According to figure, the reading of the spring balance will be: (all contacts are smooth)  $[g=10m/s^2]$

From the frame of the chamber,
 
 

$T-50-10=5a_r$

$T-60=5a_r$             ...(1)

$\left(100+20\right)\sin 30°-T=10a_r$

$60-T=10a_r$           ...(2)

From (1) and (2), we get,
$a_r=0,T=60\text{N}=6\text{kgf}$