Centripetal force in a curved road

[ShayanJ]

I read in the jearl walker's fundamentals of physics,that the centripetal force for a car moving in a curved road,is static friction.But as we now static friction is the result of another force and can't be existed alone.what is the force causing this friction?
thanks

 

[Doc Al]

But as we now static friction is the result of another force and can't be existed alone.

What do you mean? There's also a normal force pressing the car against the road due to its weight. Is that what you mean?

 

[NEILS BOHR]

sorry but i didnt get u ?what do u mean??

 

[rcgldr]

To initiate a turn, the front tires are aimed inwards of the rear tires, and this results in the tires attempting to roll in the direction of their orientation. During the initial transition, the tires increasingly deform at the contact patch, and this corresponds an increasing amount of a Newton third law pair of radial forces, the tires outwards on the pavement, coexistant with the pavement inwards on the tires. Eventually a state of equilibrium is reached where the forces correspond to the reaction force of the car to centripetal acceleration. These radial forces are limited by the maximum amount of static friction. There's also some small amount of backwards (tangental) force due the orientation of the tires and deformation of the contact patches.

 

[ShayanJ]

I understood nothing of your post.Could you explain it simpler?
thanks

 

[Doc Al]

I understood nothing of your post.Could you explain it simpler?
thanks

Why don't you explain what you meant in your original post?

 

[rcgldr]

Could you explain it simpler?

To turn left, you turn the front tires left, while the rear tires point straight. Since the front tires are turned left they want to roll to the left, while the rear tires want to roll straight. This makes the car turn left, and generates the turning forces depending on the speed of the car and how much the tires are turned left.

 

[siddharth5129]

in reference to rcgldr's original post:-
Hey thanks man that really cleared up the concept of cars turning round a bend. I just couldn't understand where that inward centripetal force came from. Just to be clear , the force by the pavement inwards on the tires is the static frictional force right?

 

[Dr Lots-o'watts]

"Frictional" yes, but "static" doesn't seem appropriate. A turn involves an acceleration (inwards), which rather makes the system a dynamic one.

 

[Doc Al]

Just to be clear , the force by the pavement inwards on the tires is the static frictional force right?

Yes.

"Frictional" yes, but "static" doesn't seem appropriate. A turn involves an acceleration (inwards), which rather makes the system a dynamic one.

The friction involved will be static friction, unless the tires slip against the road surface.

 

[rcgldr]

The force by the pavement inwards on the tires is the static frictional force right?

And so is the force by the tires outwards on the pavement. Forces only exist in equal and opposing pairs, Newtons third law.

 

[K^2]

But as we now static friction is the result of another force and can't be existed alone.

That is only true when object in question is not accelerating. A car going through the curve is accelerating. (Centripetal acceleration.) When body does accelerate, things become far more complicated.

There is a simple work around. Take a rotating reference frame in which the car is at rest. In this frame, we can apply the rule about static friction force being opposite to the sum of all other forces applied to the body. We also get a few new forces. This coordinate system is not inertial, and so we will have inertial forces, also known as fictitious forces.

In rotating reference frame, there are two fictitious forces. Centrifugal force and Coriolis force. Since the car turns at constant radius, Coriolis force is zero. We only have to compute centrifugal force, and the static friction force will be equal to it.

When we go back to an inertial reference frame, relative to which the car travels in a curve, the centrifugal force goes away. There are no inertial forces in an inertial frame. Static friction force, however, remains. It provides centripetal force necessary for the car to make a turn.

 

[Doc Al]

But as we now static friction is the result of another force and can't be existed alone.

That is only true when object in question is not accelerating.

Ah... now I understand what the OP meant by his question!

(When an object is in equilibrium, static friction only exists to resist some applied force--the 'other force' referred to. But as K^2 explains, a car going around a curve is not in equilibrium.)

