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anonymous

  • one year ago

A curve of radius 76 m is banked for a design speed of 100 km/h. If the coefficient of static friction is 0.38 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or too fast.]

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  1. rvc
    • one year ago
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    r= 76m v= 100 km/h --> 100 X 5/18 m/s \(\mu\)= 0.38

  2. anonymous
    • one year ago
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    @rvc it needs a max and a min /:

  3. Michele_Laino
    • one year ago
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    on our car are acting two forces, namely: |dw:1435475379190:dw|

  4. Michele_Laino
    • one year ago
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    the condition for v_max, is then: \[\Large m\frac{{{v^2}}}{R} = \mu mg\]

  5. Michele_Laino
    • one year ago
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    which comes from the subsequent condition: \[\Large {F_{centrifugal}} = {F_{friction}}\]

  6. Michele_Laino
    • one year ago
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    solving the last expression for v, we get: \[\Large {v_{\max }} = \sqrt {\mu Rg} \]

  7. Michele_Laino
    • one year ago
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    actually our curve is, like below: |dw:1435475861915:dw|

  8. rvc
    • one year ago
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    v^2= r g tan \(\theta\)

  9. Michele_Laino
    • one year ago
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    we have this vector decomposition: |dw:1435476024271:dw|

  10. Michele_Laino
    • one year ago
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    so we can write: \[m\frac{{{v^2}}}{R}\cos \theta = \mu mg\]

  11. Michele_Laino
    • one year ago
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    \[\Large m\frac{{{v^2}}}{R}\cos \theta = \mu mg\]

  12. rvc
    • one year ago
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    i think it should be tan

  13. Michele_Laino
    • one year ago
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    using initial data, we get: \[\Large \cos \theta = \frac{{\mu gR}}{{{v^2}}} = \frac{{0.38 \times 9.81 \times 76}}{{{{\left( {100/3.6} \right)}^2}}}\]

  14. Michele_Laino
    • one year ago
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    why, tan?

  15. Michele_Laino
    • one year ago
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    |dw:1435476363488:dw|

  16. Michele_Laino
    • one year ago
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    |dw:1435476471088:dw|

  17. Michele_Laino
    • one year ago
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    you are right! @rvc

  18. rvc
    • one year ago
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    :)

  19. Michele_Laino
    • one year ago
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    the right friction force is: \[\Large \mu \left( {mg\cos \theta + m\frac{{{v^2}}}{R}\sin \theta } \right)\]

  20. Michele_Laino
    • one year ago
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    so the right equation is: \[\large m\frac{{{v^2}}}{R}\cos \theta = m\mu \left( {g\cos \theta + \frac{{{v^2}}}{R}\sin \theta } \right)\]

  21. Michele_Laino
    • one year ago
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    oops.. I forgot the component of the weight of the car along the inclined plane: \[\large m\frac{{{v^2}}}{R}\cos \theta = m\mu \left( {g\cos \theta + \frac{{{v^2}}}{R}\sin \theta } \right) + mg\sin \theta \]

  22. rvc
    • one year ago
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    look too big

  23. Michele_Laino
    • one year ago
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    if we can neglect the weight of our car, then we can write: \[\Large \mu m\frac{{{v^2}}}{R}\sin \theta = m\frac{{{v^2}}}{R}\cos \theta \]

  24. Michele_Laino
    • one year ago
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    or: \[\Large \tan \theta = \frac{1}{\mu }\]

  25. Michele_Laino
    • one year ago
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    the external forces acting on our car, are the subsequent: |dw:1435478456892:dw| where N is the perpendicular reaction of the plane, R is the friction force3, Fc I the centrifugal force, and W is the weight of our car. So we can write the subsequent vector equation: \[\Large {{\mathbf{F}}_{\mathbf{C}}} + {\mathbf{W}} + {\mathbf{N}} + {\mathbf{R}} = {\mathbf{0}}\]

  26. Michele_Laino
    • one year ago
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    which is equivalent to these 2 scalar equations: \[\Large \left\{ \begin{gathered} m\frac{{{v^2}}}{R}\cos \theta - mg\sin \theta - \mu N = 0 \hfill \\ \hfill \\ - m\frac{{{v^2}}}{R}\sin \theta - mg\cos \theta + N = 0 \hfill \\ \end{gathered} \right.\] where N is the magnitude of the vector N

  27. Michele_Laino
    • one year ago
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    solving the second equation for N, and substituting into the first equation, we get: \[\large m\frac{{{v^2}}}{R}\cos \theta - mg\sin \theta - \mu \left( {mg\cos \theta + m\frac{{{v^2}}}{R}\sin \theta } \right) = 0\]

  28. rvc
    • one year ago
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    cool

  29. Michele_Laino
    • one year ago
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    after simplification, we get: \[\Large {v^2} = gr\frac{{\mu + \tan \theta }}{{1 - \mu \tan \theta }}\] where r is the radius of our curve

  30. rvc
    • one year ago
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    yes!

  31. Michele_Laino
    • one year ago
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    we get the min value of v, if we set: \[\Large \tan \theta = 0\]

  32. Michele_Laino
    • one year ago
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    better is if we study this function, using the mathematical analysis: \[\Large f\left( \theta \right) = \frac{{\mu + \tan \theta }}{{1 - \mu \tan \theta }}\]

  33. rvc
    • one year ago
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    thats simpler :)

  34. Michele_Laino
    • one year ago
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    I got that function, is an increasing function. Nevertheless I want to check my conjecture using this one: \[\Large {v^2} = gr\frac{{\mu \cos \theta + \sin \theta }}{{\cos \theta - \mu \sin \theta }}\]

  35. Michele_Laino
    • one year ago
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    which is equivalent

  36. Michele_Laino
    • one year ago
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    I got this: \[\Large \frac{d}{{d\theta }}\left( {\frac{{\mu \cos \theta + \sin \theta }}{{\cos \theta - \mu \sin \theta }}} \right) = \frac{{1 + {\mu ^2}}}{{{{\left( {\cos \theta - \mu \sin \theta } \right)}^2}}}\]

  37. rvc
    • one year ago
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    well now let the user find out the equation

  38. Michele_Laino
    • one year ago
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    ok! I can state this, since that function is an increasing function, then the min value is at \theta=0, namely: \[\Large \tan \theta = 0\] and therefore: \[\Large {v^2} = \mu gr\]

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