Problem involving circular movement and friction.

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Homework Help Overview

The discussion revolves around a physics problem involving circular motion and friction, specifically related to a high-speed train negotiating curves. The problem includes calculations for friction forces experienced by passengers under different conditions: when the train is not banked and when it tilts at a specific angle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations required for friction forces in both scenarios, questioning the need for coefficients of friction and the types of friction involved. There is mention of relevant equations and the role of forces in circular motion.

Discussion Status

Some participants have provided helpful insights regarding the relationship between normal force, friction, and centripetal force. There is ongoing exploration of how to approach the problem, particularly in the context of the tilted train.

Contextual Notes

Participants are considering the implications of the train's tilt and the assumptions regarding friction types and coefficients, which are not explicitly provided in the problem statement.

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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