In the pulley system shown in figure . P and Q are fixed pulleys while A, B and C are movable pulleys each of mass 1kg . The strings are vertical and inextensible . Find the tension in the string and acceleration of frictionless pulleys A,B and C

Explanation

A single string whose ends are tied to centres of A and B, passes over all the pulleys, so tension at each point of string is same equal to T Weight of each pulley A,B andC is
`mg=1g `Newton.
Let `y_(A),y_(B)` and y_(C)` be the distance of centres of pulleys A,B and C from fixed pulleys at any time t.
Following the string starting from end A and reaching upto end B, we have
`(y_(B)-y_(A))+y_(B) +2y_(A)+y_(C)+y_(C)-y_(B)=L`= constant
Differentitating twice with respect to t, we get
`(d^(2)y_(A))/(dt^(2))+(d^(2)y_(B))/(dt^(2))+2(d^(2)y_(C))/(dt^(2))=0`
i.e., `a_(A)+a_(B)+2a_(C)=0`
where `a_(A),a_(B)`, and `a_(C)` are acceleration of pulleys A,B,C respectively.
Now equations of motion of pulleys A,B and C are
`mg+T-2T=ma_(A) rArr mg-T=ma_(A) ` ....(2)
`mg+T-2T=ma_(B) rArr mg-T=ma_(B)` ....(3) and
`mg-2T=ma_(C)` ....(4)
From (2) and(3) it is obvious that
and `a_(A)=a_(B) =(g-T/M) `.....(5) and from (4), `a_(C) =g-(2T)/m`.....(6)
substituting `a_(A),a_(B)` and `a_(C) ` in (1) , we get
(g-T/m)+(g-T/m)+2(g-(2T)/m)=0`
`4g-(6T)/m=0 rArr T=2/3 mg=2/3xx1xx9.8=6.5N`
`:. a_(A)=a_(B)=(g-T/m)=9.8-(6.5)/1=3.3 m//s^(2)`
From (1), `ac=-a_(A)= -3.3m//s^(2)`