Two blocks 'A' and 'B' having masses `m_(A)` and `m_(a)`, respectively are connectd by an arrangement shown in the figure. Calculate the downward acceleration of the block B. Assume the pulleys to be massless. Under what condition will block A have downward acceleration?

Explanation

According to Newton `2^(nd)` Law
`m_(B)g-T=m_(B)a_(B)" "("From FBD of " m_(B))" " ...1`

`2T-m_(A)g=m_(A)a_(A)" "("From FBD of "m_(A))" "...2`
Constraint relation
`a_(B) = 2a_(A)=a" "...3`
Solving equation 1, 2 & 3
`a_(B)=a=(2g(2m_(B)-m_(A)))/((4m_(B)+m_(A))), a_(A)= (a)/(2)=(g(2m_(B)-m_(A)))/((4m_(B)+m_(A)))`.
Block A will have downward motion if `a_(A) lt 0 rArr (2m_(B)-m_(A)) lt 0`
`rArr m_(A) gt 2m_(B)`