Normal Force - Car over a Circular Arc

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Homework Help Overview

The discussion revolves around a physics problem involving a car passing over a circular arc, specifically focusing on the forces acting on the car at the highest point of the bump. The subject area includes concepts of forces, centripetal acceleration, and normal force.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply the equation relating normal force, gravitational force, and centripetal acceleration but encounters difficulties with the results. Some participants question the correctness of the equation used and the direction of acceleration, suggesting a need for clarification on the signs in the equations.

Discussion Status

Participants are actively engaging in clarifying the setup of the problem and the equations involved. There is a productive exchange regarding the direction of forces and the correct formulation of the equations, with some guidance provided on how to express the relationship between the forces acting on the car.

Contextual Notes

The original poster's attempts at calculations yielded results that were not correct, indicating possible misunderstandings in the application of the equations or the assumptions made. There is an ongoing exploration of the implications of the signs in the equations used.

nicknaq
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Homework Statement


An 1850 kg car passes over a bump in a road that follows the arc of a circle of radius 40.60 m as seen in the figure below.

a) What force does the road exert on the car as the car passes the highest point of the bump if the car travels at 14.9 m/s?

b) What is the maximum speed the car can have as it passes this highest point before losing contact with the road?

Homework Equations


N-mg=mv^2/r
mg=mv2/r

The Attempt at a Solution


I tried using the equation above and got 28246N, but that was not correct.
For part b) I got 6.37m/s. Also not correct.
 
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nicknaq said:
N-mg=mv^2/r
This equation is not quite correct. Hint: What's the direction of the acceleration?
 
Doc Al said:
This equation is not quite correct. Hint: What's the direction of the acceleration?

Do I need a negative sign?
So like this? mg-N=mv^2/r
 
nicknaq said:
Do I need a negative sign?
So like this? mg-N=mv^2/r
Yes.

Using your previous convention of positive for up, I'd write it as:
N - mg = - mv^2/r

The centripetal acceleration (magnitude v^2/r) points toward the center of the circle, which in this case means it points down. Thus it should have a minus sign.
 
Doc Al said:
Yes.

Using your previous convention of positive for up, I'd write it as:
N - mg = - mv^2/r

The centripetal acceleration (magnitude v^2/r) points toward the center of the circle, which in this case means it points down. Thus it should have a minus sign.

Thanks a lot for the help Doc Al!
If you have time to answer https://www.physicsforums.com/showthread.php?t=440062"too, that'd be great:
 
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