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The banked roadway

  • Thread starter roam
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  • #1
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Homework Statement



A car goes around a curved stretch of flat roadway of radius R = 96.0 m. The magnitudes of the horizontal and vertical components of force the car exerts on a securely seated passenger are, respectively, X = 230.0 N and Y = 530.0 N.

This stretch of highway is a notorious hazard during the winter months when it can be quite slippery. Accordingly they decide to bank it at an angle φ = 21.0° to the horizontal. At what speed could the car now negotiate this curve without needing to rely on any frictional force to prevent it slipping upwards or downwards on the banked surface?

[PLAIN]http://img121.imageshack.us/img121/8359/carsh.jpg [Broken]

The answer must be: 68.4

(P.S. I have for the previous parts of this question worked out that the car is travelling at 19.18 m/s, also the minimum coefficient of static friction between the tyres and the road needed to negotiate this turn without sliding out is 0.434).

The Attempt at a Solution



The mass is: [tex]m=\frac{530}{9.81} =54.02[/tex]

Now writing Newton's law for the car in the radial direction:

[tex]\sum F_r=n sin \theta = \frac{mv^2}{r}[/tex] .........(1)

[tex]\sum F_y =ncos \theta - mg = 0[/tex]

[tex]ncos \theta = mg[/tex] .........(2)

divide equation 1 by 2:

[tex]tan \theta = \frac{v^2}{rg}[/tex] .........(3)

[tex]tan 21 = \frac{v^2}{96 \times 9.81} \Rightarrow v= \sqrt{(tan \theta) (96 \times 9.81)}[/tex]

However, the answer I got did not agree with the given model answer (68.4). Can anyone show me my mistakes? Any help is appreciated. :smile:
 
Last edited by a moderator:

Answers and Replies

  • #2
ehild
Homework Helper
15,406
1,810
Your solution is correct, and if you convert the speed to km/h you get the model answer. Never use any data without its unit!!!

ehild
 

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