Counterintuitive tension: weight attached to block on table via a pulley? A problem in Halliday-Resnick's "Fundamentals of Physics" poses the question of finding the tension in a rope connecting two blocks of masses $m$ and $M$. The block with mass $m$ is hanging from a rope that goes over a pulley which is attached to a block of mass $M$ on a frictionless table. So as the block of mass $m$ falls, the rope will pull the block of mass $M$ across the table.
They calculate that
$$T={M \over {M+m}}mg.$$
Now if I set $m:=100M$, we will then have $$T={M \over {M+100M}}100Mg \approx Mg.$$
This seems counterintuitive since the hanging block is much more massive (100 times as much mass), but the the tension in the rope is being dictated by the much smaller mass on the table, which is just sitting there passively.
Can anyone resolve this apparent paradox?
 A: If the force were determined by the massive block then it would mean that any small block could slow down the large block the same. If the force were approximately proportional to $m$ then both $m=1000M$ and $m=100M$ would produce the same force. That would be bizarre. The large masses would accelerate the same but the small masses would accelerate vastly differently in the two cases.
A: For a better visualization it can be helpful to take a problem to extremes. Say the falling mass weighs 1 million kilos and the horizontally accelerating mass weighs 1 milligram, so the rope only has tension to accelerate 1 milligram to near 9.8 mps^2 despite the 1 million kilos falling at near g.
A: Let's get straight to the situation. As your hanging and subsequently, falling mass $m \gg M$ (the mass on the table) you can assume the tiny lingering constraints of the string would not effectively resist the free fall of the massive body. So, the acceleration of the massive body can be assumed to be $=g$. This acceleration would be same for the other block connected to via the string. Thus the tension would be $Mg$
