# Centripetal Force Problem

## Homework Statement

1988M1. A highway curve that has a radius of curvature of 100 meters is
banked at an angle of 15° as shown above.
a.Determine the vehicle speed for which this curve is appropriate if there is no friction between the road and the tires of the vehicle.

On a dry day when friction is present, an automobile successfully negotiates the curve at a speed of 25 m/s.

b.Draw and label all of the forces on the automobile.

## Homework Equations

Centripetal force = mv^2 / r
(I know how to do the problem, but I'm not sure why what I'm supposed to do is correct)

## The Attempt at a Solution

I know that in part a) the centripetal force is provided by the horizontal component of the normal force (Nsin(15)), but shouldn't the gravitational force also play a role? The mgsin(15) component is pushing the car down the ramp, which is pushing the car closer towards the center, albeit at an angle.

For part b) I've been told that the friction force points down the ramp and that its horizontal component contributes to the net centripetal force, but shouldn't it be pointing up the ramp to counteract mgsin(theta)? Furthermore, since friction opposes the direction of motion, shouldn't it be pointing away from the center since that's where the cart is moving towards?

Thanks!
[/B]

TSny
Homework Helper
Gold Member
Hello, welcome to PF!

The force of gravity, mg, acts vertically. So, it does not have a centripetal component. You could break up the gravity force into components along the slop (mgsinθ) and perpendicular to the slope (mgcosθ). But it is confusing to do this for this type of problem. Your are right that mgsinθ by itself has a component toward the center. But what about the mgcosθ part? Does it have a component toward or away from the center?

Friction does not always oppose the direction of motion. When you start to walk, the friction forces on your feet act in the direction you start moving.

The cart is not moving toward the center. That is, it always keeps the same distance from the center. The velocity vector is always perpendicular to the radius of the circle.

Thanks! But how do we know that friction is pointing towards the center?

TSny
Homework Helper
Gold Member
When you work out part (a) you will know the speed required when there is no friction. If the speed in (b) is faster than what you get in (a), which way would friction need to point so that you can still go around the curve? Up the slope or down the slope?

Mister T
Gold Member
I know that in part a) the centripetal force is provided by the horizontal component of the normal force (Nsin(15)), but shouldn't the gravitational force also play a role? The mgsin(15) component is pushing the car down the ramp, which is pushing the car closer towards the center, albeit at an angle.

The force mg points downward. A component of it, mg sin 15°, points down the hill. Your intuition is that it must play some role? Well, it does. It plays the same role it would play if the car were moving either up or down the hill. But as far as the car moving a horizontal direction, it plays no role.

For part b) I've been told that the friction force points down the ramp and that its horizontal component contributes to the net centripetal force, but shouldn't it be pointing up the ramp to counteract mgsin(theta)?

Again, if the car were moving up or down the hill, the friction force would oppose that motion. So if the car starts to drift up the hill, a friction force would act down the hill to oppose that motion.

Furthermore, since friction opposes the direction of motion, shouldn't it be pointing away from the center since that's where the cart is moving towards

If the car is spiraling outward we wouldn't say it's moving away from the center! As long as its path is concave towards the center we would say, for purposes of describing how it's instantaneous direction is changing, we would say it's moving towards the center.

Think of centripetal as a direction, not a force. If the car's path is concave inward, there is an inward net force. In the same way the path of a projectile is concave downwards because there is a downward net force.