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Speed at which a car on a banked curve will slide.

  • Thread starter SheldonG
  • Start date
  • #1
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Homework Statement


If a car goes around a banked curve too fast, the car will slide out of the curve. Find an expression for the car speed [tex] v_{max} [/tex] that puts the car on the verge of sliding out. What is the value for R=200, [tex] \theta = 10 [/tex], [tex] \mu_s = 0.60 [/tex]?


Homework Equations



a = v^2/r, F = ma.


The Attempt at a Solution


Place the x axis along the bank, and write the for equations for x and y:

[tex] \sum F_x = \frac{mv^2}{R}\cos\theta = mg\sin\theta + f_s [/tex]
[tex] \sum F_y = 0 = F_n - \frac{mv^2}{R}\sin\theta - mg\cos\theta [/tex]

It seems that the car should start to slide when
[tex] \frac{mv^2}{R}\cos\theta > mg\sin\theta + f_s[/tex]

Using [tex] fs = \mu_s F_n [/tex]

[tex] \frac{mv^2}{R}\cos\theta > mg\sin\theta + \mu_s\left(\frac{mv^2}{R}\sin\theta + mg\cos\theta\right) [/tex]

Solving for v:

[tex] v > \sqrt{\frac{gR(\tan\theta + \mu_s/\tan\theta)}{1-\mu_s\tan\theta} [/tex]

Unfortunately, the figure I get for v using R = 200, theta = 10, coefficient of static friction = 0.60 is much too high (88.6 m/s).

Could someone give me a hint about where I went wrong?

Thank you,
Sheldon

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
18,706
1,720
In this term [tex]\mu_s mg\cos\theta[/tex], if one divides by cos[itex]\theta[/itex], then one does not get a tan function, but simply

[tex]\mu_s mg[/tex]


So one would obtain

[tex] v > \sqrt{\frac{gR(\tan\theta + \mu_s)}{1-\mu_s\tan\theta} [/tex]
 
Last edited:
  • #3
50
0
Thank you, Astronuc. I am in your debt. What a stupid error :-)

Sheldon
 

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