Normal Force - Car over a Circular Arc

In summary, the conversation discusses the problem of an 1850 kg car passing over a bump in the road that follows the arc of a circle with a radius of 40.60 m. The first question asks for the force exerted by the road on the car at the highest point of the bump, given a speed of 14.9 m/s. The second question asks for the maximum speed the car can have at this point before losing contact with the road. After some trial and error with equations, it is determined that the correct equation is N - mg = - mv^2/r, taking into account the direction of the acceleration. The conversation ends with a thank you and a reference to another problem for further assistance.
  • #1
nicknaq
67
0

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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  • #2
nicknaq said:
N-mg=mv^2/r
This equation is not quite correct. Hint: What's the direction of the acceleration?
 
  • #3
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
 
  • #4
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.
 
  • #5
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:
 
Last edited by a moderator:

1. What is the normal force in a car traveling over a circular arc?

The normal force in a car traveling over a circular arc is the force exerted by the surface of the road on the tires of the car, perpendicular to the surface. It is also known as the support force or contact force.

2. How does the normal force affect the motion of a car over a circular arc?

The normal force helps to keep the car in equilibrium by balancing the weight of the car and providing the necessary centripetal force to keep the car moving in a circular path. If the normal force is insufficient, the car may slip or lose traction on the road.

3. What factors affect the magnitude of the normal force in a car over a circular arc?

The magnitude of the normal force depends on the weight of the car, the speed of the car, and the curvature of the circular arc. The greater the weight and speed of the car, and the tighter the curve, the greater the normal force needed to maintain the car's motion.

4. Can the normal force in a car over a circular arc ever be less than the weight of the car?

No, the normal force can never be less than the weight of the car. The normal force must always be equal or greater than the weight of the car in order for the car to stay in contact with the road and maintain its motion along the circular arc.

5. How does the normal force change as the car travels along different sections of a circular arc?

The normal force will change as the car travels along different sections of a circular arc depending on the curvature of the arc. If the curvature increases, the normal force will also increase to keep the car in equilibrium. Similarly, if the curvature decreases, the normal force will decrease.

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