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Original question

In the question above, what I don't understand is how to calculate the tension in the second case (case in which the pulley is there). I realized after reading some questions, that the arrangement in question is an Atwood machine. While I can do this question by using the formulae, I do not understand how to obtain the formulae.

Wikipedia states that (http://en.wikipedia.org/wiki/Atwood_machine#Equation_for_constant_acceleration) when T is the tension and m1 and m2 are the masses in the Atwood machine, such that m1 > m2, i.e." Forces affecting m1 = m1g - T = m1a and forces affecting m2 = T - m2g = m2a

I do not understand how we got this, and this is probably because I don't understand how the tension divides in such a case. Please help me by explaining what pulls what which leads to which equation and how these equations (from Wikipedia) are true.

EDIT: To me, I imagine this question as if the tension is exerted by the midpoint of the string of the pulley and the tension exerted for each side = Mg N. which makes the total 2Mg N tension, leading to 2l elongation. But pulleys are supposed to make life easier, and my calculation does not show anything like that. Please clarify.

EDIT 2: OK, I try to comprehend what Wikipedia says. So the first question: How can the tension be UPWARDS at BOTH sides of the pulley? By the definition of tension, it seems fine, but how does this work in the real world at the midpoint of the pulley string? The tensions are coming in opposite directions, so if they are equal, tension at midpoint = 0?

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    $\begingroup$ An illustration of this would be helpful. $\endgroup$ – Steeven Oct 12 '14 at 8:29
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    $\begingroup$ "So the first question: How can the tension be UPWARDS at BOTH sides of the pulley?" This is a puzzling question because I can't imagine how you could think it would be otherwise. Both masses are being pulled downward by gravity which means that each end of the wire is being pulled acting to stretch (lengthen) the wire. It seems intuitive that the wire would resist this by 'pulling up'. You wouldn't expect the wire to push down on the masses would you? If the wire were horizontal and being pulled at both ends, wouldn't the wire 'pull back' at both ends? $\endgroup$ – Alfred Centauri Oct 19 '14 at 12:00
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    $\begingroup$ "then I don't understand what happens at midpoint of rope?" - if the rope is static (unmoving, unaccelerating), then the net force at any point in the rope must be zero. Pick a point, the midpoint for example, on the rope and draw the forces on it; they are equal and opposing forces (otherwise the point would accelerate). However, at the ends of the rope, you and the rigid support must supply the opposing forces. Does this help? $\endgroup$ – Alfred Centauri Oct 19 '14 at 12:40
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    $\begingroup$ @Swapnil, your assumptions (or drawings) are incorrect. Consider again what I wrote in my previous comment and also, take a look at the Wikipedia article on tension: en.wikipedia.org/wiki/Tension_%28physics%29: "In physics, tension describes the pulling force exerted by each end of a string, cable, chain, or similar one-dimensional continuous object," $\endgroup$ – Alfred Centauri Oct 19 '14 at 18:07
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    $\begingroup$ I'm going with E) none of the above, since there will be lower tension (because of friction) where the rope is in touch with the pulley... Question did not state "frictionless" - just "light" (although I have no idea why that should come into it). $\endgroup$ – Floris Oct 28 '14 at 0:25