A pulley-rope-mass arrangement is shown in fig. Find the acceleration of block `m_(1)` when the masses are set free to move. Assume that the pulley and the ropes are ideal.

Explanation

Appling the equations of motion for pulleys and blocks
For `m_(1):T-m_(1)g=m_(1)a_(1)` ….(i)
For `m_(2): m_(2)g-T_(2)=m_(2)a_(2)` …(ii)
For pulley `P_(1) : 2T-T = 0 xx a_(p_1)`
For pulley `P_(2): T_(2)-T=0xxa_(p_2)` …(iv)

From Eq (iii), `T=0`, from (iv), `T_(2)=0` and from (i) and (ii). `a_(1)=-g` , also `a_(2)=g`. Hence, blocks `m_(1)` and `m_(2)` both will move down with acceleration g downward direction.
Calculating `a_(P_1)` and `a_(P_2)`
Constraint relations: Consider the reference line and the position vectors of the pulley and masses as shown in fig .write the length of the rope in terms of position vector and differentiate it to obtain the relation between acceleration of the masses and pulleys.

* For the length of the string connecting `P_(2)` and `m_(2)` not to change and for this rope not to slacken, `a_(p_2)=a_(2)=g`.
* Length of the string connecting `P_(1)` to `m_(1)` not to change and for this rope not to slacken:
`l=(x_(1)-x_(P_1))+x_(P_(1))+(x_(P_(2))-x_(P_(1)))+x_(P_(2))`
`=x_(1)-x_(P_1)+x_(P_2)`
Differentiating this equation w.r.t twice we get
`(d^(2)l)/(dt^(2))=(d^(2)x_(1))/(dt^(2))-(d^(2)x_(P_(1)))/(dt^(2)) +2(d^(2)x_(P_(2)))/(dt^(2))implies 0=a_(1)-a_(P_1)-2a_(P_2)`
`implies 0=g-a_(P_1)+2g implies a_(P_1)=3g`.