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corrected_Q5

# corrected_Q5 - z N fs N 15o 15o v 70m • r θ mg...

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Unformatted text preview: z N fs N 15o 15o v 70m • r θ mg Horizontal (a ) fs Vertical ( b) mg (c) Question 5: A curve in the road, with a radius of 70.0m, Figures (a) and (b) show the real life scenario; (c) is the is banked at an angle of 15o. If the coefficient of friction free body diagram of (a). Note that the frictional force f s is a static frictional force as there is no relative motion between the road and tires is 0.70, what is the maximum speed a car can make the corner without sliding? Hint: eliminate the normal force. between the tire and the road, i.e., no skidding or wheel spin. Take radial and vertical components in (c). Note that the net radial force is the centripetal force. In (a), 2 Radial direction: Nsin15o + f s cos15o = mv r . z-direction: N cos15o − mg − f s sin15o = 0 . But f s = µ s N. So, re-arranging these equations we get: 2 Radial direction: N sin15o + µ sN cos15o = mv r . z-direction: N cos15o − µ sN sin15o = mg. 1 Hence ( )2 N (cos15o − µ s sin15o ) = mg ... N sin15o + µ s cos15o = mv r and ... ... [1] ... ... [2]. Dividing [1] by [2] we get sin15o + µ s cos15o v 2 =. cos15o − µ s sin15o gr i.e., tan15o + µ s v 2 = gr 1 − µ s tan15o 0.268 + 0.7 = 818.2 (m/s)2. = 9.81 × 70 1 − 0.7 × 0.268 ∴ v = 28.6 m/s. At home, show that: (1) if θ = 0 and µ s = 0.70, i.e., no banking but static friction, then v = 21.9 m/s. (2) if θ = 15o and µ s = 0 , i.e., banked corner but zero static friction, then v = 13.6 m/s. 2 ...
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