0
$\begingroup$

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?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ZeroTheHero, Jon Custer, Bill N, Kyle Kanos, stafusa
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This arrangement is sometimes called a "half-Atwood's machine" and similar names. $\endgroup$ – dmckee Jan 29 at 0:17
  • $\begingroup$ 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 $a_1=-a_2$. $\endgroup$ – Bill N Jan 31 at 20:51
  • $\begingroup$ @BillN Isn’t that what I did on my answer? Or am I misinterpreting something? $\endgroup$ – Brady Dean Feb 1 at 0:54
  • $\begingroup$ 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 $m_2 g - T = m_2 a.$ $\endgroup$ – Bill N Feb 1 at 3:53
0
$\begingroup$

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

$\endgroup$

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