Figure shows a system of four pulleys with two masses `m_(A) =3kg` and `m_(B)=4 kg`. At an instant force acting on block A if block `B` is going up at an acceleration of `3 m//s^(2)` and pulley `Q` is going down at an acceleration of `1 m//s^(2)` is
.

Options

A. `7N` acting upward
B. `7N` acting downward
C. `10.5 N` acting upward
D. `10.5 N` acting downward

Answer

D

Explanation

Writing constraint for the string
`2x_(A) +2x_(B) -x_(Q) +k =l`
Differnting twice w.r.t time `2a_(A)+2a_(B) -a_(Q)=0`
Taking downward direction to be positive
`2a_(A)+2(-3) -1 =0, 2a_(A) =7, a_(A) =3.5 m//s^(2)`
Positive sign indicates that block is acceleration at `3.5m//s^(2)` in downward direaction.
So, force acting `F =3 xx 3.5 =10.5 N` (downward) .