# Newtons third law and Atwood Machines, confusion about tension

The classic way of solving these problems is like so: $T = mg - ma$ and $T = mg + ma$. However, according to Newton's third law, is this not wrong?

I know how to solve these problems, but the concepts are confusing to me. First off, apparently both tensions in the string are equal to each other. But I don't understand this, and according to Newton's third law, this should be false? Assuming $T_1$ acts on $m_1$ and $T_2$ acts on $m_2$ Newton's third law should mean that $T_1 = -m_1g$ because the force of gravity from the box $m_1$ acts on the rope, and the rope reacts with equal and opposite force.

But according to this same logic, $T_2 = -m_2*g$

And since the tension in the rope must be equal at all points, and also since $m_1 \ne m_2$ this means that this is false.

Does Newton's third law not apply to Atwood's machines? Or am I confused?

## 1 Answer

What is confusing you seems to be the rope-pulley component itself. When you have a set-up like this, it is equivalent to draw it without the pulley. Imagine instead that the same masses were attached to each other via a massless horizontal adamantium bar. Then imagine there is a force pulling $m_1$ to the left by $m_1g$ and pulling $m_2$ to the right by $m_2g$. Now how does the system react? Clearly, this would be the same as if the two masses were combined and you had the same forces: $$F_{net}=(m_2-m_1)g\\ a=\frac{m_2-m_1}{m_2+m_1}g$$

When using the rope and pulley, tension does the same thing. To keep the two masses the same distance apart and to uphold Newton's Third Law, the rope must apply the same tension force on both ends. Let's get that force from the horizontal bar model: We know the net acceleration of the system, which means the left side of the bar must pull $m_1$ with a force strong enough to overcome the leftward gravity and accelerate it rightwards $$F_{net}=T-F_g\rightarrow T=m_1g+m_1a$$ We also know the right side of the bar must apply a force to the left to cancel the rightward force on $m_2$ just enough so it accelerates at $a$ $$F_{net}=F_g-T\rightarrow T=m_2g-m_2a$$ Plug in the value for $a$ we found above and lo you find the T's are equal! Not surprising since that was a necessary outcome. But this is the same as the Atwood machine. Note though, that the Tensions aren't equal to the force of gravity on each mass because the system is accelerating. In this case, the true scope of Newton's Third Law is upheld because while the masses accelerate toward Earth, Earth imperceptibly accelerates towards the masses.