Frictionless Equilibrium of Block and Wedge System

[Guppy]

Hello, I am new member here.
I would like to know how should I handle this question:

A block of mass 2.5kg rests on the rough surface of a wedge, which in turn rests on a rough horizontal floor. What is the frictional force exerted by the floor on the wedge if the whole system is at rest?

The height of the wedge is 30cm and the length is 40cm (Side view)

 

[Doc Al]

Consider that the system of "block + wedge" is in equilibrium: the net force on it must be zero. Besides the force that the floor exerts, what forces act on that system? And what frictional force must the floor exert to couteract those forces?

 

[Guppy]

The system of Block & Wedge is in equilibrium, so the whole system is balanced, and the only force acting on the floor is the Mg of the whole block & wedge system, therefore no frictional force is exerted by the floor on the wedge as there is no frictional force exerted by the wedge on the floor, isn't it?

By the way, if it's a smooth floor, the block will also stay the same way as the above case?

 

[Doc Al]

The system of Block & Wedge is in equilibrium, so the whole system is balanced, and the only force acting on the floor is the Mg of the whole block & wedge system, therefore no frictional force is exerted by the floor on the wedge as there is no frictional force exerted by the wedge on the floor, isn't it?

This is true. Here's how I would word it. Other than the floor (which merely exerts a passive force), the only force on the system (block + wedge) is gravity. Since the weight of the system acts straight down, the only force that the floor applies to counter it is the normal force--no frictional force is needed.

By the way, if it's a smooth floor, the block will also stay the same way as the above case?

Yes. Since no frictional force is needed, it doesn't matter if the floor is smooth or not. (Things would change if the mass were able to slide down the wedge.)