Max speed of a car at a curve banked road.

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SUMMARY

The discussion focuses on the dynamics of a car navigating a banked curve, specifically addressing the forces at play, including normal force (N), gravitational force (mg), and friction (μ). The participants derive the equations governing the vertical and horizontal forces acting on the car, emphasizing the importance of centripetal acceleration in determining maximum speed without skidding. The correct vertical force equation is established as NCosθ - mg - μNSinθ = 0, while the horizontal equation is identified as mv²/r. The conversation highlights the relationship between speed, friction, and normal force in maintaining equilibrium on a banked road.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with centripetal acceleration and its formula (mv²/r)
  • Knowledge of forces acting on an inclined plane
  • Basic grasp of static friction and its coefficient (μ)
NEXT STEPS
  • Study the derivation of the maximum speed formula for banked curves
  • Explore the effects of varying the banking angle (θ) on vehicle dynamics
  • Investigate the role of friction in different road conditions and its impact on vehicle stability
  • Learn about the application of centripetal force in real-world scenarios, such as racing and vehicle design
USEFUL FOR

Physics students, automotive engineers, and anyone interested in understanding vehicle dynamics on curved paths.

azizlwl
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If given
radius=r meters
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! :smile:

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

one is horizontal radial …

what is the other? :wink:​
 
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? :wink:

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

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 :redface:

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 :smile:​
 
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
 
Last edited by a moderator:
azizlwl said:
If given
radius=r meters
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! :smile:

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?​
 
  • #10
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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  • #11
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.
 
  • #12
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 :rolleyes:

the equation in your diagram is wrong :redface:

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! :smile:

remember, the friction is not µsN …

it's ≤ µsN ! :wink:​

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 :smile:

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! :wink:​
 
  • #13
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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  • #14
azizlwl said:
And finally from Wikipedia which different from my first image. Which one is correct?

wikipedia ! :smile:
 
  • #15
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 !
 
Last edited by a moderator:

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