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Problem involving circular movement and friction.

by anightlikethis
Tags: circular, friction, involving, movement
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anightlikethis
#1
Mar2-07, 09:30 PM
P: 10
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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PhanthomJay
#2
Mar2-07, 10:45 PM
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Quote Quote by anightlikethis View Post
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.
Harmony
#3
Mar2-07, 10:58 PM
P: 209
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.

anightlikethis
#4
Mar3-07, 11:17 AM
P: 10
Problem involving circular movement and friction.

Thanks, that was very helpful


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