Reaction force at the top of a rollercoaster loop

  • Thread starter dainckp
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  • #1
dainckp
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I understand that in a rollercoaster loop, two forces combine to make the centripetal force required to keep the cart moving in a circular path - the cart's weight, and the reaction force of the track against the cart (except in the special case where the cart's weight alone is sufficient for the centripetal force)

What I don't understand is, why is there a reaction force at all? If the rails are pushing back on the cart, by Newton's Third Law the cart must be in the first place exerting an equal and opposite force on the rails, but I can't see where such a force would come from. The cart's velocity is perpendicular to the rail, and the acceleration is directed towards the centre of the loop, and I was taught there is no such thing as an outward centrifugal force, so why does the cart push on the rails in the first place?

A reaction force yes, but a reaction to what?
 

Answers and Replies

  • #2
russ_watters
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Welcome to PF!

The force is from (or, rather, causes) part of the acceleration: f=ma. Gravity causes the rest.
 
  • #3
CWatters
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If the rails are pushing back on the cart, by Newton's Third Law the cart must be in the first place exerting an equal and opposite force on the rails, but I can't see where such a force would come from.

The cart wants to go in a straight line but the rails force the cart to move in a curve. eg The rails apply a force on the cart.
 
  • #4
A.T.
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If the rails are pushing back on the cart, by Newton's Third Law the cart must be in the first place...
No, not "in the first place". The two equal and opposite forces act simultaneously, and are on the same footing. There is no cause-effect relationship between them, contrary to what the unfortunate naming "action & reaction" suggests.
 

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