I have been given the above question, with the solution,

One correction I observed from the answer was, in question c part would be acceleration of m1 and m2 are same and equal to g

Now the problem I have with the solution is that it is not in line with the string constraints. I know that T=0 so the string is not taut but just as soon as the motion will begin, it would be a problem right? They are saying in the solution that m1 comes down with g acceleration and m2 goes up with g. But if we apply string constraint then for every x distance that m1 comes down, m2 would move x/2 distance up. (in which I assumed that pulley A is fixed) and if it is movable then I am not able to apply it properly.

Also I don't understand how the motion of the pulleys would be. If m1 is coming down then the motion of pulley A should be anticlockwise by my intuition. (I might be wrong here)

Kindly explain me what is right and why

## Best Answer

If the acceleration of the two masses is $g$ downwards then the acceleration of pulley $A$ is $3g$ downwards, so pullet $A$ does move.

Let the centre of pulley $A$, $a$ move down a distance $x$.

On pulley $B$ the string on side $b1$ moves down $x$ and the string on side $b2$ moves up a distance $x$.

If pulley $C$ did not move then on side $c1$ the string would have moved up a distance $x$ and the string on side $c2$ the string would have moved down a distance $x$.

However the pulley $C$ moved down a distance $y$ so the string at $c2$ must have moved down a distance $x+2y$.

If pulley $A$ did not move then on side $a1$ the string would have moved down a distance $x+2y$ and the string on side $a2$ the string would have moved

upa distance $x+2y$.However the pulley $A$ moved

downa distance $y$ so the string at $a2$ must have moveddowna distance $2x-(x+2y) = x-2y$.This must equal $y$ as the distance moved down by the two masses is the same.

$x-2y = y \Rightarrow x = 3y$.

You will note from the right hand diagram that the acceleration of a pulley is half the vector sum of the strings.