By what acceleration the boy must go up so that 100 kg block remains stationary on the wedge. The wedge is fixed and is smooth. (g=10 m//s^(2))

By what acceleration the boy must go up so that 100 kg block remains s

1.	By what acceleration the boy must go up so that 100 kg block remains stationary on the wedge. The wedge is fixed and is smooth. (g=10 m//s^(2))
Answer» Solution :For the block to REMAIN STATIONARY, `T = mg sin THETA` `T=100xx10xx sin 53` `= 100xx ""^(2)cancel(10)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)`