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Two masses M and m, attached by a massless string. The surface they are on is frictionless. In the first case, another massless string attached to mass m, which is pulled to the right with a force F = mg

Pulley is massless and frictionless too. Everything else in the picture is self-explanatory I think.

I have studied that we can treat the two masses as a system in the first case because they have the same acceleration (in magnitude as well as in direction). While they can’t be treated as a system in the 2nd case because their acceleration is different (magnitude is same but direction is not).

My question is, why can’t we treat the two masses as a system in the $2nd$ case? I treated them as a system and got to the correct answer when working out their individual accelerations, which is of course equal because both are constrained to move with the same acceleration (in magnitude).

Since they are moving with the same acceleration (in magnitude), why can’t we treat them as a system?

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2 Answers 2

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1) You can indeed treat them as a single system, but it is important to see why. The pulley, absence of friction, and the taut string imply that there are no unbalanced internal forces between them(if there were, it would be like 2 smaller interacting subsystems).

2) The problem is exactly one dimensional now-the one dimension being dictated by the taut string. This is what allows you to use a single $a$ for both masses(if you draw the actual figure, they have different directions).

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  • $\begingroup$ Thanks for explaining it $\endgroup$
    – 4d_
    Apr 7, 2020 at 14:45
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The only reason is that we are dealing with vector equations and treating a single system in the second equation violates vector laws. True, it works in this simple case, but to describe in a general case whether it works would not be easy.

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