Centripetal force- Car making a turn

[scilearner]

Hello everyone,

 

I have got confused with this.

 

Normal the friction force acts to cancel the force. If I travel with 10 N force and friction is 5N, I travel at 5N. Why is that when you make a turn they don't create a net force, but act as separate forces?

 

EDIT: I suppose the tires are turning outward all the time in different directions to keep it on the road. when you have turn in the circle outward force is little bit greater than friction I think, which makes the car turn in the right direction, before it is stopped by friction again.

Edited June 6, 2010 by scilearner

[Physicsfan]

In the case of cars it is the friction between the wheels and the road that gets the car moving forward. it is the friction between the wheel bearings that slows down the car

[swansont]

When you turn you do create a net force, but in the case of uniform circular motion the force is toward the center of the circle. As such, it is tangent to the motion and can do no work, so you neither speed up nor slow down. The change in velocity (because you have an acceleration) is purely a change in direction.

[scilearner]

In the case of cars it is the friction between the wheels and the road that gets the car moving forward. it is the friction between the wheel bearings that slows down the car

 

Thanks for all the answers Wait I didn't know this before. So car exerts force on road, road exerts on tyres, that's how the car moves forward. So how does a car move in ice, if the road is not exerting a force on the tyres? So can you easily brake on ice because friction between the wheel bearings can be achieved. Can you elaborate on this. Upto this point I was thinking car moving forward, friction from the road stopping the car? Why is this wrong?

[swansont]

Thanks for all the answers Wait I didn't know this before. So car exerts force on road, road exerts on tyres, that's how the car moves forward. So how does a car move in ice, if the road is not exerting a force on the tyres? So can you easily brake on ice because friction between the wheel bearings can be achieved. Can you elaborate on this. Upto this point I was thinking car moving forward, friction from the road stopping the car? Why is this wrong?

 

There is friction with ice, but it is greatly reduced compared to dry asphalt. The ice still exerts a force on the tires, and vice-versa, but since it is slippery and the force involving the bearings is unaffected, it is easier to make the wheels slip and skid.

 

When the tires roll forward, there is a forward force from the road. When you brake, the force is toward the rear. Friction opposes the motion or intended motion of the object.

[Double K]

It's also important to note that with a road especially, they form a "camber" on the road so that the force acts to keep your car on the road.

Road design is fairly complex in some regards, design speeds (these are the speeds that you see signed as "safe" or "recommended" speeds on corners) and lines of sight etc are all taken into consideration when designing road alignment.

 

Most roads have a crossfall, and a camber. Camber is also referred to as superelevation.

 

Crossfall deals with road drainage, but also is factored into the camber design.

 

"For low speed rural & urban designs and high speed designs, Table 5-1 Rural Low Speed, Table 5-2 Urban Low Speed and Table 5-3 High Speed provide the basic design criteria for rates of superelevation (e) (measured in %) and Total Transition Length (TTL) (measured in feet or meters) as related to a selected design speed and curve radius. The minimum radius for each design speed is also shown. These tables are based on a maximum superelevation rate of 6%. In unusual situations, a superelevation rate less than the recommended value may be necessary for a given design speed. For these instances, the following formulas can be used to determine the Minimum Radius:

English Metric

Rmin = V2 / (15 x ((emaxx0.01) + fmax)) Rmin = V2 / (127 x ((emaxx0.01) + fmax))

Rmin - curve radius ( ft or m)

emax - superelevation rate (%)

fmax - friction factor (dimensionless)

V2 - vehicle speed (mph) (kph)"

http://www.sddot.com/pe/Roaddesign/docs/rdmanual/rdmch05.pdf

 

This is a south dakota design manual, all councils have slightly different standards, but the basics are the same across the board, having a quick read of the concepts in this manual might give you some deeper understanding.

 

This manual is a bit more easy to follow

http://safety.fhwa.dot.gov/speedmgt/ref_mats/fhwasa10001/

 

It's very rare for a road to have a reverse camber, in fact it's extremely bad design to do it, as this effect actually "throws" cars off the curve at higher speeds. Trucks especially have terrible problems with reverse cambers as they have a much higher centre of gravity.

Edited June 7, 2010 by Double K

[scilearner]

Thanks everyone My knowledge on this is very poor. I'll take this step by step. So from what you guys are saying, if there is a single tyre, I push it, it rolls rather than slide due to friction. Is that right?

 

Ok then in a car what creates the initial push? With the tyre I pushed it, in a car what gives this initial push to the car?

 

I think may be the question I should ask is why does a tyre roll? What makes it roll in one direction.

 

So is the friction making the car move forward, or is it making the tyre roll? Also since I'm not sure about this I don't understand how friction helps braking.

Isn't braking due to discs?

Edited June 8, 2010 by scilearner