# Centrifugal force

1. Apr 22, 2009

### Ry122

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?

Last edited by a moderator: May 4, 2017
2. Apr 22, 2009

### Delphi51

Looks okay to me. I got 4536 N using g = 9.81.

3. Apr 22, 2009

### Laudau

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. Apr 23, 2009

### Ry122

it says to use 9.8 for gravity and round off to the nearest tenth. What would it be then?

5. Apr 23, 2009

### ImAnEngineer

Simply 9.8? :tongue2:

6. Apr 26, 2009

### Laudau

That's exactly what I did.
I don't see anything wrong.