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I was working over some problems from my physics textbook, and I came across this one, it involves circular movement:

A car with the mass of 1t is moving over a hill with the velocity of 20 m/s. The radius of the hill (which can be viewed as a half-circle) is 100m.

When it comes to the top of the hill, with what force does it act upon the hill?

The correct answer is, apparently, 6kN. How do I solve this?

The situation can be illustrated like this I guess: enter image description here

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    $\begingroup$ What thought have you already made to this? Surely, you have done some! $\endgroup$
    – leftaroundabout
    May 22, 2012 at 21:32
  • $\begingroup$ Well, I realize that if the car was just standing there, he would act upon the hill with the force of 10kN. And it's pretty obvious to me that if it was going any faster, the hill would be "pushing him up", and once it would come to the top it would almost fly away. That's why the force is lower when it's moving. $\endgroup$
    – jcora
    May 23, 2012 at 7:43
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    $\begingroup$ Hi Bane - as described in our FAQ and homework policy, we don't allow questions that just ask for the solution to a homework-like problem. If you can edit your problem to focus on the specific physical concept that is giving you trouble, I'd be happy to reopen it. (Adding the contents of your last comment to the question would be a great start.) $\endgroup$
    – David Z
    May 23, 2012 at 15:32

2 Answers 2

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There are two forces... one is the gravitational force, and another is an apparent force arising from a moving reference frame, the centrifugal force. The gravitational force is straightforward. The centrifugal acceleration is given by a=v^2/r.

Gravity provides the centripetal acceleration (the acceleration that opposes the centrifugal acceleration) that keeps the car moving along the circular path, in contact with the ground. Because the car is moving relatively slowly, gravity provides more force (than neccessary to constrain the car to a circular path), which is opposed by the normal force of the ground.

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  • $\begingroup$ You mean centripetal, right? $\endgroup$
    – jcora
    May 23, 2012 at 7:29
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Hint: From Newton's 2nd law, $-m g + F_N = -m v^2/r$.

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