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I found this question in a book:

Is the acceleration of the block of mass m the same in part (a) and (b)?

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At first I thought that the force applied on the right side is the same in both cases so the answer should be the same. But the answer was actually that they are different.

After a bit of thinking I realised that in part (a) the tension in the wire is not $2mg$ but less than that. However in part (b) we would take tension in the wire equal to $2mg$.

That got me thinking:

What exactly does it mean (physically/experimentally) to apply a force $2mg$ on a wire and how is it different from attaching a block of mass $2m$ onto it?

Clarification: What if I am appying the 2mg force by hand .. why doesnt the mass of my hand need to be taken into account?

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It does not matter how the external forces ($mg$ and $2mg$) are generated because they have the same magnitude and direction in both cases, what is important is the total mass which they are accelerating, $m$ in one case and $3m$ in the other case.

It is perhaps easier to see what the difference is by converting the pulley arrangement into a horizontal arrangement where again the two external forces $mg$ and $2mg$ are acting.

enter image description here

The net force is the same in both cases ($mg$ to the right) but the total mass being accelerated is different.

You could say in your pulley arrangements that in (a) the force of $2mg$ has to accelerate the mass $m$ and the mass $2m$ whereas in (b) the force of $2mg$ has to accelerate only the mass $m$.

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    $\begingroup$ Perfect answer. $\endgroup$ – Mitchell May 3 '17 at 9:18
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In the first case, more mass is accelerating. More mass brought along naturally makes it all accelerate slower. From Newton's 2nd law:

\begin{align} \text{First case, big box:}&\quad \sum F=2ma\quad\Leftrightarrow\quad -T+2mg=2ma \quad\Leftrightarrow\quad T=2mg-2ma\\ \text{Second case, no box:}&\quad T=F \quad\Leftrightarrow\quad T=2mg \end{align}

The extra acceleration term $2ma$ because of the extra box accelerating as well is what makes the difference in the tension. Smaller tension causes the smaller box to accelerate slower (just as slow as the heavier other one).

Clarification: What if I am appying the 2mg force by hand .. why doesnt the mass of my hand need to be taken into account?

Your hand is not taken into account because we only consider the force it causes. We do not only consider the force, which the bigger mass causes, we also consider it's motion. (In other words, the box is a part of the system, the hand isn't.) If the hand should be considered with both force and motion, the external force would be on the hand and would be even bigger.

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Whenever we apply the force on the wire then we actually have overcome the tension and created our own tension in the wire. That's the reason why we do $T=2mg$ in the case. While on the other hand we have mass which creates its own force but

Don't overcome the tension, But produces a change in tension due to its force.

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