`(1)` Find the acceleration of centre of mass of the system shown.

`(ii)` Find the normal reaction applied by the pulley on the thread.

Explanation

`(i)` Driving force on the system `=2g=20N`
Limiting fricition `= mu mg=0.5xx3xx10=15N`
Magnitude of acceleration of both masses `a=(20N-15N)/(2+3)=(5)/(5)=1 m//s^(2)`
`m_(1)=3kg`, `a_(1)=1hatjm//s`
`m_(2)=2kg`, `a_(2)=-1hatjm//s`
`vec(a_(CM))=(m_(1)veca_(1)+m_(2)veca_(2))/(m_(1)+m_(2))`
`=(3hati-2hatj)/(3+2)=(3)/(5)hati-(2)/(5)hatj`
Magnitude of acceleration of `CM`
`a_(CM)=sqrt(((3)/(5))^(2)+((2)/(5))^(2))=(sqrt(13))/(5)m//s^(2)`
`tan alpha=(2//5)/(3//5)=(2)/(3)`
`alpha=tan^(-1)((2)/(3))` below horizontal

`(ii)` Let the two blocks and the thread is our system.

Let `N_(1)` and `N_(2)` are the vertical and horizontal component of the normal force applied by the pulley on thread.
`N_(3)=3g`
For vertical direction,
`vecF_("net external y")=Mveca_(CM y)`
`implies(3g+2g-N_(3)-N_(1))=5xx(2)/(5)`
`impliesN_(1)=20-2=18N`
Now, for horizontal direction,
`N_(2)=Ma_(CM x)`
`N_(2)=5xx(3)/(5)`
`N_(2)=3N`
Net normal reaction `=sqrt(18^(2)+3^(2))=sqrt(333)N`