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In the 1st picture we consider the tension in the rope to be the same on both the sides. However in the second picture we take it to be different. Why? If we have to consider a general case then why didn't we take $T$ to be different in the 1st case?

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    $\begingroup$ Without knowing the assumptions of the system in the second picture there is no way to know. Also you should please type out the relevant text rather than pictures of the text, and please make all figures upright. $\endgroup$ Mar 22, 2019 at 4:12
  • $\begingroup$ What is the data and problem of 2nd picture???? $\endgroup$
    – user213933
    Mar 22, 2019 at 4:22
  • $\begingroup$ Also keep in mind that some solutions keep things separate at first before setting them equal to each other. Just because two things are denoted by different variables does not mean they cannot be equal. $\endgroup$ Mar 22, 2019 at 4:50
  • $\begingroup$ Imagine a case when there is no pulley and mass 1 is tied above mass 2. The ropes and objects are in free fall and there will be no tension in the rope. Your second scenario is half way between the first scenario and the pulley-less scenario. $\endgroup$ Mar 22, 2019 at 5:10
  • $\begingroup$ Same rope, same tension. However as others said, depends on the system. $\endgroup$
    – Nikita
    Mar 22, 2019 at 9:07

1 Answer 1

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Your calculations are correct, just up to the point where you assume the cord tensions are not equal. In fact by acknowledging them being equal we get to the answer. I just type what you have in your photos, then continue with the rest.

for T2,

$$m_2g - T_2 = m_2\alpha \quad (1) $$ For T1

$$ T_1 =m_1\alpha \quad (2)$$

Now we know $ T_1=T_2 $

Substituting T from (2) into (1) we get

$$ m_2g-m_1\alpha=m_2\alpha $$

From here we get the acceleration, $ \alpha $.

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  • $\begingroup$ But how do I know when to take T as equal? $\endgroup$
    – T.James
    Apr 7, 2019 at 2:31
  • $\begingroup$ @T.James, because a free pully just changes the direction of the rope and its tension vector. If your pully had some friction or was attached to a mechanism inputting energy or taking energy out as work, then T1 would not be equal to T2. $\endgroup$
    – kamran
    Apr 7, 2019 at 4:43
  • $\begingroup$ Please elaborate what you mean by inputting energy or taking out energy as work $\endgroup$
    – T.James
    Apr 9, 2019 at 20:02
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    $\begingroup$ @T.James, if you have a source of power attached to the axis of pulley, such as an electrical motor your T1 and TWO will be different by the amount of the motor's torque divided by the pulley's radius. Or if you attach the pulley to a fan, again the tension is different. But for a frictionless pulley and a rope the will effortlessly bend around it, tension of the rope is constant. $\endgroup$
    – kamran
    Apr 9, 2019 at 20:54

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