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cheska_P

  • one year ago

The curved section of a horizontal highway is a circular unbanked arc of radius 740 m. If the coefficient of static friction between this roadway and typical tires is 0.40, what would be the maximum safe driving speed for this horizontal curved section of highway? Can someone please show me how to do this?

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  1. DanJS
    • one year ago
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    when going around the circle , the car would accelerate radially towards the center of that circle

  2. DanJS
    • one year ago
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    , the car would have a net force then [F=m*a] in the horizontal direction But the car does not want to slide off the road, so have to balance that with the friction force between the tire material and road material

  3. cheska_P
    • one year ago
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    Yes, but how do you calculate the velocity without mass?

  4. DanJS
    • one year ago
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    The radial acceleration can be a function of the velocity

  5. cheska_P
    • one year ago
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    I'm sorry I don't understand

  6. DanJS
    • one year ago
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    mass can be found in the other dimension if you need it, the normal force is to weight

  7. cheska_P
    • one year ago
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    And how would you calculate normal force?

  8. DanJS
    • one year ago
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    The acceleration towards the circle creates a Force, centripital acceleration = v^2 / r F = m*a = m*v^2/r

  9. Astrophysics
    • one year ago
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    Maybe a FBD will help, |dw:1443498860246:dw| maybe something like this, sorry for the bad drawing but notice that friction and centripetal acceleration are towards the center of the circle

  10. Astrophysics
    • one year ago
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    \[\vec a_c = \frac{ v^2 }{ r }\] so we can use @DanJS equation and find the max speed

  11. DanJS
    • one year ago
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    Normal Force is in the other direction, to get that balance those forces Sum of forces in vertical direction = Normal - Weight = Normal - m*g

  12. Astrophysics
    • one year ago
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    Yeah haha was just going to say

  13. DanJS
    • one year ago
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    Does not accelerate in vertical direction, and so F= m*a = 0 = N - mg

  14. Astrophysics
    • one year ago
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    \[\vec F_f = \mu \vec F_N = \mu mg\]

  15. cheska_P
    • one year ago
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    Okay bare with me for one moment, I want to write out what I understand and what I don't understand

  16. cheska_P
    • one year ago
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    I understand that those are the equations I have to use. But I still don't understand how I can use any of those equations to find velocity if I don't know the mass. Yes, F=m(g) but I don't know what force is, I only know what the static friction coefficient is (.040). I don't know what force is then I don't have anything to divide by 9.81 (g) to find mass (kg)

  17. Astrophysics
    • one year ago
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    The masses get cancelled out, because we are focusing on the horizontal direction and the only forces acting are the centripetal force and force of friction

  18. Astrophysics
    • one year ago
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    \[\vec F_c = \vec F_f\]

  19. cheska_P
    • one year ago
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    So in the equation F⃗ f=μF⃗ N=μmg, m=1?

  20. Astrophysics
    • one year ago
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    Your net force = centripetal

  21. Astrophysics
    • one year ago
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    \[\vec F_c = \vec F_f \implies \mu mg = \frac{ mv^2 }{ r }\] where the normal force can be found from the vertical direction

  22. Astrophysics
    • one year ago
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    \[\vec F_c = \vec F_f \implies \frac{ mv^2 }{ r } = \mu \vec F_N \implies \frac{ mv^2 }{ r }= \mu \vec mg\] \[\vec F_c = \text{centripetal force}~~~~\vec F_N = \text{Normal force}~~~~\vec F_f = \text{force of friction}\]

  23. cheska_P
    • one year ago
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    I promise I'm trying my hardest to understand what you're saying, I have an exam in 12 hours. Could you please work the problem out so I can see what steps you take, because it's not clicking and I'm getting so frustrated with myself right now

  24. ganeshie8
    • one year ago
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    While Astro composes the solution, let me ask you a question. In what direction does the frictional force acts when the car moves on a straight road ? |dw:1443500920629:dw|

  25. cheska_P
    • one year ago
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    Against the tires, so towards the center of the circle?

  26. ganeshie8
    • one year ago
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    In above picture, the car is moving on a straight road. There is no circle, yet.

