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Car banking an angle

  1. Jan 16, 2008 #1


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    1. The problem statement, all variables and given/known data

    A curve of radius 45 m is banked for a design speed of 90 km/hr. If the coefficient of static friction is .3 what is the maximum speed that the car can go around the curve safely. What is the minimum speed?

    2. Relevant equations


    3. The attempt at a solution

    θ = tan^-1((25 m/s)^2/(45m * 9.8 m/s^2) = 54.8 degrees

    Fn(sinθ) + Ff = (m)(v^2)/r

    Fn(cosθ) = mg

    Fn= (mg)/(cosθ)

    mg(tanθ) + μ(Fn) = (m)(v^2)/r

    mg(tanθ) + μ(mg/cosθ) = (m)(v^2)/r

    g(tanθ) + μ(g/cosθ) = (v^2)/r

    45m((9.8m/s^s(tan(54.8)) + .3(9.8m/s^s/cos(54.8)) = (v^2)

    v = 29.23 m/s or 105.25 km/hr.

    The key that I'm checking this with says that the above answer is not correct. I haven't tried to find the minimum speed yet because my maximum is incorrect. Could someone please let me know where I'm making a mistake or if I'm going about this the wrong way? Thanks.
  2. jcsd
  3. Jan 17, 2008 #2

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    Draw a freebody diagram. The forces acting on the car on the inclined road at a point are the force of friction inward, the normal reaction and the weight. The sum of the horizontal component of these is equal to the centripetal force.

    I’m quite sure that you have done just these, but if you plug in numbers right at the beginning, it makes things look messy and hard to check. Use symbols.
  4. Jan 20, 2008 #3


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    I tried this problem again with out putting numbers in at the beginning.

    (m)(v^2)/r = Fn(sinθ) + Ff

    (m)(v^2)/r = mg(tanθ) + Ff

    (m)(v^2)/r = (m)(v1^2)/r + Ff

    (m)(v^2)/r = (m)(v1^2)/r + μ(m)(v1^2)/(r(sinθ))

    v^2 = (v1^2) + μ(v1^2)/(sinθ)

    v^2 = (25 m/s)^2 + 0.3(25 m/s)^2/(sin(54.8))

    v = 29.23 m/s or 105.23 km

    No matter how I go about this I always get the same answer. Could someone point out what I'm doing wrong or at least steer me in the right direction? Thanks.
  5. Jan 20, 2008 #4
    I've done a similar problem like this, does your FBD look like this ...


    Last edited: Jan 20, 2008
  6. Jan 20, 2008 #5


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    My freebody diagram looks like that. But aren't the two angles equal? (alpha and beta)
  7. Jan 20, 2008 #6
    I get 36.286 m/s. What is the actual answer?

    My final equation was ...

    [tex]v=\sqrt{\frac{-Rg(\mu_s\cos \beta+\sin \alpha)}{\mu_s\sin \beta-\cos \alpha}}}[/tex]

    And yes, alpha and beta are the same b/c I used trig identities and I think you did too ...
    Last edited: Jan 20, 2008
  8. Jan 21, 2008 #7


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    That is the correct answer. I found out how you got that formula and I've also solved for the lowest speed which is about 18.6 m/s or 66.96 km/hr. Thank you for all of your help.
  9. Jan 22, 2008 #8


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    Alternatively, if you would have chosen your reference axis differently, the only force with an angle would have been gravity.
  10. Jan 22, 2008 #9
    You forgot about the acceleration towards the center.
  11. Jan 22, 2008 #10


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    both these equations are wrong they should read

    Fn(sinθ) + Ff(cosθ) = (m)(v^2)/r


    Fn(cosθ) = mg + Ff(sinθ)
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