Banked Circular Motion with Static Friction

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SUMMARY

The discussion focuses on calculating the coefficient of static friction required for a car to navigate a banked curve without skidding. Given a curve radius of 85 meters and a speed of 95 km/h, the angle of banking (theta) was determined to be 21.4 degrees. The user attempted to apply the equations of motion, specifically FNcos(theta) = mg and the centripetal force equation, but faced challenges in deriving the necessary static friction coefficient. The conversation highlights the importance of understanding the forces acting on a vehicle in banked circular motion.

PREREQUISITES
  • Understanding of centripetal force in circular motion
  • Knowledge of static friction and its role in motion
  • Familiarity with trigonometric functions and angles
  • Basic principles of dynamics and forces
NEXT STEPS
  • Study the derivation of the banking angle formula for circular motion
  • Learn how to calculate the coefficient of static friction in various scenarios
  • Explore the effects of speed on friction and banking angles
  • Investigate real-world applications of banked curves in road design
USEFUL FOR

Physics students, automotive engineers, and anyone interested in the dynamics of vehicles on curved paths will benefit from this discussion.

milky9311
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Q: "If a curve with a radius of 85m is properly banked for a car traveling 65km/h, what must be the coefficient of static friction for a car not to skid when traveling at 95km/h?"

[PLAIN]http://a4.sphotos.ak.fbcdn.net/hphotos-ak-ash4/215513_1974874979462_1470693940_32310002_1293630_n.jpg


2. The attempt at a solution

I found theta

FNcos(theta) = mg

(mgsin(theta))/cos(theta) = (mv2)/r

theta = 21.4o

But next bit i couldn't do.

Wouldnt the centripetal force of the 95km/h car be FNsin(theta) + musFN? as the static friction + the horizontal component of the normal force allows circular motion?

thanks
 

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milky9311 said:
FNcos(theta) = mg


FNcos(theta) = mg??:confused: Check that again...
 

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