# Determine the acceleration in a double Atwood machine

Trying to solve the following double Atwood machine:

Suppose there is a mass of 12kg hanging on an ideal rope that wraps around an ideal pulley, and that the other end of the rope is attached to another ideal pulley's axis. This second pulley has a rope around it which on one side holds a mass of 4kg and on the other a mass of 8kg. What is the acceleration of the first mass?

So here's what I've done so far: First I've determined that the tension in the first rope is twice the tension in the rope for the "subsystem", and that the acceleration of the first mass is the average of the accelerations of the two other masses but in the opposite direction. That is to say

$$T' = T/2$$

$$a_{1}=-\frac{a_{2}+a_{3}}{2}$$

From free-body diagrams we can determine that

$$-12a = T-12g$$

$$4a_{2} = T/2 - 4g$$

$$-8a_{3} = T/2 - 8g$$

However, when I solve this system I obtain $T=32a$ which seems impossible that the tension would be greater than the greatest force of gravity of any mass.

Part of me wonders if this has anything to do with it: How do I know which direction the third mass is traveling? The first mass is causing the subsystem to accelerate upwards but the third mass is accelerating downwards relative to the subsystem. So how do I know the direction of acceleration relative to the ground?

• Good homework question. – BMS Jul 31 '14 at 20:57

Let \begin{align} m_1 = 12\,\mathrm{kg}, \qquad m_2 = 4\,\mathrm{kg}, \qquad m_3 = 8\,\mathrm{kg} \end{align} If you solve this problem symbolically, then you'll find that the tension $T$ applied to mass $m_1$ satisfies \begin{align} T = \left(\frac{8m_1m_2m_3}{m_1m_2+m_1m_3+4m_2m_3}\right)g. \end{align} If you plug in the values given for the various masses, then you obtain \begin{align} T=\frac{192}{17}g \approx (11.30\,\mathrm{kg})g <\text{weight of mass $m_1$}, \end{align} so it seems that your claim
• @Addem I think you may have a couple of sign errors. Also, note that you want to compute $T$ as a multiple of $g$ not as a multiple of $a_1$ in order to compare to the weight of mass $m_1$. See the equations I added at the end. – joshphysics Aug 1 '14 at 0:49
• @Addem Some advice. Choose the "ground" to be $y=0$, and let "up" be the positive $y$-direction. Then write everything relative to this coordinate choice. You should never manually put in a negative sign with the acceleration $a$ when you write Newton's Second Law; the sign is contained in that symbol. – joshphysics Aug 1 '14 at 3:03
• @Addem I think that is the correct answer for $a_1$. Note that I wrote the answer for the tension which is not $m_1a_1$. It's $m_1a_1+m_1g$. – joshphysics Aug 1 '14 at 3:37