In the arrangement shown in figure, what should be the mass of block A so that the system remains at rest. Also find force exerted by string on the pulley Q.
`(g = 10 m//s^(2))`

Answer

`m=3kg30sqrt(3)N`.

Explanation

From figure
FBD of 2 kg block hanging vertically `to`

`T=20N . . . . . . . . (1)`
FBD of 2kg block on incline plane `to`

Along the plane `toT+2gsin30^(@)=T'`
`T'=20+20xx1/2=30N`
FBD of block A `to`

So `T'=M_(A)g`
`M_(A)=(T')/(g)=(30)/(10)=3kg`
`M_(A)=3kg`
FBD of pulley Q `to`

So, `R=2T'"cos"(theta)/(2)`
`R=2xx30"cos"30^(@)`
`R=2xx30xx(sqrt(3))/(2)`
`R=30sqrt(3)N`