A block of mass m is placed on a smooth wedge of inclination Q. The whole system is accelerated horizontally so that the block does not slip on the wedge. Find the
i) Acceleration of the wedge
ii) Force to be applied on the wedge
iii) Force exerted by the wedge on the block.

Explanation

(I) For observer on the ground :

`R sin theta = ma`
`R cos theta = mg rArr a = g tan theta`
(a) Force to be applied on the wedge
`F=(M+m)a=(M+m)g tan theta`
(b) Force exerted by the wedge on the block
`=R=(mg)/(cos theta)` or `R=mg sec theta`
(II) For the observer on the wedge :
Pseudo force = ma towards left.
`because` the block appears to be at rest to him
`ma cos theta = mg sin theta`
`a=g tan theta`
`R = mg cos theta + ma sin theta`
`= mg cos theta+mg tan theta sin theta`
`= mg cos theta+mg.(sin^(2))/(cos theta)`

`= mg((cos^(2)theta+sin^(2)theta))/(cos theta)`
`R=(mg)/(cos theta)=mg sec theta`