# Derivation of acceleration and tension of pulley system [closed]

On this website I am considering the question that goes along with figure 30. I am going to assume downwards is negative and rightwards is positive (not to confuse upwards as the website says).

The net forces on the object with mass $$m_1$$: $$\sum{\vec{F}}=T+m_1g-m_1g=T$$ The net forces on the object with mass $$m_2$$: $$\sum{\vec{F}}=T-m_2g$$

Using Newton's Second Law: $$T=m_1a$$ $$T-m_2g=m_2a$$

Solving for $$T$$ and $$a$$ as the website does I get: $$a=g\left(\frac{m_2}{m_1-m_2}\right)$$ $$T=g\left(\frac{m_1m_2}{m_1-m_2}\right)$$

My question is are these equations correct as the result of changing the signs of the directions? Or should I make all forces positive as done on the website?

## closed as off-topic by ZeroTheHero, Jon Custer, Bill N, Kyle Kanos, stafusaFeb 2 at 0:01

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• This arrangement is sometimes called a "half-Atwood's machine" and similar names. – dmckee Jan 29 at 0:17
• No! You should choose down as positive so that the positive acceleration of $m_1$ is coordinated with the positive acceleration with $m_2$. if you are going to use the same letter $a$ for both accelerations. Otherwise, use $a_1$ and $a_2$ with $a_1=-a_2$. – Bill N Jan 31 at 20:51
• @BillN Isn’t that what I did on my answer? Or am I misinterpreting something? – Brady Dean Feb 1 at 0:54
• If to the right is positive for $m_1$, and the string stays tight, then down must be positive for $m_2$ if you use the same variable letter for the acceleration of both masses. The N2L equation should be $m_2 g - T = m_2 a.$ – Bill N Feb 1 at 3:53

I have realized that I must set $$T-m_2g=-m_2a$$ because of my sign changes and then the equations work out.