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Old Oct 2nd 2008, 07:29 PM   #1
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Vertical Circle

You are driving down the dark, winding Northwest Georgia two-lane road at a speed of 34 miles per hour. You are driving due north when you come upon perfectly flat, circular curve. Exactly one-tenth of a mile along the curve later, you are driving due west If the car has a mass of 2308 kg, what is the force of friction holding the car on the road (in Newtons)? (Hint: make sure all units are converted appropriately; you will need to find the radius of the curve.) (Double hint: If you highlight this parenthetical clause, you will find that the curve has a radius of *102* m.

That is the question I was given. the number between the **'s is 102.

so my velocity is 34 mi/h, converted to m/s is 15.2
my mass is 2308 kg
and my radius is 102 m
the length of the curve, if important, is 320 m

i was trying to use the equation F= m(v^2/r)

but i get 5227.8, and that answer is incorrect

if you could even tell me what equation to use I would be so grateful!
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Old Oct 6th 2008, 11:40 AM   #2
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While moving in a circle, your car is undergoing acceleration directed towards the centre of this circle; the magnitude of this accln. is $\displaystyle mv^2/r.$. Now, since there is an accleration in the car, there has to be an external force. External forces generally come from the immediate external -- in your case, its the road. The road applies frictional force of magnitude $\displaystyle m\cdot v^2/r$ on the car, where $\displaystyle m$ is the car's mass, directed at the centre, which causes the acceleration, and thus the car turns in a circle.

There is no other solution. Perhaps you must read the question again?!
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