Third law of motion is so confusing
If you and your friend are interacting, when he pulls you, he will feel that you apply to him the same force as he applies to you. If both of you are in vacuum and no further forces are present, the change in momentum will be equal for both of you. If further forces like friction are present, then the total forces acting on each of you might differ. But the parts which come due to you two interacting with each other will still be equal.
In the rope example there are two things playing a role and they should be kept apart:
- Each segment of the rope acts on any of the adjacent segments with the same force as the adjacent segment acts on it. Here we are looking at forces acting on different objects interacting with each other. This is Newton's 3rd law.
- The weight and the rope (and thus each segment of the rope) are in a force equilibrium, so the total force on each segment exercised by both adjacent segments totals to zero. Here we are looking at the forces on one object exercised by all other objects it interacts with.
While the first point is always true, the second one doesn't have to be true, say, if the rope and the weight are in a free fall accelerating towards the earth, rather than in an equilibrium situation. Obviously, there will be no tension force in case that the second point is not given.
The motion of you and your friend depends on all the forces which are acting on you you and your masses.
When you pull your friend forward although the forces between you and your friend are equal in magnitude what about the force that you bother experience due to the Earth?
Assuming that you had equal masses if net force on your friend is greater than the net force on you then your friend will undergo a greater acceleration.
Mass will a play a part and so when you start walking towards the front of a moving train the change is speed is not noticeable although it possibly would be if you were in a small boat whose mass is much less than that of the train and thus closer to your mass.
The Atwood machine with one pulley can be thought of being equivalent to the following example.
Where the two masses at the ends of the rope are called $1$ and $3$ and the rope $2$ but you could extend it to being a whole string of as many masses as you wish.
The masses at the ends of the rope are subjected to gravitational attractive forces and the Newton third law force are labelled $F_{12}$ etc where $F_{12}$ is the force on mass $1$ due to mass $2$, the rope.
Taking towards the right as positive and applying Newton's second law gives.
$$m_3g-F_{32} = m_3a \;\;\; F_{23}-F_{21} = m_2a \;\;\; F_{12}-m_1g = m_1a $$
where is the acceleration of all the masses as the rope is assumed to be inextensible.
It is often the case that the rope is assumed to be massless which means that $F_{23}-F_{21} = 0 \Rightarrow F_{23} = F_{21}$ and these forces are called the tension in the rope.
In the case of unequal masses attached to a massless rope supported by a frictionless pulley, the tension in the rope is the same everywhere within the rope. Using m1 to represent the smaller mass and m2 to represent the larger mass, and a to represent the rate of acceleration, then
tension = m1 (g + a) = m2 (g - a)
where
a = (m2-m1) g / (m1+m2)
The same formulas apply for m1 >= m2, corresponding to a <= 0.