ch06-p058 - 58 We refer the reader to Sample Problem 6-10...

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with v = 60(1000/3600) = 17 m/s and R = 200 m. The banking angle is therefore θ = 8.1°. Now we consider a vehicle taking this banked curve at v' = 40(1000/3600) = 11 m/s. Its (horizontal) acceleration is 2 / av R ′′ = , which has components parallel the incline and perpendicular to it: 2 | | 2 cos cos sin sin . v aa R v R == These enter Newton’s second law as follows (choosing downhill as the + x direction and away-from-incline as + y ): || sin cos s N mg f ma
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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