Question

A block of mass m is placed on a smooth wedge of inclination θ. 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 has a magnitude
(a) mg
(b) mg/cosθ
(c) mg cosθ
(d) mg tanθ

Answer 1

(b) mg/cosθ


Free-body Diagram of the Small Block of Mass 'm'

The block is at equilibrium w.r.t. to wedge. Therefore,
mg sinθ = ma cosθ
⇒ a = gtanθ

Normal reaction on the block is
N = mg cosθ + ma sinθ

Putting the value of a, we get:
N = mg cosθ + mg tanθsinθ

$N=mg\cos \theta +mg\frac{\sin \theta }{\cos \theta }\sin \theta N=\frac{mg}{\cos \theta }$