# Centripal Force - Friction needed to keep a car tracking a curve

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1. Feb 4, 2015

### Julian3

1. The problem statement, all variables and given/known data
The question is- What is the minimum coefficient of static friction that would keep the car from sliding off the curve? The Cars mass is 13500KG and it is traveling at 50.0hm/hr(13.9m/s) and the curve has a radius of 2.00 x102 m. I know the centripetal acceleration of the car is .97m/s and the centripetal force is 3381.2N. FF=FN(coefficient of friction)

2. Relevant equations
I know you have to draw a free body diagram, which I think it would be Force Friction pointing down as well as Force weight and Force Normal pointing up and Force Velocity to the right. I know you have to make it into a equation to solve for the Coefficient of Friction.

3. The attempt at a solution
Im basically stuck at this part because I don't know what the force would have to be to keep the car on the curve, and i dont know how to set up the equation, sorry for not having an attempt at solving it but im really confused by what we are currently covering in class.

2. Feb 4, 2015

### Svein

Well, no. The forces from the road on to the car: Force Normal pointing up and Force Friction pointing towards the curve center.
The forces from the car on the road: Force Normal pointing down and Force Velocity pointing away from the curve center.
In addition: Force Frictionk(Force Normal), where k is the coefficient of static friction.

3. Feb 4, 2015

### Julian3

Is Force Normal the centripetal force on the car and if so how do you get force friction?

4. Feb 4, 2015

### Svein

Force Normal is the force from the car normal to the road surface. Call it the weight of the car. In your terms, Force Velocity is the centripetal force.

5. Feb 4, 2015

### haruspex

That doesn't seem any more accurate than the original statements.
Both of those forces are at an angle, the normal force being normal to the road surface (so up and towards the curve centre), while force of friction is parallel to the road surface. In principle, both acting up the slope and acting down the slope should be considered. (On a steeply banked road there may be a minimum speed.)
This is the 'centrifugal force' view, i.e. using non-inertial frames. It's valid, if you know what you are doing. I find it less confusing to stick with inertial frames. In that view, centripetal force is the required resultant force (not an applied force) to keep the car going around the curve.
I would recommend not calling it the weight, though maybe that's reasonable in the non-inertial frame.

6. Feb 5, 2015

### Svein

Well, without any force from the road on to the car (think slippery ice), the car is unable to turn and will continue straight ahead. So, whatever you call it, some force from the road is necessary for the car to turn.

7. Feb 5, 2015

### rcgldr

There's no mentioned of a banked road, so I would assume it's a flat road. You already know the centripetal acceleration, and you assume some constant for g, the acceleration related to gravity when in free fall. The centripetal force is mass x centripetal acceleration, and the vertical force is mass x g . How can you use this information to determine the minimal value for the coefficient of friction?