Answer to Question #92485 in Classical Mechanics for Neha

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. What is the magnitude of force exerted by the wedge on the block?

Answer 1

When the whole system is accelerated to balance the sliding of the block, horizontal force (which accelerates the system) acts on the block such that the block doesn't slide:

$$F_{ext}\cos\theta=mg\sin\theta$$

On the other hand, the block pushes the wedge with the force of:

$$F=F_{ext}\sin\theta+mg\cos\theta$$

It can be rewritten as:

$$F=mg\tan\theta\sin\theta+mg\cos\theta=\frac{mg}{\cos\theta}$$