Question

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. The force exerted by the wedge on the block(g is acceleration due to gravity) will be?

Answer 1

As there is no relative motion between block and wedge:

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

$⟹a=g\tan \theta$

The normal force by wedge on the block is

$\begin{array}{c}N=mg\cos \theta +ma\sin \theta \\ =mg\cos \theta +m\left(g\tan \theta \right)\sin \theta \\ =mg\cos +mg\frac{{\sin }^2\theta }{\cos \theta }\\ =mg\left(\frac{{\cos }^2\theta +{\sin }^2\theta }{\cos \theta }\right)\\ =\frac{mg}{\cos \theta }\end{array}$