In the arrangement shown , by what acceleration the boy must go up so that 100 kg block remains stationary on the wedge. The wedge is fixed and friction is absent everywhere . ( Take `g = 10 m s^(-2)` )

Explanation

For the block to remain stationary, `T = mg sin theta`
`T=100xx10xx sin 53`
`= 100xx ""^(2)cancel(2)xx(4)/(cancel(5))=800N`
FBD for man
`T-mg=ma`
`T=m(g+a)`
`80 cancel(0)=5cancel(0)(10+a)`
`a=6 m//s^(2)`