  27. ganeshie8
    • one year ago
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    Remember, frictional force always "opposes" the direction of motion.

  28. cheska_P
    • one year ago
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    Yes, so if the car is move to the right the frictional force would be towards the left, correct?

  29. cheska_P
    • one year ago
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    moving*

  30. ganeshie8
    • one year ago
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    Why ?

  31. Astrophysics
    • one year ago
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    It's alright, ok so first thing you want to do is make a diagram with all these kinds of questions to see what's going on and keep track of all the forces, the net force is in the horizontal direction. |dw:1443500609394:dw| the centripetal acceleration will be inside as it's always going towards the center of the circle. So lets split the forces in components, in the horizontal direction we have \[\sum F_x = ma_x~~~~\sum F_y = ma_y\] So focusing on the horizontal direction (indicated with x subscript) we have \[\sum F_x \implies F_f = ma_c \implies F_f = m \frac{ v^2 }{ r }\] since we need the normal force to solve for the force of friction we look at our vertical direction, because note that \[F_f = \mu F_N\] looking at the vertical direction we have \[\sum F_y = 0 \implies -W+F_N \implies F_N = W\] where W is the weight notice how the sum of the forces of the y - direction = 0 because there is no acceleration in the y - direction. Our net force then becomes \[F_f = ma_c \implies \mu F_N = m \frac{ v^2 }{ r } \implies \mu mg = \frac{ mv^2 }{ r } \] you should be able to do the algebra! Note that force of friction opposes motion hence it's called friction.

  32. cheska_P
    • one year ago
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    Because friction is the force trying to stop the motion opposed to it?

  33. Astrophysics
    • one year ago
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    Ah it was much longer than that, a few things got overwritten

  34. Astrophysics
    • one year ago
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    But anyways I hope that helped haha...\[\sum \] this symbol indicates the sum of the forces in the certain direction

  35. cheska_P
    • one year ago
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    The answer is suppose to be 54 m/s, @ganeshie8 do you get that when you do the math?

  36. ganeshie8
    • one year ago
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    Yes, as the car is taking a turn to the left on a curved road : 1) the tires kick the road to the right 2) Now, static friction does its job : the floor pushes back the tires to the left. When the friction is not enough, the tires start spinning and the car skids. The curved motion is made possible because of this "static" frictional force(\(\mu mg\)). Any object in circular motion experiences a centripetal force(\(\frac{mv^2}{r}\)) towards the center of circle. So the frictional force provides this centripetal force, they are same. |dw:1443502345216:dw|

  37. ganeshie8
    • one year ago
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    \(\mu mg = \dfrac{mv^2}{r}\) plugin the given numbers to find at what speed the static friction gives up

  38. ganeshie8
    • one year ago
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    Hey please read it agian, I have fixed left/right typoes..

  39. Astrophysics
    • one year ago
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    Nice post @ganeshie8 Also what do you get when you solve for v for the above equation @cheska_P

  40. cheska_P
    • one year ago
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    I get 17.04

  41. Astrophysics
    • one year ago
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    Hmm, what does your expression look like, can you show the algebra? \[\mu mg = \dfrac{mv^2}{r} \]

  42. cheska_P
    • one year ago
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    I'll try give me one second

  43. cheska_P
    • one year ago
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    |dw:1443502674089:dw|

  44. Astrophysics
    • one year ago
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    Good so, \[v = \sqrt{r \mu g} \implies \sqrt{(740m)(0.40)(9.81m/s^2)}\] you got the right numbers?

  45. cheska_P
    • one year ago
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    yes

  46. cheska_P
    • one year ago
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    wait no

  47. cheska_P
    • one year ago
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    I'm an idiot, I had .040 not .40 for my static friction coefficient

  48. cheska_P
    • one year ago
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    This whole time I've bee doing the math right, I just keep doing it with the wrong coefficient. I'm so sorry guys. You don't understand how grateful I am that you both stuck with me and tried helping me. Thank you so much and now i'm a bit embarrassed

  49. cheska_P
    • one year ago
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    been*

  50. Astrophysics
    • one year ago
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    Np, make sure you read the post above explaining unbanked curves, it's pretty great! Take care!

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