# Nature of tension in a massless rope [duplicate]

This question already has an answer here:

I'm faced with a conceptual problem, where I am supposed to describe the motions of the following system (two masses on each end of a totally inelastic rope hanged on a pulley):

I understand that there is the gravitational force acting, and also a second, opposing force in the rope. My problem is that I don't understand the nature of this second force. Where does it come from? How do I summarize it for each of the masses? From my solutions manual I know that it must be equal $|T_1|=|T_2|$ for each masses, but that sounds plain counterintuitive for me. Can you please shed some light on this?

UPDATE: the problem is formulated so that the pulley and the rope are both massless, and the bearing has no friction.

## marked as duplicate by Qmechanic♦Oct 4 '14 at 11:02

The nature of the rope force is such that the accelerations of the two masses are connected. If the two heights are $y_1$ and $y_2$, then the their sum is constant, and their derivatives are equal to zero
$$y_1 + y_2 = \ell \\ \dot{y}_1 + \dot{y}_2 = 0 \\ \ddot{y}_1 + \ddot{y}_2 = 0$$
So the nature of the force is such to enforce $\ddot{y}_1 = - \ddot{y}_2$ such that when you do a free body diagram you will get
$$m_1 \ddot{y}_1 = T - m_1 g = - m_1 \ddot{y}_2 \\ m_2 \ddot{y}_2 = T - m_2 g$$ which is 2 linear equations solved for 2 variables $T$ and $\ddot{y}_2$