1 )}{ m_2 \sin\theta \cos\theta} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.