Banked curve: question about friction and normal force

[swell9]

Hello,

From what I understand a car negotiating a banked cruve of angle \theta at a speed V0 does not experience a force of friction.

But if the car goes to a speed higher than V0, then friction acts in the direction of the cruve. i.e. inwards
And if the car goes at a speed lower than V0, then friction acts in the opposite direction of the incline.
i.e. up the incline thus preventing the car from sliding inwards.

How does friction change? is it not it a constant force? To me this all seems like friction knows what's happening and thus acts accordingly to help the car make it around the curve or prevent it from sliding


Thanks a lot
swell9

 

[jtbell]

How does friction change? is it not it a constant force?

Kinetic (sliding) friction is constant (more or less). Static friction adjusts itself as necessary to keep the two surfaces stuck together, until it reaches the limit given by \mu_s F_N. It changes the same way as when you push harder against a stationary wall, the reaction force the wall exerts on you also increases.

Friction between tires and road is static friction until you spin out and the tires start to skid.

 

[swell9]

Kinetic (sliding) friction is constant (more or less). Static friction adjusts itself as necessary to keep the two surfaces stuck together, until it reaches the limit given by \mu_s F_N. It changes the same way as when you push harder against a stationary wall, the reaction force the wall exerts on you also increases.

Friction between tires and road is static friction until you spin out and the tires start to skid.

Thank you.. I see.. but then, why does the normal force change?
does that mean it is speed dependent?

thanks

 

[log0]

The normal force is also a reaction force, keeps the car from falling through the ground. So it is going to change if you drive a banked curve (centripetal force: F = m*v²/r).

 

[swell9]

The normal force is also a reaction force, keeps the car from falling through the ground. So it is going to change if you drive a banked curve (centripetal force: F = m*v²/r).

thanks for the help.
So the normal force is dependent upon speed in this case. I guess I cannot see how speed changes the normal force...

 

[rcgldr]

There are 4 forces involved:

the weight of the car due to gravity (downwards)
the reaction force from the car related to centrpetal acceleration of the car (outwards)
the force from the road perpendicular to the road's surface (always away from the road)
the force from the road parallel to the road's surface (could be downwards+inwards or upwards+outwards)

At V₀ the force parallel to the road's surface is zero, because the force from the road perpendicular to the surface of the road is equal and opposing to the vector sum of forces related to the weight of the car and the cars reaction to centripetal acceleration.

At speeds other than V₀, then a component of force parallel to the surface is involved. At speeds less than V₀, an upwards (and outwards force) parallel to the surface is required to counter a component of weight (gravity). At speeds greather than V₀, an inwards (and downwards) force parallel to the surface is required counter the reaction force related to centripetal acceleration of the car. The force perpendicular to the road will be also be affected by speed, smaller at speeds lower than V₀, greater at speeds above V₀.

Note that the reaction force is just that, a reaction to acceleration. The rate of acceleration is the vector sum of the other 3 forces divided by the mass of the car.

 

[swell9]

There are 4 forces involved:

the weight of the car due to gravity (downwards)
the reaction force from the car related to centrpetal acceleration of the car (outwards)
the force from the road perpendicular to the road's surface (always away from the road)
the force from the road parallel to the road's surface (could be downwards+inwards or upwards+outwards)

At V₀ the force parallel to the road's surface is zero, because the force from the road perpendicular to the surface of the road is equal and opposing to the vector sum of forces related to the weight of the car and the cars reaction to centripetal acceleration.

At speeds other than V₀, then a component of force parallel to the surface is involved. At speeds less than V₀, an upwards (and outwards force) parallel to the surface is required to counter a component of weight (gravity). At speeds greather than V₀, an inwards (and downwards) force parallel to the surface is required counter the reaction force related to centripetal acceleration of the car. The force perpendicular to the road will be also be affected by speed, smaller at speeds lower than V₀, greater at speeds above V₀.

Note that the reaction force is just that, a reaction to acceleration. The rate of acceleration is the vector sum of the other 3 forces divided by the mass of the car.

Thanks a lot.
I never knew there were so many forces.

the weight of the car due to gravity (downwards)
>> This would be the weight, correct?
the reaction force from the car related to centrpetal acceleration of the car (outwards)
>> isn't this force the centrifugal force and thus is a fictitious force? meaning we should not consider it in our calculations...
the force from the road perpendicular to the road's surface (always away from the road)
>> This is just the normal force, correct?
the force from the road parallel to the road's surface (could be downwards+inwards or upwards+outwards)
>> and this would be static friction?

If you could show me a FBD, that would be great too :)

agains thank for your time

 

[log0]

There you go!

http://www.batesville.k12.in.us/physics/PHYNET/Mechanics/Circular Motion/banked_with_friction.htm

Wikipedia: http://en.wikipedia.org/wiki/Banked_turn