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Consider a system in which you have a rope connected to 1 pull above (A) which connects to a pull below (B) in which some object of mass 10kg is attached To the pully(not the rope). The rest of the rope goes to the celing which contains infinite rope.

I know that there is 2 tensions acting on this rope (the left side) and (the right side). So if I want to raise this object by say 1 meter how will I mathematically explain that I must pull down the rope with twice the distance. I cannot seem to prove this mathematically using the simple kinematic equations because if you pull with a force = to the weight of the object it rises with constant speed. i cannot find this speed and therefore cannot find the time so that I can calculate the position.enter image description here

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  • $\begingroup$ Can you provide a sketch of the system? $\endgroup$ – Steeven May 11 '15 at 19:30
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In the configuration illustrated in the image you have a "rope total length conservation law". I'll notate the part of the rope attached to the ceiling as A, the part that is hold by the hand as C, and the remaining part as B. So $ A+ B+C=l_{tot} $ . Raising the mass by $ \Delta x $ will cause parts A and B to loose each one that amout of length, and by "rope total length conservation law" the change in length of C is $ \Delta C = - 2 \Delta x $ (the minus just tells that the change in length of C is opposite to the distance of the mass to the ceiling).

$$ l_{tot} =A+B+C =A_{new}+B_{new}+C_{new}=(A- \Delta x)+(B - \Delta x) + C_{new}=l_{tot}-C - 2 \Delta x + C_{new} $$

$$ C - C_{new} = \Delta C = - 2 \Delta x $$

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