A cylinder of mass m rests on a supporting block as shown. If `beta=60^(@) " and " theta=30^(@)`, calculate the maximum acceleration 'a' which the block may be given up the incline so that the cylinder does not lose contact at B. (neglect friction anywhere).

Explanation

`(sqrt(3)N_(1))/(2)-(mg)/(2)-(N_(2)sqrt(3))/(2)=ma`
`(sqrt(3)N_(1))/(2)-(sqrt(3)N_(2))/(2)=ma+(mg)/(2)`.....(`i`)
`rArr (N_(1)+N_(2))/(2)=(mgsqrt(3))/(2)`
`rArr (sqrt(3)N_(1))/(2)+(sqrt(3)N_(2))/(2)=(3mg)/(2)`….. (`ii`)
From equation (`i`) and (`ii`), we have
`sqrt(3)N_(1)=2mg+ma`, `sqrt(3)N_(2)=mg-ma`
Since cylinder does not losse the contact, so `N_(1) ge 0`, `N_(2) ge 0`
`rArr a le g`
`rArr a_(max)=g`