Problem involving circular movement and friction.

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SUMMARY

The discussion focuses on calculating the friction force experienced by passengers on Amtrak's Acela train while negotiating a curve with a radius of 600 m at a speed of 160 km/h. For a non-banked track, the friction force required for a passenger with a mass of 75 kg is 247.31 N. When the train tilts at a maximum angle of 8° toward the center of the curve, the friction force must be calculated considering both the normal force and the friction force contributing to the centripetal force. The discussion emphasizes the importance of drawing a free body diagram (FBD) to identify all forces and their components.

PREREQUISITES
  • Understanding of centripetal acceleration and forces
  • Knowledge of normal force and friction force equations
  • Ability to draw and interpret free body diagrams (FBD)
  • Familiarity with trigonometric functions in physics
NEXT STEPS
  • Learn how to calculate centripetal force in non-banked and banked scenarios
  • Study the effects of tilt angles on friction forces in circular motion
  • Explore the relationship between normal force and friction in curved motion
  • Investigate the role of static and kinetic friction in real-world applications
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of high-speed trains and circular motion mechanics.

anightlikethis
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Amtrak's high speed train, the Acela, utilizes tilt of the cars when negotiating curves. The angle of tilt is adjusted so that the main force exerted on the passengers, to provide the centripetal acceleration, is the normal force. The passengers experience less friction force against the seat, thus feeling more comfortable. Consider an Acela train that rounds a curve with a radius of 600 m at a speed of 160 km/h (approximately 100 mi/h).
a)Calculate the friction force needed on a train passenger of mass 75 kg if the track is not banked and the train does not tilt.
b) Calculate the friction force on the passenger if the train tilts to its maximum tilt of 8° toward the center of the curve.

I figured out the answer to A which is 247.31 N, but I B is confusing me. I would think that in order to find the friction i would need the coefficient of friction. Do they want static or kinetic friction? The only equations I have related to friction are: Kinetic friction = uk*Fn and Maximum friction=us*Fn. However, I think that this equation has something to do with the equation for when no friction is required Fnsintheta=m(v^2/r)but I'm not sure how.
 
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anightlikethis said:
Amtrak's high speed train, the Acela, utilizes tilt of the cars when negotiating curves. The angle of tilt is adjusted so that the main force exerted on the passengers, to provide the centripetal acceleration, is the normal force. The passengers experience less friction force against the seat, thus feeling more comfortable. Consider an Acela train that rounds a curve with a radius of 600 m at a speed of 160 km/h (approximately 100 mi/h).
a)Calculate the friction force needed on a train passenger of mass 75 kg if the track is not banked and the train does not tilt.
b) Calculate the friction force on the passenger if the train tilts to its maximum tilt of 8° toward the center of the curve.

I figured out the answer to A which is 247.31 N, but I B is confusing me. I would think that in order to find the friction i would need the coefficient of friction. Do they want static or kinetic friction? The only equations I have related to friction are: Kinetic friction = uk*Fn and Maximum friction=us*Fn. However, I think that this equation has something to do with the equation for when no friction is required Fnsintheta=m(v^2/r)but I'm not sure how.
But you nicely calculated the friction force necessary in part a without having to know us or uk, so why do you feel you need it in part B? You're on the right track with your equation, but the friction force also contributes to the centripetal force. Draw a good FBD and identify all forces and force components.
 
The centripetal force is provided by both normal force and friction. Thus, consider the horizontal component of both normal and friction force, which the total will be the centripetal force.
 
Thanks, that was very helpful
 

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