# Max speed of a car at a curve banked road.

• azizlwl
In summary, if you have a car on a banked road and the car is not at rest, the normal force and the friction force are both acting. The friction force is larger than the normal force.
azizlwl
If given
weight=mg
Friction=μ
Banking angle=θ

For the y direction,
NCosθ-mg-µNSinθ=0

My question.
What is the x-direction equation?
I know its equal to mv^2/r.
Always mixed up between "real" force like mg and acquired force like N and friction.
If the car at rest on the banked road, N=MgCosθ.

Thank you.

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hi azizlwl!

the important thing in these problems is to find the directions in which you know the acceleration

what is the other?

thank you.
yes i know the object is accelerating towards the center.
What bother me the directions of frictional force and the weight of the object.
In a equilibrium state on an inclined plane, the gravity(MgSin[x]) is pulling the object down the plane and the friction force(UMgCos[x] pull it up.
But above situation the friction is towards the center means opposite direction when it in equilibrium state.

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the vertical acceleration is zero, isn't it?

so the extra equation you need comes from the vertical F = ma equation

Then can I use the pseudo force, centrifugal force in this equation.
Normal + friction= centrifugal force.
Newton 3rd law. Action= Centripetal and reaction=centrifugal.

Then i see the friction going out(not towards center) as in equilibrium state of object on inclined plane.

azizlwl said:
Then can I use the pseudo force, centrifugal force in this equation.
Normal + friction= centrifugal force.
Newton 3rd law. Action= Centripetal and reaction=centrifugal.

Then i see the friction going out(not towards center) as in equilibrium state of object on inclined plane.

i'm not sure what you're doing

and it's a very bad idea to use centrifugal force (unless you're in a rotating frame) …

it'll confuse you, and it's difficult to know where to put the plus and minus signs

always use centripetal acceleration and F = ma (in an inertial frame)

show us your two F = ma equations, one vertical, and the other radial horizontal

sorry for all the wandering. Try to fully understand.
Another conflicting facts. I see the frictional force only cause by mg not N or its the same.

A curve on a highway has a radius of curvature r . The curved road is banked
at θ with the horizontal. If the coefficient of static friction is μ,
(a) Obtain an expression for the maximum speed v with which a car can go
over the curve without skidding.
(b) Find v if r = 100 m, θ = 30◦, g = 9.8m/s2, μ = 0.25
http://img29.imageshack.us/img29/998/bankedz.jpg
http://img341.imageshack.us/img341/5640/bank2k.jpg

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azizlwl said:
If given
weight=mg
Friction=μ
Banking angle=θ

For the y direction,
NCosθ-mg-µNSinθ=0

My question.
What is the x-direction equation?
I know its equal to mv^2/r.
Always mixed up between "real" force like mg and acquired force like N and friction.
If the car at rest on the banked road, N=MgCosθ.

Thank you.

The statement in red is true, but unfortunately in this case the car is not at rest!

hi azizlwl!

yes, those equations (1) and (2) in your image are the two F =ma equations you need

what is it about those two equations that's worrying you?​

At rest friction =μmgCosθ
But as Mr. Peter0 reply it is only applicable at rest not for above case.

As you see from the equations from the diagram, the friction is constant=μMgCosθ. But intuitively i feel seated more deeply which shows exert more force to road which increases the friction.

I've seen different equation too.
NCosθ-mg-µNSinθ=0 ..here friction depends on N which also depends of (mv^2)/r.

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azizlwl said:
At rest friction =μmgCosθ
But as Mr. Peter0 reply it is only applicable at rest not for above case.

As you see from the equations from the diagram, the friction is constant=μMgCosθ. But intuitively i feel seated more deeply which shows exert more force to road which increases the friction.

I've seen different equation too.
NCosθ-mg-µNSinθ=0 ..here friction depends on N which also depends of (mv^2)/r.

The diagram does not show the friction force - which will be acting parallel to the slope, and down the slope.

The three forces acting are
weight - vertically down - that one is on your diagram.
Normal Force - perpendicular to the slope - that is also on your diagram.
Friction - parallel to the slope, down the slope.

Your diagram best covers the case when no friction was needed at all, as W and N would add vectorially [join the arrows head to tail] to give the F you show.

If you translate the N vector down until it is on top of the dotted green line, you will see what I mean.

You need to have F larger than in your diagram.
N is also larger, and will cross the F vector.
When you add one more vector [friction] parallel to the slope, pointing down and left, you will have the three acting vectors adding to produce the required resultant - the centripetal force.
Clearly then N will be longer than it is when the car is stationary.

The size of the friction force is of course μN.

azizlwl said:
I've seen different equation too.
NCosθ-mg-µNSinθ=0 ..here friction depends on N which also depends of $$\frac {mv^2}{r}$$

hmm … yes, that is different from the equation in your diagram

the equation in your diagram is wrong

Ncosθ - mg - µsNsinθ = 0 is correct

that's the vertical components of N, W, and µsN (at the maximum speed before slipping)
As you see from the equations from the diagram, the friction is constant=μMgCosθ. But intuitively i feel seated more deeply which shows exert more force to road which increases the friction.

i see … you feel that if we increase the speed, that increases the centripetal acceleration, and so N must also increase, to compensate?

yes, you're correct!

remember, the friction is not µsN

it's ≤ µsN !

as we increase the speed, both the friction force and N increase to match the increasing centripetal acceleration, and the ratio friction/N increases …

when that ratio reaches µs, that's the maximum speed

moral: with questions like this, don't worry about what happens at lower speeds, just answer the question as asked … that'll be much simpler!

And finally from Wikipedia which different from my first image. Which one is correct?
http://img829.imageshack.us/img829/6870/bank3y.jpg

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azizlwl said:
And finally from Wikipedia which different from my first image. Which one is correct?

wikipedia !

azizlwl said:
And finally from Wikipedia which different from my first image. Which one is correct?
http://img829.imageshack.us/img829/6870/bank3y.jpg

I suppose it depends who put the posting on wikipedia !

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## 1. What is the maximum speed for a car on a banked curve road?

The maximum speed for a car on a banked curve road is determined by the radius of the curve, the banking angle, and the coefficient of friction between the tires and the road surface. This speed can be calculated using the formula v = √(rgtanθ), where v is the speed, r is the radius of the curve, g is the acceleration due to gravity, and θ is the banking angle.

## 2. How does the banking angle of a curve affect the maximum speed of a car?

The banking angle of a curve affects the maximum speed of a car by providing a centripetal force that helps the car maintain a circular path. As the banking angle increases, the force required to keep the car on the curve decreases, allowing for a higher maximum speed.

## 3. What is the role of the coefficient of friction in determining the maximum speed of a car on a banked curve road?

The coefficient of friction between the tires and the road surface plays a crucial role in determining the maximum speed of a car on a banked curve road. It is a measure of the grip or traction between the two surfaces and a higher coefficient of friction allows for a higher maximum speed.

## 4. Are there any safety concerns when driving at the maximum speed on a banked curve road?

Yes, there are safety concerns when driving at the maximum speed on a banked curve road. If the speed is too high or the car's tires do not have enough friction with the road surface, the car may slide or skid off the road. It is important to always drive at a safe and reasonable speed, taking into account the road conditions and the capabilities of the car.

## 5. How can the maximum speed on a banked curve road be increased?

The maximum speed on a banked curve road can be increased by increasing the radius of the curve, increasing the banking angle, or using tires with a higher coefficient of friction. However, it is important to note that there are practical and safety limitations to how much the speed can be increased. It is also crucial to always follow the speed limit and drive safely on any road.

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