Banked Circular Motion with Static Friction

AI Thread Summary
To determine the coefficient of static friction for a car traveling at 95 km/h on a banked curve with a radius of 85 m, the angle of banking (theta) was calculated to be 21.4 degrees. The centripetal force needed for the car's motion is derived from both the static friction and the horizontal component of the normal force. The equation FNcos(theta) = mg was questioned for accuracy, indicating potential confusion in the calculations. The discussion highlights the importance of correctly applying the forces involved in banked circular motion. Understanding these dynamics is crucial for preventing skidding at higher speeds.
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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