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Normal Force - riding over a hill

  • Thread starter mybrohshi5
  • Start date
  • #1
365
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Homework Statement



I am studying for my physics exam on tuesday and i came across this question and i am not sure why this equation was used.

A bicycle and bicyclist of combined mass 62 kg go over a circular hump of radius 6.0 m at a speed of 5.0 m/s. What is the normal force on the bicycle at the top of the hump?


The Attempt at a Solution



The equation used and answer is..

n - mg = - (mv2)/r

n = - [(62*(52))/6] + (62*9.8)

n = 349 N

Can anyone explain why or how to come up with this equation used. I am confused on how and where this comes from. I would have never of thought to do this and would have probably gotten it wrong if it was on the test so any clarification to help me understand would be greatly appreciated.

Thank you :)
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
At the top of the hill, if you draw a FBD, the normal reaction N acts upwards and the weight mg acts downwards.

The resultant of these two must be sufficient such that they keep him in contact with the hill.

For that to happen, the resultant in the direction of the weight must be the centripetal force, else the resultant is opposite of the weight, which would mean the boy would fly off of the hill.

Hence mg-N= mv2/r

Does it make a little more sense now?
 
  • #3
rl.bhat
Homework Helper
4,433
7
Two forces act on the bicyclist.
i) Normal force N acting in the upward direction due to centripetal force.
ii) weight mg in the down ward direction.
So net centripetal force acting on him in the downward direction is
N - mg = -(mv^2)/R
 
  • #4
365
0
Yes That makes more sense. I guess thats a little easier and not as complex as i was making it out to be.

Thank you :)
 

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