 

[dgOnPhys]

I read in the jearl walker's fundamentals of physics,that the centripetal force for a car moving in a curved road,is static friction.But as we now static friction is the result of another force and can't be existed alone.what is the force causing this friction?
thanks

I guess by "another force" you meant the opposite force that has to exist because of Newton's 3rd law; this is not causing the centripetal force though, forces simply exist in pairs. In any case this force is the force the tires exert on the road (in an inertial frame) and it is obviously not part of the free-body diagram of the car.

 

[Doc Al]

I guess by "another force" you meant the opposite force that has to exist because of Newton's 3rd law;

I don't think that's what the OP meant (which I realized after reading K^2's post). Imagine a box sitting on the floor. You push it sideways with a force F. If the static friction is sufficient, the box remains in equilibrium. It's this applied force F that is "another force". Without that force, you won't have any static friction.

 

[dgOnPhys]

I don't think that's what the OP meant (which I realized after reading K^2's post). Imagine a box sitting on the floor. You push it sideways with a force F. If the static friction is sufficient, the box remains in equilibrium. It's this applied force F that is "another force". Without that force, you won't have any static friction.

You're right that's more likely what he was referring to: the OP thought was suggesting there would be another force acting on the tires to cause friction as in an equilibrium case

Let's see if Shyan ever clarifies

 

[ShayanJ]

I'm really sorry for leaving you in haze.I don't come online such often so Its the first time I see your demand for clarification.
I meant the centripetal force is static friction(according to jearl) so this friction is pointing towards the center of curve.Because friction is the result of a force which a body is exerting on the surface and is opposite to it,there should be a force that car is exerting to the road and is pointed in the opposite direction of friction,outwards the curve.The car exerts this force to the road,because there is force that is applied to its tires outwards the curve.I don't know what is this last force.

 

[Doc Al]

Because friction is the result of a force which a body is exerting on the surface and is opposite to it,there should be a force that car is exerting to the road and is pointed in the opposite direction of friction,outwards the curve.The car exerts this force to the road,because there is force that is applied to its tires outwards the curve.I don't know what is this last force.

Again, this is only true if the body in in equilibrium--not if it's accelerating. There's no need for any additional force to create the static friction in this case. The fact that it's accelerating does that.

If you wish to view things from the non-inertial frame of the car itself, then you'll have an outward centrifugal force. That force will balance the inward static friction, since in the non-inertial frame the car is not accelerating.

(See K^2's post #12.)

 

[rcgldr]

There should be a force that car is exerting to the road and is pointed in the opposite direction of friction,outwards the curve.The car exerts this force to the road,because there is force that is applied to its tires outwards the curve.I don't know what is this last force.

It's still the friction force. Friction is what allows this pair of forces to exist.

Again, this is only true if the body in in equilibrium--not if it's accelerating. There's no need for any additional force to create the static friction in this case. The fact that it's accelerating does that.

Static friction force only exists as the interface between a Newton third law pair of forces. It doesn't matter if one or both of the forces are reaction forces due to acceleration. Generally both forces are reaction forces, but the Earth is treated as if it had infinite mass as opposed to a huge amount of mass. Technically both car and Earth experience accelerations when the car is turning, but the Earth's linear and angular accelerations are very tiny because of it's relatively huge mass.

 

[Doc Al]

It looks like once again I misread the OP's question:

I meant the centripetal force is static friction(according to jearl) so this friction is pointing towards the center of curve.Because friction is the result of a force which a body is exerting on the surface and is opposite to it,there should be a force that car is exerting to the road and is pointed in the opposite direction of friction,outwards the curve.The car exerts this force to the road,because there is force that is applied to its tires outwards the curve.I don't know what is this last force.

I believe that while the OP said a force that the "car exerted on the road", what was meant was there should be an outward force exerted on the car.

If all that was meant was a missing 3rd law pair, then the question would be trivial. (The road and car exert equal and opposite forces on each other.) All real forces are parts of Newton's 3rd law pairs. So what?