A pulley fixed to the ceiling of an elevator car carries a thread whose ends are attached to the loads of masses `3m` and `m` . The car starts gong up with acceleration `g//s`. Assuming the masses of the pulley and the thread, as well as friction, to be negligible, find the force exerted by the pulley on the ceiling of car.

Explanation

Let the acceleration of block `A` with respect to pulley be `a` in the downward direction, obvisously acceleration of block `B` will be in upward direction.

Acceleration of `A` w.r.t. the ground
`a_(A//g)=a_(A//p)-a_(p//g)=(a-a_(0))` In downward direction
`a_(B//g)=a_(B//p)+a_(p//g)=(a+a_(0))` in upward direction

`3mg-T3m(a-a_(0))` (i)
`T'=mg=m(a+a_(0))` (ii)
Solving `a=(3g)(5),T=(9mg)/(5)`

`R-2T=0` (pulley is massless)
`R=2T=(18mg)/(5)`
Note`:` `1.` First, sketch free body diagram.
`2.` Decide acceleration of blocks with respect to ground
`3.` Solve the equation.
If we work from the elevator frame (non`-`inertial frame), we have to apply pseudo force on each block.

Pulling force `=3m(g+a_(0))-m(g+a_(0))`
`=2m(g+a_(0))`
Mass to be pulled `=3m+m=4m`
`a=(2m(g+a_(0)))/(4m)=(g+a_(0))/(2)`

`T-m(g+a_(0))=ma=(m)/(2)(g+a_(0))`
`T=(3m)/(2)(g+a_(0))=(3m)/(2)(g+(g)/(2))=(9mg)/(5)`

`R-2T=0`
`R=2T=(18mg)/(5)`