Max speed of a car at a curve banked road.

[azizlwl]

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.

 

[tiny-tim]

hi azizlwl!

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? ​

 

[azizlwl]

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.

 

[tiny-tim]

the vertical acceleration is zero, isn't it?

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

 

[azizlwl]

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.

 

[tiny-tim]

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 ​

 

[azizlwl]

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

 

[PeterO]

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!

 

[tiny-tim]

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?​

 

[azizlwl]

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.

 

[PeterO]

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.

 

[tiny-tim]

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 - µ_sNsinθ = 0 is correct …

that's the vertical components of N, W, and µ_sN (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 µ_sN …

it's ≤ µ_sN ! ​

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 µ_s, 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! ​

 

[azizlwl]

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

 

[tiny-tim]

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

wikipedia !

 

[PeterO]

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 !