A block of mass `m` is placed on a smooth wedge of inclination `theta`. The while system is acceleration horizontally so that the block does not slip. Find (a) horizontal acceleration, (b) normal force between the block and the wedge and (c) the horizontal force applied on the wedge.

Explanation

The wedge and the block move together in horizontal direction.
Block`:`

`uarrNcostheta-mg=0rArrN=(mg)/(costheta)`
`rarr:Nsintheta=marArra=gsintheta`
Wedge

`F=(M+m)a=(M+m)g tantheta`
This problem can be solved from the wedge frame. This frame is non`-`inertial as it has acceleration `a` with repect to an inertial frame (the ground). Newton's second law can be used is we apply a pseudo force on the block of mass `m`.

With respect to the wedge, the block is stationary along the plane`:`
`mgsintheta=macostheta=ma^(2)=0`
`a=g tantheta`
`bot^(ar)` to the plane`:`
`N=(mgcostheta+masintheta)=ma_(0)=0`
`N=mgcostheta+masintheta`
`mgcostheta+matanthetasintheta`
`=(mgcos^(2)theta+mgsin^(2)theta)/(costheta)=(mg)/(costheta)`
OR
If we resolve forces in the horozontal and vertical direction.

`uarr:Ncostheta-mg=0rArrN=(mg)/(costheta)`
`rarrNsintheta-ma=0rArra=g tantheta`