Banked Curve Algebra With Static Friction

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SUMMARY

The discussion focuses on calculating the maximum speed of a 2000 kg car on a banked circular path with a radius of 50.0 m and a bank angle of 15°. The coefficient of static friction is 0.100. The participant attempted to derive the centripetal force using components of the normal force and static friction but arrived at an incorrect velocity of 12.96 m/s. The correct approach requires accurately determining the normal force and its components to solve for the maximum speed.

PREREQUISITES
  • Understanding of centripetal force and its calculation
  • Knowledge of static friction and its role in circular motion
  • Ability to analyze free body diagrams in physics
  • Familiarity with trigonometric functions in the context of inclined planes
NEXT STEPS
  • Review the derivation of centripetal force in banked curves
  • Study the effects of static friction on motion in circular paths
  • Practice drawing and analyzing free body diagrams for inclined planes
  • Explore the relationship between angle of inclination and maximum speed in banked curves
USEFUL FOR

Students preparing for physics exams, particularly those focusing on mechanics and circular motion, as well as educators teaching concepts related to forces and motion on inclined surfaces.

Extremist223
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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.
 
Physics news on Phys.org
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.
 
Last edited:

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