A block of mass $m$ is placed on a smooth wedge of inclination $\theta$ . The whole system is accelerated horizontally so that the block does not slip on the wedge. Determine the force exerted by the wedge on the block.

Let the wedge is accelerated towards left with an acceleration ' $a$ '. The FBD of the block in the frame from is shown. The block remains at rest with respect to wedge, so along the inclined plane, we have,

$mg \sin \theta - ma \cos \theta = 0$

  or                               $a=g \tan \theta$ .          …(i)

Perpendicular to the inclined plane, the block is also at rest, therefore

$N = mg \cos \theta + ma \sin \theta$

    $=mg \cos \theta + m(g \tan \theta ) \sin \theta = \frac{mg}{\cos \theta }$ .

Thus force exerted by the wedge on the block,

$=\frac{mg}{\cos \theta }$