## 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.]

1. rvc

r= 76m v= 100 km/h --> 100 X 5/18 m/s $$\mu$$= 0.38

2. anonymous

@rvc it needs a max and a min /:

3. Michele_Laino

on our car are acting two forces, namely: |dw:1435475379190:dw|

4. Michele_Laino

the condition for v_max, is then: $\Large m\frac{{{v^2}}}{R} = \mu mg$

5. Michele_Laino

which comes from the subsequent condition: $\Large {F_{centrifugal}} = {F_{friction}}$

6. Michele_Laino

solving the last expression for v, we get: $\Large {v_{\max }} = \sqrt {\mu Rg}$

7. Michele_Laino

actually our curve is, like below: |dw:1435475861915:dw|

8. rvc

v^2= r g tan $$\theta$$

9. Michele_Laino

we have this vector decomposition: |dw:1435476024271:dw|

10. Michele_Laino

so we can write: $m\frac{{{v^2}}}{R}\cos \theta = \mu mg$

11. Michele_Laino

$\Large m\frac{{{v^2}}}{R}\cos \theta = \mu mg$

12. rvc

i think it should be tan

13. Michele_Laino

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

why, tan?

15. Michele_Laino

|dw:1435476363488:dw|

16. Michele_Laino

|dw:1435476471088:dw|

17. Michele_Laino

you are right! @rvc

18. rvc

:)

19. Michele_Laino

the right friction force is: $\Large \mu \left( {mg\cos \theta + m\frac{{{v^2}}}{R}\sin \theta } \right)$

20. Michele_Laino

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

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

look too big

23. Michele_Laino

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

or: $\Large \tan \theta = \frac{1}{\mu }$

25. Michele_Laino

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

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

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

cool

29. Michele_Laino

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

yes!

31. Michele_Laino

we get the min value of v, if we set: $\Large \tan \theta = 0$

32. Michele_Laino

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

thats simpler :)

34. Michele_Laino

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

which is equivalent

36. Michele_Laino

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

well now let the user find out the equation

38. Michele_Laino

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: $\Large {v^2} = \mu gr$