# Help in finding the magnitude of a force

I am trying but I'm having difficulty getting through to you.
Read all three of newtons laws
http://en.wikipedia.org/wiki/Newton's_laws_of_motion
What I am saying is a combination of all three of them. Let me just think a little more about to break it down so you see how each one applies in this case.

Delphi51
Homework Helper
I wouldn't say that how can a skater's feet stand such a force this sport would have been banned if it was so
Have you calculated the force?
I think we agree there is a centripetal acceleration and force for the circular path up to the point C. If you accept the force a nanometer before C, why does it suddenly become so unreasonable to have it for one more infinitesimally small distance?

what about the first poin of the circular path is there any vertical acceleration ?
the body started with no speed
and I don't know there is something which make no sense in this when i calculated the force it was 2400 N (nothing wrong ) while mg = 800 N

Delphi51
Homework Helper
At the top of the circular path, velocity = 0 so no centripetal acceleration.
The skater does accelerate downward as if in free fall. However, a nm down the path there is some velocity and a tiny bit of centripetal acceleration to the left. As the skater continues to fall, velocity and centripetal acceleration increase.

Your 2400 N calc is for centripetal force plus gravitational force, right?
I'm finding it interesting that the centripetal force is twice the gravitational force regardless of the height = radius. Since potential energy is converted entirely to kinetic energy as it falls,
PE at top = KE at bottom
mgR = ½mv²
v² = 2gR
Putting this into Fc = mv²/R
gives Fc = 2mg.
The total force pushing up on the skater at the bottom of the path is three times gravity regardless of the radius. Most interesting! Skateboarders are experiencing 2 G's of acceleration but feeling 3 times their normal weight at the bottom of a bowl.

thanks man I really apreciate it Delphi51
Homework Helper
Most welcome. Quite an interesting problem, too.

thanks a lot Delphi51. I don't think would have managed to make it that clear.

Last edited by a moderator: