Consider a situation where a block is kept on a wedge which is further kept on a surface which rotates with $\omega$ the FBD for the same
My question is why dont we consider centripetal force in fbd? Is it because we are drawing it from a frame such that it cancels?
Edit :My simple definition of fbd tells me that it is a diagram representing all the forces acting on the body when viewed from a particular frame which ensures that centripetal force must be acting only in a certain frame
Best Answer
From an inertial frame
You are considering the centripetal force. In fact you have already included it if you are using an inertial frame.
The centripetal force is not a new force. It is just a name we give to whichever force that pulls centripetally (towards the centre in circular motion).
In your case there is a component from the normal force which pulls centripetally (horizontally leftwards). So that is called the centripetal force.
From the non-inertial frame
This above explanation is from an inertial frame, as if standing on the ground and watching the scenario. If you want to look at this from a non-inertial frame, by imagining that you are sitting on the block that is in rotation, which might be the case judging from your sketch, then from that frame the block doesn't seem to rotate. Rather it looks like the surroundings are "rotating".
Thus from that frame there seems to be no (resultant) centripetal force acting on the block. This means that the horizontal component of the normal force must be balanced out by something. That "something" is what we invent with the name centrifugal force, pointing opposite in the outwards direction.
This centrifugal force is a so-called pseudo-force in that it doesn't really "exist" - but we invent it mathematically in order to make Newton's laws hold true even in non-inertial frames.