Solving Banked Curve Problem: Coefficient of Static Friction

[jonah]

Homework Statement

A town wants to build a banked curve to join to large roads. The maximum speed they would like cars travel on this curve is 51m/s. The angle of the banked curve is to be 19o and the radius is to be 1.8x102 m.

a) What does the coefficient of static friction need to be for the material being used to build this bank curve to achieve these parameters?

b) If the banked curve mentioned above is covered in ice (you can assume it is then a frictionless surface), what does a driver need to do to sure they make around they make it around a banked curve without slipping? Explain also what happens if they don’t do this (hint: there two scenarios if they slip, our answer should include a specific value relating to this exact situation and an equation for general banked curves).

Homework Equations

FN = M g = (M) (9.8 m/s2)
Fc = M v2 / r

The Attempt at a Solution

The net force must be horizontal--pointing toward the center of the circle--and only the friction force is available to provide it. The normal force and the weight simply cancel each other.

Fc = M v2 / r
= (M (51 m/s)2)/350 m
Fc = 14.45 M (m/s2)

FN = M g = (M) (9.8 m/s2)

For this flat curve, the centripetal force is supplied by the friction force, Ff,

Ff = FN
F f = M 9.8 m/s 2

Ff = Fnet = Fc

M 9.8 m/s 2 = 14.45 M (m/s2)

= 0.68

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I think my calculation of static friction is correct for part a). Although, I don't understand how to finish part b).

 

[BvU]

You sure they don't want you to take the 19 degrees as a given for part a) as well ?

 

[CWatters]

Start with a diagram.

 

[jonah]

I think you're right - because my answer was slightly off what it was supposed to be. The roadway exerts a normal force n perpendicular to its surface. And the downward force of gravity is present. Those two forces add as vectors to provide a resultant or net force Fnet which points toward the center of the circle. I used equations I am familiar with but I don't know how to incorporate the slope of the banked curve.

Thanks

 

[BvU]

Forgot to welcome you to PF: welcome!
There is one other force: along the slope pointing down is the friction force. It has a horizontal component helping to cause the circular trajectory and a vertical component in the same direction as mg.

Since there is no vertical acceleration desired and you have the desired horizontal force, you have two eqations with two unknowns (N and $\mu$).

 

[jonah]

Thanks for the reply. Although it is not clear to me how to derive those two equations.
- I am a high school physics student and this problem seems more difficult than others I have encountered in the curriculum.

 

[BvU]

Yes, this isn't easy at all. Requires a step by step approach. Fortunately, once a) is clear, b) isn't that compicated any more (in fact it's first part is easier than a!).

Did you follow Watters' suggestion and make a diagram ? Can you post it so we can check for completeness ?

As for the relevant equations: $\vec F_{\rm normal} = -m\vec g$ holds for a horizontal plane. With a slope it changes. (Just imagine a very steep slope!) Part of the normal force still comes from gravity, but the normal force now also has a horizontal component -- which is good because it helps to follow the curved trajectory.

The roadway exerts a normal force n perpendicular to its surface. And the downward force of gravity is present. Those two forces add as vectors to provide a resultant or net force Fnet which points toward the center of the circle.

is spot on(*). And its magnitude you know (why do you fill in 350 m for r if the text states 180 m ?), apart from a factor m (which will cancel later on).
(*) But don't forget you need the friction as a third force.

Since vertically there is no change, there is no acceleration, so the vertical components must add up to zero. We know there is friction and we don't really know which way it's pointing yet. In a horizontal curve it is the only force with a horizontal component, so it definitely should point towards the center of the curve, so inwards. A very steep slope might require friction to point outwards and up to prevent a drop. Make a guess and now it's really time for the diagram.

Free advice: don't fill in numbers but work with symbols as much as possible, such as: v for $|\vec v|$ is just fine, as well as g for $|\vec g|$ -- but beware of the sign! --, $\theta$, $\mu$ for $\vec F_{\rm friction, max} = \mu \ \vec F_{\rm normal}$ and the M (which , hopefully, cancels out at some point).