Banked Curve Algebra With Static Friction

AI Thread Summary
The discussion revolves around calculating the maximum speed of a car on a banked curve, considering static friction and the angle of the curve. The car's mass is 2000 kg, the radius of curvature is 50.0 m, and the coefficient of static friction is 0.100 with a curve angle of 15°. The user attempts to derive the centripetal force using components of normal force and static friction but struggles with the correct application of the normal force in this context. A suggestion is made to revisit the free body diagram to accurately determine the normal force needed for the observed acceleration. The conversation emphasizes the importance of correctly analyzing forces to solve the problem effectively.
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Homework Statement



9. A 2000 kg car is racing around a banked circular path whose radius of curvature is 50.0m. The coefficient of static friction between the car and the road is 0.100. What is the maximum possible speed for the car to remain on the road if the angle of the curve is 15°?

a) 45.7 km/h
b) 49.0 km/h
c) 60.0 km/h
d) 130 km/h
e) 197 km/h


Homework Equations


Fsmax = Us FN
FN = cos15mg
Fc = mV^2/r

The Attempt at a Solution



Once again stuck on another problem. From what I see, there is a component of the normal force that points towards the center of the radius as well as a component of the static friction that points towards the center of the circle. These two values added should give me the centripetal force, if I am right. Assuming this is true this is what I have tried.

sin15(cos15)mg= X component normal force pointing towards center
Us(cos15mg)(cos15)= x component of FsMax that points towards center

sin15(cos15)mg + Us(cos15mg)(cos15) = Fc
Fc=mv^2/r
divide the equation by mass
sin15(cos15)g + Us(cos15g)(cos15) = v^2/r
multiply the left side by r and root for velocity
Root([(sin15)(cos15)g + Us(cos15g)(cos15)]r) = v
v = 12.96m/s

What am I missing here? I have a midterm tomorrow please answer as soon as you can thank you.
 
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The normal force is (mg)cosθ only if the acceleration is parallel to the incline. In general, the normal force is what is necessary to provide the observed acceleration. Redo your free body diagram to find what the normal force is in this case.
 
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