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Centrifugal force of a car

  • Thread starter Ry122
  • Start date
  • #1
565
2
http://users.on.net/~rohanlal/cent.jpg [Broken]

my attempt:
The car becomes lighter as it passes over the arc therefore its going to exert a force on the road that is less than its normal force.
f=mv^2/r
f=1710(16.6)^2/38.5 = 12239.15844
9.8 x 1710 = 16758
16758-12239.15844
this is incorrect, what am i doing wrong?
 
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Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
Looks okay to me. I got 4536 N using g = 9.81.
 
  • #3
3
0
As a matter of fact let me do the math for ya here.(I've no idea if my expression is clear or not cuz English is not my mother tongue.)
The gravity exerted on the car is :
G=mg=16758N
And the Centripetal force needed is:
f=mv^2/R=12239.15844N
As the forces exerted on the car is the Centripetal force if we take the ground as the reference frame
so
N+f=G
N=4518.84156N
This is right the force the road exert on the car...if I'm asked to solve it.
I don't see any problems unless it's a problem of relativity or Quantunm Mechanics...
 
  • #4
565
2
it says to use 9.8 for gravity and round off to the nearest tenth. What would it be then?
 
  • #5
209
1
Simply 9.8? :tongue2:
 
  • #6
3
0
it says to use 9.8 for gravity and round off to the nearest tenth. What would it be then?
That's exactly what I did.
I don't see anything wrong.
 

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