Putting into terms.

Question

I asked this in the physics exchange, but was told this is more really of a math issue. In this problem

\begin{eqnarray} T&=&m_1 a \\T - m_2 g &=& -m_2 a \end{eqnarray} From these equations, we can express $a$ and $T$ in terms of the masses $m_1$ and $m_2$ and $g$: \begin{eqnarray} a &=& \frac{m_2}{m_1 + m_2} g \\ T &=& \frac{m_1m_2}{m_1 + m_2} g \end{eqnarray}

I'm completely confused on how they expressed $a$ and $T$. What am I solving for in $T=m_1a$ and $T-m_2g= -m_2a$? I thought I was solving for one to input into the other, but I don't think that's it because I can't put it into each other. I don't understand how they found $a$ and $T$ from the equations $T=m_1a$ and $T-m_2g=-m_2a$. How are those 2 equations being manipulated to find $a$ and $T$?

Answer 1

1. Begin with the second equation: $T - m_2g = -m_2a$
2. Insert $T = m_1a$ into it: $m_1a - m_2g = -m_2a$
3. Add $m_2a$ on both sides: $m_1a - m_2g + m_2a = 0$
4. Add $m_2g$ on both sides: $m_1a + m_2a = m_2g$
5. Now factor out $a$: $a(m_1 + m_2) = m_2g$
6. Divide both sides by $m_1 + m_2$: $a = \frac{m_2g}{m_1 + m_2}$

Now you can do the same thing to solve for $T$. Usually, it is step 5 (factoring) that is the hardest part.

Answer 2

Since $T-m_2g=-m_2a$, you know that $T=m_2g-m_2a$. But $T=m_1a$ too. Therefore$$m_2g-m_2a=m_1a$$and this is equivalent to$$a=\frac{m_2g}{m_1+m_2}=\frac{m_2}{m_1+m_2}g.$$

From this, you deduce that$$T=m_1a=m_1\frac{m_2}{_1+m_2}g=\frac{m_1m_2}{m_1+m_2}g.$$

Answer 3

You have two equations and two unknowns: $T$ and $a$. Plug the value of $T$ from the first equation into the second equation, we get $$m_1a -m_2 g = -m_2 a$$ from which we can solve for $a$ $$a= \frac{m_2g}{m_1+m_2}$$ Now that we have an expression for $a$ independent of $T$, plug it back into equation 1: $$T=m_1 a=\frac{m_1 m_2g}{m_1+m_2} .$$ For more information see https://www.mathplanet.com/education/algebra-1/systems-of-linear-equations-and-inequalities/the-substitution-method-for-solving-linear-systems