Consider an Atwood machine with two different masses $M$ and $m$ such that $M > m$. When trying to find the acceleration of that system, all solutions I've found go like this:

There are 2 forces that act on mass $M$, tension $\vec{T}$ and the gravitational force $\vec{F_g}$. According to Newton's 2nd law we then have $\vec{T} + \vec{F_g} = m \vec{a}$. Since $\vec{T}$, $\vec{F_g}$ and $\vec{a}$ are all parallel with $\vec{a}$ directed downward, it follows that $$T - mg = -ma$$

A similar thought process for $m$ with its acceleration directed upward yields $$T - mg = ma$$

Now one can solve for a. However why does the assumption we make about the sign of $\vec{a}$ hold? Empirically speaking this seems trivial, but I still think it should be possible to deduce this from Newton's axioms. But it seems like I am too inexperienced to do this.

I do know that $\vec{T}$ is the same for both masses, both $\vec{T}$s are directed upward with the same magnitude. I also know that both $\vec{F_g}$s are directed downward, but the one for mass $M$ is greater in magnitude than the one for mass $m$. But I don't see why $\vec{F_{g, M}}$ would be large enough to overcome $\vec{T}$ and cause mass $M$ to move downward. And why isn't $\vec{F_{g, m}}$ large enough to overcome $\vec{T}$?

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    $\begingroup$ You aren't supposed to assume a sign for the acceleration. Your define a coordinate system, which determines the signs of your forces. Then using Newton's second law you determine the acceleration. The sign of the acceleration tells you what direction the acceleration is in based on your previously defined coordinate system $\endgroup$ Commented Sep 14, 2018 at 18:56
  • $\begingroup$ @AaronStevens I am sorry, but I really don't see how defining my coordinate system determines that both net forces are opposite in direction. $\endgroup$
    – ook
    Commented Sep 14, 2018 at 18:57
  • $\begingroup$ You are right. The two net forces on each object are in opposite directions regardless of the coordinate system. I am talking about when you are discussing assuming a sign for the acceleration. You don't just assume a sign for the acceleration. It arises from the coordinate system. I can type out a more detailed answer right now. $\endgroup$ Commented Sep 14, 2018 at 19:00
  • $\begingroup$ @PhysicsUndergraduateStudent, I always told my students that "up" on one side of the Atwood machine is the positive direction. Also, since the pulley changes the direction of the string, "down" on the other side of the Atwood machine is also positive. Conceptually, this means that the two weights are moving "in the same direction", and they are moving at the same speed because they are connected by a string. $\endgroup$ Commented Sep 15, 2018 at 1:34
  • $\begingroup$ You can do this, and it's easy in this simple case, and in cases where there are inclines. In situations with more objects connected in complicated ways, it gets hard to figure out the "natural" direction for the positive axes. The solution by Aaron Stevens is more general and requires less thinking. $\endgroup$
    – garyp
    Commented Sep 15, 2018 at 17:26

1 Answer 1


You don't assume a sign of the acceleration. The sign of the acceleration arises from the coordinate system you choose to use and the relative magnitudes of the forces.

Let's say that upwards is positive and downwards is negative. Let's call the masses $m_1$ and $m_2$ for this, which will make keeping track of each mass easier. A diagram of this is shown below.

Diagram of problem

You are right in assuming that each mass experiences the same magnitude of tension upwards$^*$, and the weight of each mass acts downwards. It is always helpful in problems like these to draw free body diagrams of each object. Diagrams like these can help us write out the net force on each object to use in the equation for Newton's second law.


Therefore, for each block, Newton's second law looks like this: $$\sum F_1=T-m_1g=m_1a_1$$ $$\sum F_2=T-m_2g=m_2a_2$$

Now, due to how the masses are linked, it must be that $a_1=-a_2$ (since if one goes up the other goes down. This can also be "derived" if you consider the fact that the length of the rope connecting the masses must be constant). For simplicity let's say $a_1=-a_2=a$. Then we can change our equations to be $$T-m_1g=m_1a$$ $$T-m_2g=-m_2a$$

Solving for the acceleration we find that: $$a=\frac{(m_2-m_1)g}{m_1+m_2}$$

Now, let's say $m_1>m_2$. Then we see that $a<0$. Now remember, $a=a_1$ is the acceleration of block $1$. Therefore, if block $1$ is more massive, it accelerates downward (in the negative direction based on our coordinate system). You can apply a similar argument if $m_1<m_2$. Notice how we did not assume anything about the sign of the acceleration. The sign came out of applying Newton's laws to the problem with our defined coordinate system. This is the best way to approach the problem, since in more complicated systems you might not actually know which way things are going to move at first.

It seems like it would also be helpful for you if we solved for the tension in the string as well: $$T=\frac{2m_1m_2g}{m_1+m_2}$$

If $m_1>m_2$ (block $1$ goes down and block $2$ goes up), then we see that $$T=\frac{2m_1m_2g}{m_1+m_2}>\frac{2m_1m_2g}{m_1+m_1}=m_2g$$

So we do see that in fact tension is large enough to lift the smaller mass. We can do a similar argument to show that the weight of mass $1$ is larger than the tension.

Therefore, we have two different ways to show that the larger mass will move down and the smaller one will move up.

$^*$ Having the tension magnitude as the same for each mass depends on two assumptions. First, if we assume the pulley is massless and frictionless, then a net torque is not needed to rotate the pulley. Contrast this with the case of a pulley with mass where each tension needs to be different in order to rotate the pulley. Second, if we assume the rope is massless, then the rope does not need to support its own weight. This means the tension will not depend on the height of each mass.

  • $\begingroup$ Could you perhaps explain the following step a little more? "Now, due to how the masses are linked, it must be that $a_1 = - a_2$". That's essentially what I have been trying to ask. Why is $a_1 = - a_2$? Even though this seems trivial and obvious, I still feel like it's worth proving from Newton's Axioms. $\endgroup$
    – ook
    Commented Sep 14, 2018 at 21:37
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    $\begingroup$ @PhysicsUndergraduateStudent This isn't from Newton's laws. This is just from the fact that the two masses are linked in such a way as to make this true. It is a part of our system. If one mass goes up, the other goes down. If one mass is accelerating upwards, the other one must be accelerating downwards. If you want to go a little deeper, you could argue that this arises from the fact that the total lengths from each mass to the pulley must stay constant (i.e., the rope length is constant). $\endgroup$ Commented Sep 14, 2018 at 21:40
  • $\begingroup$ @PhysicsUndergraduateStudent So $y_1+y_2=L$ is constant, where $y$ means distance to the pulley. Then it must be that $\ddot y_1+ \ddot y_2 = 0$. This is not from Netwon's laws though. $\endgroup$ Commented Sep 14, 2018 at 21:41
  • $\begingroup$ Yes! I was probably over thinking this. But that "constant length" argument is what I was looking for. $\endgroup$
    – ook
    Commented Sep 14, 2018 at 21:43
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    $\begingroup$ This is a good answer. Would you consider improving it with a diagram? $\endgroup$
    – DrSheldon
    Commented Sep 15, 2018 at 1:05

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