In the arrangement shown in fig., all pulleys are smooth and massless. When the system is released from the rest, acceleration of block 2 and 3 relative to 1 are `1 ms^(-2)` downwards and `5ms^(-2)` downwards, respectively. Acceleration of block 3 relative to 4 is zero.

Find the absoulte acceleration of block 2.

Answer

`3ms^(-2)(downwards)`; `1ms^(-2)(downwards)` ; `A=2m=18kg`; `B= m=9kg` ; `C=m'=7kg`.

Explanation

Let acceleration of block B relative to pulley E be b downward, the acceleration of block C relative to pulley E will be b upward and let acceleration of block A be a upward, then acceleration of pulley E will also be a but downward. Hence, resultant accelerations of blocks B and C become `(a+b) and (a-b)` both downward, respectively.
Acceleration of B relative to A.
`a_(B,A)=(a+b)+aimplies a_(B,A)=(2a+b)=5ms^(-2)` ..(i)
And downward acceleration of C relative to A
`a_(C,A)=(a-b)+aimplies a_(C,A)=(2a-b)=3ms^(-2)` ..(ii)
from eq. (i) and (ii)
`a=2ms^(-2) and b=1ms^(-2)`
Resultant acceleration of `B=(a+b)=3ms^(-2)(downwards)`
And that of `C=(a-b)=1ms^(-2)` (downward)
Now considering free body diagrams,

For block A, `T_(1)-2mg=2ma`
`(T_(1)-2mg)=4m` ...(iii)
For block B, `mg-T_(2)=m(a+b)`
`(mg-T_(2))=3m` ..(iv)
For block C, `m'g-T_(2)=m'(a-b)`
`(m'g-T_(2))= m' xx 1 = m'` ...(v)
For pulley E, `2T_(2)+11.25g-T_(1)=11.25a`
`(T_(1)-2T_(2))=90` ..(vi)
From above equation, `m=9kg , m'=7kg`
Mass of block A=`2m=18kg`
Mass of block B= `m=9kg`
Mass of block C=`m'=7kg`.