Centripetal Acceleration Quick Question

[Speedking96]

Homework Statement

A bridge over a small river has a roadway which is in the shape of an arch having radius of curvature of 41 m. What is the maximum speed at which an automobile can travel across the bridge without leaving the bridge?

Homework Equations

Fc = (mv^2)/r

Fg= mg

The Attempt at a Solution

Fnet = ma

Fc = Fg - Fn

(mv^2) / r = mg - fn

(v^2) / r = g

root ( 41 * 9.81) = 20.06 m/s

I know that this is the right answer, but I don't understand why the normal force is zero. Isn't the bridge exerting a force equal to that of gravity on the car so that it doesn't go through the bridge?

 

[frazdaz]

The point at which the automobile leaves the bridge is when it's no longer in contact with the bridge i.e. there is no normal force. There's no force preventing it going through the bridge because it isn't in contact with it.

 

[Chestermiller]

Homework Statement

A bridge over a small river has a roadway which is in the shape of an arch having radius of curvature of 41 m. What is the maximum speed at which an automobile can travel across the bridge without leaving the bridge?

Homework Equations

Fc = (mv^2)/r

Fg= mg

The Attempt at a Solution

Fnet = ma

Fc = Fg - Fn

(mv^2) / r = mg - fn

(v^2) / r = g

root ( 41 * 9.81) = 20.06 m/s

I know that this is the right answer, but I don't understand why the normal force is zero. Isn't the bridge exerting a force equal to that of gravity on the car so that it doesn't go through the bridge?

That's only if the bridge is flat. Did you ever feel like, when you were going over a bump in the road, you were almost flying out of your seat. Part of the gravitational force is used up just trying to hold you down in the curved trajectory.

 

[haruspex]

(v^2) / r = g

I know that this is the right answer,

Actually, it isn't. That's the limiting speed for it to stay in contact at the apex of the bridge, but if it's a constant arc then the speed at start and end of the arc will need to be lower. To find out what that is you'd need to know the length of the bridge.

 

[HalfLostAndSearching]

I can explain that the normal force in this scenario is indeed zero because the car is traveling at a constant speed and direction, which means it is not accelerating. In order for the car to stay on the bridge, the centripetal force (Fc) provided by the road must be equal to the force of gravity (Fg) pulling the car downwards. This means that the net force (Fnet) acting on the car is zero, and therefore there is no need for a normal force to counteract it. The fact that the bridge is exerting a force on the car is true, but it is not a separate force from the centripetal force provided by the road. They are both part of the net force acting on the car. I hope this helps clarify any confusion.