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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):

enter image description here

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.

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

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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 $\ddot{y}_2$

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  • $\begingroup$ Thanks for your reply! But I have to say I still don't get why the forces exerted by two differing masses on a rope are the same. This is still totally counterintuitive for me, even I if imagine a massless, inextensible rope. I would think the forces are not the same, and the rope and the masses will move in the direction of the greater pull. :/ $\endgroup$ – Benjamin Márkus Oct 4 '14 at 9:33
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    $\begingroup$ The force is a constraint. It is there to ensure the fixed length of the rope. It the rope was straight you would see it as equal and opposite forces (action/reaction), but the pulley transforms the direction. $\endgroup$ – ja72 Oct 4 '14 at 14:43
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Ok, so since I asked the question I figured out that this setup is called Atwood's machine, and has a very nice and easy to understand explanation here: Tension in an Atwoods machine conceptual?

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