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The acceleration for this system is given by

$$a = \frac{F_{\text{net}}}{m_{\text{tot}}} = \frac{m_2 g - m_1 g}{m_1 + m_g}.$$

My question is, why do we neglect tension here when it is clearly acting on the masses and will contribute to the net force on this system?

enter image description here

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To obtain the solution the system with the pulley was transformed into the following horizontal system where the massless inextensible string could just as well be a massless rigid rod.

enter image description here

The system of the two masses and the string has two external forces acting on it $m_1g$ and $m_2g$ with the forces labelled $T$ as internal forces and which are equal in magnitude and oppose in direction.

taking to the right as positive the net external force, $m_2g-m_1g$, is the force which accelerates the two masses.

If one needed to consider a system of only one of the masses then the force $T$ would then be considered to be one of the external force.

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  • $\begingroup$ How did you define them to be system?Is it like a convention for such setups to be considered system $\endgroup$ – user184271 Apr 14 '18 at 6:34
  • $\begingroup$ @magemaro How you define a system is up to you but once you have done so you can then decide which forces are external and which forces are internal. Internal forces will always be in Newton's third law pairs. $\endgroup$ – Farcher Apr 14 '18 at 6:55
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The solution doesn’t neglect tension. To see that:

Consider the case of equal masses. In that case they don’t move. What holds them up? The tension T equals their weight. The pulley provides 2T to hold both up.

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  • $\begingroup$ Okay.So what happens to tension here...did it get cancelled due to any opposite force $\endgroup$ – user184272 Apr 7 '18 at 2:44
  • $\begingroup$ The action of tension is responsible for the minus sign in the numerator. $\endgroup$ – probably_someone Apr 7 '18 at 2:45
  • $\begingroup$ When I make force equations for both the masses ans solve it then indeed the tension gets cancelled but what I don't understand that even though they act in same direction,they are taken to be opposite $\endgroup$ – user184272 Apr 7 '18 at 2:55
  • $\begingroup$ The pulley changes the direction of the string. When one side goes up, the other goes down. Hence different signs. $\endgroup$ – Bob Jacobsen Apr 7 '18 at 2:58
  • $\begingroup$ So they get cancelled in the numerator.That makes sense now $\endgroup$ – user184272 Apr 7 '18 at 2:59
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You can just sum all the gravitational forces, and than divide with tota mass. You will get the acceleration. For example,if mass one is 2 kg and mass 2 is 4 kg total mass is 6 kg and total force (neglecting the string) is 40 N -20 N which equals 20 N as resultant force. So acceleration is 20 N / 6 kg. So in general for the pulley you can just add up external forces and total mass and you get the result.

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If the system is the pulley, the two blocks and the string then tension in the string can be ignored because it is an internal force. Internal forces do not affect the motion of the centre of mass of the system.

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