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centrifugal force |
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| Apr22-09, 08:49 PM | #1 |
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centrifugal force![]() 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? |
| Apr22-09, 11:17 PM | #2 |
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Recognitions:
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Looks okay to me. I got 4536 N using g = 9.81.
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| Apr22-09, 11:19 PM | #3 |
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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... |
| Apr23-09, 12:52 AM | #4 |
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centrifugal force
it says to use 9.8 for gravity and round off to the nearest tenth. What would it be then?
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| Apr23-09, 05:05 AM | #5 |
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Simply 9.8?
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| Apr26-09, 10:52 AM | #6 |
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I don't see anything wrong. |
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