Centrifugal forces don't exist in reality?

[Riichiro Mizoguchi]

You claim that if O is exerting an outward force on something then it follows that something else must be exerting an outward force on O.

No. I mean if O is exerting a reactive outward force, ...

The proof you attempt sub-divides O into a bunch of little pieces looks at the outward forces between the various pieces, ignores the inward forces between the various pieces, and concludes that the sum of the forces that are examined is as desired.

No. I don't ignore them because my goal is to show the existence of a particular kind of force. There is no cancelation of forces when we are not talking about motion caused by forces. Imagine you are pushed by two persons in opposite directions of the same strength. You would say these two forces cancel each other. You don't move but it is different from a situation where no one is pushing you. You are exerted on two forces.

By the way, what we see at every boundary between pieces of objects is the same at the boundary between O and the other object generating centripetal force both of which would form action/reaction pairs?

Or, are you going to say, in the case of a book on the table, my proof doesn't work because the downward force exerted on the book is canceled by the upward force exerted on the book by the table so that there is no downward force exerted on the book?

 

[Dale]

Suppose you slice O into two pieces O1 and O2 so that |F1|=|F2| where -(F1+F2)=-F and -F1 is exerting on O2 by O1 and -F2 is on the wall by O2.

I could really use a drawing here because from your description I don’t think that this combination of facts is possible. If F2 is the force on the wall then F2=F (cutting O into pieces internally doesn’t change anything externally). So then you cannot have both F1+F2=F and |F1|=|F2|

You would say these two forces cancel each other. You don't move but it is different from a situation where no one is pushing you. You are exerted on two forces.

The force does cancel. What is different is the stress, not the force. When no one is pushing you the net force is zero and the stress is zero. When two people are pushing in opposite directions the net force is still zero but the stress is non-zero.

O does have a nonzero stress. It does not have an additional force.

 

[Riichiro Mizoguchi]

I could really use a drawing here because from your description I don’t think that this combination of facts is possible. If F2 is the force on the wall then F2=F (cutting O into pieces internally doesn’t change anything externally). So then you cannot have both F1+F2=F and |F1|=|F2|

The force does cancel. What is different is the stress, not the force. When no one is pushing you the net force is zero and the stress is zero. When two people are pushing in opposite directions the net force is still zero but the stress is non-zero.

O does have a nonzero stress. It does not have an additional force.

I'm afraid there might be a misunderstanding. Not F but -F. F is the force the wall pushes O and -F is the force O pushes the wall.

The notion of the net force is defined for describing motion caused by forces. It doesn't affect the existence of those forces. In the case of two persons push me, even if the two forces are canceled, the existence of those two forces is never canceled. They are there.

Allow me to attach a detailed proof of my claim. I hope my translation from Japanese to English works well and hope this clarifies my proof.

 

[Nugatory]

hope this clarifies my proof.

It does, and you are repeating the same error. The only outwards forces in your diagram are the second law partners of the inwards centripetal force - there is no centrifugal force. The best way to see this is to consider the forces acting on the innermost object: there is an inwards-directed force pushing it onto a curved path instead of the straight-line path it would follow in the absence of any net force, and no outward force acting on it at all.

This thread has reached the point where we're propagating misinformation under the guise of a proof, so it is closed.